# Help me! This power series is driving me crazy!

• Jun 29th 2008, 03:18 PM
sgares
Help me! This power series is driving me crazy!
The function http://i29.tinypic.com/2cpezvr.png is represented as a power series
http://i28.tinypic.com/rbk0fp.png
Find the first few coefficients in the power series.
C0-C5

I've obtained C0 and the radius of convergence but every other answer I put in keeps coming back as incorrect.
I'm going by 1/(1-x) = 1 + x + x^2 + ...
So for C0 I get 1x3=3
This is where it goes downhill... For C1 I get 3[-100x^2]^1 so C1=-300?? and I sort of repeat this method except each time the final exponent gets larger.
I found the radius of convergence by doing |100x^2| < 1 so x < 1/10
Please help with the rest! I've been doing this problem FOREVER and I don't understand why I at very least haven't guessed the answers! :(
• Jun 29th 2008, 03:20 PM
Mathstud28
Quote:

Originally Posted by sgares
The function http://i29.tinypic.com/2cpezvr.png is represented as a power series
http://i28.tinypic.com/rbk0fp.png
Find the first few coefficients in the power series.
C0-C5

I've obtained C0 and the radius of convergence but every other answer I put in keeps coming back as incorrect.
I'm going by 1/(1-x) = 1 + x + x^2 + ...
So for C0 I get 1x3=3
This is where it goes downhill... For C1 I get 3[-100x^2]^1 so C1=-300?? and I sort of repeat this method except each time the final exponent gets larger.
I found the radius of convergence by doing |100x^2| < 1 so x < 1/10
Please help with the rest! I've been doing this problem FOREVER and I don't understand why I at very least haven't guessed the answers! :(

$\displaystyle \frac{3}{1+100x^2}=3\cdot\frac{1}{1+(10x)^2}=3\cdo t\sum_{n=0}^{\infty}(-1)^n(10x)^{2n}$

use the Root test
• Jun 29th 2008, 03:38 PM
sgares
What do you mean by "use the root test"? How will that help me find the coefficients?
Also, using your formula and simply plugging in n=1, n=2, n=3, and n=4 I received the coefficients 3, -300, 30000, -30000000, 300000000... these are the same answers I was receiving prior to this, and they were all wrong besides the first one.
• Jun 29th 2008, 03:42 PM
Mathstud28
Quote:

Originally Posted by sgares
What do you mean by "use the root test"? How will that help me find the coefficients?
Also, using your formula and simply plugging in n=1, n=2, n=3, and n=4 I received the coefficients 3, -300, 30000, -30000000, 300000000... these are the same answers I was receiving prior to this, and they were all wrong besides the first one.

You asked for the radius of convergence, and that is the right power series.
• Jun 29th 2008, 03:55 PM
sgares
I believe you that it is the correct power series, but why am I getting incorrect answers for the coefficients then? Is it something wrong with my calculations or with my webwork? Also, I stated that I already calculated the radius of convergence, and it was 1/10.

As I just stated, I received the coefficients 3, -300, 30000, -30000000, 300000000, and the final four of these are incorrect. Why though? :\
• Jun 29th 2008, 03:56 PM
Mathstud28
Quote:

Originally Posted by sgares
I believe you that it is the correct power series, but why am I getting incorrect answers for the coefficients then? Is it something wrong with my calculations or with my webwork? Also, I stated that I already calculated the radius of convergence, and it was 1/10.

As I just stated, I received the coefficients 3, -300, 30000, -30000000, 300000000, and the final four of these are incorrect. Why though? :\

That should be right...why do you say it is wrong?
• Jun 29th 2008, 04:05 PM
sgares
I don't know man... I'm so flustered at this point from this problem! Grrr!
http://i32.tinypic.com/k54wb5.jpg
• Jun 29th 2008, 04:31 PM
topsquark
It is probably because you have found the first five nonzero coefficients. I suspect the answer is
3, 0, -300, 0, 30000

-Dan
• Jun 29th 2008, 04:58 PM
sgares
Woah, that was correct topsquark... thank you so much for that lol. I still have no idea why that is correct though, so if anyone could explain this to me that would make my day a lot better!

Seriously though, if I could I would sit here thanking you all day for coming up with that answer. That just made me feel so much better.
• Jun 29th 2008, 05:40 PM
Mathstud28
Quote:

Originally Posted by sgares
Woah, that was correct topsquark... thank you so much for that lol. I still have no idea why that is correct though, so if anyone could explain this to me that would make my day a lot better!

Seriously though, if I could I would sit here thanking you all day for coming up with that answer. That just made me feel so much better.

Do you see how the exponent of $\displaystyle x^{2n}$ is going to be even since $\displaystyle n\in\mathbb{N}$?

Well did you wonder what happened to the even terms?

The reason is that if we let

$\displaystyle f(x)=\frac{1}{1+100x^2}$

and let $\displaystyle f^{(n)}(0)$

We see that

$\displaystyle f^{(n)}(0)=\left\{ \begin{array}{cc} (-1)^n\cdot{10}^{2n} & \mbox{ if } {\text{n is even}}\\ 0 & \mbox{ if } {\text{n is odd}}\end{array}\right.$

So since two (three if excluding zero) of the first terms are odd numbers, we can see that they are zero.
• Jun 30th 2008, 02:34 AM
topsquark
Another piece of advice. The problem asked you to find the Taylor series about x = 0. What Mathstud (and presumably you) derived was based on a geometric series summation. The two give identical answers, but if you did the actual work with the Taylor series you would have seen the coefficients that were 0 directly.

That being said I would probably have given the same answer as you as such problems typically ask for only the non-zero coefficients.

-Dan