Question

7. Use the Taylor expansions for $f(x + h)$, $f(x - h)$, $f(x + 2h)$, and $f(x - 2h)$ and derive the central-difference formula: $f^{(3)}(x) \approx \frac{f(x + 2h) - 2f(x + h) + 2f(x - h) - f(x - 2h)}{2h^3}$ 8. Use the Taylor expansions for $f(x + h)$, $f(x - h)$, $f(x + 2h)$, and $f(x - 2h)$ and derive the central-difference formula: $f^{(4)}(x) \approx \frac{f(x + 2h) - 4f(x + h) + 6f(x) - 4f(x - h) + f(x - 2h)}{h^4}$

          7. Use the Taylor expansions for $f(x + h)$, $f(x - h)$, $f(x + 2h)$, and $f(x - 2h)$ and
derive the central-difference formula:
$f^{(3)}(x) \approx \frac{f(x + 2h) - 2f(x + h) + 2f(x - h) - f(x - 2h)}{2h^3}$
8. Use the Taylor expansions for $f(x + h)$, $f(x - h)$, $f(x + 2h)$, and $f(x - 2h)$ and
derive the central-difference formula:
$f^{(4)}(x) \approx \frac{f(x + 2h) - 4f(x + h) + 6f(x) - 4f(x - h) + f(x - 2h)}{h^4}$

7. Use the Taylor expansions for f(x + h), f(x - h), f(x + 2h), and f(x - 2h) and
derive the central-difference formula:
f^(3)(x) ≈(f(x + 2h) - 2f(x + h) + 2f(x - h) - f(x - 2h))/(2h^3)
8. Use the Taylor expansions for f(x + h), f(x - h), f(x + 2h), and f(x - 2h) and
derive the central-difference formula:
f^(4)(x) ≈(f(x + 2h) - 4f(x + h) + 6f(x) - 4f(x - h) + f(x - 2h))/(h^4)

Added by Juan M.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Mukesh Devi

Step 1

First, we need to write out the Taylor expansions for f(x + h), f(x - h), f(x + 2h), and f(x - 2h): f(x + h) = f(x) + hf'(x) + (h^2/2)f''(x) + (h^3/6)f'''(x) + O(h^4) f(x - h) = f(x) - hf'(x) + (h^2/2)f''(x) - (h^3/6)f'''(x) + O(h^4) f(x + 2h) = f(x) + 2hf'(x) + Show more…

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Use the Taylor expansions for f(x + h), f(x - h), f(x + 2h), and f(x - 2h) and derive the central-difference formula: f(x + 2h) - 2f(x + h) + 2f(x - h) - f(x - 2h). Use the Taylor expansions for f(x + h), f(x - h), f(x + 2h), and f(x - 2h) and derive the central-difference formula: f(x + h) - 4f(x) + 6f(x - h) - 4f(x - 2h) + f(x - 3h).