# Thread: antiderivative of a power series

1. ## antiderivative of a power series

Hi everyone. My final is tomorrow, and I'm going over a lot of stuff that plans to be covered on the test. My professor had to leave town suddenly before he could fully cover everything, but chose not to change the test. He never went over differentiation or antiderivatives of a power series. I was wondering if anyone could help me out with both of these things for this given problem, and Im sure I could apply these techniques to all future problems.

Differentiation and antiderivative of the following power series:

this would be a great help! thanks so much!

2. Originally Posted by etha
Hi everyone. My final is tomorrow, and I'm going over a lot of stuff that plans to be covered on the test. My professor had to leave town suddenly before he could fully cover everything, but chose not to change the test. He never went over differentiation or antiderivatives of a power series. I was wondering if anyone could help me out with both of these things for this given problem, and Im sure I could apply these techniques to all future problems.

Differentiation and antiderivative of the following power series:

this would be a great help! thanks so much!
you have the formula $\displaystyle \frac {2n}{2n + 1} \cdot x^n$ ............$\displaystyle \frac {2n}{2n + 1}$ is a constant. you can integrate and differentiate using the power rules

3. Originally Posted by etha
Hi everyone. My final is tomorrow, and I'm going over a lot of stuff that plans to be covered on the test. My professor had to leave town suddenly before he could fully cover everything, but chose not to change the test. He never went over differentiation or antiderivatives of a power series. I was wondering if anyone could help me out with both of these things for this given problem, and Im sure I could apply these techniques to all future problems.

Differentiation and antiderivative of the following power series:

this would be a great help! thanks so much!
Power series integrate and differentiate term wise (like a polynomial)

$\displaystyle \frac{d}{dx}\sum_{n=0}^{\infty}\frac{2nx^n}{2n+1}= \sum_{n=0}^{\infty}\frac{2n(n)x^{n-1}}{2n+1}=\sum_{n=0}^{\infty}\frac{2n^2x^{n-1}}{2n+1}$

Now for the integral

$\displaystyle \int \sum_{n=0}^{\infty}\frac{2nx^n}{2n+1}= C+\sum_{n=0}^{\infty}\frac{2nx^{n+1}}{(2n+1)(n+1)} =$

4. what exactly are the power rules? When I mean my professor skipped town, he literally skipped town a week before finals and expected us to learn the material on our own. The book I have is really confusing, and the help from this forum has much easier to understand.

5. Originally Posted by TheEmptySet
Power series integrate and differentiate term wise (like a polynomial)

$\displaystyle \frac{d}{dx}\sum_{n=0}^{\infty}\frac{2nx^n}{2n+1}= \sum_{n=0}^{\infty}\frac{2n(n)x^{n-1}}{2n+1}=\sum_{n=0}^{\infty}\frac{2n^2x^{n-1}}{2n+1}$

Now for the integral

$\displaystyle \int \sum_{n=0}^{\infty}\frac{2nx^n}{2n+1}= C+\sum_{n=0}^{\infty}\frac{2nx^{n+1}}{(2n+1)(n+1)} =$
i see, that helps a lot. I have a question though, when you differentiate, why does the 2n multiply by n and become n^2

6. Originally Posted by etha
what exactly are the power rules? When I mean my professor skipped town, he literally skipped town a week before finals and expected us to learn the material on our own. The book I have is really confusing, and the help from this forum has much easier to understand.
As you may have figured out from EmptySet's post...

The Power Rules for differentiation and integration are respectively as follows:

$\displaystyle \frac d{dx} x^n = n x^{n - 1}$

$\displaystyle \int x^n~dx = \frac {x^{n + 1}}{n + 1} + C$ for $\displaystyle n \ne -1$. (if $\displaystyle n = -1$, $\displaystyle \int x^{-1}~dx = \ln x + C$)

7. Originally Posted by etha
i see, that helps a lot. I have a question though, when you differentiate, why does the 2n multiply by n and become n^2
$\displaystyle n \cdot n = n^2$ because $\displaystyle x^a \cdot x^b = x^{a + b}$

8. Originally Posted by etha
i see, that helps a lot. I have a question though, when you differentiate, why does the 2n multiply by n and become n^2
It is just simplification

$\displaystyle 2n(n)=2n^2$ n is just a parameter.

this is just like$\displaystyle 2x(x)=2x^2$

Good luck.

Edit: Dang you are speedy.
I need to get faster

9. Originally Posted by TheEmptySet
Edit: Dang you are speedy.
I need to get faster
Hehe, i'm not nearly fast enough to be around here. I was just lucky. anyway, i'll leave you to run the forum, i'm logging off

10. Originally Posted by TheEmptySet

$\displaystyle 2n(n)=2n^2$ n is just a parameter.

this is just like$\displaystyle 2x(x)=2x^2$
well i know that when you multiply the n in it becomes n^2 lol. I just don't understand why you mutiply it by n. I'm sorry if it seems like I'm a slow learner!

11. Originally Posted by etha
well i know that when you multiply the n in it becomes n^2 lol. I just don't understand why you mutiply it by n. I'm sorry if it seems like I'm a slow learner!
It is what the power rule says we should do. see post #6