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www.bookspar.com | Website for students | VTU NOTES ENGINEERING MATHEMATICS IV SUBJECT CODE: 06MAT41 | 10MAT41 Author: Dr. K. GURURAJAN Designation: Professor Department: MATHEMATICS Malnad College of Engineering, Hassan UNIT 7: Hypothesis Testing Introduction: Statistical Inference is a branch of Statistics which uses probability concepts to deal with uncertainty in decision making. There are a number of situations where in we come across problems involving decision making. For example, consider the problem of buying 1 kilogram of rice, when we visit the shop, we do not check each and every rice grains stored in a gunny bag; rather we put our hand inside the bag and collect a sample of rice grains. Then analysis takes place. Based on this, we decide to buy or not. Thus, the problem involves studying whole rice stored in a bag using only a sample of rice grains. Another example is consider the experiment of finding how far this e – learning program started by VTU yielding results. Usually the authorities visit the colleges and collect the information about the programme from a sample of students. Later study will be done, based on this sample study, suitable actions will be taken. Like this, one can give a number of examples. With these in view, this chapter focuses a detailed discussion of statistical inference. This topic considers two different classes of problems 1. Hypothesis testing – we test a statement about the population parameter from which the sample is drawn. 2. Estimation – A statistic obtained from the sample collected is used to estimate the population parameter. (Dr. K. Gururajan, MCE, Hassan Page 1 - www.bookspar.com | Website for students | VTU NOTES 2/5/2016) www.bookspar.com | Website for students | VTU NOTES First what is meant by hypothesis testing? This means that testing of hypothetical statement about a parameter of population. Conventional approach to testing: The procedure involves the following: 1. First we set up a definite statement about the population parameter which we call it as null hypothesis, denoted by H0 . According to Professor R. A. Fisher, Null Hypothesis is the statement which is tested for possible rejection under the assumption that it is true. Next we set up another hypothesis called alternate statement which is just opposite of null statement; denoted by H1 which is just complimentary to the null hypothesis. Therefore, if we start with H0 : 0 then alternate hypothesis may be considered as either one of the following statements; H1 : 0 , or H1 : 0 or H1 : 0 . As we are studying population parameter based on some sample study, one can not do the job with 100% accuracy since sample is drawn from the population and possible sample may not represent the whole population. Therefore, usually we conduct analysis at certain level of significance (lower than 100%. The possible choices include 99%, or 95% or 98% or 90%. Usually we conduct analysis at 99% or 95% level of significance, denoted by the symbol . We test H0 against H1 at certain level of significance. The confidence with which a person rejects or accepts H0 depends upon the significance level adopted. It is usually expressed in percentage forms such as 5% or 1% etc. Note that when is set as 5%, then probability of rejecting null hypothesis when it is true is only 5%. It also means that when the hypothesis in question is accepted at 5% level of significance, then statistician runs the risk of taking wrong decisions, in the long run, is only 5%. The above is called II step of hypothesis testing. (Dr. K. Gururajan, MCE, Hassan Page 2 www.bookspar.com | Website for students | VTU NOTES 2/5/2016) www.bookspar.com | Website for students | VTU NOTES Critical values or Fiducial limit values for a two tailed test: Sl. No Level of significance Theoretical Value 1 1% 2.58 2 2% 2.33 3 5% 1.96 Critical values or Fiducial limit values for a single tailed test (right and test) 5% 10% Tabulated value 1% Right – tailed test 2.33 1.645 1.28 Left tailed test -2.33 -1.645 -1.28 Setting a test criterion: The third step in hypothesis testing procedure is to construct a test criterion. This involves selecting an appropriate probability distribution for the particular test i.e. a proper probability distribution function to be chosen. Some of the distribution functions used are t, F, when the sample size is small (size lower than 30). However, for large samples, normal distribution function is preferred. Next step is the computation of statistic using the sample items drawn from the population. Usually, samples are drawn from the population by a procedure called random, where in each and every data of the population has the same chance of being included in the sample. Then the computed value of the test criterion is compared with the tabular value; as long the calculated value is lower then or equal to tabulated value, we accept the null hypothesis, otherwise, we reject null hypothesis and accept the alternate hypothesis. Decisions are valid only at the particular level significance of level adopted. During the course of analysis, there are two types of errors bound to occur. These are (i) Type – I error and (ii) Type – II error. (Dr. K. Gururajan, MCE, Hassan Page 3 www.bookspar.com | Website for students | VTU NOTES 2/5/2016) www.bookspar.com | Website for students | VTU NOTES Type – I error: This error usually occurs in a situation, when the null hypothesis is true, but we reject it i.e. rejection of a correct/true hypothesis constitute type I error. Type – II error: Here, null hypothesis is actually false, but we accept it. Equivalently, accepting a hypothesis which is wrong results in a type – II error. The probability of committing a type – I error is denoted by where = Probability of making type I error = Probability [Rejecting H0 | H0 is true] On the other hand, type – II error is committed by not rejecting a hypothesis when it is false. The probability of committing this error is denoted by . Note that = Probability of making type II error = Probability [Accepting H1 | H1 is false] Critical region: A region in a sample space S which amounts to Rejection of H0 is termed as critical region. One tailed test and two tailed test: This depends upon the setting up of both null and alternative hypothesis. A note on computed test criterion value: 1. When the sampling distribution is based on population of proportions/Means, then test criterion may be given as Zcal Expected results - Observed results Standard error of the distribution Application of standard error: 1. S.E. enables us to determine the probable limit within which the population parameter may be expected to lie. For example, the probable limits for population of proportion are given by p 3 pqn . Here, p represents the chance of achieving a success in a single trial, q stands for the chance that there is a failure in the trial and n refers to the size of the sample. 2. The magnitude of standard error gives an index of the precision of the parameter. (Dr. K. Gururajan, MCE, Hassan Page 4 - www.bookspar.com | Website for students | VTU NOTES 2/5/2016) www.bookspar.com | Website for students | VTU NOTES ILLUSTRATIVE EXAMPLES 1. A coin is tossed 400 times and the head turned up 216 times. Test the hypothesis that the coin is un– biased? Solution: First we construct null and alternate hypotheses set up H0 : The coin is not a biased one. Set up H1 : Yes, the coin is biased. As the coin is assumed be fair and it is tossed 400 times, clearly we must expect 200 times heads occurring and 200 times tails. Thus, expected number of heads is 200. But the observed result is 216. There is a difference of 16. Further, standard error is npq . With p = ½, q = ½ and n = 400, clearly 10 . The test criterion is zcal difference 216 240 1.6 standard error 10 If we choose 5% , then the tabulated value for a two tailed test is 1.96. Since, the calculated value is lower than the tabulated value; we accept the null hypothesis that coin is un – biased. 2. A person throws a 10 dice 500 times and obtains 2560 times 4, 5, or 6. Can this be attributed to fluctuations in sampling? Solution: As in the previous problem first we shall set up H0 : The die is fair and H1 : The die is unfair. We consider that problem is based on a two – tailed test. Let us choose level of significance as 5% then, the tabulated value is 1.96. Consider computing test criterion, zcal Expected value - observed result ; here, as the dice is tossed by a person standard error 5000 times, and on the basis that die is fair, then chance of getting any of the 6 numbers is 1/6. Thus, chance of getting either 4 or 5, or 6 is p = ½. Also, q = ½. With n = 5000, standard error, npq = 35.36. Further, expected value of obtaining 4 or 5 or 6 is 2500. Hence, zcal 2500 - 2560 1.7 which is lower than 1.96. Hence, we conclude 35.36 that die is a fair one. (Dr. K. Gururajan, MCE, Hassan Page 5 - www.bookspar.com | Website for students | VTU NOTES 2/5/2016) www.bookspar.com | Website for students | VTU NOTES 3. A sample of 1000 days is taken from meteorological records of a certain district and 120 of them are found to be foggy. What are the probable limits to the percentage of foggy days in the district? Solution: Let p denote the probability that a day is foggy in nature in a district as reported by meteorological records. Clearly, p 120 0.12 and q = 0.88. With n = 1000 1000, the probable limits to the percentage of foggy days is given by p 3 pqn . Using the data available in this problem, one obtains the answer as 0.12 3 0.12 88 1000 . Equivalently, 8.91% to 15.07%. 4. A die was thrown 9000 times and a throw of 5 or 6 was obtained 3240 times. On the assumption of random throwing, do the data indicate that die is biased? (Model Question Paper Problem) Solution: We set up the null hypothesis as H1 : Die is biased. . H0 : D i e Let us take level of significance as i s u n Also, - b i a s e d . 5% . Based on the assumption that distribution is normally distributed, the tabulated value is 1.96. The chance of getting each of the 6 numbers is same and it equals to 1/6 therefore chance of getting either 5 or 6 is 1/3. In a throw of 9000 times, getting the numbers either 5 or 6 is 1 ×9000 = 3000 . Now the difference in these two results is 240. With p = 1/3, q = 2/3, 3 n = 9000, S.E. npq = 44.72. Now consider the test criterion zcal Difference = S.E. 240 5.367 which is again more than the tabulated value. Therefore, we reject null 44.72 hypothesis and accept the alternate that die is highly biased. (Dr. K. Gururajan, MCE, Hassan Page 6 - www.bookspar.com | Website for students | VTU NOTES 2/5/2016) www.bookspar.com | Website for students | VTU NOTES Tests of significance for large samples: In the previous section, we discussed problems pertaining to sampling of attributes. It is time to think of sampling of other variables one may come across in a practical situation such as height weight etc. We say that a sample is small when the size is usually lower than 30, otherwise it is called a large one. The study here is based on the following assumptions: (i) the random sampling distribution of a statistic is approximately normal and (ii) values given by the samples are sufficiently close to the population value and can be used in its place for calculating standard error. When the standard deviation of population is known, then S.E (X) = p n where p denotes the standard deviation of population . When the standard deviation of the population is unknown, then S.E (X) = n where is the standard deviation of the sample. Fiducial limits of population mean are: 95% fiducial limits of population mean are X 1.96 99% fiducial limits of population mean are X 2.58 Further, test criterion zcal n n . x - S.E. LLUSTRATIVE EXAMPLES 1. A sample of 100 tyres is taken from a lot. The mean life of tyres is found to be 39, 350 kilo meters with a standard deviation of 3, 260. Could the sample come from a population with mean life of 40, 000 kilometers? Establish 99% confidence limits within which the mean life of tyres is expected to lie. Solution: First we shall set up null hypothesis, H0 : 40,000 , alternate hypothesis as H1 : 40, 000 . We consider that the problem follows a two tailed test and (Dr. K. Gururajan, MCE, Hassan Page 7 - www.bookspar.com | Website for students | VTU NOTES 2/5/2016) www.bookspar.com | Website for students | VTU NOTES chose 5% . Then corresponding to this, tabulated value is 1.96. expression for finding test criterion, zcal 3, 260 , n = 100. S.E. = n x - . S.E. Consider the Here, =40, 000, x 39, 350 and 3,260 326 . Thus, zcal 1.994. As this value is 100 slightly greater than 1.96, we reject the null hypothesis and conclude that sample has not come from a population of 40, 000 kilometers. The 99% confidence limits within which population mean is expected to lie is given as x 2.58×S.E. i.e. 39,350±2.58×326 = (38, 509, 40, 191) . 2. the mean life time of a sample of 400 fluorescent light bulbs produced by a company is found to be 1, 570 hours with a standard deviation of 150 hours. Test the hypothesis that the mean life time of bulbs is 1600 hours against the alternative hypothesis that it is greater than 1, 600 hours at 1% and 5% level of significance. Solution: First we shall set up null hypothesis, H0 : 1, 600 hours , alternate hypothesis as H1 : 1, 600 hours . We consider that the problem follows a two tailed test and chose 5% . Then corresponding to this, tabulated value is 1.96. expression for finding test criterion, zcal x - . S.E. Consider the Here, =1, 600, x 1, 570, n = 400 , 150 hours so that using all these values above, it can be seen that zcal 4.0 which is really greater than 1.96. Hence, we have to reject null hypothesis and to accept the alternate hypothesis. Test of significance of difference between the means of two samples Consider two populations P1 and P2. Let S1 and S2 be two samples drawn at random from these two different populations. Suppose we have the following data about these two samples, say (Dr. K. Gururajan, MCE, Hassan Page 8 - www.bookspar.com | Website for students | VTU NOTES 2/5/2016) www.bookspar.com | Website for students | VTU NOTES Samples/Data Sample size Mean Standard Deviation S1 n1 x1 1 S2 n2 x2 2 then standard error of difference between the means of two samples 2 22 S1 and S2 Difference of sample means . The Standard error n1 n2 rest of the analysis is same as in the preceding sections. is S.E = 1 and the test criterion is Zcal When the two samples are drawn from the same population, then standard error is S.E = Difference of sample means 1 1 and test criterion is Zcal . Standard error n1 n2 When the standard deviations are un – known, then standard deviations of the two s12 s22 where n1 n2 deviations of the two samples considered in the problem. samples must be replaced. s1 and s2 are standard Thus, S.E = ILLUSTRATIVE EXAMPLES 1. Intelligence test on two groups of boys and girls gave the following data: Data Mean Standard deviation Sample size Boys 75 15 150 Girls 70 20 250 Is there a significant difference in the mean scores obtained by boys and girls? (Dr. K. Gururajan, MCE, Hassan Page 9 - www.bookspar.com | Website for students | VTU NOTES 2/5/2016) www.bookspar.com | Website for students | VTU NOTES Solution: We set up null hypothesis as H0 : there is no significant difference between the mean scores obtained by boys and girls. The alternate hypothesis is considered as H1 : Yes, there is a significant difference in the mean scores obtained by boys and girls. We choose level of significance as 5% so that tabulated value is 1.96. Difference of means Consider z cal = . The standard error may be calculated as Standard Error 75 - 70 152 202 2.84 . As 2.84 is more + =1.761 , The test criterion is z cal = 1.761 150 250 than 1.96, we have to reject null hypothesis and to accept alternate hypothesis that there are some significant difference in the mean marks scored by boys and girls. S.E = 2. A man buys 50 electric bulbs of “Philips” and 50 bulbs of “Surya”. He finds that Philips bulbs give an average life of 1,500 hours with a standard deviation of 60 hours and Surya bulbs gave an average life of 1, 512 hours with a standard deviation of 80 hours. Is there a significant difference in the mean life of the two makes of bulbs? Solution: we set up null hypothesis, H0 : there is no significant difference between the bulbs made by the two companies, the alternate hypothesis can be set as H1 : Yes, and there could be some significant difference in the mean life of bulbs. Taking Consider 1% and =5% , the respective tabulated values are 2.58 and 1.96. 602 802 1512 - 1500 0.849 . Since the + =14.14 so that z cal = 14.14 50 50 calculated value is certainly lower than the two tabulated values, we accept the hypothesis there is no significant difference in the make of the two bulbs produced by the companies. standard error is S.E = A discussion on tests of significance for small samples So far the problem of testing a hypothesis about a population parameter was based on the assumption that sample drawn from population is large in size (more than 30) and the probability distribution is normally distributed. However, when the size of the sample is small, (say < 30) tests considered above are not suitable because the assumptions on which they are based generally do not hold good in the case of small samples. IN particular, here one cannot assume that the problem follows a normal distribution function and those values given by sample data are sufficiently close to the population values and can be used in their place for the calculation of standard error. Thus, it is a necessity to develop some alternative strategies to deal with problems having sample size (Dr. K. Gururajan, MCE, Hassan Page 10 - www.bookspar.com | Website for students | VTU NOTES 2/5/2016) www.bookspar.com | Website for students | VTU NOTES relatively small. Also, we do see a number of problems involving small samples. With these in view, here, we will initiate a detailed discussion on the same. Here, too, the problem is about testing a statement about population parameter; i.e. in ascertaining whether observed values could have arisen by sampling fluctuations from some value given in advance. For example, if a sample of 15 gives a correlation coefficient of +0.4, we shall be interested not so much in the value of the correlation in the parent population, but more generally this value could have come from an un – correlated population, i.e. whether it is significant in the parent population. It is widely accepted that when we work with small samples, estimates will vary from sample to sample. Further, in the theory of small samples also, we begin study by making an assumption that parent population is normally distributed unless otherwise stated. Strictly, whatever the decision one takes in hypothesis testing problems is valid only for normal populations. Sir William Gosset and R. A. Fisher have contributed a lot to theory of small samples. Sir W. Gosset published his findings in the year 1905 under the pen name “student”. He gave a test popularly known as “t – test” and Fisher gave another test known as “z – test”. These tests are based on “t distribution and “z – distribution”. Student’s t - distribution function Gosset was employed by the Guinness and Son, Dublin bravery, Ireland which did not permit employees to publish research work under their own names. So Gosset adopted the pen name “student” and published his findings under this name. Thereafter, the t – distribution commonly called student’s t – distribution or simply student’s distribution. The t – distribution to be used in a situation when the sample drawn from a population is of size lower than 30 and population standard deviation is un – known. The t – statistic, i n x t cal n where S n 1 12 S sample mean, n is the sample size, and x i are the data items. t cal is defined as x i 1 i x n 1 2 , x is the The t – distribution function has been derived mathematically under the assumption of a normally distributed population; it has the following form (Dr. K. Gururajan, MCE, Hassan Page 11 - www.bookspar.com | Website for students | VTU NOTES 2/5/2016) www.bookspar.com | Website for students | VTU NOTES 1 t2 2 where C is a constant term and = n - 1 denotes the number of f (t ) C 1 degrees of freedom. As the p.d.f. of a t – distribution is not suitable for analytical treatment. Therefore, the function is evaluated numerically for various values of t, and for particular values of . The t – distribution table normally given in statistics text books gives, over a range of values of , the probability values of exceeding by chance value of t at different levels of significance. The t – distribution function has a different value for each degree of freedom and when degrees of freedom approach a large value, t – distribution is equivalent to normal distribution function. The application of t – distribution includes (i) testing the significance of the mean of a random sample i.e. determining whether the mean of a sample drawn from drawn from a normal population deviates significantly from a stated value (i.e. hypothetical value of the populations mean) and (ii) testing whether difference between means of two independent samples is significant or not i.e. ascertaining whether the two samples comes from the same normal population? (iii) Testing difference between means of two dependent samples is significant? (iv) Testing the significance of on observed correlation coefficient. Procedures to be followed in testing a hypothesis made about the population parameter using student’s t - distribution: As usual first set up null hypothesis, Then, set up alternate hypothesis, Choose a suitable level of significance, • Note down the sample size, n and the number of degrees of freedom, • • Compute the theoretical value, t tab by using t – distribution table. t tab value is to be obtained as follows: If we set up 5% 0.05 , suppose 9 then, t tab is to be obtained by looking in 9th row and in the column 0.025 (i.e. half of = 0.05) . x The test criterion is then calculated using the formula, t cal n S Later, the calculated value above is compared with tabulated value. As long as the calculated value matches with the tabulated value, we as usual accept the null hypothesis and on the other hand, when the calculated value becomes more than tabulated value, we reject the null hypothesis and accept the alternate hypothesis. (Dr. K. Gururajan, MCE, Hassan Page 12 - www.bookspar.com | Website for students | VTU NOTES 2/5/2016) www.bookspar.com | Website for students | VTU NOTES ILLUSTRATIVE EXAMPLES 1. The manufacturer of a certain make of electric bulbs claims that his bulbs have a mean life of 25 months with a standard deviation of 5 months. Random samples of 6 such bulbs have the following values: Life of bulbs in months: 24, 20, 30, 20, 20, and 18. Can you regard the producer’s claim to valid at 1% level of significance? (Given that t tab 4.032 corresponding to 5 ). Solution: To solve the problem, we first set up the null hypothesis H0 : 25 months , alternate hypothesis may be treated as H0 : 25 months . To set up 1% , then tabulated value corresponding to this level of significance is th t tab | 1% and =5 4.032 (4.032 value has been got by looking in the 5 row ) . The test i n x criterion is given by t cal n where S S x i 1 i x n 1 2 . Consider xi x x x 24 1 1 26 3 9 30 7 49 -3 9 20 -3 9 18 -5 25 Total = 138 - Total = 102 xi 20 x 23 2 i 23 25 102 6 1.084 . Since the calculated 20.4 4.517 and t cal 4.517 5 value, 1.084 is lower than the tabulated value of 4.032; we accept the null hypothesis as mean life of bulbs could be about 25 hours. Thus, S (Dr. K. Gururajan, MCE, Hassan Page 13 www.bookspar.com | Website for students | VTU NOTES 2/5/2016) www.bookspar.com | Website for students | VTU NOTES 2. A certain stimulus administered to each of the 13 patients resulted in the following increase of blood pressure: 5, 2, 8,-1, 3, 0, -2, 1, 5, 0, 4, 6, 8. Can it be concluded that the stimulus, in general, be accompanied by an increase in the blood pressure? (Model Question Paper Problem) Solution: We shall set up H0 : before after i.e. there is no significant difference in the blood pressure readings before and after the injection of the drug. The alternate hypothesis is H0 : before after i.e. the stimulus resulted in an increase in the blood pressure of the patients. Taking 1% and 5% , as n = 13, n 1 12 , respective tabulated values are t tab | 1% and =12 3.055 and t tab | 5% and =12 2.179 . Now, we compute the value of test criterion. For this, consider xi x x x 5 2 4 2 -1 1 8 5 25 -1 -4 16 3 0 0 0 -3 9 -2 -5 25 -2 4 5 2 4 0 -3 9 4 1 1 6 3 9 8 5 25 Total = 39 - Total = 132 xi 1 x 3 (Dr. K. Gururajan, MCE, Hassan Page 14 - www.bookspar.com | Website for students | VTU NOTES 2 i 2/5/2016) www.bookspar.com | Website for students | VTU NOTES i n Consider S x i 1 i x n 1 2 x 132 n may 11 3.317 . Therefore, t cal S 12 03 13 3.2614 . As the calculated value 3.2614 is more than 3.317 the tabulated values of 3.055 and 2.179, we accept the alternate hypothesis that after the drug is given to patients, there is an increase in the blood pressure level. be obtained as t cal 3. the life time of electric bulbs for a random sample of 10 from a large consignment gave the following data: 4.2, 4.6, 3.9, 4.1, 5.2, 3.8, 3.9, 4.3, 4.4, 5.6 (in ‘000 hours). Can we accept the hypothesis that the average life time of bulbs is 4, 000 hours? Solution: Set up H0 : 4,000 hours , H1 : 4, 000 hours . Let us choose that 5% . Then tabulated value is t tab | 5% and =9 2.262 . To find the test criterion, consider xi x x x 4.2 -0.2 0.04 4.6 0.2 0.04 3.9 -0.5 0.25 4.1 -0.3 0.09 0.8 0.64 3.8 -0.6 0.36 3.9 -0.5 0.25 4.3 -0.1 0.01 4.4 0.0 0.0 5.6 1.2 1.44 Total = 44 - Total = 3.12 xi x 5.2 4.4 (Dr. K. Gururajan, MCE, Hassan Page 15 - www.bookspar.com | Website for students | VTU NOTES 2 i 2/5/2016) www.bookspar.com | Website for students | VTU NOTES i n x Consider S i 1 i x n 1 2 x 3.12 n is computed 0.589 . Therefore, t cal S 9 4.4 4.0 10 = 2.148. As the computed value is lower than the tabulated 0.589 value of 2.262, we conclude that mean life of time bulbs is about 4, 000 hours. as t cal A discussion on 2 test and Goodness of Fit Recently, we have discussed t – distribution function (i.e. t – test). The study was based on the assumption that the samples were drawn from normally distributed populations, or, more accurately that the sample means were normally distributed. Since test required such an assumption about population parameters. For this reason, A test of this kind is called parametric test. There are situations in which it may not be possible to make any rigid assumption about the distribution of population from which one has to draw a sample. Thus, there is a need to develop some non – parametric tests which does not require any assumptions about the population parameters. With this in view, now we shall consider a discussion on distribution which does not require any assumption with regard to the population. The test criterion 2 Oi Ei corresponding to this distribution may be given as 2 i where Ei RT CT Oi : Observed values , Ei : Expected values . Ei : N 2 When Expected values are not given, one can calculate these by using the following RT CT relation; E i : . Here, RT means the row total for the cell containing the row, CT N is for the column total for the cell containing columns, and N represent the total number of observations in the problem. The calculated 2 value (i.e. test criterion value or calculated value) is compared with the tabular value of value for given degree of freedom at a certain prefixed level of significance. Whenever the calculated value is lower than the tabular value, we continue to accept the fact that there is not much significant difference between expected and observed results. 2 (Dr. K. Gururajan, MCE, Hassan Page 16 - www.bookspar.com | Website for students | VTU NOTES 2/5/2016) www.bookspar.com | Website for students | VTU NOTES On the other hand, if the calculated value is found to be more than the value suggested in the table, then we have to conclude that there is a significant difference between observed and expected frequencies. As usual, degrees of freedom are n k where k denotes the number of independent constraints. Usually, it is 1 as we will be always testing null hypothesis against only one hypothesis, namely, alternate hypothesis. This is an approximate test for relatively a large population. For the usage of test, the following conditions must checked before employing the test. These are: 1. The sample observations should be independent. 2. Constraints on the cell frequencies, if any, must be linear. 3. i.e. the sum of all the observed values must match with the sum of all the expected values. 4. N, total frequency should be reasonably large 5. No theoretical frequency should be lower than 5. 2 6. It may be recalled this test is depends on test: The set of observed and expected frequencies and on the degrees of freedom, it does not make any assumptions regarding the population. ILLUSTRATIVE EXAMPLES 1. From the data given below about the treatment of 250 patients suffering from a disease, state whether new treatment is superior to the conventional test. Data Number of patients Favourable Not favorable Total New one 140 30 170 Conventional 60 20 80 Total 200 50 280 Solution: We set up null hypothesis as there is no significance in results due to the two procedures adopted. The alternate hypothesis may be assumed as there could be some difference in the results. Set up level of significance as www.bookspar.com | Website for students | VTU NOTES www.bookspar.com | Website for students | VTU NOTES 112 100 2 71 50 2 32 10 2 + 5% 100 50 10 2 | 0.05, 1 3.841 . (Dr. K. Gururajan, MCE, Hassan then tabulated Page 17 - value is 2/5/2016) RT CT N where RT means that the row total for the row containing the cell, CT means that the total for the column containing the cell and N, total number of frequencies. Keeping these in view, we find that expected frequencies are Consider finding expected values given by the formula, Expectation(AB) B A Note: Oi 170 200 136 ; 250 Ei 136 34 170 64 16 80 200 50 250 170 50 80 200 34 , 64 and 250 250 Oi E i Oi E i 2 80 50 16 . 250 Oi Ei 2 Ei 140 136 4 16 0.118 60 64 -4 16 0.250 30 34 -4 16 0.471 20 16 4 16 1.000 Total 1.839 As the calculated value 1.839 is lower than the tabulated value 2 | 0.05, 1 3.841 , we accept the null hypothesis, namely, that there is not much significant difference between the two procedures. (Dr. K. Gururajan, MCE, Hassan Page 18 www.bookspar.com | Website for students | VTU NOTES 2/5/2016) www.bookspar.com | Website for students | VTU NOTES 2. A set of five similar coins is tossed 320 times and the result is No. of heads Frequency 0 6 1 27 2 72 3 112 4 71 5 32 Test the hypothesis that the data follow a binomial distribution function. Solution: We shall set up the null hypothesis that data actually follows a binomial distribution. Then alternate hypothesis is, namely, data does not follow binomial distribution. Next, to set up a suitable level of significance, 5% , with n = 6, degrees of freedom is 5. Therefore, the tabulated value is 2 | 0.05, 5 11.07 . Before proceeding to finding test criterion, first we compute the various expected frequencies. As the data is set to be following binomial distribution, clearly probability density n function is b n, p, k p k q n k . Here, n 320, p 0.5, q 0.5 , and k k takes the values right from 0 up to 5. Hence, the expected frequencies of getting 0, 1, 2, 5 3, 4, 5 heads are the successive terms of the binomial expansion of 320 p q . Thus, expected frequencies E i are 10, 50, 100, 100, 50, 10. Consider the test criterion given by 2 |cal O i Ei 2 i ; Ei Here, observed values are: The expected values Oi : 6, 27, 72, 112, 71, 32 are: Ei : 10, 50, 100, 100, 50, 10 . Consider 6 10 2 27 50 2 72 100 2 |cal + 10 50 100 2 112 100 2 71 50 2 32 10 2 + 100 50 10 = 78.68. As the calculated value is very much higher than the tabulated value of 3.841, we reject the null hypothesis and accept the alternate hypothesis that data does not follow the binomial distribution. (Dr. K. Gururajan, MCE, Hassan - page 19, 23 – 05 – 2008) www.bookspar.com | Website for students | VTU NOTES www.bookspar.com | Website for students | VTU NOTES Subject: Engineering Mathematics IV Subject Code: 06MAT41 Author: Dr. K. Gururajan Assistant Professor Department of Mathematics Malnad College of Engineering, Hassan – 573 201 PART – B CLASS NOTES OF THE TOPIC RANDOM VARIABLE AND PROBABILITY DISTRIBUTION FUNCTIONS UNIT 6: Introduction: The previous chapter was devoted to a detailed study of basic probability concepts. The following were taught to you: Random Experiment, Sample Space. Events in a Sample Space Definition of probability, Axioms Addition Law, Multiplication Rule Independent and Exclusive Events Bayes Rule Mainly computation of probability of events From these analyses, it is clear that calculation of probability of events to be treated individually. However, in a practical situation, one may be interested in finding the probabilities of all the events and may wishes to have the results in a tabular form for any future reference. Since for an experiment having n outcomes, totally, there are 2 n totally events; finding probabilities of each of these and keeping them in a tabular form may be an interesting problem. Thus, if we develop a procedure, using which if it is possible to compute the probability of all the events, is certainly an improvement. The aim of this chapter is to initiate a discussion on the above. (Dr. K. Gururajan, MCE, Hassan www.bookspar.com | Website for students | VTU NOTES page 1) www.bookspar.com | Website for students | VTU NOTES Also, in many random experiments, outcomes may not involve a numerical value. In such a situation, to employ mathematical treatment, there is a need to bring in numbers into the problem. Further, probability theory must be supported and supplemented by other concepts to make application oriented. In many problems, we usually do not show interest on finding the chance of occurrence of an event, but, rather we work on an experiment with lot of expectations Considering these in view, the present chapter is dedicated to a discussion of random variables which will address these problems. First what is a random variable? Let S denote the sample space of a random experiment. A random variable means it is a rule which assigns a numerical value to each and every outcome of the experiment. Thus, random variable may be viewed as a function from the sample space S to the set of f :S R. all real numbers; denoted as For example, consider the random experiment of tossing three fair coins up. Then S = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}. Define f as the number of heads that appear. Hence, f ( HHH ) 3 , f ( HHT ) 2 , f ( HTH ) 2 , f (THH ) 2 , f ( HTT ) 1 , f (THT ) 1 , f (TTH ) 1 and f (TTT ) 0 . The same can be explained by means of a table as given below: HHH HHT HTH THH TTH THT HTT TTT 3 2 2 2 1 1 1 1 Note that all the outcomes of the experiment are associated with a unique number. Therefore, f is an example of a random variable. Usually a random variable is denoted by using upper case letters such as X, Y, Z etc. The image set of the random variable may be written as f ( S ) {0, 1, 2, 3} . (Dr. K. Gururajan, MCE, Hassan A random variable is divided into www.bookspar.com | Website for students | VTU NOTES page 2) www.bookspar.com | Website for students | VTU NOTES Discrete Random Variable (DRV) Continuous Random Variable (CRV). If the image set, X(S), is either finite or countable, then X is called as a discrete random variable, otherwise, it is referred to as a continuous random variable i.e. if X is a CRV, then X(S) is infinite and un – countable. Example of Discrete Random Variables: (i) In the experiment of throwing a die, define X as the number that is obtained. Then X takes any of the values 1 – 6. Thus, X(S) = {1, 2, 3. . . 6} which is a finite set and hence X is a DRV. (ii) Let X denotes the number of attempts required for an engineering graduate to obtain a satisfactory job in a firm? Then X(S) = {1, 2, 3,. . . . }. Clearly X is a DRV but having a image set countably infinite. (iii) If X denote the random variable equals to the number of marks scored by a student in a subject of an examination, then X(S) = {0, 1, 2, 3, . . . . 100}. Thus, X is a DRV, Discrete Random Variable. (iv) In an experiment, if the results turned to be a subset of the non – zero integers, Then it may be treated as a Discrete Random Variable. Examples of Continuous Random Variable: 1. Let X denote the random variable equals the speed of a moving car, say, from a destination A to another location B, then it is known that speedometer indicates the speed of the car continuously over a range from 0 up to 160 KM per hour. Therefore, X is a CRV, Continuously Varying Random Variable. (Dr. K. Gururajan, MCE, Hassan page 3) 2. Let X denotes the monitoring index of a patient admitted in ICU in a good hospital. Then it is a known fact that patient’s condition will be watched by the doctors continuously over a range of time. Thus, X is a CRV. www.bookspar.com | Website for students | VTU NOTES www.bookspar.com | Website for students | VTU NOTES 3. Let X denote the number of minutes a person has to wait at a bus stop in Bangalore to catch a bus, then it is true that the person has to wait any where from 0 up to 20 minutes (say). Will you agree with me? Since waiting to be done continuously, random variable in this case is called as CRV. 4. Results of any experiments accompanied by continuous changes at random over a range of values may be classified as a continuous random variable. Probability function/probability mass function f xi P X xi of a discrete random variable: Let X be a random variable taking the values, say X : x1 x2 x3 . . . xn then f xi P X xi is called as probability mass function or just probability function of the discrete random variable, X. Usually, this is described in a tabular form: X xi f xi x2 x1 P (2 X 5) f x1 f x2 x3 . . . xn f x3 . . . f xn Note: When X is a discrete random variable, it is necessary to compute f xi P X xi for each i = 1, 2, 3. . n. This function has the following properties: (Dr. K. Gururajan, MCE, Hassan f xi 0 0 f xi 1 f x 1 i i www.bookspar.com | Website for students | VTU NOTES page 4) www.bookspar.com | Website for students | VTU NOTES On the other hand, X is a continuous random variable, then its probability function will be usually given or has a closed form, given as f ( x ) P ( X x ) where x is defined over a range of values., it is called as probability density function usually has some standard form. This function too has the following properties: f ( x) 0 0 f ( x) 1 f ( x) 1 . To begin with we shall discuss in detail, discrete random variables and its distribution functions. Consider a discrete random variable, X with the distribution function as given below: X xi x1 f xi f x1 f x2 x2 x3 . . . xn f x3 . . . f xn Using this table, one can find probability of various events associated with X. For example, P xi X x j P X xi P X xi 1 up to + P X x j f xi f xi 1 f xi 2 + up to +f x j 1 +f x j P xi X x j P X xi 1 P X xi 2 . . . + P X x j 1 f xi 1 f xi 2 + up to +f x j 1 (Dr. K. Gururajan, MCE, Hassan P X x j 1 P X x j 1 page 5) 1 P X x1 P X x2 up to +P X x j -1 The probability distribution function or cumulative distribution function is given as www.bookspar.com | Website for students | VTU NOTES www.bookspar.com | Website for students | VTU NOTES F ( xt ) P ( X xt ) P X x1 P X x2 up to +P X xt It has the following properties: F ( x) 0 0 F ( x) 1 When xi x j then F xi F x j i.e. it is a strictly monotonic increasing function. when x , F ( x ) approaches 1 when x , F ( x ) approaches 0 A brief note on Expectation, Variance, Standard Deviation of a Discrete Random Variable: i n E ( X ) xi f xi i 1 i n E X 2 xi2 f xi i 1 Var( X ) E X 2 E ( X ) 2 (Dr. K. Gururajan, MCE, Hassan page 6) ILLUSTRATIVE EXAMPLES: 1. The probability density function of a discrete below: random variable X is given X: 0 1 2 3 4 5 6 f xi : k 3k 5k 7k 9k 11k 13k www.bookspar.com | Website for students | VTU NOTES www.bookspar.com | Website for students | VTU NOTES Find (i) k; (ii) F (4) ; (iii) P ( X 5) ; (iv) P (2 X 5) (v) E(X) and (vi) Var (X). Solution: To find the value of k, consider the sum of all the probabilities which equals to 49k. Equating this to 1, we obtain k 1 . 