Taylor's inequality problem

**crymorenoobs** · March 28th 2010, 08:16 PM

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.

**Anonymous1** · March 28th 2010, 09:21 PM

$\int_0^1 \cos(x^2)\approx\int_0^1 (1-\frac{x^4}{2}) = \int_0^1 1-\int_0^1 \frac{x^4}{2}$

Also, you need to find the remainder term to find the bound on error. You know... $R_n(x)= \frac{f^{n+1}(x)}{n!}$

**crymorenoobs** · March 28th 2010, 09:33 PM

Ok thank you, I don't know why the integral threw me off...seems straight forward enough

.

**crymorenoobs** · March 28th 2010, 09:41 PM

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.

**Anonymous1** · March 28th 2010, 10:02 PM

Yes, this is correct. Note that if you want $equality$ to $cos(x^2),$ use the $\infty ^{th}$ order expansion, which is what I implied with $+...$

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

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