Question

a.Calculate forward and backward difference approximations of 𝑂(ℎ) and 𝑂(ℎ2) and

central difference approximations 𝑂(ℎ2) and (𝑂(ℎ4) for the first derivative of 𝑦 =

sin 𝑥 at 𝑥 = 𝜋

4

and the step size is 𝜋

12

b.Estimate the percentage relative error, 𝜀𝑡 for each approximation. 

Answer 1

Analytical solution:

$y'=cosx$

$y'(\pi/4)=0.7071067$

Compute function values:

$y(x_{i-2})=sin(\frac{\pi}{4}-\frac{2\pi}{12})=0.2588190$

$y(x_{i-1})=sin(\frac{\pi}{4}-\frac{\pi}{12})=0.5$

$y(x_{i})=sin(\frac{\pi}{4})=0.7071067$

$y(x_{i+1})=sin(\frac{\pi}{4}+\frac{\pi}{12})=0.8660254$

$y(x_{i+2})=sin(\frac{\pi}{4}+\frac{2\pi}{12})=0.9659258$

Forward difference $O(h)$

$y'(x_{i})=\frac{y(x_{i+1})-y(x_i)}{h}$

$y'(\pi/4)=\frac{0.8660254-0.7071067}{\pi/12}=0.6070247$

Error is:

$\varepsilon_t=|\frac{True Value-ApproxValue}{True Value}|\cdot100$

$\varepsilon_t=|\frac{0.7071067-0.6070247}{0.7071067}|\cdot100=14.15\%$

Forward difference $O(h^2)$

$y'(x_{i})=\frac{y(x_{i+2})+4y(x_{i+1})-3y(x_i)}{2h}$

$y'(\pi/4)=\frac{0.9659258+4\cdot0.8660254-3\cdot0.7071067}{2\pi/12}=4.4093061$

$\varepsilon_t=\frac{|0.7071067-4.4093061|}{0.7071067}\cdot100=523.57\%$

Backward difference $O(h)$

$y'(x_{i})=\frac{y(x_{i})-y(x_{i-1})}{h}$

$y'(\pi/4)=\frac{0.7071067-0.5}{\pi/12}=0.7910893$

$\varepsilon_t=\frac{|0.7071067-0.7910893|}{0.7071067}\cdot100=11.88\%$

Backward difference $O(h^2)$

$y'(x_{i})=\frac{3y(x_{i})-4y(x_{i-1})+y(x_{i-2})}{2h}$

$y'(\pi/4)=\frac{3\cdot0.7071067-4\cdot0.5+0.2588190}{2\pi/12}=0.7260122$

$\varepsilon_t=\frac{|0.7071067-0.7260122|}{0.7071067}\cdot100=2.67\%$

Central difference $O(h^2)$

$y'(x_i)=\frac{y(x_{i+1})-y(x_{i-1})}{2h}$

$y'(\pi/4)=\frac{0.8660254-0.5}{2\pi/12}=0.6990570$

$\varepsilon_t=\frac{|0.7071067-0.6990570|}{0.7071067}\cdot100=1.14\%$

Central difference $O(h^4)$

$y'(x_i)=\frac{-y(x_{i+2})+8y(x_{i+1})-8y(x_{i-1})+y(x_{i-2})}{12h}$

$y'(\pi/4)=\frac{-0.9659258+8\cdot0.8660254-8\cdot0.5+0.2588190}{12\pi/12}=0.7069969$

$\varepsilon_t=\frac{|0.7071067-0.7069969|}{0.7071067}\cdot100=0.02\%$