Question

Use the forward finite difference method to estimate the first derivative of f(x) = 5.4x^4 -0.12 x^3 + 2x^2 – 0.5x +1.7 at x =0.25, using a step size of h =0.5 and then with a step size of 0.125. Explicitly calculate the derivative. Then calculate the absolute relative true percent error for each step size. 5 points

Answer 1

Answer:

$\frac{\Delta_{0.5}[f](0.25)}{0.5}=4.7775$

$\frac{\Delta_{0.125}[f](0.25)}{0.125}=1.3999$

$f'(x)=0.815$

For h=0.5:

$E_{absolute}=3.962$

$E_{relative}=4.8619$

$E_{percent}=486\%$

For h=0.125:

$E_{absolute}=1.3999$

$E_{relative}=0.7176$

$E_{percent}=71.7\%$

Step-by-step explanation:

The forward finite difference has the next formula:

$\frac{\Delta_{h}[f](x)}{h} =\frac{f(x+h)-f(x)}{h}$

With x=0.25 and h=0.5:

x+h=0.75:

$f(x)=5.4(x)^{4}-0.12(x)^{3}+2(x)^{2}-0.5(x)+1.7\\f(x+h)=5.4(x+h)^{4}-0.12(x+h)^{3}+2(x+h)^{2}-0.5(x+h)+1.7$

then:

$f(0.25)=5.4(0.25)^{4}-0.12(0.25)^{3}+2(0.25)^{2}-0.5(0.25)+1.7\\f(0.25)=1.7192\\f(0.75)=5.4(0.75)^{4}-0.12(0.75)^{3}+2(0.75)^{2}-0.5(0.75)+1.7\\f(0.75)=4.1079$

$\frac{\Delta_{h}[f](x)}{h} = \frac{f(x+h)-f(x)}{h}$

$\frac{\Delta_{h}[f](x)}{h} = \frac{4.1079-1.7192}{0.5}=4.7775$

Now with x=0.25 and h=0.125:

x+h=0.375:

$f(x)=5.4(x)^{4}-0.12(x)^{3}+2(x)^{2}-0.5(x)+1.7\\f(x+h)=5.4(x+h)^{4}-0.12(x+h)^{3}+2(x+h)^{2}-0.5(x+h)+1.7$

then:

$f(0.25)=5.4(0.25)^{4}-0.12(0.25)^{3}+2(0.25)^{2}-0.5(0.25)+1.7\\f(0.25)=1.7192\\f(0.375)=5.4(0.375)^{4}-0.12(0.375)^{3}+2(0.375)^{2}-0.5(0.375)+1.7\\f(0.375)=1.8942$

$\frac{\Delta_{h}[f](x)}{h} = \frac{f(x+h)-f(x)}{h}$

$\frac{\Delta_{h}[f](x)}{h} = \frac{1.8942-1.7192}{0.125}=1.3999$

We have to find the derivative at x=0.25:

$f(x)=ax^{n}\\\\f'(x)=anx^{n-1}$

$f'(x)=5.4(4)x^{(4-1)}-0.12(3)x^{(3-1)}+2(2)x^{(2-1)}-0.5(1)x^{(1-1)}\\\\f'(x) =21.6x^{3}-0.36x^{2}+4x-0.5\\\\f'(0.25)=21.6(0.25)^{3}-0.36(0.25)^{2}+4(0.25)-0.5\\\\f'(0.25)=0.815$

Now the errors:

For h=0.5:

$E_{absolute}= V_{finite}-V_{derivative}\\\\E_{absolute}= 4.7775-0.815=3.9625\\\\E_{relative}=\frac{E_{absolute}}{V_{derivative}}\\\\E_{relative}=\frac{3.9625}{0.815}=4.8619\\\\Percent=E_{relative}*100=486\%$

For h=0.125:

$E_{absolute}= V_{finite}-V_{derivative}\\\\E_{absolute}= 1.3999-0.815=0.5849\\\\E_{relative}=\frac{E_{absolute}}{V_{derivative}}\\\\E_{relative}=\frac{0.5849}{0.815}=0.7176\\\\Percent=E_{relative}*100=71.7\%$