49 Therefore, distribution of X may now be written as X: f xi : 0 1 49 1 3 49 2 5 49 3 7 49 4 9 49 5 11 49 6 13 49 Using this, we may solve the other problems in hand. F (4) P[ X 4] P[ X 0] P[ X 1] P[ X 2] P[ X 3] P[ X 4] P[ X 5] P[ X 5] P[ X 6] 24 49 P[2 X 5] P[ X 2] P[ X 3] P[ X 4] E ( X ) xi f xi i 21 . Next to find E(X), consider 49 203 . To obtain Variance, it is necessary to compute 49 (Dr. K. Gururajan, MCE, Hassan E X 2 xi 2 f xi i page 7) 973 . 49 Thus, Variance of X is obtained by using the 2 E( X ) relation, Var( X ) E X 25 . 49 2 2 973 203 . 49 49 2. A random variable, X, has the following distribution function. X: -2 -1 0 1 2 3 f xi : 0.1 k 0.2 2k 0.3 k Find (i) k, (ii) F (2) , (iii) P (2 X 2) , (iv) P (1 X 2) , (v) E(X) , Variance. Solution: Consider the result, namely, sum of all the probabilities equals 1, www.bookspar.com | Website for students | VTU NOTES www.bookspar.com | Website for students | VTU NOTES 0.1+ k + 0.2 + 2k 0.3 + k 1 Yields k = 0.1. In view of this, distribution function of X may be formulated as X: -2 -1 0 1 2 3 f xi : 0.1 0.1 0.2 0.2 0.3 0.1 Note that F (2) P[ X 2] P[ X 2] P[ X 1] P[ X 0] P[ X 1] P[ X 2] 0.9 . The same also be obtained using the result, F (2) P[ X 2] 1 - P[ X 1] 1 P[ X 2] P[ X 1] P[ X 0] = 0.6. Next, P (2 X 2) P[ X 1] P[ X 0] P[ X 1] 0.5 . Clearly, P (1 X 2) = 0.7. Now, consider E ( X ) xi f xi = 0.8. i Then E X 2 xi 2 f xi = 2.8. Var( X ) E X 2 E ( X ) = 2.8 - 0.64 = 2.16. 2 i (Dr. K. Gururajan, MCE, Hassan page 8) A DISCUSSION ON A CONTINUOUS RANDOM VARIABLE AND IT’S DENSITY FUNCTION: Consider a continuous random variable, X. Then its probability density is usually given in the form of a function f ( x) with the following properties. (i) f ( x ) 0, (ii) 0 f ( x) 1 and ( iii ) f ( x) dx 1 . Using the definition of f ( x ) , it is possible to compute the probabilities of various events associated with X. b P (a X b) f ( x ) dx , a b P (a X b) f ( x ) dx a www.bookspar.com | Website for students | VTU NOTES www.bookspar.com | Website for students | VTU NOTES b x P (a X b) f ( x ) dx , F ( x) P ( X x) E( X ) f ( x ) dx a EX x f ( x) dx , 2 x 2 f ( x ) dx Var( X ) E X 2 E ( X ) P (a X b ) F (b ) F (a ) f ( x) 2 dF ( x ) ,if the derivative exists dx SOME STANDARD DISTRIBUTIONS OF A DISCRETE RANDOM VARIABLE: Binomial distribution function: Consider a random experiment having only two outcomes, say success (S) and failure (F). Suppose that trial is conducted, say, n number of times. One might be interested in knowing how many number of times success was achieved. Let p denotes the probability of obtaining a success in a single trial and q stands for the chance of getting a failure in one attempt implying that p + q = 1. If the experiment has the following characteristics; (Dr. K. Gururajan, MCE, Hassan page 9) the probability of obtaining failure or success is same for each and every trial trials are independent of one another probability of having a success is a finite number, then We say that the problem is based on the binomial distribution. In a problem like this, we define X as the random variable equals the number of successes obtained in n trials. Then X takes the values 0, 1, 2, 3 . . . up to n. Therefore, one can view X as a discrete random variable. Since number of ways of obtaining k successes in n trials may be n n! achieved in , therefore, binomial probability function may be formulated k k ! n k ! n as b(n, p, k ) p k q n k . k www.bookspar.com | Website for students | VTU NOTES www.bookspar.com | Website for students | VTU NOTES Illustrative examples: 1. It is known that among the 10 telephone lines available in an office, the chance that any telephone is busy at an instant of time is 0.2. Find the probability that (i) exactly 3 lines are busy, (ii) What is the most probable number of busy lines and compute its probability, and (iii) What is the probability that all the telephones are busy? Solution: Here, the experiment about finding the number of busy telephone lines at an instant of time. Let X denotes the number of telephones which are active at a point of time, as there are n = 10 telephones available; clearly X takes the values right from 0 up to 10. Let p denotes the chance of a telephone being busy, then it is given that p = 0.2, a finite value. The chance that a telephone line is free is q = 0.8. Since a telephone line being free or working is independent of one another, and since this value being same for each and every telephone line, we consider that this problem is based on binomial distribution. Therefore, the required probability mass function is (Dr. K. Gururajan, MCE, Hassan 10 b(10, 0.2, k ) (0.2) k (0.8)(10 k ) Where k = 0, 1, 2 . . . 10. k (i) page 10) To find the chance that 3 lines are busy i.e. P[X = 3] = 10 b(10, 0.2,3) (0.2)3 (0.8) 7 3 (ii) With p = 0.2, most probable number of busy lines is n p 10 0.2 2 . The 10 probability of this number equals b(10, 0.2, 2) (0.2) 2 (0.8)8 . 2 (iii) The chance that all the telephone lines are busy = (0.2)10 . 2. The chance that a bomb dropped from an airplane will strike a target is 0.4. 6 bombs are dropped from the airplane. Find the probability that (i) exactly 2 bombs strike the target? (ii) At least 1 strikes the target. (iii) None of the bombs hits the target? www.bookspar.com | Website for students | VTU NOTES www.bookspar.com | Website for students | VTU NOTES Solution: Here, the experiment about finding the number of bombs hitting a target. Let X denotes the number of bombs hitting a target. As n = 6 bombs are dropped from an airplane, clearly X takes the values right from 0 up to 6. Let p denotes the chance that a bomb hits a target, then it is given that p = 0.4, a finite value. The chance that a telephone line is free is q = 0.6. Since a bomb dropped from airplane hitting a target or not is an independent event, and the probability of striking a target is same for all the bombs dropped from the plane, therefore one may consider that hat this problem is based on binomial distribution. Therefore, the required probability 10 mass function is b(10, 0.4, k ) (0.4) k (0.8) 6 k . k (i) To find the chance that exactly 2 bombs hits a target, 10 i.e. P[X = 2] b(10, 0.4, 2) (0.4) 2 (0.8) 4 2 (Dr. K. Gururajan, MCE, Hassan page 11) (ii) Next to find the chance of the event, namely, at least 1 bomb hitting the target; i.e. P[ X 1] 1 P[ X 1] 1 P[ X 0] 1 (0.6)6 . (iii) The chance that none of the bombs are going to hit the target is P[X=0] = (0.6)6 . A discussion on Mean and Variance of Binomial Distribution Function Let X be a discrete random variable following a binomial distribution function with the n probability mass function given by b(n, p, k ) p k q n k . Consider the expectation of k X, namely, k n n E ( X ) k p k q n k k 0 k k n k k 0 n! pk q n k k ! n k ! www.bookspar.com | Website for students | VTU NOTES www.bookspar.com | Website for students | VTU NOTES n(n 1)! pp k 1q ( n11 k ) k 0 ( k 1)! n 1 1 k ! k n (n 1)! p k 1q[( n1 ( k 1)] ( k 1)![( n 1)! ( k 1)!] k 0 k n np (n 1)! p k 1q[( n1 ( k 1)] k 1 ( k 1)![( n 1)! ( k 1)!] k n np k n n 1 k 1 [( n1 ( k 1)] np p q k 2 k 1 np ( p q)n1 np as p q 1 Thus, expected value of binomial distribution function is np . (Dr. K. Gururajan, MCE, Hassan To find variance of X, consider EX k n 2 k k 0 2 n p k q n k k n k (k 1 1) p k q n k k 0 k k n k n n n! k n k k (k 1) p q k p k q n k k ! n k ! k 0 k 0 k k n n(n 1)(n 2)! p 2 p k 2q[( n 2 ( k 2)] E ( X ) k 0 ( k 2)![( n 2)! ( k 2)!] k n (n 2)! p k 2q[( n 2 ( k 2)] np k 0 ( k 2)![( n 2)! ( k 2)!] k n n(n 1) p 2 (n 2)! p k 2q[( n 2 ( k 2)] np k 2 ( k 2)![( n 2)! ( k 2)!] k n n(n 1) p 2 k n n 2 k 2 [( n 2 ( k 2)] n(n 1) p 2 np p q k 2 k 2 n(n 1) p2 ( p q)n 2 np . Since p +q = 1, it follows that www.bookspar.com | Website for students | VTU NOTES page 12) www.bookspar.com | Website for students | VTU NOTES n(n 1) p2 np Therefore, Var( X ) E X 2 - E ( X ) 2 n(n 1) p2 np np 2 n2 p2 np2 np n2 p2 np np2 np(1 p) npq . Hence, standard deviation of binomially distributed random variable is Var( X ) npq . (Dr. K. Gururajan, MCE, Hassan page 13) A DISCUSSION ON POISSON DISTRIBUTION FUNCTION This is a limiting case of the binomial distribution function. It is obtained by considering that the number of trials conducted is large and the probability of achieving a success in a single trial is very small i.e. here n is large and p is a small value. Therefore, Poisson distribution may be derived on the assumption that n and p 0 . It is found that Poisson distribution function is e k p ( , k ) . Here, np and k 0, 1, 2, 3, . . . . . k! Expectation and Variance of a Poisson distribution function e k Consider E ( X ) k p( , k ) k k! k 0 k 0 k k e k 1 k 0 ( k 1)! k e k k 1 k 1 (k 1)! k . But k 1 (k 1)! e , therefore it follows that for k 1 a Poisson distribution function, E ( X ) . Next to find Variance of X, first consider www.bookspar.com | Website for students | VTU NOTES www.bookspar.com | Website for students | VTU NOTES EX2 k k 2 p ( , k ) k 0 k k2 k 0 e k k! e k k (k 1 1) k! k 0 k k k (k 1) k 0 k k (k 1) k 0 e k k e k k k! k! k 0 e k E ( X ) k! (Dr. K. Gururajan, MCE, Hassan k 2 k 2e k 0 e 2 (k 2)! k page 14) k 2 (k 2)! k 2 2 e e . Thus, E X 2 2 . Hence, Variance of the Poisson distribution function is Var( X ) E X 2 - E ( X ) . 2 The standard deviation is Var( X ) Illustrative Examples: It is known that the chance of an error in the transmission of a message through a communication channel is 0.002. 1000 messages are sent through the channel; find the probability that at least 3 messages will be received incorrectly. Solution: Here, the random experiment consists of finding an error in the transmission of a message. It is given that n = 1000 messages are sent, a very large number, if p denote the probability of error in the transmission, we have p = 0.002, relatively a small number, therefore, this problem may be viewed as Poisson oriented. Thus, average number of messages with an error is np 2 . Therefore, required probability function www.bookspar.com | Website for students | VTU NOTES www.bookspar.com | Website for students | VTU NOTES e 2 2k is . p(2, k ) , k 0, 1, 2, 3, . . . . k! Here, the problem is about finding the probability of the event, namely, P ( X 3) 1 P ( X 3) 1 {P[ X 0] P[ X 1] P[ X 2]} k 2 e 2 2 k 1 k 0 k ! 1 e 2 1 2 2 1 5e 2 (Dr. K. Gururajan, MCE, Hassan page 15) 2. A car hire –firm has two cars which it hires out on a day to day basis. The number of demands for a car is known to be Poisson distributed with mean 1.5. Find the proportion of days on which (i) There is no demand for the car and (ii) The demand is rejected. Solution: Here, let us consider that random variable X as the number of persons or demands for a car to be hired. Then X assumes the values 0, 1, 2, 3. … . . . It is given that problem follows a Poisson distribution with mean, 1.5 . Thus, required probability mass function may be written as p(1.5, k ) e 1.5 (1.5)k . k! (i) Solution to I problem consists of finding the probability of the event, namely P[X = 0] = e 1.5 . (ii) The demand for a car will have to be rejected, when 3 or more persons approaches the firm seeking a car on hire. Thus, to find the probability of the event P[ X 3]. Hence, (1.5) 2 P[ X 3] 1 P{ X 3] 1 P[ X 0,1, 2] = e 1.5 1 1.5 . 2 Illustrative examples based on Continuous Random Variable and it’s Probability Density Function www.bookspar.com | Website for students | VTU NOTES www.bookspar.com | Website for students | VTU NOTES Suppose that the error in the reaction temperature, in oC, for a controlled laboratory experiment is a R.V. X having the p.d.f x2 , 1 x 2 f ( x) 3 0 elsewhere. Find (i) F(x) and (ii) use it to evaluate P (0<X1). (Dr. K. Gururajan, MCE, Hassan page 16) x Solution: Consider F ( x ) P ( X x ) f ( t )dt x x F ( x) Case (i) x -1 f ( t )dt 0dt 0 Case (ii) -1< x < 2 1 x F ( x) f ( t )dt x f ( t )dt 1 t3 t2 f ( t )dt 0 dt 9 3 1 1 x Case (iii) x = 2 F ( x ) x f ( t )dt 2 x 1 2 x 1 2 x3 1 . 9 f ( t )dt f ( t )dt f ( t )dt t3 t2 0 dt 0 9 3 1 2 1 81 1 . Therefore, 9 x 1 0, 3 x 1 F ( x) , 1 x 2. 9 1, x 2. 2kxe x , for x 0 2. If the p.d.f of a R.V. X having is given by f ( x ) 0, for x 0. 2 Find (a) the value of k and (b) distribution function F(X) for X. www.bookspar.com | Website for students | VTU NOTES www.bookspar.com | Website for students | VTU NOTES 2kxe WKT x2 dx 1 0 ke t dt 1(put x 2 t ) 0 ke t 0 1 (0 k ) 1 k 1 (Dr. K. Gururajan, MCE, Hassan page 17) x F ( x) P( X x) f ( t )dt 0, if x 0 0 x f ( t )dt f ( t )dt , if x 0 0 x 0 2te t dt ( e z )0x (1 e x ) . 2 2 2 0 1 e x , for x 0 F ( x) 0, otherwise. 2 3. Find the C.D.F of the R.V. whose P.D.F is given by x 2 , for 0<x 1 1 , for 1<x 2 f ( x) 2 3 x , for 2<x 3 2 0, otherwise x Solution: Case (i) x 0 F ( x ) 0 f ( t )dt 0dt 0 x Case (ii) 0 < x 1 F ( x) 0 f ( t )dt x t x2 0 dt dt 0 2 4 www.bookspar.com | Website for students | VTU NOTES www.bookspar.com | Website for students | VTU NOTES 0 Case (iii) 1 < x 2 F ( x) 1 x 0 1 f ( t )dt f ( t )dt f ( t )dt t 1 2x 1 0 dt dt 2 2 4 0 1 1 0 Case (iv) 2 < x 3 F ( x ) x 1 2 x 0 1 2 f ( t )dt f ( t )dt f ( t )dt f ( t )dt (Dr. K. Gururajan, MCE, Hassan 6 x x2 5 F ( x) 4 page 18) Case (v) for x >3, F(x) = 1. Therefore, if x 0 0, 2 x , if 0<x 1 4 2 x 1 F ( x) , if 1<x 2 4 6x x2 5 , if 2<x 3 4 if x>3 1, 4. The trouble shooting of an I.C. is a R.V. X whose distribution function is given by 0, for x 3 F ( x) 9 1 2 , for x 3. x If X denotes the number of years, find the probability that the I.C. will work properly (a) less than 8 years (b) beyond 8 years (c) anywhere from 5 to 7 years (d) Anywhere from 2 to 5 years. www.bookspar.com | Website for students | VTU NOTES www.bookspar.com | Website for students | VTU NOTES 0, for x 3 Solution: We have F ( x ) f ( t )dt 9 0 1 x 2 , for x 3. 8 9 For (a): P ( x 8) f ( t )dt 1 2 0.8594 8 0 x For Case (b): P(x > 8) = 1 – P(x 8) = 0.1406 (Dr. K. Gururajan, MCE, Hassan page 19) 2 2 For Case (c): P (5 x 7) = F (7) – F (5) = (1-9/7 ) – (1-9/5 ) = 0.1763 2 For Case (d): P (2 x 5) = F (5) – F (2) = (1-9/5 ) – (0) = 0.64 5. A continuous R.V. X has the distribution function is given by x1 0, 4 F ( x ) c( x 1) , 1 x 3 1, x 3. Find c and the probability density function. d [F ( x )] dx x1 0, 3 f ( x ) 4c( x 1) , 1 x 3 0, x 3. Solution: We know that f ( x ) 4c( x 1)3 , 1 x 3 f ( x) 0, otherwise Since we must have f ( x )dx 1, 3 3 3 4 4c ( x 1) dx 1 c ( x 1) 1 1 1 1 16c 1 c 16 www.bookspar.com | Website for students | VTU NOTES www.bookspar.com | Website for students | VTU NOTES Using this, one can give the probability function just by substituting the value of c above. (Dr. K. Gururajan, MCE, Hassan page 20) A discussion on some standard distribution functions of continuously distributed random variable: This distribution, sometimes called the negative exponential distribution, occurs in applications such as reliability theory and queuing theory. Reasons for its use include its memory less (Markov) property (and resulting analytical tractability) and its relation to the (discrete) Poisson distribution. Thus, the following random variables may be modeled as exponential: Time between two successive job arrivals to a computing center (often called inter-arrival time) Service time at a server in a queuing network; the server could be a resource such as CPU, I/O device, or a communication channel Time to failure of a component i.e. life time of a component Time required repairing a component that has malfunctioned. The exponential distribution function is given by, x e f ( x) 0, The probability distribution function may be written as computed as x >0, otherwise. x F ( x ) f ( x )dx 1 e x , if 0 x F( x ) . otherwise. 0 , www.bookspar.com | Website for students | VTU NOTES which may be www.bookspar.com | Website for students | VTU NOTES (Dr. K. Gururajan, MCE, Hassan page 21) Mean and Variance of Exponential distribution function Consider mean ( ) x f ( x )dx xe x dx 0 x e x Consider E X2 x 2 f ( x )dx x 2 e = x Var ( X ) E X 2 e x 1 1 1 - 0 2 2 0 x 2 e x dx 0 e x 2 x 2 e x 2 3 0 2 2 1 2 - E( X ) . 2 The standard deviation is Var( X ) 1 . Illustrative examples based on Exponential distribution function 1. The duration of telephone conservation has been found to have an exponential distribution with mean 2 minutes. Find the probabilities that the conservation may last (i) more than 3 minutes, (ii) less than 4 minutes and (iii) between 3 and 5 minutes. www.bookspar.com | Website for students | VTU NOTES www.bookspar.com | Website for students | VTU NOTES Solution: Let X denotes the random variable equals number of minute’s conversation may last. It is given that X is exponentially distributed with mean 3 minutes. Since for an exponential distribution function, mean is known to be 1 1 2 or =0.5 . The , so (Dr. K. Gururajan, MCE, Hassan page 22) 0.5 x , if x 0 , Probability density function can now be written as f ( x ) 0.5e . 0 , otherwise. (i) To find the probability of the event, namely, 3 P [ X 3] 1 P[ X 3] 1 0.5e 0.5 x dx 0 4 (ii) To find the probability of the event, namely P [ X 4] 0.5e 0.5 x dx . 0 5 (iii) To find the probability of the event P [3 X 5] 0.5e 0.5 x dx . 3 2. in a town, the duration of a rain is exponentially distributed with mean equal to 5 minutes. What is the probability that (i) the rain will last not more than 10 minutes (ii) between 4 and 7 minutes and (iii) between 5 and 8 minutes? Solution: An identical problem to the previous one. Thus, may be solved on similar lines. Discussion on Gaussian or Normal Distribution Function Among all the distribution of a continuous random variable, the most popular and widely used one is normal distribution function. Most of the work in correlation and regression analysis, testing of hypothesis, has been done based on the assumption that problem follows a normal distribution function or just everything normal. Also, this distribution www.bookspar.com | Website for students | VTU NOTES www.bookspar.com | Website for students | VTU NOTES is extremely important in statistical applications because of the central limit theorem, which states that under very general assumptions, the mean of a sample of n mutually (Dr. K. Gururajan, MCE, Hassan page 23) Independent random variables (having finite mean and variance) are normally distributed in the limit n . It has been observed that errors of measurement often possess this distribution. Experience also shows that during the wear – out phase, component life time follows a normal distribution. The purpose of today’s lecture is to have a detailed discussion on the same. The normal density function has well known bell shaped curve which will be shown on the board and it may be given as 1 x 1 f ( x) e2 2 2 , - x where and 0 . It will be shown that and are respectively denotes mean and variance of the normal distribution. As the probability or cumulative x F ( x) P( X x) f ( x) dx has no closed form, distribution function, namely, evaluation of integral in an interval is difficult. Therefore, results relating to probabilities are computed numerically and recorded in special table called normal distribution table. However, It pertain to the standard normal distribution function by choosing Fz ( z ) shown that and 1 2 z e t 2 2 and their entries are values of the function, dt. Since the standard normal distribution is symmetric, it can be 1 Fz ( z ) 2 z f (t ) dt =1 - F ( z ) . z www.bookspar.com | Website for students | VTU NOTES www.bookspar.com | Website for students | VTU NOTES (Dr. K. Gururajan, MCE, Hassan page 24) Thus, tabulations are done for positive values of z only. From this it is clear that Normal distribution table, please refer table 3, in page number 591, Probability and Statistics, Reliability, Queuing and Computer Science Applications” by K. S. Trivedi, a PHI Publications. It is clear that P(a X b) F (b) - F (a) P(a X b) F (b) - F (a) P(a X ) 1 P( X a) 1 F ( a ) Note: Let X be a normally distributed random variable taking a particular value, x, the x corresponding value of the standardized variable is given by z . Hence, F ( x) P( X x) Fz x . Illustrative Examples based on Normal Distribution function: 1. In a test on 2000 electric bulbs, it was found that the life of a particular make was normally distributed with an average life of 2040 hours and standard deviation of 60 hours. Estimate the number of bulbs likely to burn for (a) more than 2150 hours, (b) less than 1940 hours and (c) more than 1920 hours and but less than 2060 hours. Solution: Here, the experiment consists of finding the life of electric bulbs of a particular make (measured in hours) from a lot of 2000 bulbs. Let X denotes the random variable equals the life of an electric bulb measured in hours. distribution with mean It is given that X follows normal 2040 hours and 60 hours . First to calculate P ( X 2150 hours) 1 P ( X 2150) = 1 Fz 1.8333 1 0.9664 0.0336 www.bookspar.com | Website for students | VTU NOTES www.bookspar.com | Website for students | VTU NOTES Therefore, number of electrical bulbs with life expectancy more than 2150 hours is 0.0336 2000 67 . (Dr. K. Gururajan, MCE, Hassan page 25) 1950 2040 60 Next to compute the probability of the event; P ( X 1950 hours)=Fz =F 1.5 1 F (1.5) 1 0.9332 0.0668 z z Therefore, in a lot of 2000 bulbs, number of bulbs with life expectancy less than 1950 hours is 0.0668 * 2000 = 134 bulbs. Finally, to find the probability of the event, namely, P (1920 X 2060) F (2060) F (1920) 2060 2040 1920 2040 F Fz z 60 60 F 0.3333 F 2 z z F 0.3333 1 F 2 z z 0.6293 1 0.9774 = 0.6065 . Therefore, number of bulbs having life any where in between 1920 hours and 2060 hours is 0.6065 * 2000 = 1213. 