# Integration of taylor series

1. Mar 30, 2013

### Zondrina

1. The problem statement, all variables and given/known data

A problem from advanced calculus by Taylor :

http://gyazo.com/5d52ea79420c8998a668fab0010857cf

2. Relevant equations

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

$sin(3x) = \sum_{n=0}^{∞} (-1)^n \frac{3^{2n+1}x^{2n+1}}{(2n+1)!}$

3. The attempt at a solution

I'm not quite seeing what the question is trying to ask me here. It wants me to find a series for the integral of f(x)?

I wrote out the sin(px) in terms of their power series and then integrated f(x) from 0 to π/2. So I have :

$\int_{0}^{π/2} f(x) dx = \frac{\sum_{n=0}^{∞} \frac{(-1)^n 3^{2n+1} (π/2)^{2n+2}}{(2n+1)!(2n+2)}}{1 * 2} + \frac{\sum_{n=0}^{∞} \frac{(-1)^n 5^{2n+1} (π/2)^{2n+2}}{(2n+1)!(2n+2)}}{3 * 4} + ......$

I don't see where to go from here or what I'm even supposed to be doing with this monster.

2. Mar 30, 2013

### voko

What makes you think you are required to mess with power series? Can you integrate the series term by term? Why?

3. Mar 30, 2013

### Zondrina

Well I don't HAVE to use the power series I suppose.

Yes I can integrate the series term by term. Hm, what if I write :

$f(x) = \sum_{n=1}^{∞} \frac{sin((2n+1)x)}{not sure} = \sum f_n(x)$

Having a bit of trouble getting the denominator on that one. If I can show that the series is uniformly convergent on $[0, π/2]$ and each $f_n(x)$ is continuous on the interval, then I can integrate the terms of the series term by term which will give me the series I'm looking for.

$\int_{0}^{π/2} f(x) dx = \int_{0}^{π/2} \frac{sin(3x)}{1*2} + \int_{0}^{π/2} \frac{sin(5x)}{3*4} + \int_{0}^{π/2} \frac{sin(7x)}{5*6} + ...$

4. Mar 30, 2013

### voko

Does (2n - 1)(2n) work for the denominator?

5. Mar 30, 2013

### Zondrina

I was just thinking (2n+1)(2n), but (2n-1) works since it goes oddeven oddeven. I was actually getting shafted by (2n+1) screwing everything.

So the series would be uniformly bounded by $M_n = \frac{1}{n^2}$. Since $\sum M_n$ converges by p-comparison, we know $\sum f_n(x)$ will uniformly converge by the M-Test.

Since each term in the series is also continuous I integrate :

$\int_{0}^{π/2} f(x) dx = \sum_{n=1}^{∞} \int_{0}^{π/2} \frac{(2n+1)x}{4n^2 - 2n} = \sum_{n=1}^{∞} \frac{(2n+1)x^2}{8n^2 - 4n}$

and... there's my series !

6. Mar 30, 2013

### voko

I do not understand your last step. You should be integrating $\frac {\sin (2n + 1)x} {(2n - 1)(2n)}$, no?

7. Mar 30, 2013

### Zondrina

Oh whoops, I jumped a little too quickly and forgot the sin.

$\int_{0}^{π/2} f(x) dx = \sum_{n=1}^{∞} \int_{0}^{π/2} \frac{sin((2n+1)x)}{4n^2 - 2n}dx$

Then using $u = (2n+1)x$, I get $\frac{1}{2n+1}du = dx$ and so on and so forth.

I'm sure I got this now :). Thanks voko.