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Chapter 4 Exponential and Logarithmic Functions (指数函数和对数函数) In this Chapter, we will encounter some important concepts Exponential Functions(指数函数) Logarithmic Functions(对数函数) Differentiation of Logarithmic and Exponential Functions 1 Section 4.1 Exponential Functions Exponential function(指数函数): If b is a positive number other than 1 (b>0, b≠1), there is a unique function called the exponential function with base b that is defined by f(x)=bx for every real number x NOTE: Such function can be used to describe exponential and logistic growth and a variety of other important quantities. 2 Definition of b n for Rational Values of n (and b>0) Integer powers: If n is a positive integer, bn b b b n facters Fractional powers: If n and m are positive integers, b where m n/m b m n b m n b denotes the positive mth root. Negative powers: b n 1 n b 0 b 1 Zero power: 3 Example 1 Solution: 4 Figure below shows graphs of various members of the x family of exponential functions y b b p ( x ) x NOTE: Students often confuse the power function x f ( x ) b with the exponential function 5 6 7 Definition: The natural exponential function is Where n 10 100 2.59374 2.70481 1000 2.71692 10,000 2.711815 100,000 2.71827 8 Continuous Compounding of Interest(连续利息) If P is the initial investment (the principal) and r is the interest rate (expressed as a decimal), the balance B after the interest is added will be B=P+Pr=P(1+r) dollars 9 10 11 12 Example 2 Suppose $1,000 is invested at an annual interest rate of 6%. Compute the balance after 10 years if the interest is compounded a. Quarterly b. Monthly c. Daily d. Continuously Solution: a. To compute the balance after 10 years if the interest is compounded quarterly, using the formula r B (t ) p1 k kt with t=10, p=1,000, r=0.06, and k=4: 40 0.06 B(10) 1,0001 $1,814.02 4 to be continued 13 b. This time, take t=10, p=1,000, r=0.06, and k=12 to get 120 0.06 B(10) 1,0001 12 $1,819.40 c. Take t=10, p=1,000, r=0.06, and k=365 to obtain 0.06 B(10) 1,0001 365 3, 650 $1,822.03 d. For continuously compounded interest use the formula B(t ) pert with t=10, p=1,000, and r=0.06: B(10) 1,000e0.6 $1,822.12 This value, $1,822.12, is an upper bound for the possible balance. No matter how often interest is compounded, $1,000 invested at an annual interest rate of 6% can not grow to more than $1,822.12 in 10 years. 14 Present Value(现值) 15 16 Example 3 Sue is about to enter college. When she graduates 4 years from now, she wants to take a trip to Europe that she estimates will cost $5,000. How much should she invest now at 7% to have enough for the trip if interest is compounded: a. Quarterly b. Continuously Solution: The required future value is F=$5,000 in t=4 years with r=0.07. a. If the compounding is quarterly, then k=4 and the present value is to be continued 17 0.07 P 5,0001 4 4 ( 4 ) $3,788.08 b. For continuous compounding, the present value is P 5,000e 0.07( 4) $3,778.92 Thus, Sue would have to invest about $9 more if interest is compounded quarterly than if the compounding is continuous. 18 指数增长和指数下降 19 Section 4.2 Logarithmic Functions(对数函数) 20 21 Example 4 Use logarithm rules to rewrite each of the following expressions in terms of log 5 2 and log. 5 3 5 log a. b. log 5 8 c. log 5 36 5 3 Solution: a. 5 log 5 log 5 5 log 5 3 3 1 log 5 3 quotient rule since log 5 5 1 b. log 5 8 log 5 23 3 log 5 2 c. power rule log 5 36 log 5 (2232 ) log 5 22 log 5 32 2log 5 2 2 log 5 3 product rule power rule 22 23 24 Example 5 Solution: 25 Exponential function: series representation Let n=k/r Expand and take the limit, we get 26 Exponential & Logarithmic functions: important limits Lim (x0) (exp(x) – 1 ) / x = 1 Or exp(x) 1 + x as x0 , and Lim (x0) ln(1+x) / x = 1 From these, we can prove d/dx ( exp(x) ) = exp(x) , d/dx ( ln(x) ) = 1/x 27 Section 4.3 Differentiation of Logarithmic and Exponential Function Example 6 Solution: 28 Example 7 29 Solution: 30 Differentiate both sides of the equation 31 Example 8 Solution: 32 33 Example 9 Solution: 34 Taking the derivatives of some complicated functions can be simplified by using logarithms. This is called logarithmic differentiation. Example 10 Solution: to be continued 35 36 Example 11 Differentiate each of these function a. y 2x b. y log 3 x Solution: a. To differentiate follows: y 2 x , we use logarithmic differentiation as to be continued 37 b. 38 39 40 The relative rate of change of a quantity Q(x) can be computed by finding the derivative of lnQ. d Q '( x) (ln Q) dx Q( x) 41 Example 12 Solution: to be continued 42 43 Summary Exponential Functions, Basic Properties of Exponential Functions, The Natural Exponential Base e. Compound Interest, Continuously Compounded Interest, Present Value. Exponential Growth and Decay. Logarithmic Functions, The Natural Logarithm. Differentiation of logarithmic and Exponential Functions. Optimal holding time. 44 Trigonometric functions: a very important limit From this we can prove: 45 Derivatives of trigonometric functions Using the previous results we can derive d/dx sin(x) = cos(x) d/dx cos(x) = - sin(x) d/dx tan(x) = ? d/dx sec(x) = ? 