2. Assume that the reduction of a person’s oxygen consumption during a period of Transcendenta Meditation (T.M.) is a continuous random variable X normally distributed with mean 37.6 cc/min and S.D. 4.6 cc/min. Determine the probability that during a period of T.M. a person’s oxygen consumption will be reduced by (a) at least 44.5 cc/min (b) at most 35.0 cc/min and (c) anywhere from 30.0 to 40.0 cc/min. Solution: Here, X a random variable is given to be following normal distribution function with mean .P 37.6 and 4.6 . Let us consider that X as the random equals the rejection of oxygen consumption during T M period and measured in cc/min. www.bookspar.com | Website for students | VTU NOTES www.bookspar.com | Website for students | VTU NOTES (Dr. K. Gururajan, MCE, Hassan page 26) (i) To find the probability of the event P[ X 44.5] 1 F (44.5) 44.5 37.6 1 Fz 4.6 1 Fz 1.5 1 0.9332 0.0668 . (ii) To find the probability of the event, P[ X 35.0] F (33.5) 35.0 37.6 Fz 4.6 Fz 0.5652 1 Fz 0.5652 1 0.7123 0.2877 . (iii) Consider the probability of the event P[30.0 X 40.0] F (40) F (30) 40 37.6 30 37.6 Fz Fz 4.6 4.6 Fz (0.5217) Fz (1.6522) 0.6985 1 0.9505 0.6490 3. An analog signal received at a detector (measured in micro volts) may be modeled as a Gaussian random variable N (200, 256) at a fixed point in time. What is the probability that the signal will exceed 240 micro volts? What is the probability that the signal is larger than 240 micro volts, given that it is larger than 210 micro volts? (Dr. K. Gururajan, MCE, Hassan www.bookspar.com | Website for students | VTU NOTES page 27) www.bookspar.com | Website for students | VTU NOTES Solution: Let X be a CRV denotes the signal as detected by a detector in terms of micro volts. Given that X is normally distributed with mean 200 micro volts and variance 256 micro volts. To find the probability of the events, namely, (i) P (X > 240 micro volts] and (ii) P[X > 240 micro volts | X > 210 micro volts]. Consider P[ X 240] 1 P[ X 240] 1 F (240) 240 200 1 Fz 16 1 Fz 2.5 = 1 – 0.9938 = 0.00621 Next consider P[X > 240 | X > 210] P[ X 240 and X > 210] P[ X 210] P[ X 240 ] 1 P[ X 240 ] P[ X 210] 1 P[ X 210] 240 200 1 Fz 1 Fz (2.5) 16 210 200 1 Fz (0.625) 1 Fz 16 1 0.9939 0.2335 1 0.73401 (Dr. K. Gururajan, MCE, Hassan www.bookspar.com | Website for students | VTU NOTES page 28) www.bookspar.com | Website for students | VTU NOTES Author: Dr. K. Gururajan Assistant Professor Department of Mathematics Malnad College of Engineering, Hassan – 573 201 UNIT 8: Jointly Distributed Random Variables Introduction: So far the study was restricted to one dimensional random variable, distribution function, and other related aspects. However, in real world, one come across a number of situations where we do find the presence of two or correlated random variables. For example, consider the experiment of finding how far e - learning programme initiated by VTU Belgaum has become popular among the engineering students? To find this say, authorities collect feed back from the students by visiting their institutions. Let the problem be about finding the opinion of students regarding two parameters; (i) quality of transmission from studio situated at Bangalore which we call it as X and (ii) student’s interest in this kind of programs which we shall refer it to as Y. For convenience of discussion, authorities visit seven colleges located at different parts of the state and results are in the following table. We assume that these are given in terms of percentages. Engg. Colleges X Y PGA SSIT BVB GSSIT AIT SBMJCE KLE X1 x2 x3 x4 x5 x6 x7 y1 y2 y3 y4 y5 y6 y7 In problems like this, we/authorities are certainly interested to learn the mood of the students/teachers about the e – learning programme initiated by us of course with huge cost. It is known to you that one satellite channel has been completely dedicated for this purpose in India. Many people are involved in this programme to reach the un – reached and needy. One comes across many illustrative examples like that. Therefore, there is a necessity to extend the study beyond single random variable. This chapter is devoted to a discussion on jointly related variables, their distribution functions and other important characteristics. First we shall have a discussion on discrete case. (Dr. K. Gururajan, MCE, Hassan www.bookspar.com | Website for students | VTU NOTES page 1) www.bookspar.com | Website for students | VTU NOTES Consider a random experiment and let S denotes its sample space. Let X and Y be two discrete random variables defined on S. Let the image set of these be X : x1 x2 x3 . . . xm Y: y1 y2 y3 . . . yn Suppose that there exists a correlation between the random variables X and Y. Then X and Y are jointly related/distributed variables Also, note that X and Y together assumes values. The same can be shown by means of a matrix or a table. Y y1 y2 y3 . . . . yn .. . . . . . . x1 , yn x2 , yn . . . . x3 , yn X x2 x1 , y1 x1 , y2 x2 , y1 x2 , y2 x3 x3 , y1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . xm , yn x1 x1 , y3 x2 , y3 x3 , y2 x3 , y3 xm , y1 xm , y2 xm , y3 xm Joint Probability Function/Joint probability mass function: The probability of the event X xi , Y y j is called joint probability function, denoted by h xi , y j P X xi , Y y j . Here, i take the values from 1 to m and j assumes the values right from 1 to n. The function h xi , y j has the following properties: 1. h x i , y j 0 2. 0 h xi , y j 1 3. h x , y 1 i i j j (Dr. K. Gururajan, MCE, Hassan www.bookspar.com | Website for students | VTU NOTES page 2) www.bookspar.com | Website for students | VTU NOTES Note: One caution with discrete random variables is that probabilities of events must be calculated individually. From the preceding sections, it is clear that in the current problem, there totally m n events. Thus, it is necessary to compute the probability of each and every event. This can also be shown by means of table. Y y1 y2 y3 . . . . yn X x1 h x1 , y1 h x1 , y2 h x1 , y3 .. . . h x1 , yn x2 h x2 , y1 h x2 , y2 h x2 , y3 . . . . h x2 , yn x3 h x3 , y1 h x3 , y2 h x3 , y3 . . . . h x3 , yn . . . . . . . . . . . . . . . . . . . . . . . . . . . . xm h xm , y1 h xm , y2 h xm , y3 . . . . h xm , yn Note: When the problem is based on discrete random variables, it is necessary to compute the probability and each and every event separately. Illustrative examples: A fair coin is tossed 3 times. Let X denote the random variable equals to 0 or 1 accordingly as a head or a tail occurs on the first toss, and let Y be the random variable representing the total number of heads that occurs. Find the joint distribution function of (X, Y). Solution: S = {HHT, HHT, HTH, THH, THT, TTH, HTT, TTT}. Thus, |S| = 8. Here, X takes the values 0 or 1 accordingly as a H appears on the I toss or a Tail appears on the I toss, while Y takes the values 0, 1, 2, 3 where these numbers represent the number of heads appearing in the experiment. Observe that joint variables(X, Y) assume eight values. Thus, there are (0, 0), (0, 1), (0, 2), (0, 3), (1, 0), (1, 1), (1, 2), (1, 3) – totally 8 events. Thus, we need to find the probabilities of all these 8 events. First we shall list the respective events corresponding to X and Y. (Dr. K. Gururajan, MCE, Hassan www.bookspar.com | Website for students | VTU NOTES page 3) www.bookspar.com | Website for students | VTU NOTES [X = 0] = {HHH, HHT, HTH, HTT}, [X = 1] = {THH, TTH, THT, TTT} [Y = 0] = {TTT}, [Y = 1] = {HTT, HTH, TTH], [Y = 2] = {HHT, HTH, THH} [Y = 3] = {HHH} Therefore, [X = 0, Y = 0] = { }, a null event, hence P[ X 0, Y 0] h(0, 0) 0 . Similarly, [ X 0, Y 1] = {HTT}, so P[ X 0, Y 1] h(0, 1) 1 . Note that 8 [ X 0, Y 2] = {HHT, HTH}, thus P[ X 0, Y 2] h(0, 2) 2 . Another example, 8 consider the event [X = 1, Y = 2] = {THH} implying that P[ X 1, Y 2] h(1, 2) 1 . 8 Continuing like this, one can compute the probabilities of all the events. The results are shown in the form of the following table; Y 0 1 2 3 X 0 h(0, 0) 0 h(0, 1) 1 1 h(1, 0) 1 h(1, 1) 2 8 8 8 h(0, 2) 2 h(1, 2) 1 8 8 h(0, 3) 1 8 h(1, 3) 0 Two cards are selected at random from a box which contains 5 cards numbered 1, 1, 2, 2, and 3. Let X denote the sum selected, and Y the maximum of two numbers drawn. Determine the joint probability function of X and Y. Solution: It is known that 2 cards can be selected from the available 5 in C (5, 2) = 10 ways. With cards being numbered as 1, 1, 2, 2, and 3, clearly, S = {(1, 1), (1, 2), (1, 2), (1, 2), (1, 2), (1, 3), (1, 3), (2, 2), (2, 3), (2, 3)}. As X denote the sum of numbered chosen, X takes the values 2, 3, 4, and 5. Y is defined as the maximum of the two numbers, so Y takes the values 1, 2, 3. Therefore, (X, Y) assumes totally 12 values. It can be shown as (Dr. K. Gururajan, MCE, Hassan page 4) www.bookspar.com | Website for students | VTU NOTES www.bookspar.com | Website for students | VTU NOTES X 1 2 3 2 (2, 1) (2, 2) (2, 3) 3 (3, 1) (3, 2) (3, 2) 4 5 (4, 1) (5, 1) (4, 2) (5, 2) (4, 3) (5, 3) Let h(x, y) denotes the joint probability function. Now, [X = 2] = {(1, 1)}, [Y = 1] = {(1, 1)}, therefore, h(2, 1) = 1/10 = 0.1; Observe that [Y = 2] = {(1, 2), (1, 2), (1, 2), (1, 2), (2, 2)}, thus, [X = 2, Y = 2] = { }, so h(2, 2) = 0.0, a similar argument will show that h(2, 3) = 0.0 since [y = 3] = {(1, 3), (1, 3), (2, 3), ( 2, 3)}. Next, [X = 3] = {(1, 2), (1, 2), (1, 2), (1, 2)}, clearly, [X = 3, Y = 1] = { }, so h (3, 1) = 0.0, [X = 3, Y = 2] = {(1, 2), (1, 2), (1, 2), (1, 2)}, thus, h(3, 2) = 4/10 = 0.4. However, h (3, 3) = 0.0. This is because, [Y = 3] = {(1, 3), (1, 3), (2, 3), (2, 3) and nothing is common between the events [X = 3] and [Y = 3]. By following above arguments, it may be seen that h (4, 1) = 0, h (4, 2) = 0.1, h (4, 3) = 0.2, h (5, 1) = 0.0, h (5, 2) = 0.0 and h (5, 3) = 0.2. Thus, the joint distribution function of X and Y may be given as Y X 1 2 3 2 0.1 0.0 0.0 3 0.0 0.4 0.0 4 0.0 0.1 0.2 5 0.0 0.0 0.2 The joint probability distribution function may be written as H ( xt , yu ) P[ X xt , Y yu ] x x t y yu h( x, y ) . The expectation of X and Y is x x1 y y1 i m j n E ( XY ) xi y j h xi , y j . From the joint probability mass function, h(x, y) it i 1 j 1 is possible to obtain distribution functions of X and Y. These are called as marginal distribution functions. The distribution of X may be obtained by adding all the probabilities row wise. Similarly, Y’s distribution function is determined by summing the probabilities column wise. For example, consider the joint probability function h (x, y) as given in the table (Dr. K. Gururajan, MCE, Hassan www.bookspar.com | Website for students | VTU NOTES page 5) www.bookspar.com | Website for students | VTU NOTES Y Y 0 1 2 3 X 0 h(0, 0) 0 h(0, 1) 1 1 h(1, 0) 1 h(1, 1) 2 8 8 8 h(0, 2) 2 h(1, 2) 1 h(0, 3) 1 8 8 h(1, 3) 0 8 The distribution function of X is got by adding the probabilities row wise. Thus, xi : 0 1 1 1 2 2 and the marginal distribution function of Y is to be obtained by the adding the probabilities column wise. Therefore, f xi : yi g yj 0 1 2 3 1 3 3 1 8 8 8 8 Cov( X , Y ) E ( XY ) E ( X ) E (Y ) . The Cov( X , Y ) regression coefficient between X and Y can be written as ( x, y) . Here, x y The Co – Variance of X and Y is given as x and σ y are standard deviation of X and Y series respectively. ILLUSTRATIVE EXAMPLES: Consider the joint distribution of X and Y Y X -4 2 7 1 1/8 ¼ 1/8 5 2/8 1/8 1/8 Compute E(X), E(Y), E (XY), Covar (X, Y), x , y and ( x, y ). Solution: First, we obtain the distribution functions of X and Y. To get the distribution of X, it is sufficient to add the probabilities row wise. Thus, (Dr. K. Gururajan, MCE, Hassan page 6) www.bookspar.com | Website for students | VTU NOTES www.bookspar.com | Website for students | VTU NOTES X f xi 1 5 1/2 1/2 E ( X ) 1 (1 / 2) 5 (1 / 2) 3 E X 2 1 (1 / 2) 25 (1 / 2) 13 Var( X ) E X 2 E ( X ) 4 and x 2 . 2 Similarly, Y -4 2 7 g yj 3/8 3/8 2/8 E (Y ) 4 (3 / 8) 2 (3 / 8) 7 (2 / 8) 1.0 E Y 2 16 (3 / 8) 4 (3 / 8) 49 (2 / 8) 39.5 Var(Y ) E Y 2 E (Y ) = 38.5 2 y 38.5 = 6.2048. Consider E ( XY ) x y h( x , y ) x = (2)(1)(0.2) + (3)(2)(0.4) + (4)(2)(0.1) + y (4)(3)(0.2) + (5)(3)(0.2) = 8.8. With E(X) = 3, and E(Y) = 1, it follows that Cov( X , Y ) E ( XY ) E ( X ) E (Y ) = 5.8 and ( x , y ) Cov( X ,Y ) 5.8 0.4673 . x y ( 2.0)(6.2048) As Cov( X , Y ) is not zero, we conclude that X and Y are not Independent. It may be noted that so far the study on probability theory was concentrated on either a single random variable or a two random variables (correlated variables) defined on the same sample space, S of some random experiment. It is interesting to see that in a practical situation, we come across numerous occasions involving a family or a collection (Dr. K. Gururajan, MCE, Hassan www.bookspar.com | Website for students | VTU NOTES page 7) www.bookspar.com | Website for students | VTU NOTES of random variables. This is especially true in problems of queuing theory and in networking. As today’s problems are mainly based on theory of queues, thus, there is a strong necessity to study the same. Also, this has enormous amount of applications in engineering and technology. With these in view, we shall consider a brief discussion on stochastic processes. Among various types of stochastic processes available in literature, Markov processes have found significant applications in many fields of science and engineering; therefore, current study will be restricted to this only. A discussion on Independent Random Variables Let X and Y be two discrete random variables. One says that X and Y are independent whenever the joint distribution function is the product of the respective marginal distribution functions. Equivalently, suppose that f xi , g y j and h xi , y j denote respectively, the distribution function of X, Y and that of X and Y, and h xi , y j f xi g y j , then we say that X and Y are independent random variables. Note: If it is known that X and Y are independent random variables, then it is very easy to compute the joint distribution function. This may be obtained by just multiplying the respective probabilities in the corresponding row and column of the table. For example, consider X f xi 0 0.2 y g yj 1 0.2 3 0.2 2 0.5 5 0.5 3 0.1 6 0.3 Then, joint distribution function of X and Y may be obtained as follows: h xi , y j 0 1 2 3 3 5 6 0.04 0.04 0.10 0.02 0.10 0.10 0.25 0.05 0.06 0.06 0.15 0.03 (Dr. K. Gururajan, MCE, Hassan www.bookspar.com | Website for students | VTU NOTES page 8) www.bookspar.com | Website for students | VTU NOTES A discussion on Stochastic Process and Queuing Theory First What is a Stochastic Process? It is a family of random variables {x (t ) | t T } , defined on a given probability space, indexed by the parameter, t, where t varies over an Index set, T usually, t is considered as time parameter. If T {t1 , t 2 , t 3 . . . } , then one can have a family of random variables like X t1 , X t 2 , X t 3 . . . etc. For a particular value of parameter, t, the values X ( t ) are called states. The set of all possible states is referred to as state taken by space, denoted by I. As it is known that a random variable is a function from the sample space to the set of all real pij( n ) pij (n) P x m n j | x m i numbers. Thus, one can define a stochastic process as a function X t , s where t T and Here, we can have many options. • One can fix t, and vary s. • t may be varied, but s could be fixed. • Both can be fixed to some specific values. • Both t and s may be allowed to vary. s I . When both s and t are varied, we generate a family of random variables constituting a stochastic process. Type of stochastic processes: 1. If the state space I and index set T of a stochastic process is discrete, then it is called Discrete – state – discrete – parameter (time) process. 2. On the other hand, if state space is continuous, and index set T is discrete, then we have a continuous – state – discrete – parameter process. 3. Similarly, one have discrete state – continuous parameter process and 4 Continuous state – continuous parameter process. Theory of queues provides a number of examples of stochastic processes. Among the various processes, Markov process is seen to be more useful. Markov Process (Memory less process): A stochastic process {x (t ) | t T } is called a Markov process if for any t 0 t1 t 2 < . . . <t n < t The conditional distribution of X (t ) for given values depends only on X (t n ) ; that is probability of occurrence of an event in future is completely dependent on the chance of occurrence of the event in the present state but (Dr. K. Gururajan, MCE, Hassan www.bookspar.com | Website for students | VTU NOTES page 9) www.bookspar.com | Website for students | VTU NOTES not on the previous or past records. Mathematically, this can be explained as P X (t ) x | X t n x n , X t n 1 x n 1 , . . . X t 0 x o P X (t ) x | X t n x n i.e. here, behavior of the stochastic process is such that probability distributions for its future development depend only the present state and not on how the process arrived in that state. Also, state space, I is discrete in nature. Equivalently, in a Markov chain, (a Markov process where state space, I takes discrete values), the past history is completely summarized in the current state, and, future is independent of its past but depends only on present state. First we shall discuss few basic concepts pertaining to Markov chains. A vector a a1 , a2 , a3 . . . , an is called a probability vector if all the components are non– negative and their sum is 1. A matrix, A such that each row is a probability vector, then A is called stochastic matrix. If A and B are stochastic matrices, then AB is a stochastic matrix. In fact, any power of A is a a stochastic matrix. A stochastic matrix A is said to be regular if all the entries of A are positive for some power Am of A where m is a positive integer. Result 1: Let A be a regular stochastic matrix. Then A has a unique fixed probability vector, t. i.e. One can find a unique fixed vector t such that t t A . Result 2: The sequence A, A2 , A3 , . . . of powers of A approaches the matrix T whose rows are each the fixed point t. Result 3: If q is any vector, then q A, q A2 , q A3 , . approaches the fixed point t. Transition Matrix of a Markov Chain: Consider a Markov chain – a fine stochastic process consisting of a finite sequence of trials, whose outcomes say, x1 , x 2 , x 3 . . . satisfy the following properties: (Dr. K. Gururajan, MCE, Hassan – page 10) www.bookspar.com | Website for students | VTU NOTES www.bookspar.com | Website for students | VTU NOTES Each outcome belongs to the state space, I {a1 , a2 , a3 . . .ma . }The outcome of any trial depends, at most, on the outcome of the trial and not on any other previous outcomes. If the outcome in n th trial is ai then we say that the system is in ai state at the n th stage. Thus, with each pair of states ai , a j , we associate a probability value pij which indicate probability of system reaching the state ai from the state a j in n steps. These probabilities form a matrix called transition probability matrix or just transition matrix. This may be written as p11 p12 p13 . . . p1n p21 p22 p23 . . . p2 n M . