1. Let f(x)=sin(x) where x is measured in radians. (a) Calculate approximations to f'(0.8) using formula (3 ) with h=0.1,h=0.01 and h=0.001 Carry eight or nine decimal places. (b) Compare with the value f'(0.8)=cos(0.8) (c) Compute bounds for the truncation error (4). Use vert f^(3)(c)vert leqslant cos(0.7)approx 0.764842187 for all cases.

Question

Soru

1. Let f(x)=sin(x) where x is measured in radians. (a) Calculate approximations to f'(0.8) using formula (3 ) with h=0.1,h=0.01 and h=0.001 Carry eight or nine decimal places. (b) Compare with the value f'(0.8)=cos(0.8) (c) Compute bounds for the truncation error (4). Use vert f^(3)(c)vert leqslant cos(0.7)approx 0.764842187 for all cases.

Answer 1

(a) To calculate approximations to $f'(0.8)$ using the formula (3), we can use the forward difference formula: $f'(x) \approx \frac{f(x+h) - f(x)}{h}$ Let's calculate the approximations for $h=0.1$, $h=0.01$, and $h=0.001$: For $h=0.1$: $f'(0.8) \approx \frac{f(0.9) - f(0.8)}{0.1} = \frac{\sin(0.9) - \sin(0.8)}{0.1} \approx 0.145520000$ For $h=0.01$: $f'(0.8) \approx \frac{f(0.81) - f(0.8)}{0.01} = \frac{\sin(0.81) - \sin(0.8)}{0.01} \approx 0.145520000$ For $h=0.001$: $f'(0.8) \approx \frac{f(0.801) - f(0.8)}{0.001} = \frac{\sin(0.801) - \sin(0.8)}{0.001} \approx 0.145520000$ (b) Comparing the approximations with the value $f'(0.8)=\cos(0.8)$, we have: $f'(0.8) \approx \cos(0.8) \approx 0.696708} (c) To compute bounds for the truncation error, we can use the error bound formula (4): $\vert f'(x) - f'(x)\vert \leq \frac{h^2}{6}\vert f^{(3)}(c)\vert$ where $f'(x)$ is the approximation, $f'(x)$ is the exact value, $h$ is the step size, and $c$ is some value in the interval $[x, x+h]$. Given that $\vert f^{(3)}(c)\vert \leqslant \cos(0.7) \approx 0.764842187$ for all cases, we can substitute this value into the error bound formula: $\vert f'(0.8) - \cos(0.8)\vert \leq \frac{h^2}{6}\cdot 0.764842187$ For $h=0.1$: $\vert f'(0.8) - \cos(0.8)\vert \leq \frac{(0.1)^2}{6}\cdot 0.764842187 \approx 0.001271$ For $h=0.01$: $\vert f'(0.8) - \cos(0.8)\vert \leq \frac{(0.01)^2}{6}\cdot 0.764842187 \approx 0.000127$ For $h=0.001$: $\vert f'(0.8) - \cos(0.8)\vert \leq \frac{(0.001)^2}{6}\cdot 0.764842187 \approx 0.0000127$

Soru

Çözüm

Cevap