4.7 Use forward and backward difference approximations of O(h)and a centered difference approximation of O(h²) to estimate thefirst derivative of the function examined in Prob. 4.5. Evaluate thederivative at x = 2 using a step size of h = 0.2. Compare your resultswith the true value of the derivative. Interpret your results on thebasis of the remainder term of the Taylor series expansion.

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Answered: 4.7 Use forward and backward difference…

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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Find the first four nonzero terms of the Taylor series about 0 for the function f(x) = √1 + x cos(6x). Note that you may want to find these in a manner other than by direct differentiation of the function. √1 + x cos(6x) = = + .. 1 + 1/2x +-145/4x^2 +
3. (a) Use derivatives and the definition of the Taylor series to write out the degree three Taylor polynomial T3(x) expanded about X-value a=1 for the functionff(x) = ln (2 - x). (Caution: use a = 1; do not use a = 0.) (b)Use the degree three Taylor polynomial found in part (a) to estimate In(1.5).
Write the first three terms of the Taylor series for f(x) = Vx at a = 1. Show your reasoning
(part 1 of 4) Find the coefficient of 6 in the Taylor series expansion centered at the origin for the function f(x) = 6+2 sin(x²). 1. coefficient 2. coefficient ofx6 3. 4. 5. coefficient ofx6 = 2 on (-∞, ∞). = 1 coefficient ofx6 = 3 coefficient ofx6 (part 2 of 4) Find a power series representation for the function f(x) = x sin(2x) 1. f(x) = Σ (−1)n-1. n=1 5. f(x) ∞ 2. f(x) = Σ (-1)″-1. n=1 6. f(x) of 6 = −1 -1 -2 1. 2. ∞ 3. f(x) = Σ (−1)″-1. n=1 3. ∞ 4. f(x) = Σ (−1)n-1. n=1 == = 4. = 5. ∞ 1 3 n=1 Σ(-1)-1 22n 2n 1 2n-1 n - 1 Σ (1)n-1. n=1 f(x) = f(x) = Σn=0 f(x) = Σn=0_n! f(x) = Σn=0 (-1)" 2n n! xn+1 2n (-1)" (2n)! (part 3 of 4) Find the Taylor series centered at the origin for the function - 22n-1 (2n-1)! x = x cos(2x). f(x) = En=0 22n f(x) = Σ=0 (2n)! 2n (n − 1)! 1 (2n)! 1 xn (n − 1)! x²n xn xn+1 x²n+1 (-1)" 22n (2n)! x²n+1 x²n 2n x²n xn x²n+1
Approximate ln 0.5 using three terms of the Taylor series expansion of ln(1−x)about 0.
Find the first 3 terms of the Taylor series for f(x) = ln(x) centered at 4.
The Taylor series for f(x) = cos(x) at a = is Cn (x - )". 4 n=0 Find the first few coefficients. CO = C2 = C3 = C4 =
What is the coefficient of 23 in the Taylor series for f(x) = centered at # = should have at least 6 decimal places. 2 ? Your answer
1.6. By manipulating Taylor series, determine the constant C for an error expansion of (1.3) of the form wj−u' (xj) ~ Ch¼u (5) (x;), where u (5) denotes the fifth derivative. Based on this value of C and on the formula for u(5) (x) with u(x) = esin(x), determine the leading term in the expansion for w; - u'(x;) for u(x) = esin(x). (You will have to find maxε[-T,T] |u(5) (x)| numerically.) Modify Program 1 so that it plots the dashed line corresponding to this leading term rather than just N-4. This adjusted dashed line should fit the data almost perfectly. Plot the difference between the two on a log-log scale and verify that it shrinks at the rate O(h6).
1/2 Give an upper bound for the error in using a Taylor polynomial of degree 3 to approximate e" dt. You may assume the Maclaurin series of e* as well as the fact that e < 3.
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