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . p n1 pn 2 pn 3 . . . pnn Note: Here, i th row of M represents the probabilities of that system will change from ai to a1 , a2 , a3 , . . . an . Equivalently, pij P xn j | x n1 i . n – step transition probabilities: The probability that a Markov chain will move from state I to state j in exactly n – steps is denoted by pij( n ) pij (n) P x m n j | x m i Evaluation of n–step transition probability matrix If M is the transition matrix of a Markov chain, then the n – step transition matrix may be obtained by taking nth power of M. Suppose that a system at time t = 0 is in the state a a1 , a2 , a3 . . . a n where the process begins, then corresponding probability vector denotes the initial probability distribution. Similarly, the a ( 0) a1( 0) , a 2( 0) , a (30) . . . a (0) n step transition probability matrix may be given as a ( n ) a1( n ) , a 2( n ) , a 3( n ) . . . a (n) . Now, n the (marginal) probability mass function of the random variable may be obtained from the nth step transition probabilities and the initial distribution function as follows: a ( n ) a ( 0) M ( n ) a ( 0) M n . Thus, the probability distribution of a homogeneous Markov chain is completely determined from the one step transition probability matrix M and the initial probability distribution a (0) . Note: The same may be explained as (Dr. K. Gururajan, MCE, Hassan – page 11) www.bookspar.com | Website for students | VTU NOTES www.bookspar.com | Website for students | VTU NOTES a (1 ) a ( 0 ) M a ( 2 ) a (1 ) M a ( 0 ) M 2 a ( 3) a ( 2) M a ( 0) M 3 Some illustrative examples on Markov chains: 1. Determine which of the following are stochastic matrices? 1 1 1 1 1 3 3 - 3 3 3 4 4 4 4 (i) (ii) (iii) 2 2 2 1 1 1 0 3 3 2 2 3 3 Solution: (i) is a stochastic matrix as all the entries are non – negative and the sum of all values in both the rows equals 1. (ii) This is not a stochastic matrix, since the second row is such that sum exceeding 1 (iii) this is again not a stochastic matrix, because one of the entry is negative. 2. Determine which of the following are regular stochastic matrices? 1 1 1 1 1 1 2 4 4 2 4 4 1 0 0 1 (i) A 0 (ii) B 0 1 0 1 1 0 0 2 2 Solution: A is not a regular stochastic matrix, as it has a 1 on the main diagonal. B is a regular stochastic matrix, Compute B3 , you would notice that all the entries are positive. 3. Find the unique fixed probability vector of the following stochastic matrices. 3 1 0 0 1 0 4 4 2 1 1 1 1 1 1 (i) A 3 (ii) (iii) A 0 3 2 2 6 2 3 0 1 1 0 2 1 0 0 3 3 (Dr. K. Gururajan, MCE, Hassan – page 12) www.bookspar.com | Website for students | VTU NOTES www.bookspar.com | Website for students | VTU NOTES Solution: Here, to find a unique fixed probability vector t ( x , y ) such that t t A . As t is a probability vector, one can take t ( x , 1 x ) . Consider the matrix equation, 2 1 (x,1 x ) (x,1 x ) 3 3 ; A multiplication of RHS matrices, yields, 0 1 1 2 x x (1 x ) and (1 x ) x . Thus, one can find that x 0.6 , hence the 3 3 required unique fixed probability vector t (0.6, 0.4) . (ii) We shall set up t ( x , y , z ) as the unique fixed probability vector. Again this is a probability vector, therefore, t ( x , y , 1 x y ) . Now, consider the matrix equation, 3 1 0 4 4 1 1 (x, y , 1 x y ) (x, y , 1 x y ) 0 2 2 1 0 0 Multiplying the matrices on the right hand side gives, 1 x y 2 3 1 y x y (1 x y ) 4 2 Solving these two equations, one obtains the unique fixed probability vector. (iii) Here, the unique fixed probability vector may be obtained as t (0.1, 0.6, 0.3) . 5. A boy’s studying habits are as follows. If he studies one night, he is 70 percent sure not to study the next night. On the other hand, if he does not study one night, he is only 60 percent sure not to study the next night as well. In the long run, how often does the boy studies? Solution: First we construct transition probability matrix. The problem here is about whether the boy studies or not on consecutive nights. As his study is given to be dependent on study in one night or not, therefore, it may be modeled in terms of Markov chain. Thus, first we construct the required transition probability matrix. The following table indicates the pattern of the study of the boy: (Dr. K. Gururajan, MCE, Hassan – page 13) www.bookspar.com | Website for students | VTU NOTES www.bookspar.com | Website for students | VTU NOTES I night Transition probability II night Boy Studying Boy not studying Boy Studying 0.3 0.7 Boy not studying 0.4 0.6 0.7 0.3 Therefore, it is clear that required stochastic matrix is M . To solve the 0.6 0.4 given problem, it is sufficient to compute a unique fixed probability vector, t ( x , 1 x ) such that t t M . Using the above, one obtains x 0.3 x 0.4(1 x ) . From this 4 equation, we can calculate x . Thus, a chance of boy studying in the long run is 11 36.7%. 6. A person’s playing chess game or tennis is as follows: If he plays chess game one week then he switches to playing tennis the next week with a probability 0.2. On the other hand, if he plays tennis one week, then there is a probability of 0.7 that he will play tennis only in the next week as well. In the long run, how often does he play chess game? Solution: Like in the previous case, this one too is based on Markov process. Here, parameters are about a player playing either chess or tennis game. Also, playing the game in the next week is treated to be dependent on which game player plays in the previous week. As usual, first we obtain transition probability matrix. II Week Transition probabilities I Week Player Playing Chess Player Playing Tennis Player Playing Chess 0.8 0.2 Player Playing Tennis 0.3 0.7 (Dr. K. Gururajan, MCE, Hassan, page number 14) www.bookspar.com | Website for students | VTU NOTES www.bookspar.com | Website for students | VTU NOTES 0.2 0.8 Clearly, the transition probability matrix is M . Here, too, the problem is 0.7 0.3 to find a unique fixed probability vector, t ( x , 1 x ) such that t t M . Consider the 0.2 0.8 matrix equation; x 1 - x ) ( x 1 - x ) . A simple multiplication work 0.7 0.3 results in x 0.8 x 0.7(1 x ) from which we get x 0.6 . Hence, we conclude that in the long run, person playing chess game is about 60%. 7. A sales man S sells in only three cities, A, B, and C. Suppose that S never sells in the same city on successive days. If S sells in city A, then the next day S sells in the city B. However, if S sells in either B or C, then the next day S is twice likely to sell in city A as in the other city. Find out how often, in the long run, S sells in each city. Solution: First we shall obtain transition probability matrix which will have 3 rows and 3 columns. As a salesman does not sell in a particular city on consecutive days, clearly, main diagonal is zero. It is given that if S sells in city A on first day, then next day he will sell in city B, therefore, first row is 0, 1, 0. next, one can consider two cases: (i) S selling city B or (ii) S selling in city C. It is also given that if S sells in city B with probability p, then his chances of selling in city A, next day is 2p. There is no way, he will sell in city C. Thus, we have p + 2p + 0 = 1 implying that p = 1/3. Therefore middle row of the matrix is 2/3, 1/3, 0. Similarly, if S sells in city C with probability q, then his chances of selling in city A, next day is 2q. Again, a chance of selling city B is 0. Thus, last row is 2/3, 0 , 1/3. The probability transition matrix is Next Day I day A A B C 0 1 0 0 1 B 2 C 2 3 3 1 3 3 0 (Dr. K. Gururajan, MCE, Hassan, page number 15 , 24 – 05 2-008) www.bookspar.com | Website for students | VTU NOTES www.bookspar.com | Website for students | VTU NOTES 0 1 0 1 . As in the previous two The matrix of the Markov process is M 2 0 3 3 2 1 0 3 3 problems, we need to find a unique fixed probability vector, t ( x , y , z ) such that 0 1 0 x y 1 - x - y x y 1 - x - y 2 3 0 13 Or 2 1 0 3 3 x 2 y 2 (1 x y ) and y x 1 (1 x y ) . Solving these two equations, 3 3 3 one obtains x 0.4, y 0.45 and z 0.15 . Hence, chances of a sales man selling in each of the cities is 40%, 45% and 15% respectively. 8. Mary’s gambling luck follows a pattern. If she wins a game, the probability of winning the next game is 0.6. However, if she loses a game, the probability of losing the next game is 0.7. There is an even chance that she wins the first game. (a) Find the transition matrix M of the Markov process. (b) Find the probability that she wins the second game? (c) Find the probability that she wins the third game? Find out how often, in the long run, she wins? 10. Each year Rohith trades his car for a new car. If he has a Maruti, he trades it in for a Santro. If he has a Santro car, then he trades it in for a Ford. However, is he has a Ford car; he is just as likely to trade it in for a new Ford as to trade it in for a Maruti or for a Santro. In 1995, he bought his first car which was a Ford. (a) Find the probability that he has bought (i) a 1997 Buick, (ii) a 1998 Plymouth, (iii) a 1998 Ford? (b) Find out how often, in the long run, he will have a Ford? Definition of various states: Transient state: A state i is said to be transient (or non – recurrent) if and only if there is a positive probability that process will not return to this state. For example, if we model a program as a Markov chain all but the final state will be transient. Otherwise program ends in an infinite loop. Recurrent state: A state i is said to be recurrent if and only if process returns to this state with probability one. Periodic state: A recurrent state which returns to the state i at regular periods of time. Absorbing state: A state i is said to be absorbing if and only if pii 1 . Here, once a Markov chain enters a state, and it remains there itself. (Dr. K. Gururajan, MCE, Hassan, page number 16, 24 – 05 2-008) www.bookspar.com | Website for students | VTU NOTES www.bookspar.com | Website for students | VTU NOTES PLEDGE I commit to excel, in all I do. I will apply myself actively to making a difference at Malnad College of Engineering, Hassan. I will use every opportunity presented to me by my superiors from this moment to make that difference. For myself, for colleagues, and for my students (Dr. K. Gururajan) www.bookspar.com | Website for students | VTU NOTES www.bookspar.com | Website for students | VTU NOTES Unit 1: Numerical Methods Numerical Solution of First Order and First Degree Ordinary Differential Equations Motivation: The fundamental laws of physics, mechanics, electricity and thermodynamics are usually based on empirical observations that explain variations in physical properties and states of the systems. Rather than describing the state of physical system directly, the laws are usually couched in terms of spatial and temporal changes. The following table gives a few examples of such fundamental laws that are written in terms of the rate of change of variables (t = time and x = position) Physical Law Mathematical Expression Variables and Parameters Newton’s second law of motion Fourier’s Law of Heat Conduction dv F dt m dT q k dx Faraday’s Law (Voltage drop across an inductor) VL L Velocity(v), force (F) and mass (M) Heat flux (q), thermal conductivity(k) and temperature (T) Voltage drop ( VL ), inductance4 (L) and current (i) di dt The above laws define mechanism of change. When combined with continuity laws for energy, mass or momentum, differential equation arises. The mathematical expression in the above table is an example of the Conversion of a Fundamental law to an Ordinary Differential Equation. Subsequent integration of these differential equations results in mathematical functions that describe the spatial and temporal state of a system in terms of energy, mass or velocity variations. In fact, such mathematical relationships are the basis of the solution for a great number of engineering problems. But, many ordinary differential equations arising in real-world applications and having lot of practical significance cannot be solved exactly using the classical analytical methods. These ode can be analyized qualitatively. However, qualitative analysis may not be able to give accurate answers. A numerical method can be used to get an accurate approximate solution to a differential equation. There are many programs and packages available for solving these differential equations. With today's computer, an accurate solution can be obtained rapidly. In this chapter we focus on basic numerical methods for solving initial value problems. Analytical methods, when available, generally enable to find the value of y for all values of x. Numerical methods, on the other hand, lead to the values of y corresponding only to some finite set of values of x. That is the solution is obtained as a table of values, www.bookspar.com | Website for students | VTU NOTES www.bookspar.com | Website for students | VTU NOTES rather than as continuous function. Moreover, analytical solution, if it can be found, is exact, whereas a numerical solution inevitably involves an error which should be small but may, if it is not controlled, swamp the true solution. Therefore we must be concerned with two aspects of numerical solutions of ODEs: both the method itself and its accuracy. In this chapter some methods for the numerical solution of ODEs are described. The general form of first order differential equation, in implicit form, is dy f ( x, y ) . An Initial Value Problem (IVP) F ( x, y, y / ) 0 and in the explicit form is dx consists of a differential equation and a condition which the solution much satisfies (or several conditions referring to the same value of x if the differential equation is of higher order). In this chapter we shall consider IVPs of the form dy f ( x, y ), y ( x0 ) y0 . (1) dx Assuming f to be such that the problem has a unique solution in some interval containing x0, we shall discuss the methods for computing numerical values of the solution. These methods are step-by-step methods. That is, we start from y0 y( x0 ) and proceed stepwise. In the first step, we compute an approximate value y1 of the solution y of (1) at x = x1 = x0 + h. In the second step we compute an approximate value y2 of the solution y at x = x2 = x0 + 2h, etc. Here h is fixed number for example 0.1 or 0.001 or 0.5 depends on the requirement of the problem. In each step the computations are done by the same formula. The following methods are used to solve the IVP (1). 1. 2. 3. 4. 5. 1. Taylor’s Series Method Euler and Modified Euler Method Runge – Kutta Method Milne’s Method Adams – Bashforth Method Taylor’s Series Method dy f ( x, y ), y ( x0 ) y0 . Let us approximate the exact solution y(x) dx to a power series in ( x x0 ) using Taylor’s series. The Taylor’s series expansion of y(x) Consider an IVP about the point x x0 is ( x x0 ) / ( x x0 ) 2 // ( x x0 ) 3 /// ( x x0 ) 4 / V y ( x0 ) y ( x0 ) y ( x0 ) y ( x0 ) 1! 2! 3! 4! (1) dy f ( x, y ) . Differentiating this From the differential equation, we have y / ( x) dx successively, we can get y // ( x), y /// ( x), y / V ( x), etc. Putting x x0 and y y 0 , the values y ( x) y ( x0 ) www.bookspar.com | Website for students | VTU NOTES www.bookspar.com | Website for students | VTU NOTES of y // ( x0 ), y /// ( x0 ), y / V ( x0 ), etc. can be obtained. Hence the Taylor’s series (1) gives the values of y for every value of x for which (1) converges. On finding the value of y1 for x x1 from (1), y // ( x), y /// ( x), y / V ( x), etc. can be evaluated at x x1 from the differential equation Problems: 1. Find by Taylor’s series method the value of y at x = 0.1 and 0.2 five places of decimals dy x 2 y 1, y (0) 1 . for the IVP dx Solution: Given x0 0, y 0 1 and f ( x, y ) x 2 y 1 Taylor’s series expansion about the point x 0( x0 ) is ( x 0) / ( x 0) 2 // ( x 0) 3 /// ( x 0) 4 / V y (0) y (0) y (0) y (0) 1! 2! 3! 4! x 2 // x 3 /// x 4 /V y( x) y(0) xy / (0) y (0) y (0) y (0) 2 6 24 y ( x) y (0) i.e. It is given that dy y / ( x) x 2 y 1 dx (1) y ( 0) 1 y / (0) 1 Differentiating y / ( x) x 2 y 1 successively three times and putting x = 0 & y = 1, we get y // ( x) 2 xy x 2 y / y // (0) 0 y /// ( x) 2 y 4 xy / x 2 y // y /// (0) 2 y iv ( x) 6 y / 6 xy // x 2 y /// y iv (0) 6 Putting the values of y(0), y / (0), y // (0), y /// (0), y iv (0) in (1), we get x2 x3 x4 (0) (2) (6) 2 6 24 x3 x4 y ( x) 1 x 3 4 y ( x) 1 x(1) Hence y(0.1) = 0.90033 and y(0.2) = 0.80227. www.bookspar.com | Website for students | VTU NOTES www.bookspar.com | Website for students | VTU NOTES 2. Employ Taylor’s series method to obtain approximate value of y at x = 0.1 and 0.2 for dy 2 y 3e x , y (0) 0. Compare the numerical solution the differential equation dx obtained with the exact solution. Solution: Given x0 0, y 0 0 and f ( x, y ) 2 y 3e x Taylor’s series expansion about the point x 0 is y( x) y (0) xy / (0) x 2 // x 3 /// x 4 /V y (0) y (0) y (0) 2 6 24 It is given that dy y / ( x) 2 y 3e x dx (2) y (0) 0 y / (0) 2 y(0) 3e 0 3 Differentiating y / ( x) 2 y 3e x successively three times and putting x = y = 0, we get y // ( x) 2 y / 3e x y // (0) 2 y / (0) 3 9 y /// ( x) 2 y // 3e x y /// (0) 2 y // (0) 3 21 y iv ( x) 2 y /// 3e x y iv (0) 2 y /// (0) 3 45 Putting the values of y(0), y / (0), y // (0), y /// (0), y iv (0) in (2), we get 9 2 21 3 45 4 x x x 2 6 24 9 7 15 3x x 2 x 3 x 4 2 2 8 y ( x) 0 3x Hence, y(0.1) = 3(0.1)+4.5(0.1)2+3.5(0.1)3+1.875(0.1)4 = 0.3486875 and y(0.2) = 3(0.2)+4.5(0.2)2+3.5(0.2)3+1.875(0.2)4 = 0.8110. www.bookspar.com | Website for students | VTU NOTES www.bookspar.com | Website for students | VTU NOTES Exact Solution: The given differential equation can be written as dy 2 y 3e x which is Leibnitz’s linear dx differential equation. 2d x Its I.F. is I.F = e e 2 x Therefore the general solution is, ye 2 x 3e x (e 2 x )dx c 3e x c y 3e x ce 2 x (3) Using the given initial condition y = 0 when x = 0 in (3) we get c = 3. Thus the exact solution is y 3 e 2 x e x When x = 0.1, the exact solution is y(0.1) = 0.348695 When x = 0.2, the exact solution is y(0.2) = 0.811266 The above solutions are tabulated as follows: x Numerical Exact Absolute Error Value 0.1 0.3486875 0.348695 0. 75x10 -5 0.2 0.8110 0.811266 0. 266x10 -3 www.bookspar.com | Website for students | VTU NOTES www.bookspar.com | Website for students | VTU NOTES 3. Using Taylor’s series method solve dy x 2 y, dx y (0) 1 at 0.1 ≤ x ≤ 0.4. Compare the values with the exact solution. Solution: Given x0 0, y 0 1 and f ( x, y ) x 2 y Taylor’s series expansion about the point x 0 is y( x) y (0) xy / (0) x 2 // x 3 /// x 4 /V y (0) y (0) y (0) 2 6 24 y (0) 0 It is given that dy y / ( x) x 2 y dx y / (0) (0) 2 1 1 Differentiating y / ( x) x 2 y successively and putting x =0, y = 1, we get y // ( x) 2 x y / y // (0) 0 y / (0) 0 (1) 1 y /// ( x) 2 y // y /// (0) 2 y // (0) 1 y iv ( x) y /// y iv (0) y /// (0) 1 Putting the values of y(0), y / (0), y // (0), y /// (0), y iv (0) in (4), we get y ( x) 1 x x2 x3 x4 2 6 24 Hence, y (0.1) 1 (0.1) (0.1) 2 (0.1) 3 (0.1) 4 2 6 24 = 0.9051625 y (0.2) 1 (0.2) (0.2) 2 (0.2) 3 (0.2) 4 2 6 24 = 0.8212667. www.bookspar.com | Website for students | VTU NOTES (4) www.bookspar.com | Website for students | VTU NOTES y (0.3) 1 (0.3) (0.3) 2 (0.3) 3 (0.3) 4 2 6 24 = 0.7491625. y (0.4) 1 (0.4) (0.4) 2 (0.4) 3 (0.4) 4 2 6 24 = 0.6896. Exact Solution: The given differential equation can be written as dy y x 2 a linear differential dx equation. dx Its I.F. is I.F = e e x Therefore the general solution is, ye x e x ( x 2 )dx c ( x 2 2 x 2)e x c y ( x 2 2 x 2) ce x (5) Using the given condition y(0) = 1 in (5) we get 1 = 2 + c or c = -1. Hence the exact solution is y ( x 2 2 x 2) e x The exact solution at x = 0.1, 0.2, 0.3 and 0.4 are y(0.1) = 0.9051625, y(0.2) = 0.8212692, y(0.3) = 0.7491817 and y(0.4) = 0.6896799 www.bookspar.com | Website for students | VTU NOTES www.bookspar.com | Website for students | VTU NOTES The above solutions are tabulated as follows: x Numerical Exact Absolute Error Value 0.1 0.9051625 0.9051625, 0 0.2 0.8212667 0.8212692, 0. 