# Maclaurin Series

1. Oct 10, 2009

### .....

1. The problem statement, all variables and given/known data

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

2. Relevant equations

Given in the problem.

3. 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

2. Oct 10, 2009

### Dick

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.

3. Oct 10, 2009

### .....

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

4. Oct 10, 2009

### Dick

Alright. So now you've got 1/(1+1-x^2+x^4/3+...). That's 1/(2-x^2/+x^4/3...). Write it as (1/2)*1/(1-(x^2/2-x^4/6...)). Now use 1/(1-a)=1+a+a^2+... with a=x^2/2-x^4/6+... If you only want a quadratic approximation then you didn't need to keep the x^4 stuff, right?