LaGrange Error and power series to approximate 1/(1-x^2)

[SoaringQuail]

There's a homework problem that I've been struggling over:

Find a formula for the truncation error if we use 1 + x^2 + x^4 +x^6 to approximate 1/(1-x^2) over the interval (-1, 1).

Now, I assume that you need to use LaGrange error but I'm not sure how to proceed. Any help would be greatly appreciated.

 

[Count Iblis]

Re: LaGrange Error and power series

There's a homework problem that I've been struggling over:

Find a formula for the truncation error if we use 1 + x^2 + x^4 +x^6 to approximate 1/(1-x^2) over the interval (-1, 1).

Now, I assume that you need to use LaGrange error but I'm not sure how to proceed. Any help would be greatly appreciated.

This problem can be solved by the most powerful method that exists in mathematics: cheating.

You know that in the interval (-1, 1) you have:


\(\displaystyle \L \frac{1}{1-x^2}=\sum_{n=0}^{\infty}x^{2n}\)

This means that the error \(\displaystyle e\) is given by:

\(\displaystyle \L e=\sum_{n=8}^{\infty}x^{2n} = \frac{x^{8}}{1-x^{2}}\)

 

[SoaringQuail]

that actually makes sense. Thanks