25x10 -5 0.3 0.7491625 0.7491817 0. 192x10 -4 0.4 0.6896 0.6896799 0. 799x10 -4 Assignments: 1. Use Taylor’s series method to find an approximate value of y at x = 0.1 of the IVP dy x y 2 , y (0) 1 Answer : 0.9138 dx 2. Evaluate y(0.1) correct to six decimal places by Taylor’s series method if y(x) satisfies Answer : 1.1053425 y / xy 1, y(0) 1 dy x y2, dx and y(0.2) 3. Solve y 1 at x 0 using Taylor’s series method and compute y(0.1) 4. Using Taylor’s series method, compute the solution of point x = 0.2 correct to three decimal places. 2. Answer : 1.1164 and 1.2725 dy x y, dx y (0) 1 at the Answer : 1.243 Modified Euler’s Method dy f ( x, y ), y ( x0 ) y0 . The following two methods can be used to dx determine the solution at a point x xn x0 nh . Consider the IVP www.bookspar.com | Website for students | VTU NOTES www.bookspar.com | Website for students | VTU NOTES Euler’s Method : y nE1 y n hf ( x n , y n ), n 0,1,2,3, (1) Modified Euler’s Method : y n 1 y n h f ( x n , y n ) f ( x n 1 , y nE1 ) , 2 n 0,1,2,3, (2) Remark: 1. The formulae (1) and (2) are also known as Euler’s Predictor – Corrector formula. 2. When Modified Euler’s method is applied to find the solution at a give point, we first find the solution at that point by using Euler’s method and the same will be used in the calculation of Modified Euler’s method. Also Modified Euler’s method has to be applied repeatedly until the solution is stationary. Problems : dy y2 , y (0) 1 by Euler’s method by choosing h = 0.1 and h = 0.05. dx 1 x Also solve the same problem by modified Euler’s method by choosing h = 0.05. Compare the numerical solution with analytical solution. 1. Solve Solution : Analytical solution is : dy dx 2 1 x y 1 log( 1 x) c y Using the condition y(0) = 1, we get c = 1. 1 Hence the analytical solution is y y(0.2) = 0.84579 1 log( 1 x) y2 Now by Euler’s method, we have y n1 y n 0.1 n 1 xn 2 (1) 0.9 y1 y(0.1) 1 0.1 1 0 (0.9) 2 0.82636 y 2 y(0.2) 0.9 0.1 1 0.1 www.bookspar.com | Website for students | VTU NOTES www.bookspar.com | Website for students | VTU NOTES Error = 0.84579 – 0.82636 = 0.01943 y2 Now taking h = 0.05, Euler’s method is y n1 y n 0.05 n 1 xn (1) 2 0.95 y1 y(0.05) 1.0 0.05 1 0 (0.95) 2 0.90702 y 2 y(0.1) 0.95 0.05 1 0 . 05 (0.90702) 2 0.86963 y3 y(0.15) 0.90702 0.05 1 0.1 (0.86963) 2 0.83675 y 4 y(0.2) 0.86963 0.05 1 0.15 Error = 0.84579 – 0.83675 = 0.00904 Note that when h = 0.1, Error was 0.01943, WHICH IS MORE. Now we use modified Euler’s method to find y(0.2) with h = 0.05 Euler’s Formula is y n1 y n2 y n 0.05 1 xn Modified Euler Formula is y n1 , n = 0,1,2 and 3 y2 yE y n 0.025 n n1 , n = 0,1,2 and 3 1 xn 1 xn1 Stage – I: Finding y1 = y(0.05) From Euler’s formula (for n = 0), (1) 2 0.95 y1E y(0.05) 1.0 0.05 1 0 From Modified Euler’s formula, we have (1) 2 (0.95) 2 0.95351 y1(1) y(0.05) 1.0 0.025 1 0 1 0.05 (1) 2 (0.95351) 2 0.95335 y1( 2) y(0.05) 1.0 0.025 1 0.05 1 0 www.bookspar.com | Website for students | VTU NOTES www.bookspar.com | Website for students | VTU NOTES (1) 2 (0.95335) 2 0.95336 y1(3) y(0.05) 1.0 0.025 1 0.05 1 0 (1) 2 (0.95336) 2 ( 4) 0.95336 y1 y(0.05) 1.0 0.025 1 0 1 0 . 05 Hence y1 y(0.05) 0.95336 Stage – II: Finding y2 = y(0.1) From Euler’s formula (for n = 1), we get (0.95336) 2 0.91008 y y(0.1) 0.95336 0.05 1 0 . 05 From Modified Euler’s formula, we have E 2 (0.95336) 2 (0.91008) 2 0.91286 y 2(1) y(0.1) 0.95336 0.025 1 0.1 1 0.05 (0.95336) 2 (0.91286) 2 0.91278 y 2( 2) y(0.1) 0.95336 0.025 1 0 . 05 1 0 . 1 (0.95336) 2 (0.91278) 2 0.91278 y 2( 2) y(0.1) 0.95336 0.025 1 0.1 1 0.05 Hence y2 y(0.1) 0.91278 Stage – III: Finding y3 = y(0.15) From Euler’s formula (for n = 2), we get (0.91278) 2 0.87491 y3E y(0.15) 0.91278 0.05 1 0 . 1 From Modified Euler’s formula (for n = 2), we have (0.91278) 2 (0.87491) 2 0.87720 y3(1) y(0.15) 0.91278 0.025 1 0.15 1 0.1 (0.91278) 2 (0.87720) 2 0.87712 y3( 2) y(0.15) 0.91278 0.025 1 0.15 1 0.1 (0.91278) 2 (0.87712) 2 0.87712 y3(3) y(0.15) 0.91278 0.025 1 0.15 1 0.1 www.bookspar.com | Website for students | VTU NOTES www.bookspar.com | Website for students | VTU NOTES Hence y3 y(0.15) 0.87712 Stage – IV: Finding y4 = y(0.2) From Euler’s formula (for n = 3), we get (0.87712) 2 0.84367 y 4E y(0.2) 0.87712 0.05 1 0.15 From Modified Euler’s formula(for n = 3), we have (0.87712) 2 (0.84367) 2 0.84557 y y(0.2) 0.87712 0.025 1 0.2 1 0.15 (0.87712) 2 (0.84557) 2 0.84550 y 4( 2) y(0.2) 0.87712 0.025 1 0.2 1 0.15 (1) 4 y ( 2) 4 (0.87712) 2 (0.84550) 2 0.84550 y(0.2) 0.87712 0.025 1 0.2 1 0.15 Hence y4 y(0.2) 0.84550 Error = 0.84579 – 0.84550 = 0.00029 Recall that the error from Euler’s Method is 0.00904 2. Solve the following IVP by Euler’s modified method at 0.2 x 0.8 with h = 0.2: dy log 10 ( x y ), y (0) 2 . dx Solution: Given Data is : x0 0, y0 2, h 0.2 and f ( x, y) log 10 ( x y) www.bookspar.com | Website for students | VTU NOTES www.bookspar.com | Website for students | VTU NOTES To Find: y1 y( x1 ) y(0.2), y 2 y( x2 ) y(0.4), y3 y( x3 ) y(0.6) & y4 y( x4 ) y(0.8) Stage – I: Finding y1 = y(0.2) From Euler’s formula (for n = 0), y1E y (0.2) 2.0 0.2 log 10 0 2 2.0602 Now from Modified Euler’s formula (for n = 0), we have y1(1) y (0.2) 2.0 0.1log 10 (0 2) log 10 (0.2 2.0602 2.0655 y1( 2) y (0.2) 2.0 0.1log 10 (0 2) log 10 (0.2 2.0655 2.0656 y1(3) y (0.2) 2.0 0.1log 10 (0 2) log 10 (0.2 2.0656 2.0656 Hence y1 y(0.2) 2.0656 Stage – II: Finding y2 = y(0.4) From Euler’s formula (for n = 1), y 2E y (0.4) 2.0656 0.2 log 10 0.2 2.0656 2.1366 Now from Modified Euler’s formula (for n = 1), we have y 2(1) y (0.4) 2.0656 0.1log 10 (0.2 2.0656) log 10 (0.4 2.1366 2.1415 y 2( 2) y (0.4) 2.0656 0.1log 10 (0.2 2.0656) log 10 (0.4 2.1415 2.1416 y 2(3) y (0.4) 2.0656 0.1log 10 (0.2 2.0656) log 10 (0.4 2.1416 2.1416 Hence y2 y(0.4) 2.1416 Stage – III: Finding y3 = y(0.6) From Euler’s formula (for n = 2), www.bookspar.com | Website for students | VTU NOTES www.bookspar.com | Website for students | VTU NOTES y3E y (0.6) 2.1416 0.2 log 10 0.4 2.1416 2.2226 Now from Modified Euler’s formula (for n = 2), we have y 3(1) y (0.6) 2.1416 0.1log 10 (0.4 2.1416) log 10 (0.6 2.2226 2.2272 y 3( 2) y (0.6) 2.1416 0.1log 10 (0.4 2.1416) log 10 (0.6 2.2272 2.2272 Hence y3 y(0.6) 2.2272 Stage – IV: Finding y4 = y(0.8) From Euler’s formula (for n = 3), y 3E y (0.8) 2.2272 0.2 log 10 0.6 2.2272 2.3175 Now from Modified Euler’s formula (for n = 3), we have y 4(1) y (0.8) 2.2272 0.1log 10 (0.6 2.2272) log 10 (0.8 2.3175 2.3217 y 4( 2) y (0.8) 2.2272 0.1log 10 (0.6 2.2272) log 10 (0.8 2.3217 2.3217 Hence y4 y(0.8) 2.3217 The solutions at 0.2 x 0.8 are tabulated as follows: xn yn 0.2 y1 y(0.2) 2.0656 0.4 y2 y(0.4) 2.1416 0.6 y3 y(0.6) 2.2272 0.8 y4 y(0.8) 2.3217 www.bookspar.com | Website for students | VTU NOTES www.bookspar.com | Website for students | VTU NOTES dy sin x cos y, y (2.5) 0 dx at x = 3.5 in two steps, modifying the solution thrice at each stages. Here x is in radians. 3. Using modified Euler’s method solve the IVP Solution: Given x0 2.5, y0 0, h 0.5 and f ( x, y) sin x cos y To Find: y1 y( x1 ) y(3.0) and y2 y( x2 ) y(3.5) Stage – I: Finding y1 = y(3.0) From Euler’s formula (for n = 0), y1E y(3.0) 0.0 0.5sin( 2.5) cos(0) 0.7992 Now from Modified Euler’s formula (for n = 0), we have y1(1) y(3.0) 0.0 0.25sin( 2.5) cos(0) sin( 3.0) cos(0.7992) 0.6092 y1( 2) y(3.0) 0.0 0.25sin( 2.5) cos(0) sin( 3.0) cos(0.6092) 0.6399 y1(3) y(3.0) 0.0 0.25sin( 2.5) cos(0) sin( 3.0) cos(0.6399) 0.6354 Hence y1 y(3.0) 0.6354 Stage – II: Finding y2 = y(3.5) From Euler’s formula (for n = 1), y 2E y(3.5) 0.6354 0.5sin( 3.0) cos(0.6354) 1.10837 Now from Modified Euler’s formula (for n = 1), we have y 2(1) y(3.5) 0.6354 0.25sin( 3.0) cos(0.6354) sin( 3.5) cos(1.10837) 0.89572 y 2( 2) y(3.5) 0.6354 0.25sin( 3.0) cos(0.6354) sin( 3.5) cos(0.89572) 0.94043 y 2(3) y(3.5) 0.6354 0.25sin( 3.0) cos(0.6354) sin( 3.5) cos(0.94043) 0.93155 Hence y2 y(3.5) 0.93155 www.bookspar.com | Website for students | VTU NOTES www.bookspar.com | Website for students | VTU NOTES 4. Using modified Euler’s method obtain the solution of the differential equation dy x y with the initial condition y = 1 at x = 0 for the range 0 < x 0.6 in dx steps of 0.2. Solution: Given x0 0, y 0 1, h 0.2 and f ( x, y) x To Find: y y1 y( x1 ) y(0.2), y 2 y( x2 ) y(0.4) and y3 y( x3 ) y(0.6) The Entire Calculations can be put in the following Tabular Form y n hf ( xn , y n ) Y2 f ( xn1 , y n1 ) 0.2 1 1.2 0.4 1.3094 1.4927 0.6 1.6350 1.8523 1.2954 1.3088 1.3094 1.6218 1.6345 1.6350 1.9610 1.9729 1.9734 x Y1 f ( xn , y n ) The solution is: y n 1 y n h Y1 Y2 2 1.2295 1.2309 1.2309 1.5240 1.5253 1.5253 1.8849 1.8861 1.8861 y(0.2) 1.2309, y(0.4) 1.5253 and y(0.6) 1.8861 Assignments : 1. Use Modified Euler’s method to find an approximate value of y at x = 0.3, in three dy x y , y ( 0) 1 . steps, of the IVP Answer : 1.1105,1.2432,1.4004 dx 2. Evaluate y at x=1.2 and x=1.4 by Euler’s modified method if y(x) satisfies dy y 1 , y (1) 2 . Compare the numerical solution with the analytical solution. dx x Answer : 2.6182, 3.2699 dy xy 2 , y 2 at x 0 using Euler’s modified method 3. Compute y(0.2) for the IVP dx in two steps. Answer : 1.9227 www.bookspar.com | Website for students | VTU NOTES www.bookspar.com | Website for students | VTU NOTES 4. Using modified Euler’s method, compute the solution of point x = 2 with h = 0.2. 5. Using modified Euler’s method, solve the IVP 3. dy 2 xy , y (1) 1 at the dx Answer : 5.051 dy x y 2 , y (0) 1 at the point x=0.5. dx Answer : 2.2352 Runge – Kutta Method The Taylor’s series method to solve IVPs is restricted by the difficulty in finding the higher order derivatives. However, Runge – Kutta method do not require the calculations of higher order derivatives. Euler’s method and modified Euler’s method are Runge – Kutta methods of first and second order respectively. dy f ( x, y ), y ( x0 ) y0 . Let us find the approximate value of y at dx x xn1 , n = 0,1,2,3,….. of this numerically, using Runge – Kutta method, as follows: Consider the IVP First let us calculate the quantities k1 , k 2 , k 3 and k 4 using the following formulae. k1 hf ( xn , y n ) k h k 2 hf x n , y n 1 2 2 k h k 3 hf x n , y n 2 2 2 k 4 hf xn h, y n k 3 Finally, the required solution y is given by y n 1 y n 1 k1 2k 2 2k 3 k 4 6 Problems: 1. Apply Runge – Kutta method, to find an approximate value of y when x = 0.2 given dy x y , y ( 0) 1 . that dx Solution: Given: x0 0, y0 1, h 0.2 and f ( x, y) x y www.bookspar.com | Website for students | VTU NOTES www.bookspar.com | Website for students | VTU NOTES R – K method (for n = 0) is: y1 y (0.2) y 0 1 k1 2k 2 2k 3 k 4 6 -------- (1) Now k1 hf ( x0 , y0 ) 0.2 [0 1] = 0.2 k h 0.2 0.2 = 0.2400 k 2 hf x0 , y 0 1 0.2 0 1 2 2 2 2 k h 0.2 0.24 k 3 hf x0 , y 0 2 0.2 0 1 = 0.2440 2 2 2 2 k 4 hf x0 h, y0 k 3 0.2 0 0.2 1 0.2440 = 0.2888 Using the values of k1 , k 2 , k 3 and k 4 in (1), we get 1 0.2 0.24 0.244 0.2888 = 1.2468 6 Hence the required approximate value of y is 1.2468. y1 y (0.2) 1 2. Using Runge – Kutta method of fourth order, solve dy y 2 x 2 , dx y 2 x 2 y (0) 1 at x = 0.2 & 0.4. Solution: Given: x0 0, y 0 1, h 0.2 and f ( x, y ) y2 x2 y2 x2 Stage – I: Finding y1 y(0.2) R – K method (for n = 0) is: y1 y (0.2) y 0 1 k1 2k 2 2k 3 k 4 6 k1 hf ( x0 , y0 ) 0.2 f (0,1) k h k 2 hf x0 , y 0 1 0.2 f 0.1,1.1 2 2 k h k 3 hf x0 , y 0 2 0.2 f 0.1,1.0936 2 2 www.bookspar.com | Website for students | VTU NOTES = 0.2 = 0.19672 = 0.1967 -------- (2) www.bookspar.com | Website for students | VTU NOTES k 4 hf x0 h, y0 k 3 0.2 f 0.2,1.1967 = 0.1891 Using the values of k1 , k 2 , k 3 and k 4 in (2), we get y1 y (0.2) 1 1 0.2 2(0.19672) 2(0.1967) 0.1891 6 = 1+0.19599 = 1.19599 Hence the required approximate value of y is 1.19599. Stage – II: Finding y 2 y(0.4) We have x1 0.1, y1 1.19599 and h 0.2 R – K method (for n = 1) is: y 2 y (0.4) y1 1 k1 2k 2 2k 3 k 4 6 k1 hf ( x1 , y1 ) 0.2 f (0.2, 1.19599) k h k 2 hf x1 , y1 1 0.2 f 0.3, 1.2906 2 2 -------- (3) = 0.1891 = 0.1795 k h k 3 hf x1 , y1 2 0.2 f 0.3, 1.2858 2 2 = 0.1793 k 4 hf x1 h, y1 k 3 0.2 f 0.4, 1.3753 = 0.1688 Using the values of k1 , k 2 , k 3 and k 4 in (3), we get 1 0.1891 2(0.1795) 2(0.1793) 0.1688 6 = 1.19599 + 0.1792 y 2 y (0.4) 1.19599 = 1.37519 Hence the required approximate value of y is 1.37519. 3. Apply Runge – Kutta method to find an approximate value of y when x = 0.2 with dy y 3 x , y (0) 1 . Also find the Analytical solution and h = 0.1 for the IVP dx 2 compare with the Numerical solution. www.bookspar.com | Website for students | VTU NOTES www.bookspar.com | Website for students | VTU NOTES Solution: Given: x0 0, y 0 1, h 0.1 and f ( x, y ) 3 x y 2 Stage – I: Finding y1 y(0.1) R – K method (for n = 0) is: y1 y (0.1) y 0 1 k1 2k 2 2k 3 k 4 6 k1 hf ( x0 , y0 ) 0.1 f (0,1) -------- (4) = 0.05 k h k 2 hf x0 , y 0 1 0.1 f 0.05, 1.025 2 2 k h k 3 hf x0 , y 0 1 0.1 f 0.05, 1.033125 2 2 k h k 4 hf x0 , y 0 1 0.1 f 0.1, 1.0666563 2 2 = 0.06625 = 0.0666563 = 0.0833328 Using the values of k1 , k 2 , k 3 and k 4 , we get 1 0.05 2(0.06625) 2(0.0666563) 0.0833328 6 = 1.0 + 0.0665242 y1 y (0.1) 1.0 = 1.0665242 Hence the required approximate value of y is 1.0665242. Stage – II: Finding y 2 y(0.2) We have x1 0.1, y1 1.0665242 and h 0.1 R – K method (for n = 1) is: y 2 y (0.2) y1 1 k1 2k 2 2k 3 k 4 6 k1 hf ( x1 , y1 ) 0.1 f (0.1, 1.0665242) k h k 2 hf x1 , y1 1 0.1 f 0.15, 1.04 2 2 = 0.0833262 = 0.1004094 k h k 3 hf x1 , y1 2 0.1 f 0.15, 1.0485 2 2 = 0.1008364 k 4 hf x1 h, y1 k 3 0.1 f 0.2, 1.097425 = 0.1183680 www.bookspar.com | Website for students | VTU NOTES -------- (5) www.bookspar.com | Website for students | VTU NOTES Using the values of k1 , k 2 , k 3 and k 4 in (5), we get 1 y 2 y (0.2) 1.0665242 0.0833262 2(0.1004094) 2(0.1008364) 0.1006976 6 = 1.0665242 + 0.1006976 = 1.1672218 Hence the required approximate value of y is 1.1672218. Exact Solution The given DE can be written as dy y 3 x which is a linear equation whose solution is: dx 2 x 2 y 6 x 12 13e . The Exact solution at x = 0.1 is y(0.1) = 1.0665242 and at x = 0.2 is y(0.2) = 1.1672218 Both the solutions and the error between them are tabulated as follows: x 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 xn yn (Exact) yn (Numerical) Absolute Error 0.1 1.0665243 1.0665242 0.0000001 0.2 1.1672219 1.1672218 0.0000001 y y (Exact) (Numerical) 1.0665243 1.0665242 1.1672219 1.1672218 1.3038452 1.3038450 1.4782359 1.4782357 1.6923304 1.6923302 1.9481645 1.9481643 2.2478782 2.2478779 2.5937211 2.5937207 2.9880585 2.9880580 3.4333766 3.4333761 www.bookspar.com | Website for students | VTU NOTES www.bookspar.com | Website for students | VTU NOTES Assignments: 1. Apply Runge – Kutta method to compute an approximate value of y (0.2) and y(0.4), by dy x 2 y 2 , y ( 0) 1 . taking h = 0.1 for the IVP 10 Answer: 1.0207, 1.0438 dx 2.Use Runge – Kutta method to find y when x = 0.2 in steps of 0.1 for the differential dy y x equation Answer: 1.0911278, 1.1678430 , y (0) 1 dx y x dy ( y x) , y (0.4) 0.41 by Runge – Kutta method at x = 0.8 in two steps. 3. Solve dx Answer: 0.6103476, 0.8489914 4.Using Runge – Kutta method, find an approximate value of y for x = 0.2 in steps of 0.1 dy x y 2 , given that y = 1 when x = 0, if Answer: 1.1165, 1.2736 dx Multi-step Methods: To solve a differential equation over an interval (xn, xn+1), using previous single-step methods, only the values of y at the beginning of interval is required. However, in the following methods, four prior values are needed for finding the value of yn+1 at a given value of x. Also the solution at yn+1 is obtained in two stages. This method of refining an initially crude estimate of yn+1 by means of a more accurate formula is known as Predictor–Corrector method. A Predictor Formula is used to predict the value of yn+1 and then a Corrector Formula is applied to calculate a still better approximation of yn+1. Now we study two such methods namely (i) Milne’s method and (ii) Adams – Bashforth method. (I) Milne’s Method: dy f ( x, y ), y ( x0 ) y 0 . To find an approximate value of y at x=x0+nh dx by Milne’s method, we proceed as follows: Using the given value of y( x0 ) y 0 , we Given compute y( x1 ) y( x0 h) y1 , y( x2 ) y( x0 2h) y 2 and using Taylor’s series method. y( x3 ) y( x0 3h) y3 Next, we calculate f1 f ( x1 , y1 ), f 2 f ( x2 , y 2 ) and f 3 f ( x3 , y3 ) . Then, the value of y at x = x4 = x0+4h can be found in the following two stages. I Stage : Predictor Method y 4( P ) y 0 4h 2 f1 f 2 2 f 3 3 www.bookspar.com | Website for students | VTU NOTES www.bookspar.com | Website for students | VTU NOTES Then we compute f 4 f ( x4 , y 4( P ) ) II Stage : Corrector Method y 4(C ) y 2 h f2 4 f3 f 4 3 Then, an improved value of f4 is computed and again, corrector formula is applied to find a better value of y4. We repeat the step until y4 remains unchanged. (II) Adams – Bashforth Method: dy f ( x, y ), y ( x0 ) y 0 . Using the given value of y( x0 ) y 0 , we first compute dx y( x1 ) y( x0 h) y 1 , y( x2 ) y( x0 2h) y 2 and y( x3 ) y( x0 3h) y 3 using Taylor’s series method. Given Next, we calculate f 1 f ( x1 , y 1 ), f 2 f ( x2 , y 2 ) and value of y at x = x1 (or y1) can be determined in two stages: f 3 f ( x3 , y 3 ) . Now, the I Stage : Predictor Method y1( P ) y 0 h 55 f 0 59 f 1 37 f 2 9 f 3 24 Next, we compute f1 f ( x1 , y1( P ) ) . To find a better approximation to y1, the following corrector formula is used. II Stage : Corrector Method y1(C ) y 0 h 9 f1 19 f 0 5 f 1 f 2 24 Then, an improved value of f1 f ( x1 , y1(C ) ) is calculated and again, corrector formula is applied to find a better value of y1. This step is repeated until y1 remains unchanged and then proceeds to calculate y2 as above. Problems: 1. Use Milne’s method to find y(0.3) for the IVP dy x2 y2 , dx Solution: www.bookspar.com | Website for students | VTU NOTES y ( 0) 1 www.bookspar.com | Website for students | VTU NOTES First, let us find the values of y at the points x = -0.1, x = 0.1 and x = 0.2 by using Taylor’s series method for the given IVP. Taylor’s expansion of y(x) about the point x = 0( = x0) is y ( x) y (0) xy / (0) x 2 // x 3 /// y (0) y (0) 2 6 -------- (1) Given y / ( x) x 2 y 2 y / (0) 0 1 1 y // ( x) 2 x 2 yy / y // (0) 2 1 1 2 y /// ( x) 2 2 yy // 2( y / ) 2 y /// (0) 2 4 2 8 Using the values of y(0), y/(0), y//(0) and y///(0) in (1), we get y ( x) 1 x x 2 4x 3 3 Putting x = -0.1, x = 0.1 and x = 0.2 in the above expression, we get y(-0.1) = 0.9087, y(0.1) = 1.1113 and y(0.2) = 1.2507 Given: x0 0.1, y0 0.9087 and f 0 0.8357 x1 0, y1 1 and f1 1 x2 0.1, y2 1.1113 and f 2 1.2449 x3 0.2, y3 1.2507 and f 3 1.6043 To Find : y4 y( x4 ) y(0.3) I Stage : Predictor Method 4h 2 f1 f 2 2 f 3 3 4(0.1) (2 1) 1.2449 2 1.6043 0.9087 3 = 1.4372 Now we compute f 4 f (0.3,1.4372) 2.1555 y 4( P ) y (0.3) y 0 www.bookspar.com | Website for students | VTU NOTES www.bookspar.com | Website for students | VTU NOTES II Stage : Corrector Method h f2 4 f3 f4 3 0.1 1.2449 (4 1.6043) 2.1555 y 4(C ) y (0.3) 1.1113 3 = 1.4386 Now, we compute f 4 f (0.3,1.4386) 2.1596 y 4(C ) y (0.3) y 2 y 4(C ,1) y (0.3) 1.1113 0.1 1.2449 (4 1.6043) 2.1596 3 y(0.3) = 1.43869 Hence, the approximate solution is y(0.3) = 1.43869 dy x y 2 , y (0) 0 , y(0.2) = 0.02, y(0.4) = 0.0795 and y(0.6) = 0.1762. dx Compute y(1) using Milne’s Method. 2. Given Solution: Stage - I : Finding y(0.8) Given: x0 0, y0 0 and f 0 f ( x0 , y0 ) 0 x1 0.2, y1 0.02 and f1 f ( x1 , y1 ) 0.1996 x2 0.4, y2 0.0795 and f 2 f ( x2 , y2 ) 0.3937 x3 0.6, y3 0.1762 and f 3 f ( x3 , y3 ) 0.56895 To Find : y4 y( x4 ) y(0.8) I Stage : Predictor Method y 4( P ) y (0.8) y 0 4h 2 f1 f 2 2 f 3 3 4(0.2) (2 0.1996) 0.3937 2 2 0.56895 3 = 0.30491 0 www.bookspar.com | Website for students | VTU NOTES www.bookspar.com | Website for students | VTU NOTES Now we compute f 4 f (0.8,0.30491) 0.7070 II Stage : Corrector Method h f2 4 f3 f 4 3 0.2 0.3937 4 0.56895 0.7070 y (0.8) 0.0795 3 = 0.3046 y 4(C ) y (0.8) y 2 y 4(C ) Now f 4 f (0.8,0.3046) 0.7072 Again applying corrector formula with new f4, we get y 4(C ,1) y (0.8) 0.0795 0.2 0.3937 4 0.56895 0.7072 3 y(0.8) = 0.3046 Stage - II : Finding y(1.0) Given: x1 0.2, y1 0.02 and f1 f ( x1 , y1 ) 0.1996 x2 0.4, y2 0.0795 and f 2 f ( x2 , y2 ) 0.3937 x3 0.6, y3 0.1762 and f 3 f ( x3 , y3 ) 0.56895 x4 0.8, y4 0.3046 and f 4 f ( x4 , y4 ) 0.7072 To Find : y5 y( x5 ) y(1.0) I Stage : Predictor Method y 5( P ) y (1.0) y1 4h 2 f 2 f 3 2 f 4 3 4(0.2) (2 0.3937) 0.56895 2 0.7072 3 = 0.45544 Now we compute f 5 f (1.0,0.45544) 0.7926 0.02 II Stage : Corrector Method y 5(C ) y (1.0) y 3 h f3 4 f4 f5 3 www.bookspar.com | Website for students | VTU NOTES www.bookspar.com | Website for students | VTU NOTES y 5(C ) y (1.0) 0.56895 0.2 0.56895 4 0.7072 0.7926 3 = 0.4556 Now f 5 f (1.0,0.4556) 0.7024 Again applying corrector formula with new f5, we get 0.2 0.56895 4 0.7072 0.7924 y 5(C ,1) y (1.0) 0.56895 3 y(1.0) = 0.4556 dy x 2 1 y , y (1) 1, y (1.1) 1.233, y (1.2) 1.548, y (1.3) 1.979 .Evaluate dx y(1.4) by Adam’s – Bashforth method. 