1. ## Another Power Series

The function is represented as a power series

Find the first few coefficients in the power series.

I've found c0 to be 6 in this case, and the radius of convergence to be 0.1. Can someone help me find c1, c2, c3, and c4? Please help!

2. Originally Posted by Del
The function is represented as a power series

Find the first few coefficients in the power series.

I've found c0 to be 6 in this case, and the radius of convergence to be 0.1. Can someone help me find c1, c2, c3, and c4? Please help!
$6\cdot\frac{1}{1+(10x)^2}$
and we know that $\frac{1}{1+x^2}=\sum_{n=0}^{\infty}(-1)^{n}x^{2n}$

CAn you gofrom there?

3. Series problems are my worst enemies. I don't know how to find coefficients.

4. Originally Posted by Del
Series problems are my worst enemies. I don't know how to find coefficients.
Put the 10x where the x^{2n} and expand

5. Substitute 10x in the place of x^(2n)? Or just use x = 10x?

6. Originally Posted by Del
Substitute 10x in the place of x^(2n)? Or just use x = 10x?
x=10x

7. I still can't figure this one out. What series can I use to find the coefficients?

8. Originally Posted by Del
I still can't figure this one out. What series can I use to find the coefficients?
$\frac{1}{1+x^2}=\sum_{n=0}^{\infty}(-1)^{n}x^{2n}$

So,
$\frac{1}{1+(10x)^2}=\sum_{n=0}^{\infty}(-1)^{n}(10x)^{2n}$

$\frac{6}{1+(10x)^2}=\sum_{n=0}^{\infty}(-1)^{n}6(10x)^{2n}$

$\text{But }\sum_{n=0}^{\infty}(-1)^{n}6(10x)^{2n} = \sum_{n=0}^{n=\infty} c_n x^n$

$\text{Thus }\sum_{n=0}^{\infty}6(-100)^{n}(x)^{2n} = \sum_{n=0}^{n=\infty} c_n x^n$

So clearly for an even n $c_n = 6(-100)^{\frac{n}2}$ and for an odd n $c_n = 0$