46 Hyperbolic functions sinh (x) = ( exp(x) – exp(-x) )/ 2 cosh (x) = ( exp(x) + exp(-x) )/ 2 d/dx sinh(x) = cosh(x) d/dx cosh(x) = sinh(x) 47 Taylor’s Series/Expansion 1 2 1 n e 1 x x x 2! n! x x ( , ) If then the series is convergent to ex. Proof: 48 Suppose that the function f(x) has continuous derivatives of orders 1, 2, ..., k at x=0. Can we approximate f(x) by a power series? 2 k f ( x ) c0 c1 x c2 x ck x Expand We get Thus, f (n) n ! cn , n k , (0) n k, 0, f (0) 2 f ( x ) f (0) f (0) x x 2! c0 c1 x c2 x 2 f ( k ) (0) k x k! ck x k , Suppose f(x) is a function for which you know the value and the value of all its derivatives at some point x = a. What is the value of f at some nearby point, say x = a+h, where h is sufficiently small? For example, you are studying the motion of an object, and you have measured its current position and velocity, and maybe even its acceleration. If we take f(t) to be the position of the object at time t, f’ is its velocity, and f’’ is its acceleration, then we would like to determine how much we can know about the motion a short time later. Approximation by Increment 50 Taylor's Theorem 51 Taylor Series/Expansion The remainder can be written in a number of ways. The most useful form is that derived by J. L. Lagrange (1736-1813), resulting in the Taylor formula with derivative remainder: where is some number between x and a 52 Taylor's formula with integral remainder, derived by Brook Taylor (1685-1731) By repeating this integration by parts process on the remaining integral p times, one has the result: The last term is referred to as the remainder, Rn(x) 53 Expand f ( x ) e x into Taylor series of x. Sulution thus Since f ( n) ( x ) e , f ( n ) (0) 1 x 1 2 1 n e 1 x x x 2! n! x (n 0,1,2,) x ( , ) Lagrange Residual Term: y Obviously, 6 For any real x, e| x| lim | x |n1 0, n ( n 1)! y ex 4 (n 2) 2 (n 0) thus Therefore, the expansion is convergent to f(x)! (n 3) 1 O 2 1 2 54 x Useful expansion 55 Euler's formula named after Leonhard Euler, is a mathematical formula in complex analysis that establishes the deep relationship between the trigonometric functions and the complex exponential function. Euler's formula states that, for any real number x, Hence we can get: 56 Proof of Euler's formula In the last step we have simply recognized the Taylor series for sin(x) and cos(x). The rearrangement of terms is justified because each series is absolutely convergent. 57 Indeterminate forms: application of Taylor Expansion Expressions of the form 0/0, ∞/∞, 0 ×∞, ∞−∞, 0^∞ and ∞^0 are called indeterminate forms. 58 L’Hospital’s Rule – for indeterminate forms f’(x), g’(x) exist, 59 60 Examples: 61 Example: Find the limit of as x0 Since therefore 62 What is Power Series? Power Series(幂级数)：The expression as n 2 a ( x x ) a a ( x x ) a ( x x ) ... n 0 0 1 0 2 0 n 0 Special case：as x0=0 ，i.e., expending at 0, it becomes n 2 a x a a x a x ... n 0 1 2 n 0 Convergence (收敛性) of power series： Power series an ( x x0 ) will be absolutely convergent within n n 0 interval x x0 R , and divergent within x x0 R , i.e., its convergent interval is [x0-R，x0+R]. Radius of Convergence 63 Absolute convergence and Conditional convergence A series is said to converge absolutely if the series of absolute values converges. A series of real or complex numbers is said to be conditionally convergent (or semi-convergent) if it is convergent but not absolutely convergent. A famous example is the alternating series: 64 Test of convergence: absolute ratio test Given the series: and r Then, if r < 1, the series converges absolutely. if r > 1, the series diverges. if r = 1, no conclusions can be drawn - series may converge or diverge. 65 Test of convergence: comparison test # Comparison test 1: If ∑b(n) is an absolutely convergent series such that |a(n) | ≤ C |b(n) | for some number C and for sufficiently large n , then ∑a(n) converges absolutely as well. If ∑|b(n) | diverges, and |a(n) | ≥ |b(n) | for all sufficiently large n , then ∑a(n) also fails to converge absolutely. # Comparison test 2: If ∑b(n) is an absolutely convergent series such that |an+1 /an | ≤ |b(n+1) /b(n) | for sufficiently large n , then ∑a(n) converges absolutely as well. If ∑|b(n) | diverges, and |a(n+1) /a(n) | ≥ |b(n+1) /b(n) | for all sufficiently large n , then ∑a(n) also fails to converge absolutely (though it could still be conditionally convergent, e.g. if the a(n) alternate in sign). 66 Example: r= r 67