Use the Taylor polynomial of degree up to 3 for y = cos x along with Taylor's inequality to estimate (integral from 0 to 1) of cos(x^2) and give a bound on the error that your estimate makes.

I'm not quite sure how to do this.

Printable View

- Mar 28th 2010, 08:16 PMcrymorenoobsTaylor's inequality problem
Use the Taylor polynomial of degree up to 3 for y = cos x along with Taylor's inequality to estimate (integral from 0 to 1) of cos(x^2) and give a bound on the error that your estimate makes.

I'm not quite sure how to do this. - Mar 28th 2010, 09:21 PMAnonymous1
$\displaystyle \int_0^1 \cos(x^2)\approx\int_0^1 (1-\frac{x^4}{2}) = \int_0^1 1-\int_0^1 \frac{x^4}{2}$

Also, you need to find the remainder term to find the bound on error. You know... $\displaystyle R_n(x)= \frac{f^{n+1}(x)}{n!}$ - Mar 28th 2010, 09:33 PMcrymorenoobs
Ok thank you, I don't know why the integral threw me off...seems straight forward enough :).

- Mar 28th 2010, 09:41 PMcrymorenoobs
Incase anyone else is interested in this answer in the future, the x^8 term should not be included (since it is of degree 3 and the x^4 term isn't used). Thanks again for the reply though.

- Mar 28th 2010, 10:02 PMAnonymous1
Yes, this is correct. Note that if you want $\displaystyle equality$ to $\displaystyle cos(x^2),$ use the $\displaystyle \infty ^{th}$ order expansion, which is what I implied with $\displaystyle +...$

I've edited the post to show we are finding an $\displaystyle approximation.$