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2.14 Example Solve 𝑥 4 – 11 x + 8 modified Newton-Raphson method. Solution: f (x) = 𝑥 4 – 11x + 8 f '(x) = 4x3 – 11 f "(x) = 12x2 The modified Newton-Rapson’s formula is The calculations are given as Hence, the root is 1.891876. MULLER’S METHOD: Muller’s method is an iterative method and free from the evaluation of derivative. It requires three starting points (𝑥𝑛−2 , 𝑓𝑛−2 ), (𝑥𝑛−1 , 𝑓𝑛−1 ) and (𝑥𝑛 , 𝑓𝑛 ). A parabola is constructed that passes through these points then the quadratic formula is employed to find a root for the next approximation. We assume that 𝑥𝑛 is the best approximation to the root and consider the parabola through the three starting values as shown in Fig. Let the quadratic polynomial be f (x) = a𝑥 2 + bx + c [1] 2.15 Passing through the points (𝑥𝑛−2 , 𝑓𝑛−2 ), (𝑥𝑛−1 , 𝑓𝑛−1 ) and (𝑥𝑛 , 𝑓𝑛 ), then [2] Eliminating a, b, c from Eq. 2, we obtain the following determinant [3] By expanding this determinant in Eq. 3, the function f (x) can be written as [4] Equation 4 is a quadratic polynomial passing through the three given points. Let Now, Eq. 4 becomes [5] Noting f(x)=0, Let Eq. 5 will take the form [6] [7] 2.16 Where first part of [7] can be written as [8] Solving Eq.(3.35) for 1/𝝀, we obtain [9] The sign in the denominator of Eq. 9 is ± according as 𝑔𝑛 > 0 or 𝑔𝑛 < 0. Hence [10] Now, replacing x on left hand side by 𝑥𝑛+1 in Eq. 10, we obtain Summary Example Find a root of the equation 𝑥 3 – 3x – 7 = 0 using the Miller’s method where the root lies between 2 and 3. Solution: Let 𝑥0 = 2, 𝑥1 = 2.5 and 𝑥2 = 3. The calculations are shown in Tables 1 and 2. Table 1 2.17 Table 2 Hence one root is 2.42599 correct up to five decimal places. AITKEN’S 𝚫𝟐 METHOD: Suppose we have an equation f (x) = 0 whose roots are to be determined. Let I be an interval containing the point x = 𝛼. Now, can be written as x = 𝜑(x) such that 𝜑 (x) and 𝜑‘(x) are continuous in I and | 𝜑 ‘(x)| < 1 for all x in I. Denoting 𝑥𝑖−1 , 𝑥𝑖 and 𝑥𝑖 +1 as the three successive approximations to the desired root 𝛼, we can write 𝛼 -𝑥𝑖 = 𝜆(𝛼 – xi –1) (1) 𝛼 – xi +1 = 𝜆 (𝛼 – xi) (2) where 𝜆 is a constant so that | 𝜑 (x)| ≤ 𝜆 ≤1 for all i. Dividing Eq.1 with Eq. 2, we obtain (3) (4) Now, And (5) Using equation 5 and 4 , we get (6) Equation (6) gives the successive approximation to the root 𝛼 and method is known as the Aitken’s𝚫𝟐 method. 2.18 Example: 1+𝑐𝑜𝑠𝑥 Find the root of the function 𝑥 = ( 3 )correct to four decimal places using Aitken’s iteration method. Solution: f (x) = cos x – 3x + 1 f (0) = 1 f (𝝅/2) = cos(𝝅 /2) – 3(𝝅 /2) + 1 = – 8.42857 Hence f (0) > 0 and f (𝝅 /2) < 0 Also, f (0) f (𝝅 /2) = 1(– 8.42857) = – 8.42857 < 0 Therefore, a root exists between 0 and 𝝅 /2. The given equation can be written as 1 + 𝑐𝑜𝑠𝑥 𝑥=( ) 3 Now, Equation shows that Aitken’s method can be employed. Let 𝑥0 = 0 be an initial approximation to the root Now, drawing the table 2.19 Therefore, Hence, the root is 0.6070 correct up to four decimal places. SECANT METHOD: The secant method is very similar to the Newton-Raphson method. The main disadvantage of the Newton-Raphson method is that the method requires the determination of the derivatives of the function at several points. Often, the calculation of these derivatives takes too much time. In some cases, a closed-form expression for f ′(x) may difficult to obtain or may not be available. Now, from the Newton-Raphson method, Convergence of the Iteration Methods: We now study the rate at which the iteration methods converge to the exact root, if the initial approximation is sufficiently close to the desired root. Defining the error of approximation at the kth iterate as 𝜀𝑘 = 𝑥𝑘 – , k = 0, 1, 2,... Definition An iterative method is said to be of order p or has the rate of convergence p, if p is the largest positive real number for which there exists a finite constant C ≠ 0 , such that |𝜀𝑘+1 | ≤ C |𝜀𝑘 |𝑝 . 