# Percentage Errors using the Maclaurin Series

• May 28th 2009, 04:06 AM
Percentage Errors using the Maclaurin Series
Hi, I'm having trouble trying to find the percentage errors of P3(x) and P4(x).

Here's the question;
When x=0.7891.

f(x)=x/(sin(x)+2)
f(x)=0.2912

P3(x)=1/2*x-1/4*x^2
P3(x) =0.2389

P4(x)=1/2*x-1/4*x^2+1/8*x^3
P4(x) =0.3003

Find the percentage errors in using P3(x), and P4(x) to approximate f(x).

Thanks, DLL.
• May 28th 2009, 08:56 AM
Spec
The formula for relative error is simply: $\eta=\left | \frac{\Delta f(x)}{f(x)}\right |$ where $\Delta f(x)=f(x)-f(x)_{approximation}$ for some value of $x$.
• May 29th 2009, 03:20 AM
I'm not sure how to use this formula as I have never come across it before. I was thinking of the Maclarin Remaider Therom. I'm not sure how I use these errors to approximate f(x) however.
• May 29th 2009, 11:17 PM
I don't need to use the relative error formula. I need to find the percentage errors of in usig P3(x) and P4(x) to approximate f(x).

f(x)=x/(sins(x)=2) and the Maclurin polynomials are P3(x)=1/2*x-1/4*x^2 and P4(x)=1/2*x-1/4*x^2+1/8*x^3, when x=0.7891.

I'm not sure how I use the percentage errors or use them to approximate f(x).
• May 30th 2009, 05:56 AM
Spec
If you express the error as percentages, then it's relative to the true value. You can find the error of using a Maclaurin series to approximate a function, by using the Lagrange remainder theorem, but the answer you get is going to be the absolute error.

The remainder is $r(x)=\frac{f^{(n+1)}(\xi)}{(n+1)!}x^{n+1}$ for some $\xi \in [0,x]$

If you want to the find the absolute error by using the Lagrange remainder theorem, then you first need to find $f^{4}(\xi)$ for the first polynomial ( $P_3$) and $f^{5}(\xi)$ for the second ( $P_4$).
• May 30th 2009, 06:09 AM