Please help answer this question QUESTION 71To derive finite difference formulas and compute their errors we usually use Taylor's expansions. The error in the finite difference approximation f'(x) ~—- −[4ƒ (x+ h) − 3f (x) − f (x+2h) ] is2ha. 1·h²ƒ''' (§)6b. 1-h²ƒ'' (5)3C. 1d.·h²ƒ ''' (§)31-h²f (4) (§)12

The Taylor series expansion of fx+h is ∑i=0n-1hii!fix…

Answered: To derive finite difference formulas…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

See similar textbooks

Related questions

Question

Please help answer this question

Expert Solution

Step 1: Finding the Taylor series expansion

The Taylor series expansion of $f (x + h)$ is $\sum_{i = 0}^{n - 1} \frac{h^{i}}{i!} f^{i} (x)$

$∴ f (x + h) = f (x) + h f' (x) + \frac{h^{2}}{2!} f " (x) + . . . . + \frac{h^{n - 1}}{(n - 1)!} + R_{n} f (x + 2 h) = f (x) + 2 h f' (x) + \frac{{(2 h)}^{2}}{2!} f " (x) + . . . . + \frac{h^{n - 1}}{(n - 1)!} + R_{n}$

Now

$4 f (x + h) - 3 f (x) - f (x + 2 h) = 4 (f (x) + h f' (x) + \frac{h^{2}}{2!} f " (x) + \frac{h^{3}}{3!} f''' (x) . . .) - 3 f (x) - (f (x) + 2 h f' (x) + \frac{{(2 h)}^{2}}{2!} f " (x) + \frac{{(2 h)}^{3}}{3!} f''' (x)) = 4 f (x) + 4 h f' (x) + 2 h^{2} f " (x) + \frac{2 h^{3}}{3} f''' (x) - 3 f (x) - f (x) - 2 h f' (x) - 2 h^{2} f " (x) - \frac{4 {(h)}^{3}}{3} f''' (x) = 2 h f' (x) - \frac{2 {(h)}^{3}}{3} f''' (x) + . . .$

Given

$f' (x) \approx \frac{1}{2 h} [4 f (x + h) - 3 f (x) - f (x + 2 h)]$

$∴ \frac{1}{2 h} [4 f (x + h) - 3 f (x) - f (x + 2 h)] = \frac{1}{2 h} [2 h f' (x) - \frac{2 {(h)}^{3}}{3} f''' (x) + . . .] = f' (x) - \frac{1}{3} h^{2} f''' (x) + . . . . ∴ |f' (x) - \frac{1}{2 h} [4 f (x + h) - 3 f (x) - f (x + 2 h)]| \leq \frac{1}{3} h^{2} f''' (x)$