So I have the following question here:
Find the Macluarin series of $\displaystyle F(x) = \int_{0}^{x} (1+t^2)\cos(t^2)dt$. Use this series to Evaluate $F(\frac{\pi}{2})$ with an error less than $0.001$.
Now, I know the basic idea. The Maclaurin series of $\displaystyle \cos(x)=\sum_{n=0}^{\infty} \frac{(-1)^n(t^{2n})}{(2n)!}$. So then I would just expand the Integral like so:
$\displaystyle F(x) = \int_{0}^{x} (1+t^2)\cos(t^2)dt$
$\displaystyle F(x) = \int_{0}^{x} (1+t^2)\sum_{n=0}^{\infty} \frac{(-1)^n(t^{4n})}{(2n)!}dt$
$\displaystyle F(x) = \int_{0}^{x}\sum_{n=0}^{\infty} \frac{(-1)^n(t^{4n})}{(2n)!}dt + \int_{0}^{x}\sum_{n=0}^{\infty} \frac{(-1)^n(t^{4n+2})}{(2n)!}dt$
$\displaystyle F(x) = \sum_{n=0}^{\infty} \frac{(-1)^n(x^{4n+1})}{(2n)!(4n+1)} + \sum_{n=0}^{\infty} \frac{(-1)^n(x^{4n+3})}{(2n)!(4n+3)}$
As far as I know, I would have to combine both of these into a single sum to get my maclaurin series.
Now I know that because these series alternate, I have to use the alternating series estimation theorem and make the error less than $0.001$.
Here's where I'm stuck… How do I do that? This would be fine if I had a single sum. However I have two sums here. How do I deal with that?
I could do this by adding up terms if I wanted to I guess. This would require me to integrate $10$ terms as such:
$\displaystyle \int_{0}^{\frac{\pi}{2}}\left(1+x^2-\frac{x^4}{2}-\frac{x^6}{2}+\frac{x^8}{24}+\frac{x^{10}}{24}-\frac{x^{12}}{720}-\frac{x^{14}}{720}+\frac{x^{16}}{40320}+\frac{x^{18}}{40320}\right)dx \approx 0.9259$ which gives me the desired ammount I want to that the error does not exceed $0.001$. However, this requires me to know the value of the integral which I can't find using elementary methods.
Is there a way I could do it with my original method or using a series + the Alternating series estimation theorem? Help would be appreciated. Thank you very much.
EDIT: Corrected to account for the $t^2$ for the maclaurin series of cosine.
Best Answer
It could be easier to use the fundamental theorem of calculus $$F(x) = \int_{0}^{x} (1+t^2)\cos(t^2)\,dt \implies F'(x)=(1+x^2)\cos(x^2)$$ Now, let $y=x^2$ and you should arrive at $$F'(x)=1+\sum_{n=1}^\infty\frac{n \sin \left(\frac{\pi n}{2}\right)+\cos \left(\frac{\pi n}{2}\right)}{n!} x^{2n}$$ and integrating termwise $$F(x)=x+\sum_{n=1}^\infty\frac{n \sin \left(\frac{\pi n}{2}\right)+\cos \left(\frac{\pi n}{2}\right)}{(2n+1)n!} x^{2n+1}$$ This is an alernating series. So, if you write $$F(x)=x+\sum_{n=1}^{p-1}\frac{n \sin \left(\frac{\pi n}{2}\right)+\cos \left(\frac{\pi n}{2}\right)}{(2n+1)n!} x^{2n+1}+\sum_{n=p}^\infty\frac{n \sin \left(\frac{\pi n}{2}\right)+\cos \left(\frac{\pi n}{2}\right)}{(2n+1)n!} x^{2n+1}$$ The first neglected term is $$R_p=\frac{p \sin \left(\frac{\pi p}{2}\right)+\cos \left(\frac{\pi p}{2}\right)}{(2p+1)p!} x^{2p+1} $$ which makes $$R_{2p}=\frac{x^{4 p+1}}{(4 p+1) (2 p)!}\sim \frac{x^{4 p+1}}{2 (2 p+1)!} \qquad\text{and}\qquad R_{2p+1}=\frac{x^{4 p+3}}{(4 p+3) (2p)!}\sim \frac{x^{4 p+3}}{2 (2 p+1)!}$$
So, depending on the value of $x$ we need to solve either $$(2p+1)!=\frac 1{2x} (x^2)^{(2p+1)} 10^k\qquad\text{or}\qquad (2p+1)!=\frac x{2} (x^2)^{(2p+1)} 10^k $$ in order to have $R \leq 10^{-k}$.
Have a look at this question of mine; you will find a magnificent approximation provided by @robjohn, an eminent user on this site. Adapted to your problem, this would give $$2p+1 \sim e x^2 \exp\Big[{W\left(2 \log \left(\frac{10^k}{8 \pi x^3}\right)\right) }\Big]-\frac 12$$ $$2p+1 \sim e x^2 \exp\Big[{W\left(2 \log \left(\frac{10^k}{8 \pi x}\right)\right) }\Big]-\frac 12$$ where $W(.)$ is Lambert function.
Applied to $x=1$ and $k=3$, both formulae will give $p=5.68784$, that is to say $p=6$.