Question

derive a suitable numerical differentiation formula of 0(h2) to find f''(2. 4) with h =0.1 given the table f(0.1)=3.41, f(1.2) =2.68, f(2.4)=1. 37, f(3.9) =-1. 48.

Answer 1

Question #74146

Derive a suitable numerical differentiation formula of $0(h2)$ to find $f'(2.4)$ with $h = 0.1$ given the table $f(0.1) = 3.41$, $f(1.2) = 2.68$, $f(2.4) = 1.37$, $f(3.9) = -1.48$.

Answer:

Using Taylor Expansion technique:

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

Thus

$$f(x + h) + f(x - h) = 2f(x) + h^2 f''(x) + \frac{h^4}{12} f^{(4)}(x) + \dots,$$ $$f(x + h) + f(x - h) - 2f(x) = h^2 f''(x) + \frac{h^4}{12} f^{(4)}(x) + \dots,$$ $$\frac{f(x + h) + f(x - h) - 2f(x)}{h^2} = f''(x) + \frac{h^2}{12} f^{(4)}(x) + \dots,$$ $$\frac{f(x + h) + f(x - h) - 2f(x)}{h^2} = f''(x) + O(h^2).$$

So

$$f''(x) \approx \frac{f(x + h) + f(x - h) - 2f(x)}{h^2}$$

with truncation error of order $O(h^2)$.

For $x = 2.4$ and $h = 0.1$

$$f(2.4) = 1.37,$$ $$f(2.4 + 0.1) = f(2.5) = 1.37 - (1.37 + 1.48) \frac{2.5 - 2.4}{3.9 - 2.4} = 1.18,$$ $$f(2.4 - 0.1) = f(2.3) = 1.37 + (2.68 - 1.37) \frac{2.4 - 2.3}{2.4 - 1.2} = 1.48,$$ $$f''(x) = \frac{1.18 + 1.48 - 2 \cdot 1.37}{0.1^2} = -8.$$

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