# Math Help - Taylor/Fourier series, Imaginary numbers, is this right?

1. ## Taylor/Fourier series, Imaginary numbers, is this right?

Hello everyone! I figured it would be easier if I just uploaded my work pages and posted them instead of trying to type everything out (because I don't know how to use those math tags ) so please find that attached. My calculus final is tomorrow, and I really am just hoping someone can review my work and help me out with a couple of questions I have.

I did everything except #3, 7, and 10.

On #3, can I use a comparison test with an is my original series listed, and with bn being 1/n^2.3? Or do I need to do a limit comparison test, and will bn = 1/n^2.3 work for that? OR did I pick the wrong comparison function? I did try doing the limit comparison test, and I believe I got c=0, so.. yeah I probably did this wrong

On #4, I realize I forgot a negative sign on one of the 1/10. Is it correct, or do I have to write it as a-1/10 < x < a+1/10? Or did I just do it completely wrong?

On #5, I think I may have gotten the wrong taylor series... and for the endpoint x=-8, I I can't use the alternating series test, right? would I be able to use a limit comparison test instead? Actually, upon further review, I'm not even sure what I would compare it to! Maybe... -1/n! ?

On #7, I hate integration by parts, so I skipped over it when I got to that point to make sure I could do everything else. I guess I should ask to confirm, anyway. d/dx [sin (kx)]/k = cos (kx) dx? I know, basic differenatiaon, I'm not very good at cos/sin functions.

And #10, I haven't gotten around to that one yet.

Thanks for the help!

2. ok for the 3th question

$\sum_{n=2}^{\infty}\frac{n^2+log(n)}{n^{4.3}-3(log(log(n))^4}$

you can use comparative test

$\frac{n^2+log(n)}{n^{4.3}-3(log(log(n)))^4}<\frac{n^2+log(n)}{n^{4.3}}$

and the second one is converge by integral test or any other test

4) where dose the $\sum_{n=0}^{\infty}c_n(x-a)$ converge
if we know

$lim_{n\rightarrow\infty}\mid\frac{c_{n+1}}{c_n}\mi d=10$

$\mid 10(x-a)\mid<1$
$-1<10(x-a)<1$
$-1/10 +a

I will continue ....

3. the question is what is the taylor series for xsin(x/4) centered at zero what is the radius of convergence of this series ?
ok you know that

$sin(x)=\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n+1}}{(2n+1)!}
$

so for

$sin\left(\frac{x}{4}\right)=\sum_{n=0}^{\infty}\fr ac{(-1)^{n}\left(\frac{x}{4}\right)^{2n+1}}{(2n+1)!}$

so

$xsin\left(\frac{x}{4}\right)=x\sum_{n=0}^{\infty}\ frac{(-1)^{n}\left(\frac{x}{4}\right)^{2n+1}}{(2n+1)!}=\s um_{n=0}^{\infty}\frac{(-1)^{n}x^{2n+2}}{4^{2n+1}(2n+1)!}$

what is the radius of convergence I do not really know ??

4. now the next one

7)
$let..f(x)=x^2$ for $-\pi\geq x\geq\pi$ and then be periodic of period $2\pi$ can you compute a couple Fourier series coefficients ? what do you already know about $b_n$ ?

first $b_n=0$ since f(x) is even

$a_n=\frac{1}{\pi}\int_{-\pi}^{\pi} x^2 cos(xn) dx$

you can solve it by parts twice let $u=x^2...du=2x dx$ and $dv=cos(xn)....v=\frac{sin(xn)}{n}$

$\frac{1}{\pi}\int_{-\pi}^{\pi} x^2 cos(nx) dx={\color{red}x^2\left(\frac{sin(nx)}{n}\right)}-\int_{-\pi}^{\pi}2x\left(\frac{sin(nx)}{n}\right) dx$

the red term is zero because [math sin(n\pi)=0...and....sin(-n\pi)=0[/tex]
by parts again let $u=x$
and
$dv=sin(nx)....v=-\frac{cos(nx)}{n}$

$\frac{1}{\pi}\int_{-\pi}^{\pi} x^2 cos(nx) dx={\color{red}x^2\left(\frac{sin(nx)}{n}\right)}-\left(\frac{-2xcos(nx)}{n^2}+{\color{red}\int_{n=0}^{\infty} \frac{2cos(nx)}{n^2}}\right) dx$

the red terms is zero

sub the limits of the integral
you continue

5. the 10th question I will write it just so anyone can help you

10)give the parametrization of the line connecting A=(2,3,4) and B=(-2,3,7) so the velocity of the parametrization is constant the point A corresponds to the line at t=0 and the point B is given by t=2

Best wishes