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Direction. Answer the following: Finds the roots of equations using required relative error dk < 0.001 1. Using Newton's Method, find two of the roots of x³ – 4e* + 3 = x. Use xº = –1and 1.

Direction. Answer the following: Finds the roots of equations using required relative error dk < 0.001 1. Using Newton's Method, find two of the roots of x³ – 4e* + 3 = x. Use xº = –1and 1.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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answer this and I provide some examples and the table of NEWTON'S method  (pls follow the attachment that i gave)

EXAMPLES Example 5.) Finds the roots of equations using required relative error dk < 0.001 NEWTON'S METHOD TABLE!!!!! a.) Using Newton's Method, calculate sin-1 0.52 f(x*) dk Solution: k f'(x*) rk+1 Remarks let: x = sin- 0.52 sin x = 0.52 f(x) = sin x - 0.52 f'(x) = cos x Newton's Method, also known as Newton Raphson Method, is important because it's an iterative process that can approximate solutions to an equation with incredible accuracy. And it's a method to approximate numerical solutions (i.e., x-intercepts, zeros, or roots) to equations that are too hard for us to solve by hand. Using radian: f'(x) = cos x, x° = 30° or n/6, therefore f'(x) = 0.866 x*+1 0.546 This is the fastest method, but requires analytical computation of the derivative off(x). k f(xk) f'(x*) Remarks Also, the method may not always converge to the desired root. "/ 0.546 -0.020 0.866 0.042 Continue 1 -0.000 0.854 0.546 Stop We can derive Newton's Method graphically, or by a Taylor series. We again want to construct a sequence xo, X1, X2, -... that converges to the root x = r. Consider the Xn+1 member of this sequence, and Taylor series expand f(xn+1) about the point xn. We have 0.546 (180/n) = 31.283° root = 0.546 = 31.28° f(Xn+1) = f(xn) + (xn+1 – Xn) f'(xn)+.... to check: To determine xn+1; we drop the higher-order terms in the Taylor series, and assume f(xn+1) = 0. Solving for xn+1, we have f(x) = 0 = sin 31.28 – 0.52 f(xn) Xn+1 = Xn -T(xn) b.) f (x) — х3-Зх + 1 Solution: Starting Newton's Method requires a guess for xo, hopefully close to the root x =r. Req'd: three roots A simple algorithm for the bisection calculation is listed below: f'(x) = 3x² - 3 i. 1st root; use x° = 0.5, therefore f'(x) = -2.25 Taylor series expansion y = f(x) expand x f(x) = f(x*)+ L (x*)(x-x*)+ !"(x*)(x-x*)*, = x* S(x*) -0.375 L'(x*) dk 0.501 k x* x*+1 Remarks Step 1 0.500 -2.250 0.333 Continue "(x*)(x-x*)" 0.037 0.002 0.0 +..+ 1 О.333 0.037 -2.667 0.346 Continue 1! 2! n! 0.003 0.0007 0.347 Continue Stop 0.346 -2.640 Delete terms beyond 1st derivative 3 0.347 -2.638 0.347 Step 2 f'(xk)(x – x*) f(x) = f(x*) + therefore, 1st root = 0.347 = x* 1! Estimate x = ĩ and f(x) = 0 2nd root; use x° = -1.5, therefore f'(x) = 3.75 f(xk) ii. f'(x*)(x – x*) 1! Step 3 k xk f'(x*) xk+1 dk Remarks f(x) = 0 = f(xk) + -2.066 0.473 -1.500 2.125 3.750 Continue 1 -2.066 -1.561 9.730 -1.899 -1.899 -О.151 7.810 -1.879 0.010 Continue Approximate x = x*+1 then solve for x*+1 3. -1.879 0.002 7.590 -1.879 0.0 Stop f(x*) f'(xk) Step 4 xk+1 = xk %3D therefore, 2nd root = -1.879 = x* Condition for convergence 3rd root; use x° = 1.25, therefore f'(x) = 1.68 iii. Step 5 k L'(xk) xk+1 Remarks f'(x) is not too close to zero 1.250 1.721 1.562 -0.796 1.687 1.721 Continue Check the following: (a) If dk > 0.001, set x*+1 to be the new xk and return to step 2. (b) If dk < 0.001, stop the computation. 1 0.934 5.885 1.562 Step 6 2 0.125 4.319 1.533 0.018 Continue 1.533 0.003 4.050 1.532 0.000 Stop therefore, 3rd root = 1.532 = x*