3. Given Solution: Given: x3 1, y 3 1 and f 3 2 x2 1.1, y2 1.233 and f 2 2.70193 x1 1.2, y1 1.548 and f 1 3.66912 x0 1.3, y0 1.979 and f 0 5.03451 To Find : y1 y( x1 ) y(1.4) I Stage : Predictor Method y1( P ) y (1.4) y 0 1.979 h 55 f 0 59 f 1 37 f 2 9 f 3 24 (0.1) (55 5.03451) (59 3.66912) (37 2.70193) (9 2) 24 = 2.57229 Now we compute f1 f (1.4,2.57229) 7.0017 II Stage : Corrector Method www.bookspar.com | Website for students | VTU NOTES www.bookspar.com | Website for students | VTU NOTES y1(C ) y (1.4) y 0 h 9 f1 19 f 0 5 f 1 f 2 24 0.1 (9 7.0017) (19 5.03451) (5 3.66912) 2.70193 24 y(1.4) = 2.57495 1.979 Now, let us compute f1 f (1.4,2.57495) 7.0069 0.1 (9 7.0069) (19 5.03451) (5 3.66912) 2.70193 24 = 2.57514 Again f1 f (1.4, 2.57514) 7.0073 0.1 (9 7.0073) (19 5.03451) (5 3.66912) 2.70193 y1(C , 2) 1.979 24 y1( C ,1) 1.979 y(1.4) = 2.57514 4. Given dy x 2 y, dx y (0) 1 . Find y(0.4) by Adam’s method. Solution: First, let us find the values of y at the points x = 0.1, x = 0.2 and x = 0.3 by using Taylor’s series method for the given IVP. Taylor’s expansion of y(x) about the point x = 0( = x0) is y ( x) y (0) xy / (0) x 2 // x 3 /// y (0) y (0) 2 6 -------- (2) Given y / ( x) x 2 y y / (0) 0 1 1 y // ( x) 2 x y / y // (0) 0 (1) 1 y /// ( x) 2 2 y // y /// (0) 2 1 1 Using the values of y(0), y/(0), y//(0) and y///(0) in (2), we get y ( x) 1 x x2 x3 2 3 www.bookspar.com | Website for students | VTU NOTES www.bookspar.com | Website for students | VTU NOTES Putting x = 0.1, x = 0.2 and x = 0.3 in the above expression, we get y(0.1) = 0.9051, y(0.2) = 0.8212 and y(0.3) = 0.7492 Let x3 0, y 3 1 and f 3 1 x2 0.1, y2 0.9051 and f 2 0.8951 x1 0.2, y1 0.8212 and f 1 0.7812 x0 0.3, y0 0.7492 and f 0 0.6592 y1 y( x1 ) y(0.4) To Find : I Stage : Predictor Method h y1( P ) y (0.4) y 0 55 f 0 59 f 1 37 f 2 9 f 3 24 (0.1) 55 (0.6592) 59 (0.7812) 37 (0.8951) 9 (1) 0.7492 24 = 0.6896507 Now we compute f1 f (0.4,0.6896507) 0.5296507 II Stage : Corrector Method y1(C ) y (0.4) y 0 0.7492 h 9 f1 19 f 0 5 f 1 f 2 24 0.1 9 (0.5297) 19 (0.6592) 5 (0.7812) 0.895125 24 y(0.4) = 0.6896522 Assignments: 1. Using Adam’s method find y(1.4) for the IVP x2y/ + xy = 1; y(1) = 1, y(1.1) = 0.996, y(1.2) = 0.986, y(1.3) = 0.972. dy 0.5 xy ; y (0) 1 . Find y(0.4) using Adam’s – Bashforth predictor – 2. Given dx corrector formulae. 3. Solve by Milne’s predictor – corrector method, the differential equation dy y x 2 with the following starting values : y(0) = 1, y(0.2) = 1.12186, dx y(0.4) = 1.4682, y(0.6) = 1.7379 to find the value of y when x = 0.8. 4. Using Milne’s method, obtain the solution of the equation 1 y / 1 x 2 y 2 ; y (0) 1 at x = 0.4. 2 Answers : (1) 0.94934, (2) 1.0408, (3) 2.0111 & (4) 1.2797 www.bookspar.com | Website for students | VTU NOTES www.bookspar.com | Website for students | VTU NOTES LECTURE NOTES OF ENGINEERING MATHEMATICS–IV (Sub Code: 06 MAT41) Text Book: Higher Engineering Mathematics by Dr. B.S.Grewal (36th Edition – 2002) Khanna Publishers,New Delhi Reference Book: Advanced Engineering Mathematics by E. Kreyszig (8th Edition – 2001) John Wiley & Sons, INC. New York SPECIAL FUNCTIONS Prepared by Dr. M. SANKAR Professor and Head Department of Mathematics Sapthagiri College of Engineering Bangalore – 560 057 Introduction Many Differential equations arising from physical problems are linear but have variable coefficients and do not permit a general analytical solution in terms of known functions. Such equations can be solved by numerical methods (Unit – I), but in many cases it is easier to find a solution in the form of an infinite convergent series. The series solution of certain differential equations give rise to special functions such as Bessel’s function, Legendre’s polynomial. These special functions have many applications in engineering. Series solution of the Bessel Differential Equation Consider the Bessel Differential equation of order n in the form d2y dy x ( x2 n2 ) y 0 2 dx dx where n is a non negative real constant or parameter. x2 (i) We assume the series solution of (i) in the form y a r x k r where a0 0 r 0 www.bookspar.com | Website for students | VTU NOTES (ii) www.bookspar.com | Website for students | VTU NOTES dy a r (k r ) x k r 1 dx r 0 Hence, d2y ar (k r )(k r 1) x k r 2 2 dx r 0 Substituting these in (i) we get, x 2 a r (k r )( k r 1) x k r 2 x a r (k r ) x k r 1 x 2 n 2 a r x k r 0 r 0 r 0 r 0 r 0 r 0 r 0 r 0 i.e., a r (k r )( k r 1) x k r a r (k r ) x k r a r x k r 2 n 2 a r x k r 0 Grouping the like powers, we get ar (k r )(k r 1) (k r ) n 2 x k r ar x k r 2 0 r 0 r 0 a r (k r ) 2 n 2 x k r a r x k r 2 0 r 0 (iii) r 0 Now we shall equate the coefficient of various powers of x to zero Equating the coefficient of xk from the first term and equating it to zero, we get a 0 k 2 n 2 0. Since a 0 0, we get k 2 n 2 0, k n Coefficient of xk+1 is got by putting r = 1 in the first term and equating it to zero, we get i.e., a1 (k 1) 2 n 2 0. This gives a1 0, since (k 1) 2 n 2 0 gives , k 1 n which is a contradiction to k = n. Let us consider the coefficient of xk+r from (iii) and equate it to zero. i.e, ar (k r ) 2 n 2 ar 2 0. ar 2 (k r ) 2 n 2 If k = +n, (iv) becomes ar 2 a ar 2 r 2 2 2 (n r ) n r 2nr ar Now putting r = 1,3,5, ….., (odd vales of n) we obtain, www.bookspar.com | Website for students | VTU NOTES (iv) www.bookspar.com | Website for students | VTU NOTES a3 a1 0 , a1 0 6n 9 Similarly a5, a7, ….. are equal to zero. i.e., a1 = a5 = a7 = …… = 0 Now, putting r = 2,4,6, ……( even values of n) we get, a0 a0 a0 a2 a2 ; a4 ; 4n 4 4(n 1) 8n 16 32(n 1)( n 2) Similarly we can obtain a6, a8, … We shall substitute the values of a1 , a2 , a3 , a4 , in the assumed series solution, we get y a r x k r x k ( a0 a1 x a 2 x 2 a 3 x 3 a 4 x 4 ) r 0 Let y1 be the solution for k = +n a0 a0 y1 x n a0 x2 x 4 4(n 1) 32(n 1)( n 2) 2 4 x x i.e., y1 a0 x n 1 2 5 2 (n 1) 2 (n 1)( n 2) This is a solution of the Bessel’s equation. (v) Let y2 be the solution corresponding to k = - n. Replacing n be – n in (v) we get x2 x4 (vi) y 2 a0 x n 1 2 5 2 (n 1) 2 (n 1)( n 2) The complete or general solution of the Bessel’s differential equation is y = c1y1 + c2y2, where c1, c2 are arbitrary constants. Now we will proceed to find the solution in terms of Bessel’s function by choosing 1 and let us denote it as Y1. a0 n 2 (n 1) 4 x 2 1 1 x i.e., Y1 n 1 2 (n 1) 2 (n 1) 2 (n 1)( n 2) 2 xn www.bookspar.com | Website for students | VTU NOTES www.bookspar.com | Website for students | VTU NOTES 2 4 1 x 1 x 1 (n 1) 2 (n 1) (n 1) 2 (n 1)( n 2) (n 1) 2 We have the result (n) = (n – 1) (n – 1) from Gamma function x 2 n Hence, (n + 2) = (n + 1) (n + 1) and (n + 3) = (n + 2) (n + 2) = (n + 2) (n + 1) (n + 1) Using the above results in Y1, we get 2 4 1 1 1 x x (n 1) 2 (n 2) 2 (n 3) 2 which can be further put in the following form n 0 2 4 0 (1)1 x (1) 2 x x (1) x Y1 (n 2) 1! 2 (n 3) 2! 2 2 (n 1) 0! 2 x Y1 2 x 2 n n r 0 (1) r x (n r 1) r! 2 x (1) 2 r 0 r n2r 2r 1 (n r 1) r! This function is called the Bessel function of the first kind of order n and is denoted by Jn(x). x Thus J n ( x) (1) r 2 r 0 n2r 1 (n r 1) r! Further the particular solution for k = -n ( replacing n by –n ) be denoted as J-n(x). Hence the general solution of the Bessel’s equation is given by y = AJn(x) + BJ-n(x), where A and B are arbitrary constants. Properties of Bessel’s function 1. J n ( x ) ( 1 ) n J n ( x ) , where n is a positive integer. Proof: By definition of Bessel’s function, we have x J n ( x) (1) 2 r 0 n2r r x Hence, J n ( x ) ( 1 ) r r 0 2 n2r 1 (n r 1) r! 1 ( n r 1 ) r! www.bookspar.com | Website for students | VTU NOTES ……….(1) ……….(2) www.bookspar.com | Website for students | VTU NOTES But gamma function is defined only for a positive real number. Thus we write (2) in the following from x J n ( x ) ( 1 ) r r n 2 n2r 1 ………..(3) ( n r 1 ) r! Let r – n = s or r = s + n. Then (3) becomes n2s2n x J n ( x ) ( 1 ) s n s 0 2 1 ( s 1 ) ( s n )! We know that (s+1) = s! and (s + n)! = (s+n+1) n2 s x ( 1 ) s n s 0 2 1 x ( 1 ) n ( 1 ) s s 0 2 ( s n 1 ) s! n2s 1 ( s n 1 ) s! Comparing the above summation with (1), we note that the RHS is Jn(x). Thus, J n (x) ( 1)n J n (x) 2. J n ( x ) ( 1 ) n J n ( x ) J n ( x ) , where n is a positive integer x Proof : By definition, J n ( x) (1) 2 r 0 n2r r x J n ( x ) ( 1 ) r r 0 2 i.e., n2r Thus, (n r 1) r! 1 ( n r 1 ) r! x ( 1 ) r 1n 2 r r 0 2 x 1n ( 1 ) r r 0 2 1 n2r 1 ( n r 1 ) r! n2r 1 ( n r 1 ) r! J n (-x) ( 1) n J n (x) Since, ( 1 ) n J n ( x ) J n ( x ) , we have J n ( x) ( 1)n J n (x) J n (x) Recurrence Relations: Recurrence Relations are relations between Bessel’s functions of different order. Recurrence Relations 1: d n x J n ( x ) x n J n 1 ( x ) dx From definition, x x n J n ( x ) x n ( 1 ) r r 0 2 n2r x ( 1 ) r r 0 2 ( n r 1 ) r! 1 www.bookspar.com | Website for students | VTU NOTES 2( n r ) 1 ( n r 1 ) r! www.bookspar.com | Website for students | VTU NOTES 2( n r )x 2( n r )1 d n x J n ( x ) ( 1 ) r dx r 0 2 n 2 r ( n r 1 ) r! ( n r )x n 2 r 1 x ( 1 ) n r 2 n 2 r 1 ( n r ) ( n r ) r! r 0 x n ( 1 ) r r 0 ( x / 2 ) n 1 2 r ( n 1 r 1 ) r! x n J n 1 ( x ) d n x J n ( x ) x n J n 1 ( x ) dx d n x J n ( x ) x n J n 1 ( x ) Recurrence Relations 2: dx Thus, --------(1) From definition, x x n J n ( x ) x n ( 1 ) r r 0 2 x ( 1 ) r r 0 2 n2r 2r 1 ( n r 1 ) r! 1 ( n r 1 ) r! 2 r x 2 r 1 d n x J n ( x ) ( 1 ) r dx r 0 2 n 2 r ( n r 1 ) r! x n 1 2( r 1 ) r 1 2 n 1 2( r 1 ) ( n r 1 ) ( r 1 )! x n ( 1 ) r 1 Let k = r – 1 x n 1 2 k k 0 2 n 1 2 k ( n 1 k 1 ) k ! x n ( 1 ) k d n x J n ( x ) x n J n 1 ( x ) dx x Recurrence Relations 3: J n ( x ) J n 1 ( x ) J n 1 ( x ) 2n d n x J n ( x ) x n J n 1 ( x ) We know that dx Thus, x n J n 1 ( x ) --------(2) Applying product rule on LHS, we get x n J n/ ( x ) nx n1 J n ( x ) x n J n1 ( x ) Dividing by xn we get J n/ ( x ) ( n / x )J n ( x ) J n1 ( x ) --------(3) Also differentiating LHS of d n x J n ( x ) x n J n 1 ( x ) , dx we get x n J n/ ( x ) nx n1 J n ( x ) x n J n1 ( x ) Dividing by –x–n we get J n/ ( x ) ( n / x )J n ( x ) J n1 ( x ) --------(4) Adding (3) and (4), we obtain 2nJ n ( x ) xJ n1 ( x ) J n1 ( x ) x J n 1 ( x ) J n 1 ( x ) 2n 1 Recurrence Relations 4: J n/ ( x ) J n 1 ( x ) J n1 ( x ) 2 i.e., Jn( x ) www.bookspar.com | Website for students | VTU NOTES www.bookspar.com | Website for students | VTU NOTES Subtracting (4) from (3), we obtain 2 J n/ ( x ) J n1 ( x ) J n1 ( x ) i.e., J n/ ( x ) 1 J n1 ( x ) J n1 ( x ) 2 n x Recurrence Relations 5: J n/ ( x ) J n ( x ) J n 1 ( x ) This recurrence relation is another way of writing the Recurrence relation 2. n x Recurrence Relations 6: J n/ ( x ) J n 1 ( x ) J n ( x ) This recurrence relation is another way of writing the Recurrence relation 1. Recurrence Relations 7: J n 1 ( x ) 2n J n ( x ) J n 1 ( x ) x This recurrence relation is another way of writing the Recurrence relation 3. Problems: 2 Prove that ( a ) J 1 / 2 ( x ) sin x ( b ) x J 1 / 2 ( x ) 2 x cos x By definition, x J n ( x ) ( 1 ) r r 0 2 n2r 1 ( n r 1 ) r! Putting n = ½, we get x J 1 / 2 ( x ) ( 1 ) r r 0 2 J1/ 2( x ) 1 / 22r 1 ( r 3 / 2 ) r! 2 4 x 1 1 1 x x 2 ( 3 / 2 ) 2 ( 5 / 2 )1! 2 ( 7 / 2 )2! --------(1) Using the results (1/2) = and (n) = (n – 1) (n–1), we get (3 / 2) 2 , ( 5 / 2 ) 3 15 and so on. , (7 / 2 ) 4 8 Using these values in (1), we get J1/ 2( x ) J1/ 2( x ) x 2 x2 4 x 4 8 2 4 3 16 15 .2 x 2 x3 x 5 x 2 x 6 120 2 x x3 x5 x 5! 3! 2 sin x x Putting n = - 1/2, we get x J 1 / 2 ( x ) ( 1 ) r r 0 2 J 1 / 2 ( x ) 1 / 2 2 r 1 ( r 1 / 2 ) r! 2 4 x 1 1 1 x x 2 ( 1 / 2 ) 2 ( 3 / 2 )1! 2 ( 5 / 2 )2! --------(2) Using the results (1/2) = and (n) = (n – 1) (n–1) in (2), we get www.bookspar.com | Website for students | VTU NOTES www.bookspar.com | Website for students | VTU NOTES J 1 / 2 ( x ) 2 1 x2 2 x 4 4 x 4 16 3 .2 2 x J 1 / 2 ( x ) x2 x4 1 4! 2! 2 cos x x 2. Prove the following results : 2 3 x2 3 2 sin x cos x and x x x ( a ) J5/ 2( x ) 2 3 x2 3 2 cos x sin x x x x ( b ) J 5 / 2 ( x ) Solution : We prove this result using the recurrence relation J n ( x ) x J n 1 ( x ) J n 1 ( x ) 2n 3 x Putting n = 3/2 in (1), we get J 1 / 2 ( x ) J 5 / 2 ( x ) J 3 / 2 ( x ) J5/ 2( x ) 3 J3 / 2( x ) J1/ 2( x ) x i.e., J 5 / 2 ( x ) J5/ 2( x ) 3 2 sin x x cos x 2 x sin x x x x 2 3 sin x 3 x cos x x 2 sin x 2 ( 3 x 2 ) 3 sin x cos x 2 2 x x x x x Also putting n = - 3/2 in (1), we get J 5 / 2 ( x ) J 1 / 2 ( x ) 3 J 3 / 2 ( x ) x 3 2 x sin x cos x 2 3 J 3 / 2 ( x ) J 1 / 2 ( x ) cos x x x x x x 2 3 x sin x 3 cos x x 2 cos x 2 3 3 x2 i.e., J 5 / 2 ( x ) sin x cos x 2 2 x x x x x J 5 / 2 ( x ) 3. Show that Solution: L.H.S = d 2 2 J n ( x ) J n21 ( x ) nJ n2 ( x ) ( n 1 )J n21 ( x ) dx x d 2 J n ( x ) J n21 ( x ) 2 J n ( x )J n/ ( x ) 2 J n 1 ( x )J n/ 1 ( x ) ------dx (1) We know the recurrence relations xJ n/ ( x ) nJ n ( x ) xJ n1 ( x ) xJ n/ 1 ( x ) xJ n ( x ) ( n 1 )J n1 ( x ) ------- (2) ------- (3) Relation (3) is obtained by replacing n by n+1 in xJ n/ ( x ) xJ n1 ( x ) nJ n ( x ) Now using (2) and (3) in (1), we get www.bookspar.com | Website for students | VTU NOTES ------ (1). www.bookspar.com | Website for students | VTU NOTES d 2 n1 n J n ( x ) J n21 ( x ) 2 J n ( x ) J n ( x ) J n 1 ( x ) 2 J n 1 ( x ) J n ( x ) J n 1 ( x ) dx x x L.H.S = 2n 2 n1 2 J n ( x ) 2 J n ( x ) J n 1 ( x ) 2 J n 1 ( x ) J n ( x ) 2 J n 1 ( x ) x x d 2 2 J n ( x ) J n21 ( x ) nJ n2 ( x ) ( n 1 )J n21 ( x ) Hence, dx x 1 4. Prove that J 0// ( x ) J 2 ( x ) J 0 ( x ) 2 Solution : We have the recurrence relation J n/ ( x ) J n 1 ( x ) J n1 ( x ) -------(1) 1 2 Putting n = 0 in (1), we get J 0/ ( x ) J 1 ( x ) J 1 ( x ) J 1 ( x ) J 1 ( x ) J 1 ( x ) 1 2 1 2 Thus, J 0/ ( x ) J 1 ( x ) . Differentiating this w.r.t. x we get, J 0// ( x ) J 1/ ( x ) ----- (2) Now, from (1), for n = 1, we get J 1/ ( x ) J 0 ( x ) J 2 ( x ) . 1 2 Using (2), the above equation becomes 1 J 0 ( x ) J 2 ( x )orJ 0// ( x ) 1 J 2 ( x ) J 0 ( x ) . 2 2 1 // Thus we have proved that, J 0 ( x ) J 2 ( x ) J 0 ( x ) 2 J 0// ( x ) 2 x 5. Show that (a) J 3 ( x )dx c J 2 ( x ) J 1 ( x ) (b) xJ 02 ( x )dx x 2 J 02 ( x ) J 12 ( x ) 1 2 Solution : (a) We know that d n x J n ( x ) x n J n 1 ( x ) dx or x n J n1 ( x )dx x n J n ( x ) ------ (1) Now, J 3 ( x )dx x 2 x 2 J 3 ( x )dx c x 2 x 2 J 3 ( x )dx 2 x x 2 J 3 ( x )dx dx c x 2 x 2 J 2 ( x ) 2 x x 2 J 2 ( x ) dx c ( from (1) when n = 2) 2 2 c J 2 ( x ) J 2 ( x )dx c J 2 ( x ) J 1 ( x ) ( from (1) when n = 1) x x 2 Hence, J 3 ( x )dx c J 2 ( x ) J 1 ( x ) x 1 1 (b) xJ 02 ( x )dx J 02 ( x ) x 2 2 J 0 ( x ) J 0/ ( x ). x 2 dx (Integrate by parts) 2 2 1 2 2 2 x J 0 ( x ) x J 0 ( x ) J 1 ( x )dx (From (1) for n = 0) 2 d 1 d xJ 1 ( x ) xJ 0 ( x ) from recurrence relation (1) x 2 J 02 ( x ) xJ 1 ( x ) xJ 1 ( x )dx dx 2 dx 1 2 2 1 1 x J 0 ( x ) xJ 1 ( x )2 x 2 J 02 ( x ) J 12 ( x ) 2 2 2 www.bookspar.com | Website for students | VTU NOTES www.bookspar.com | Website for students | VTU NOTES Generating Function for Jn(x) x To prove that e 2 ( t 1 / t ) tnJn( x ) n or x If n is an integer then Jn(x) is the coefficient of tn in the expansion of e 2 ( t 1 / t ) . Proof: x We have e 2 ( t 1 / t ) e xt / 2 e x / 2t ( xt / 2 ) ( xt / 2 ) 2 ( xt / 2 ) 3 ( xt / 2 ) ( xt / 2 ) 2 ( xt / 2 ) 3 1 1 1! 2! 3! 1! 2! 3! (using the expansion of exponential function) ( 1 ) n x n ( 1 ) n 1 x n 1 xt x 2t 2 x nt n x n 1 t n 1 x x2 1 2 n n 1 1 2 2 n n n 1 n 1 2 n! 2 ( n 1 )! 2 t n! 2 t ( n 1 )! 2 1! 2 2! 2t 1! 2 t 2! If we collect the coefficient of tn in the product, they are xn 2n n! xn 2 2n 2 ( n 1 )!1! n 1 x 1 x n! 2 ( n 1 )!1! 2 xn4 2n 4 ( n 2 )! 2! n2 1 x ( n 2 )!2! 2 n4 x ( 1 )r r 0 2 n 2r 1 ( n r 1 )r! Jn( x ) Similarly, if we collect the coefficients of t–n in the product, we get J–n(x). Thus, x ( t 1 / t ) 2 e Result: tnJn( x ) n x ( t 1 / t ) 2 e J 0 ( x ) t n ( 1 ) n t n J n ( x ) n 1 Proof : x e2 ( t 1 / t ) 1 n n t nJn( x ) t nJn( x ) tnJn( x ) n 0 n 1 n 1 n 1 t n J n ( x ) J 0 ( x ) t n J n ( x ) J 0 ( x ) t n ( 1 ) n J n ( x ) t n J n ( x ) { J n ( x ) ( 1 ) n J n ( x )} n 1 Thus, x ( t 1 / t ) e2 J 0 ( x ) t n ( 1 ) n t n J n ( x ) n 1 Problem 6: Show that (a) J n ( x ) 1 cos( n x sin )d , n being an integer 0 1 (b) J 0 ( x ) cos( x cos )d 0 (c) 2 J 22 J 32 1 Solution : J 02 2 J 12 www.bookspar.com | Website for students | VTU NOTES www.bookspar.com | Website for students | VTU NOTES x We know that e 2 ( t 1 / t ) J 0 ( x ) t n ( 1 ) n t n J n ( x ) n 1 J 0 ( x ) tJ 1 ( x ) t 2 J 2 ( x ) t 3 J 3 ( x ) t 1 J 1 ( x ) t 2 J 2 ( x ) t 3 J 3 ( x ) Since J n ( x ) ( 1 )n J n ( x ) , we have x ( t 1 / t ) 2 e J 0 ( x ) J 1 ( x )t 1 / t J 2 ( x ) t 2 1 / t 2 J 3 ( x ) t 3 1 / t 3 p ----- (1) p Let t = cos + i sin so that t = cosp + i sinp and 1/t = cosp - i sinp. From this we get, tp + 1/tp = 2cosp and tp – 1/tp = 2i sinp Using these results in (1), we get x e2 ( 2i sin ) e ix sin J 0 ( x ) 2J 2 ( x ) cos 2 J 4 ( x ) cos 4 2iJ 1 ( x ) sin J 3 ( x ) sin 3 -----(2) Since e = cos(xsin) + i sin(xsin), equating real and imaginary parts in (2) we get, cos( x sin ) J 0 ( x ) 2J 2 ( x ) cos 2 J 4 ( x ) cos 4 ----- (3) ----- (4) sin( x sin ) 2J 1 ( x ) sin J 3 ( x ) sin 3 These series are known as Jacobi Series. ixsin Now multiplying both sides of (3) by cos n and both sides of (4) by sin n and integrating each of the resulting expression between 0 and , we obtain J n ( x ), n is even or zero 1 cos( x sin ) cos nd 0 , n is odd 0 and n is even 0, 1 sin( x sin ) sin nd 0 J n ( x ), n is odd Here we used the standard result cos p cos qd sin p sin qd 2 , if p q 0 0 , if p q 0 From the above two expression, in general, if n is a positive integer, we get Jn( x ) 1 1 cos( x sin ) cos n sin( x sin ) sin n d cos( n x sin )d 0 0 (b) Changing to (/2) in (3), we get cos( x cos ) J 0 ( x ) 2J 2 ( x ) cos( 2 ) J 4 ( x ) cos( 4 ) cos( x cos ) J 0 ( x ) 2 J 2 ( x ) cos 2 2 J 4 ( x ) cos 4 Integrating the above equation w.r.t from 0 to , we get 0 0 cos( x cos )d J 0 ( x ) 2 J 2 ( x ) cos 2 2 J 4 ( x ) cos 4 cos( x cos )d J 0 ( x ) 2 J 2 ( x ) 0 Thus, J 0 ( x ) sin 2 sin 4 2J 4 ( x ) J 0 ( x ) 2 4 0 1 cos( x cos )d 0 (c) Squaring (3) and (4) and integrating w.r.t. from 0 to and noting that m and n being integers www.bookspar.com | Website for students | VTU NOTES www.bookspar.com | Website for students | VTU NOTES 0 2 2 2 2 cos ( x sin )d J 0 ( x ) 4J 2 ( x ) 0 2 2 2 sin ( x sin )d 4J 1 ( x ) 4J 3 ( x )2 2 4J 4 ( x )2 2 Adding, d J 02 ( x ) 2 J 12 ( x ) 2 J 22 ( x ) J 32 ( x ) Hence, 0 J 02 2 J 12 2 J 22 J 32 1 Orthogonality of Bessel Functions If and are the two distinct roots of Jn(x) = 0, then if 0, xJ n ( x )J n ( x )dx 1 J / ( ) 2 1 J ( )2 , if n n 1 0 2 2 Proof: We know that the solution of the equation x2u// + xu/ + (2x2 – n2)u = 0 -------- (1) x2v// + xv/ + (2x2 – n2)v = 0 -------- (2) are u = Jn(x) and v = Jn(x) respectively. Multiplying (1) by v/x and (2) by u/x and subtracting, we get x(u// v - u v//)+ (u/ v – uv/)+ (2 –2)xuv = 0 or d x u / v uv / 2 2 xuv dx Now integrating both sides from 0 to 1, we get 2 1 2 xuvdx x u / v uv / u v uv 1 0 / / x 1 ------- (3) 0 Since u = Jn(x), u / d J n ( x ) d J n ( x ) d ( x ) J n/ ( x ) dx d ( x ) dx Similarly v = Jn(x) gives v / d J n ( x ) J n/ ( x ) . dx Substituting these values in (3), we get 1 J n/ ( )J n ( ) J n ( )J n/ ( ) 0 2 2 xJ n ( x )J n ( x )dx ------- (4) If and are the two distinct roots of Jn(x) = 0, then Jn() = 0 and Jn() = 0, and hence (4) reduces to xJ n ( x )J n ( x )dx 0 . 