# Maclaurin Series

## Homework Statement

How can I calculate it for $$\frac{1}{1+cos^2(x)}$$ by using the fact that $$\frac{1}{1+x^2} = 1 - x^2 + x^4 - ...$$?

## Homework Equations

Given in the problem.

## The Attempt at a Solution

I tried letting u = cos(x), then

$$\frac{1}{1+cos^2(x)} = \frac{1}{1+u^2} = 1 - u^2 + u^4 - ... = 1 - cos^2(x) + cos^4(x) - ...$$

But I don't think this is right because the first term should be 0.5, not 1... and I don't see how a -0.5 might pop out of this series of cos terms....and even if it somehow does, I think this question is not meant to be that difficult...

The answer I am trying to get is $$\frac{1}{2} + \frac{1}{4}x^2 + ...$$

Any ideas?

Thanks

Dick
Homework Helper
Use cos(x)=1-x^2/2!+x^4/4!-... to express the first few terms of the series for 1+cos(x)^2. Then, sure, use your geometric series. Feel free to drop terms that you know won't be important.

I don't follow, could you spell it out a bit more?

So $$cos^2(x) = 1-x^2+\frac{1}{3}x^4-...$$

Now what?

Thanks

Dick