2.20 The constant C, which is independent of k, is called the asymptotic error constant and it depends on the derivatives of f(x) at x = . Method of false position If the root lies initially in the interval (x0, x1), then one of the end points is fixed for all iterations. If the left end point 𝑥0 is fixed and the right end point moves towards the required root, the method behaves like Subtitiuting we expand each term in Taylor’s series and simplify using the fact that f(α) = 0. We obtain the error equation as Since 𝜀0 is finite and fixed, the error equation becomes |𝜀𝑘+1 | ≤ |𝐶 ∗ | |𝜀𝑘 | where C*=C𝜀0 Hence, the method of false position has order 1 or has linear rate of convergence. Method of successive approximations or fixed point iteration method We have Subtracting we get Therefore, Hence, the fixed point iteration method has order 1 or has linear rate of convergence. Newton-Raphson method The method is given by Subtitiuting we obtain 2.21 Expanding each term in Taylor’s series and simplify using the fact that f(α) = 0. Neglecting the terms containing 𝜀𝑘 3 and higher powers of 𝜀𝑘 , we get |𝜀𝑘+1 | ≤ |𝐶| |𝜀𝑘 |2 And Therefore, Newton’s method is of order 2 or has quadratic rate of convergence. Convergence of Newton-Raphson Method The Newton-Raphson iteration formula is given by (1) The general form of Eq. is given by x = φ (x) (2) The Newton-Raphson iteration method given by Eq. converges if |φ′(x)| < 1. 2.22 Hence Newton-Raphson method converges if If α denotes the actual root of f (x) = 0, then we can select a small interval in which f (x), f ′(x) and f ″(x) are all continuous and the condition given is satisfied. Therefore, Newton-Raphson method always converges provided the initial approximation 𝑥0 is taken very close to the actual root α. Summing up of all methods The Bisection method and the method of False Position always converge to an answer, provided a root is bracketed in the interval (a, b) to start with. Since the root lies in the interval (a, b), on every iteration the width of the interval is reduced until the solution is obtained. The Newton-Raphson method and the method of Successive Approximations require only one initial guess and on every iteration it approaches to the true solution or the exact root. The Bisection method is guaranteed to converge. The Bisection method may fail when the function is tangent to the axis and does not cross the x-axis at f (x) = 0. The Bisection method, the method of False Position, and the method of Successive Approximations converge linearly while the Newton-Raphson method converges quadratically. Newton-Raphson method requires less number of iterations than the other three methods. One disadvantage with Newton-Raphson method is that when the derivative f ′(xi) is zero, a new starting or initial value of x must be selected to Continue with the iterative procedure. The Successive Approximation method converges only when the condition |φ′(x)| < 1 is satisfied. Futhermore, Muller method and aitken’s method can find real or complex roots of a polynomial or an arbitrary function. Question arises What is the importance of defining the order or rate of convergence of a method? Suppose that we are using Newton’s method for computing a root of f(x) = 0. Let us assume that at a particular stage of iteration, the error in magnitude in computing the root is 10−1 =0.1. We observe from (1.31), that in the next iteration, the error behaves like 𝐶(0.1)2 = C(10−2 ). That is, we may possibly get an accuracy of two decimal places. Because of the quadratic convergence of the method, we may possibly get an accuracy of four 2.23 decimal places in the next iteration. However, it also depends on the value of C. From this discussion, we conclude that both fixed point iteration and regula-falsi methods converge slowly as they have only linear rate of convergence. Further, Newton’s method converges at least twice as fast as the fixed point iteration and regula-falsi methods.