0 This is known as Orthogonality relation of Bessel functions. When = , the RHS of (4) takes 0/0 form. Its value can be found by considering as a root of Jn(x) = 0 and as a variable approaching to . Then (4) gives www.bookspar.com | Website for students | VTU NOTES www.bookspar.com | Website for students | VTU NOTES 1 Lt xJ n ( x )J n ( x )dx Lt 0 J n/ ( )J n ( ) 2 2 Applying L’Hospital rule, we get 2 J n/ ( )J n/ ( ) 1 / J n ( ) --------(5) 2 2 1 Lt xJ n ( x )J n ( x )dx Lt 0 n x We have the recurrence relation J n/ ( x ) J n ( x ) J n 1 ( x ) . J n/ ( ) n J n ( ) J n 1 ( ). Since 1 J n ( ) 0 , we have J n/ ( ) J n 1 ( ) Thus, (5) becomes Lt xJ n ( x )J n ( x )dx 0 2 1 / 1 J n ( ) J n1 ( )2 2 2 www.bookspar.com | Website for students | VTU NOTES www.bookspar.com | Website for students | VTU NOTES LECTURE NOTES OF ENGINEERING MATHEMATICS–IV (Sub Code: 06 MAT41) Text Book: Higher Engineering Mathematics by Dr. B.S.Grewal (36th Edition – 2002) Khanna Publishers,New Delhi Reference Book: Advanced Engineering Mathematics by E. Kreyszig (8th Edition – 2001) John Wiley & Sons, INC. New York SPECIAL FUNCTIONS Prepared by Dr. M. SANKAR Professor and Head Department of Mathematics Sapthagiri College of Engineering Bangalore – 560 057 Introduction Many Differential equations arising from physical problems are linear but have variable coefficients and do not permit a general analytical solution in terms of known functions. Such equations can be solved by numerical methods (Unit – I), but in many cases it is easier to find a solution in the form of an infinite convergent series. The series solution of certain differential equations give rise to special functions such as Bessel’s function, Legendre’s polynomial. These special functions have many applications in engineering. Series solution of ODE d2y dy g ( x) h( x) y 0 , where f(x), g(x) and h(x) 2 dx dx are functions in x and f(x) 0. The series solution of above type of DE is explained as follows: Consider the 2nd order ODE f ( x) 1. Assume the solution of (1) in the form y a r x r r 0 / // 2. Find the derivatives y and y from the assumed solution and substitute in to the given DE which results in an infinite series with various powers of x equal to zero. www.bookspar.com | Website for students | VTU NOTES www.bookspar.com | Website for students | VTU NOTES 3. Now equate the coefficients of various powers of x to zero and try to obtaina recurrence relation from which the constants a0, a1, a2,…. can be determined. 4. When substituted the values of a0, a1, a2,…. in to the assumed solution, we get the power series solution of the given DE in the form y = Ay1(x) + By2(x), where A and B are arbitrary constants. In general, The above type of DE can be solved by the following two methods: Type – I (Frobenius Method) Suppose f(x) = 0 at x= 0 in the above differential equation, we assume solution in the form y a r x k r r 0 where k, a0, a1, a2,…. are all constants to be determined and a0 ≠ 0. Type – II (Power Series Method) Suppose f(x) ≠ 0 at x= 0 in the above differential equation, we assume solution in the form y a r x r where a0, a1, a2,…. are all r 0 constants to be determined. Here all the constants ar’s will be expressed in terms of a0 and a1 only. Problems : 1. Obtain the series solution of the equation d2y y0 dx 2 ----- (1) Let y a r x r ----- (2) be the series solution of (1). r 0 Hence, y / a r r x r 1 , r 0 y // a r r (r 1) x r 2 , r 0 Now (1) becomes a r 2 r r (r 1) x r 2 ar x r 0 r 1 Equating the coefficients of various powers of x to zero, we get Coefficient of x–2 : a0(0)(-1) = 0 Coefficient of x–1 : a1(1)(0) = 0 Equating the coefficient of xr (r >=0) and a0 ≠ 0. and a1 ≠ 0. ar ( r 0) (r 2)( r 1) Putting r = 0,1, 2, 3, …… in (3) we obtain, ar 2 (r 2)(r 1) ar 0 or a r 2 www.bookspar.com | Website for students | VTU NOTES -------(3) www.bookspar.com | Website for students | VTU NOTES a0 a3 a1 a 2 a0 a ; a3 ; a4 ; a5 1 ; 2 6 12 24 20 120 a5 a 4 a0 a1 a6 ; a7 ; and so on. 30 720 42 5040 Substituting these values in the expanded form of (1), we get, y a0 a1 x a 2 x 2 a3 x 3 a 4 x 4 a2 x2 x4 x6 x3 x5 x7 i.e., y a0 1 a1 x 2 24 720 6 120 5040 x2 x4 x6 x3 x5 x7 y a0 1 a1 x is the required 2! 4! 6! 3! 5! 7! solution of the given DE. Hence Series solution of the Bessel Differential Equation Consider the Bessel Differential equation of order n in the form d2y dy x (x2 n2 ) y 0 2 dx dx where n is a non negative real constant or parameter. x2 (i) We assume the series solution of (i) in the form y a r x k r where a0 0 (ii) r 0 Hence, dy a r (k r ) x k r 1 dx r 0 d2y ar (k r )(k r 1) x k r 2 2 dx r 0 Substituting these in (i) we get, x 2 a r (k r )( k r 1) x k r 2 x a r (k r ) x k r 1 x 2 n 2 a r x k r 0 r 0 r 0 r 0 r 0 r 0 r 0 r 0 i.e., a r (k r )( k r 1) x k r a r (k r ) x k r a r x k r 2 n 2 a r x k r 0 Grouping the like powers, we get ar (k r )(k r 1) (k r ) n 2 x k r ar x k r 2 0 r 0 r 0 www.bookspar.com | Website for students | VTU NOTES www.bookspar.com | Website for students | VTU NOTES a r (k r ) 2 n 2 x k r a r x k r 2 0 r 0 (iii) r 0 Now we shall equate the coefficient of various powers of x to zero Equating the coefficient of xk from the first term and equating it to zero, we get a 0 k 2 n 2 0. Since a 0 0, we get k 2 n 2 0, k n Coefficient of xk+1 is got by putting r = 1 in the first term and equating it to zero, we get i.e., a1 (k 1) 2 n 2 0. This gives a1 0, since (k 1) 2 n 2 0 gives , k 1 n which is a contradiction to k = n. Let us consider the coefficient of xk+r from (iii) and equate it to zero. i.e, ar (k r ) 2 n 2 ar 2 0. ar 2 (k r ) 2 n 2 If k = +n, (iv) becomes ar 2 a ar 2 r 2 2 2 (n r ) n r 2nr ar (iv) Now putting r = 1,3,5, ….., (odd vales of n) we obtain, a3 a1 0 , a1 0 6n 9 Similarly a5, a7, ….. are equal to zero. i.e., a1 = a5 = a7 = …… = 0 Now, putting r = 2,4,6, ……( even values of n) we get, a0 a0 a0 a2 a2 ; a4 ; 4n 4 4(n 1) 8n 16 32(n 1)( n 2) Similarly we can obtain a6, a8, … We shall substitute the values of a1 , a2 , a3 , a4 , in the assumed series solution, we get y a r x k r x k ( a0 a1 x a 2 x 2 a 3 x 3 a 4 x 4 ) r 0 Let y1 be the solution for k = +n www.bookspar.com | Website for students | VTU NOTES www.bookspar.com | Website for students | VTU NOTES a0 a0 y1 x n a0 x2 x 4 4(n 1) 32(n 1)( n 2) 2 4 x x i.e., y1 a0 x n 1 2 5 2 (n 1) 2 (n 1)( n 2) This is a solution of the Bessel’s equation. (v) Let y2 be the solution corresponding to k = - n. Replacing n be – n in (v) we get x2 x4 (vi) y 2 a0 x n 1 2 5 2 (n 1) 2 (n 1)( n 2) The complete or general solution of the Bessel’s differential equation is y = c1y1 + c2y2, where c1, c2 are arbitrary constants. Now we will proceed to find the solution in terms of Bessel’s function by choosing 1 and let us denote it as Y1. a0 n 2 (n 1) i.e., Y1 4 x 2 1 1 x 1 2 n (n 1) 2 (n 1) 2 (n 1)( n 2) 2 xn 2 4 1 x 1 x 1 (n 1) 2 (n 1) (n 1) 2 (n 1)( n 2) (n 1) 2 We have the result (n) = (n – 1) (n – 1) from Gamma function x 2 n Hence, (n + 2) = (n + 1) (n + 1) and (n + 3) = (n + 2) (n + 2) = (n + 2) (n + 1) (n + 1) Using the above results in Y1, we get 2 4 1 x 1 x 1 (n 1) 2 (n 2) 2 (n 3) 2 which can be further put in the following form n 0 2 4 0 (1)1 x (1) 2 x x (1) x Y1 (n 2) 1! 2 (n 3) 2! 2 2 (n 1) 0! 2 x Y1 2 x 2 n n r 0 (1) r x (n r 1) r! 2 2r www.bookspar.com | Website for students | VTU NOTES www.bookspar.com | Website for students | VTU NOTES x (1) r 2 r 0 n2r 1 (n r 1) r! This function is called the Bessel function of the first kind of order n and is denoted by Jn(x). x Thus J n ( x) (1) 2 r 0 n2r 1 r (n r 1) r! Further the particular solution for k = -n ( replacing n by –n ) be denoted as J-n(x). Hence the general solution of the Bessel’s equation is given by y = AJn(x) + BJ-n(x), where A and B are arbitrary constants. Properties of Bessel’s function 1. J n ( x ) ( 1 ) n J n ( x ) , where n is a positive integer. Proof: By definition of Bessel’s function, we have x J n ( x) (1) 2 r 0 n2r x Hence, J n ( x ) ( 1 ) r r 0 2 1 r n2r ……….(1) (n r 1) r! 1 ……….(2) ( n r 1 ) r! But gamma function is defined only for a positive real number. Thus we write (2) in the following from x J n ( x ) ( 1 ) r r n 2 n2r 1 ( n r 1 ) r! ………..(3) Let r – n = s or r = s + n. Then (3) becomes x J n ( x ) ( 1 ) s n s 0 2 n2s2n 1 ( s 1 ) ( s n )! We know that (s+1) = s! and (s + n)! = (s+n+1) x ( 1 ) s n s 0 2 n2 s x ( 1 ) n ( 1 ) s s 0 2 1 ( s n 1 ) s! n2s 1 ( s n 1 ) s! Comparing the above summation with (1), we note that the RHS is Jn(x). Thus, J n (x) ( 1)n J n (x) 2. J n ( x ) ( 1 ) n J n ( x ) J n ( x ) , where n is a positive integer www.bookspar.com | Website for students | VTU NOTES www.bookspar.com | Website for students | VTU NOTES x Proof : By definition, J n ( x) (1) r 2 r 0 x J n ( x ) ( 1 ) r r 0 2 n2r x 1n ( 1 ) r r 0 2 1 (n r 1) r! 1 ( n r 1 ) r! x ( 1 ) r 1n 2 r r 0 2 i.e., n2r n2r 1 ( n r 1 ) r! n2r 1 ( n r 1 ) r! J n (-x) ( 1) n J n (x) Thus, Since, ( 1 ) n J n ( x ) J n ( x ) , we have J n ( x) ( 1)n J n (x) J n (x) Recurrence Relations: Recurrence Relations are relations between Bessel’s functions of different order. d n x J n ( x ) x n J n 1 ( x ) dx Recurrence Relations 1: From definition, x x n J n ( x ) x n ( 1 ) r r 0 2 n2r x ( 1 ) r r 0 2 ( n r 1 ) r! 1 2( n r ) 2( n r )x 2( n r )1 d n x J n ( x ) ( 1 ) r dx r 0 2 n 2 r ( n r 1 ) r! x ( 1 ) n r r 0 x n ( 1 ) r r 0 ( n r )x n 2 r 1 2 n 2 r 1 ( n r ) ( n r ) r! ( x / 2 ) n 1 2 r ( n 1 r 1 ) r! x n J n 1 ( x ) d n x J n ( x ) x n J n 1 ( x ) dx d n x J n ( x ) x n J n 1 ( x ) Recurrence Relations 2: dx Thus, From definition, x x n J n ( x ) x n ( 1 ) r r 0 2 x ( 1 ) r r 0 2 2r n2r 1 ( n r 1 ) r! 1 ( n r 1 ) r! 2 r x 2 r 1 d n x J n ( x ) ( 1 ) r dx r 0 2 n 2 r ( n r 1 ) r! www.bookspar.com | Website for students | VTU NOTES --------(1) 1 ( n r 1 ) r! www.bookspar.com | Website for students | VTU NOTES x n 1 2( r 1 ) r 1 2 n 1 2( r 1 ) ( n r 1 ) ( r 1 )! x n ( 1 ) r 1 Let k = r – 1 x n 1 2 k k 0 2 n 1 2 k ( n 1 k 1 ) k ! x n ( 1 ) k d n x J n ( x ) x n J n 1 ( x ) dx x Recurrence Relations 3: J n ( x ) J n 1 ( x ) J n 1 ( x ) 2n d n x J n ( x ) x n J n 1 ( x ) We know that dx Thus, x n J n 1 ( x ) --------(2) Applying product rule on LHS, we get x n J n/ ( x ) nx n1 J n ( x ) x n J n1 ( x ) Dividing by xn we get J n/ ( x ) ( n / x )J n ( x ) J n1 ( x ) --------(3) Also differentiating LHS of d n x J n ( x ) x n J n 1 ( x ) , dx we get x n J n/ ( x ) nx n1 J n ( x ) x n J n1 ( x ) Dividing by –x–n we get J n/ ( x ) ( n / x )J n ( x ) J n1 ( x ) --------(4) Adding (3) and (4), we obtain 2nJ n ( x ) xJ n1 ( x ) J n1 ( x ) x J n 1 ( x ) J n 1 ( x ) 2n 1 Recurrence Relations 4: J n/ ( x ) J n 1 ( x ) J n1 ( x ) 2 Subtracting (4) from (3), we obtain 2 J n/ ( x ) J n1 ( x ) J n1 ( x ) i.e., Jn( x ) i.e., J n/ ( x ) 1 J n1 ( x ) J n1 ( x ) 2 n x Recurrence Relations 5: J n/ ( x ) J n ( x ) J n 1 ( x ) This recurrence relation is another way of writing the Recurrence relation 2. n x Recurrence Relations 6: J n/ ( x ) J n 1 ( x ) J n ( x ) This recurrence relation is another way of writing the Recurrence relation 1. Recurrence Relations 7: J n 1 ( x ) 2n J n ( x ) J n 1 ( x ) x This recurrence relation is another way of writing the Recurrence relation 3. Problems: Prove that ( a ) J 1 / 2 ( x ) 2 sin x ( b ) x J 1 / 2 ( x ) 2 cos x x By definition, x J n ( x ) ( 1 ) r r 0 2 n2r 1 ( n r 1 ) r! Putting n = ½, we get www.bookspar.com | Website for students | VTU NOTES www.bookspar.com | Website for students | VTU NOTES x J 1 / 2 ( x ) ( 1 ) r r 0 2 J1/ 2( x ) 1 / 22r 1 ( r 3 / 2 ) r! 2 4 x 1 1 1 x x 2 ( 3 / 2 ) 2 ( 5 / 2 )1! 2 ( 7 / 2 )2! --------(1) Using the results (1/2) = and (n) = (n – 1) (n–1), we get (3 / 2) 2 , ( 5 / 2 ) 3 15 and so on. , (7 / 2 ) 4 8 Using these values in (1), we get J1/ 2( x ) J1/ 2( x ) x 2 x2 4 x 4 8 2 4 3 16 15 .2 x 2 x3 x 5 x 2 x 6 120 2 x x3 x5 x 5! 3! 2 sin x x Putting n = - 1/2, we get x J 1 / 2 ( x ) ( 1 ) r r 0 2 J 1 / 2 ( x ) 1 / 2 2 r 1 ( r 1 / 2 ) r! 2 4 x 1 1 1 x x 2 ( 1 / 2 ) 2 ( 3 / 2 )1! 2 ( 5 / 2 )2! --------(2) Using the results (1/2) = and (n) = (n – 1) (n–1) in (2), we get J 1 / 2 ( x ) J 1 / 2 ( x ) 2 1 x2 2 x 4 4 x 4 16 3 .2 2 x x2 x4 1 4! 2! 2 cos x x 2. Prove the following results : ( a ) J5/ 2( x ) ( b ) J 5 / 2 ( x ) 2 3 x2 3 2 sin x cos x and x x x 2 3 x2 3 2 cos x sin x x x x Solution : We prove this result using the recurrence relation J n ( x ) 3 x Putting n = 3/2 in (1), we get J 1 / 2 ( x ) J 5 / 2 ( x ) J 3 / 2 ( x ) J5/ 2( x ) 3 J3 / 2( x ) J1/ 2( x ) x www.bookspar.com | Website for students | VTU NOTES x J n 1 ( x ) J n 1 ( x ) 2n ------ (1). www.bookspar.com | Website for students | VTU NOTES i.e., J 5 / 2 ( x ) J5/ 2( x ) 3 2 sin x x cos x 2 x sin x x x x 2 3 sin x 3 x cos x x 2 sin x 2 ( 3 x 2 ) 3 sin x cos x x x x 2 x x2 J 5 / 2 ( x ) J 1 / 2 ( x ) Also putting n = - 3/2 in (1), we get 3 J 3 / 2 ( x ) x 3 2 x sin x cos x 2 3 J 3 / 2 ( x ) J 1 / 2 ( x ) cos x x x x x x 2 3 x sin x 3 cos x x 2 cos x 2 3 3 x2 i.e., J 5 / 2 ( x ) sin x cos x x x x x2 x2 J 5 / 2 ( x ) Solution: L.H.S = d 2 2 J n ( x ) J n21 ( x ) nJ n2 ( x ) ( n 1 )J n21 ( x ) dx x 3. Show that d 2 J n ( x ) J n21 ( x ) 2 J n ( x )J n/ ( x ) 2 J n 1 ( x )J n/ 1 ( x ) ------dx (1) We know the recurrence relations xJ n/ ( x ) nJ n ( x ) xJ n1 ( x ) xJ n/ 1 ( x ) ------- (2) ------- (3) xJ n ( x ) ( n 1 )J n1 ( x ) Relation (3) is obtained by replacing n by n+1 in xJ n/ ( x ) xJ n1 ( x ) nJ n ( x ) Now using (2) and (3) in (1), we get L.H.S = d 2 n1 n J n ( x ) J n21 ( x ) 2 J n ( x ) J n ( x ) J n 1 ( x ) 2 J n 1 ( x ) J n ( x ) J n 1 ( x ) dx x x 2n 2 n1 2 J n ( x ) 2 J n ( x ) J n 1 ( x ) 2 J n 1 ( x ) J n ( x ) 2 J n 1 ( x ) x x d 2 2 J n ( x ) J n21 ( x ) nJ n2 ( x ) ( n 1 )J n21 ( x ) Hence, dx x 1 4. Prove that J 0// ( x ) J 2 ( x ) J 0 ( x ) 2 Solution : We have the recurrence relation J n/ ( x ) J n 1 ( x ) J n1 ( x ) -------(1) 1 2 Putting n = 0 in (1), we get J 0/ ( x ) J 1 ( x ) J 1 ( x ) J 1 ( x ) J 1 ( x ) J 1 ( x ) 1 2 1 2 Thus, J 0/ ( x ) J 1 ( x ) . Differentiating this w.r.t. x we get, J 0// ( x ) J 1/ ( x ) ----- (2) Now, from (1), for n = 1, we get J 1/ ( x ) J 0 ( x ) J 2 ( x ) . 1 2 Using (2), the above equation becomes 1 J 0 ( x ) J 2 ( x )orJ 0// ( x ) 1 J 2 ( x ) J 0 ( x ) . 2 2 1 Thus we have proved that, J 0// ( x ) J 2 ( x ) J 0 ( x ) 2 J 0// ( x ) www.bookspar.com | Website for students | VTU NOTES www.bookspar.com | Website for students | VTU NOTES 2 x 5. Show that (a) J 3 ( x )dx c J 2 ( x ) J 1 ( x ) (b) xJ 02 ( x )dx x 2 J 02 ( x ) J 12 ( x ) 1 2 Solution : d n x J n ( x ) x n J n 1 ( x ) dx (a) We know that or x n J n1 ( x )dx x n J n ( x ) ------ (1) Now, J 3 ( x )dx x 2 x 2 J 3 ( x )dx c x 2 x 2 J 3 ( x )dx 2 x x 2 J 3 ( x )dx dx c x 2 x 2 J 2 ( x ) 2 x x 2 J 2 ( x ) dx c ( from (1) when n = 2) 2 2 c J 2 ( x ) J 2 ( x )dx c J 2 ( x ) J 1 ( x ) ( from (1) when n = 1) x x 2 Hence, J 3 ( x )dx c J 2 ( x ) J 1 ( x ) x 1 1 (b) xJ 02 ( x )dx J 02 ( x ) x 2 2 J 0 ( x ) J 0/ ( x ). x 2 dx (Integrate by parts) 2 2 1 x 2 J 02 ( x ) x 2 J 0 ( x ) J 1 ( x )dx (From (1) for n = 0) 2 d 1 d xJ 1 ( x ) xJ 0 ( x ) from recurrence relation (1) x 2 J 02 ( x ) xJ 1 ( x ) xJ 1 ( x )dx dx 2 dx 1 2 2 1 1 x J 0 ( x ) xJ 1 ( x )2 x 2 J 02 ( x ) J 12 ( x ) 2 2 2 Generating Function for Jn(x) x To prove that e 2 ( t 1 / t ) tnJn( x ) n or x If n is an integer then Jn(x) is the coefficient of tn in the expansion of e 2 ( t 1 / t ) . Proof: x We have e 2 ( t 1 / t ) e xt / 2 e x / 2t ( xt / 2 ) ( xt / 2 ) 2 ( xt / 2 ) 3 ( xt / 2 ) ( xt / 2 ) 2 ( xt / 2 ) 3 1 1 1! 2! 3! 1! 2! 3! (using the expansion of exponential function) ( 1 ) n x n ( 1 ) n 1 x n 1 xt x 2t 2 x nt n x n 1 t n 1 x x2 1 2 n n 1 1 2 2 n n n 1 n 1 2 n! 2 ( n 1 )! 2 t n! 2 t ( n 1 )! 2 1! 2 2! 2t 1! 2 t 2! If we collect the coefficient of tn in the product, they are xn n 2 n! xn 2 2 n2 ( n 1 )!1! xn4 2 n4 ( n 2 )! 2! www.bookspar.com | Website for students | VTU NOTES www.bookspar.com | Website for students | VTU NOTES n 1 x 1 x n! 2 ( n 1 )!1! 2 n2 1 x ( n 2 )!2! 2 n4 x ( 1 )r r 0 2 n 2r 1 ( n r 1 )r! Jn( x ) Similarly, if we collect the coefficients of t–n in the product, we get J–n(x). Thus, x ( t 1 / t ) 2 e Result: tnJn( x ) n x ( t 1 / t ) 2 e J 0 ( x ) t n ( 1 ) n t n J n ( x ) n 1 Proof : x e2 ( t 1 / t ) 1 n n t nJn( x ) t nJn( x ) tnJn( x ) n 0 n 1 n 1 n 1 t n J n ( x ) J 0 ( x ) t n J n ( x ) J 0 ( x ) t n ( 1 ) n J n ( x ) t n J n ( x ) { J n ( x ) ( 1 ) n J n ( x )} n 1 Thus, x ( t 1 / t ) e2 J 0 ( x ) t n ( 1 ) n t n J n ( x ) n 1 Problem 6: Show that (a) J n ( x ) 1 cos( n x sin )d , n being an integer 0 1 (b) J 0 ( x ) cos( x cos )d 0 (c) 2 J 22 J 32 1 Solution : J 02 2 J 12 x We know that e 2 ( t 1 / t ) J 0 ( x ) t n ( 1 ) n t n J n ( x ) n 1 J 0 ( x ) tJ 1 ( x ) t 2 J 2 ( x ) t 3 J 3 ( x ) t 1 J 1 ( x ) t 2 J 2 ( x ) t 3 J 3 ( x ) Since J n ( x ) ( 1 )n J n ( x ) , we have x e2 ( t 1 / t ) J 0 ( x ) J 1 ( x )t 1 / t J 2 ( x ) t 2 1 / t 2 J 3 ( x ) t 3 1 / t 3 p ----- (1) p Let t = cos + i sin so that t = cosp + i sinp and 1/t = cosp - i sinp. From this we get, tp + 1/tp = 2cosp and tp – 1/tp = 2i sinp Using these results in (1), we get x ( 2i sin ) 2 e e ix sin J 0 ( x ) 2J 2 ( x ) cos 2 J 4 ( x ) cos 4 2iJ 1 ( x ) sin J 3 ( x ) sin 3 -----(2) Since eixsin = cos(xsin) + i sin(xsin), equating real and imaginary parts in (2) we get, cos( x sin ) J 0 ( x ) 2J 2 ( x ) cos 2 J 4 ( x ) cos 4 ----- (3) ----- (4) sin( x sin ) 2J 1 ( x ) sin J 3 ( x ) sin 3 These series are known as Jacobi Series. Now multiplying both sides of (3) by cos n and both sides of (4) by sin n and integrating each of the resulting expression between 0 and , we obtain www.bookspar.com | Website for students | VTU NOTES www.bookspar.com | Website for students | VTU NOTES and J n ( x ), n is even or zero 1 cos( x sin ) cos nd 0 , n is odd 0 n is even 0, 1 sin( x sin ) sin nd 0 J n ( x ), n is odd Here we used the standard result cos p cos qd sin p sin qd 2 , if p q 0 0 , if p q 0 From the above two expression, in general, if n is a positive integer, we get Jn( x ) 1 1 cos( x sin ) cos n sin( x sin ) sin n d cos( n x sin )d 0 0 (b) Changing to (/2) in (3), we get cos( x cos ) J 0 ( x ) 2J 2 ( x ) cos( 2 ) J 4 ( x ) cos( 4 ) cos( x cos ) J 0 ( x ) 2 J 2 ( x ) cos 2 2 J 4 ( x ) cos 4 Integrating the above equation w.r.t from 0 to , we get 0 0 cos( x cos )d J 0 ( x ) 2 J 2 ( x ) cos 2 2 J 4 ( x ) cos 4 sin 2 sin 4 2J 4 ( x ) J 0 ( x ) cos( x cos )d J 0 ( x ) 2 J 2 ( x ) 2 4 0 0 Thus, J 0 ( x ) 1 cos( x cos )d 0 (c) Squaring (3) and (4) and integrating w.r.t. from 0 to and noting that m and n being integers 0 2 2 2 2 cos ( x sin )d J 0 ( x ) 4J 2 ( x ) 0 2 2 2 sin ( x sin )d 4J 1 ( x ) 4J 3 ( x )2 2 4J 4 ( x )2 2 Adding, d J 02 ( x ) 2 J 12 ( x ) 2 J 22 ( x ) J 32 ( x ) Hence, 0 J 02 2 J 12 2 J 22 J 32 1 Orthogonality of Bessel Functions If and are the two distinct roots of Jn(x) = 0, then if 0, xJ n ( x )J n ( x )dx 1 J / ( ) 2 1 J ( )2 , if n n 1 0 2 2 Proof: We know that the solution of the equation x2u// + xu/ + (2x2 – n2)u = 0 -------- (1) x2v// + xv/ + (2x2 – n2)v = 0 -------- (2) are u = Jn(x) and v = Jn(x) respectively. www.bookspar.com | Website for students | VTU NOTES www.bookspar.com | Website for students | VTU NOTES Multiplying (1) by v/x and (2) by u/x and subtracting, we get x(u// v - u v//)+ (u/ v – uv/)+ (2 –2)xuv = 0 or d x u / v uv / 2 2 xuv dx Now integrating both sides from 0 to 1, we get 2 1 2 xuvdx x u / v uv / u v uv 1 0 / / ------- (3) x 1 0 Since u = Jn(x), u / d J n ( x ) d J n ( x ) d ( x ) J n/ ( x ) dx d ( x ) dx Similarly v = Jn(x) gives v / d J n ( x ) J n/ ( x ) . dx Substituting these values in (3), we get 1 J n/ ( )J n ( ) J n ( )J n/ ( ) 0 2 2 xJ n ( x )J n ( x )dx ------- (4) If and are the two distinct roots of Jn(x) = 0, then Jn() = 0 and Jn() = 0, and hence (4) reduces to xJ n ( x )J n ( x )dx 0 . 0 This is known as Orthogonality relation of Bessel functions. When = , the RHS of (4) takes 0/0 form. Its value can be found by considering as a root of Jn(x) = 0 and as a variable approaching to . Then (4) gives 1 Lt xJ n ( x )J n ( x )dx Lt 0 J n/ ( )J n ( ) 2 2 Applying L’Hospital rule, we get 2 J n/ ( )J n/ ( ) 1 / J n ( ) --------(5) 2 2 1 Lt xJ n ( x )J n ( x )dx Lt 0 n x We have the recurrence relation J n/ ( x ) J n ( x ) J n 1 ( x ) . J n/ ( ) n J n ( ) J n 1 ( ). Since 1 J n ( ) 0 , we have J n/ ( ) J n 1 ( ) Thus, (5) becomes Lt xJ n ( x )J n ( x )dx 0 2 1 / 1 J n ( ) J n1 ( )2 2 2 www.bookspar.com | Website for students | VTU NOTES