Solving Power Series for 9/25: Find x When y = 9/25

[Anewk]

I was asked to find sums equal to 9/25 by using the power series of \(\displaystyle y=\frac{1}{1+x^2}\). First thing I did was to find the power series representation of the function:

\(\displaystyle \sum_{n=0}^{\infty }(-x^2)^n\)

Next I figured out the interval of convergence:

\(\displaystyle \left \| -x^2 \right \|< 1\)

This meant that \(\displaystyle x\) had to be less than 1 and more than -1 for the series to converge.

Now this is where I became a little confused... when I solved the function for 9/25, I found that \(\displaystyle x\) has to be 4/3 to get \(\displaystyle y\) to equal 9/25. But if I let the power series equal to this value then x>1 and the series would not converge. What do I do?

 

[I like Serena]

I was asked to find sums equal to 9/25 by using the power series of \(\displaystyle y=\frac{1}{1+x^2}\). First thing I did was to find the power series representation of the function:

\(\displaystyle \sum_{n=0}^{\infty }(-x^2)^n\)

Next I figured out the interval of convergence:

\(\displaystyle \left \| -x^2 \right \|< 1\)

This meant that \(\displaystyle x\) had to be less than 1 and more than -1 for the series to converge.

Now this is where I became a little confused... when I solved the function for 9/25, I found that \(\displaystyle x\) has to be 4/3 to get \(\displaystyle y\) to equal 9/25. But if I let the power series equal to this value then x>1 and the series would not converge. What do I do?

Hi Anewk! :)

Your result indeed means that the expansion around $x=0$ does not converge.

To find an expansion that does converge, you'll need a different point to expand from.
How about expanding the following instead (at $x=\infty$)?
$$y=\frac{1}{1+x^2} = \frac{x^{-2}}{x^{-2} + 1} = 1 - \frac{1}{1 + x^{-2}}$$