Transcribed Image Text:EXAMPLES Example 5.) Finds the roots of equations using required relative error dk < 0.001 NEWTON'S METHOD TABLE!!!!! a.) Using Newton's Method, calculate sin-1 0.52 f(x*) dk Solution: k f'(x*) rk+1 Remarks let: x = sin- 0.52 sin x = 0.52 f(x) = sin x - 0.52 f'(x) = cos x Newton's Method, also known as Newton Raphson Method, is important because it's an iterative process that can approximate solutions to an equation with incredible accuracy. And it's a method to approximate numerical solutions (i.e., x-intercepts, zeros, or roots) to equations that are too hard for us to solve by hand. Using radian: f'(x) = cos x, x° = 30° or n/6, therefore f'(x) = 0.866 x*+1 0.546 This is the fastest method, but requires analytical computation of the derivative off(x). k f(xk) f'(x*) Remarks Also, the method may not always converge to the desired root. "/ 0.546 -0.020 0.866 0.042 Continue 1 -0.000 0.854 0.546 Stop We can derive Newton's Method graphically, or by a Taylor series. We again want to construct a sequence xo, X1, X2, -... that converges to the root x = r. Consider the Xn+1 member of this sequence, and Taylor series expand f(xn+1) about the point xn. We have 0.546 (180/n) = 31.283° root = 0.546 = 31.28° f(Xn+1) = f(xn) + (xn+1 – Xn) f'(xn)+.... to check: To determine xn+1; we drop the higher-order terms in the Taylor series, and assume f(xn+1) = 0. Solving for xn+1, we have f(x) = 0 = sin 31.28 – 0.52 f(xn) Xn+1 = Xn -T(xn) b.) f (x) — х3-Зх + 1 Solution: Starting Newton's Method requires a guess for xo, hopefully close to the root x =r. Req'd: three roots A simple algorithm for the bisection calculation is listed below: f'(x) = 3x² - 3 i. 1st root; use x° = 0.5, therefore f'(x) = -2.25 Taylor series expansion y = f(x) expand x f(x) = f(x*)+ L (x*)(x-x*)+ !"(x*)(x-x*)*, = x* S(x*) -0.375 L'(x*) dk 0.501 k x* x*+1 Remarks Step 1 0.500 -2.250 0.333 Continue "(x*)(x-x*)" 0.037 0.002 0.0 +..+ 1 О.333 0.037 -2.667 0.346 Continue 1! 2! n! 0.003 0.0007 0.347 Continue Stop 0.346 -2.640 Delete terms beyond 1st derivative 3 0.347 -2.638 0.347 Step 2 f'(xk)(x – x*) f(x) = f(x*) + therefore, 1st root = 0.347 = x* 1! Estimate x = ĩ and f(x) = 0 2nd root; use x° = -1.5, therefore f'(x) = 3.75 f(xk) ii. f'(x*)(x – x*) 1! Step 3 k xk f'(x*) xk+1 dk Remarks f(x) = 0 = f(xk) + -2.066 0.473 -1.500 2.125 3.750 Continue 1 -2.066 -1.561 9.730 -1.899 -1.899 -О.151 7.810 -1.879 0.010 Continue Approximate x = x*+1 then solve for x*+1 3. -1.879 0.002 7.590 -1.879 0.0 Stop f(x*) f'(xk) Step 4 xk+1 = xk %3D therefore, 2nd root = -1.879 = x* Condition for convergence 3rd root; use x° = 1.25, therefore f'(x) = 1.68 iii. Step 5 k L'(xk) xk+1 Remarks f'(x) is not too close to zero 1.250 1.721 1.562 -0.796 1.687 1.721 Continue Check the following: (a) If dk > 0.001, set x*+1 to be the new xk and return to step 2. (b) If dk < 0.001, stop the computation. 1 0.934 5.885 1.562 Step 6 2 0.125 4.319 1.533 0.018 Continue 1.533 0.003 4.050 1.532 0.000 Stop therefore, 3rd root = 1.532 = x*
Direction. Answer the following: Finds the roots of equations using required relative error dk < 0.001 1. Using Newton's Method, find two of the roots of x3 - 4e* + 3 = x. Use x° = -1and 1.
Transcribed Image Text:Direction. Answer the following: Finds the roots of equations using required relative error dk < 0.001 1. Using Newton's Method, find two of the roots of x3 - 4e* + 3 = x. Use x° = -1and 1.
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