1. Maclaurin Series Expansion

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

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...

Any ideas?

Thanks

2. Originally Posted by HD09
How can I calculate it for $\frac{1}{1+cos^2(x)}$ by using the fact that $\frac{1}{1+x^2} = 1 - x^2 + x^4 - ...$?

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...

Any ideas?

Thanks
What about the identity $\cos^2 x = \frac {1 + \cos 2x}{2}$? There's a convenient "divide by 2" in there which should get you your 0.5 you're looking for.

3. Originally Posted by Matt Westwood
What about the identity $\cos^2 x = \frac {1 + \cos 2x}{2}$? There's a convenient "divide by 2" in there which should get you your 0.5 you're looking for.

Hmmm... but then $cos^4(x) = \frac{1+2cos(2x)+cos^2(2x)}{2}$ puts it right back in again

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

4. Maclaurin Series

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 work towards is $\frac{1}{2} + \frac{1}{4}x^2+...$

5. Originally Posted by HD09
How can I calculate it for $\frac{1}{1+cos^2(x)}$ by using the fact that $\frac{1}{1+x^2} = 1 - x^2 + x^4 - ...$?

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...

Any ideas?

Thanks
I don't see any simple way of doing this by using the result from an infinite geometric series. Is this how you were told to do it?

6. Well, I was told to use my previous answer... I guess I can just use the derivatives I calculated for it and the ones for u(x) to get the derivatives for f(x) if there's no simpler way to do it. Thanks for replying.

7. Originally Posted by HD09
Well, I was told to use my previous answer... I guess I can just use the derivatives I calculated for it and the ones for u(x) to get the derivatives for f(x) if there's no simpler way to do it. Thanks for replying.

Well, of course that 1 - cos^2(x) + cos^4(x)-.... cannot be a valid answer since you're not giving a McClaurin series IN X but in cos x, and that's not, apparently, what you were asked to do.

Sometimes , in some maths forum, there had been discussions as whether the infinite series 1 -1 + 1 - 1 +.... """converges""" to 1/2, and one of the reasons is precisely this developement: if you use the usual McClaurin series for cos(x), you get:

== cos x = 1 - x^2/2 + x^4/4 -....

== cox^2(x) = (1 - x^2/2 +...)^2 = 1 - x^2 +...

== cox^4(x) = (1 - x^2/2 +...)^4 = 1 - 2x^2 +....

etc.

Now, if you want to know the value of f(x) = 1/[1 + cox^2(x)] at x = 0, which clearly is 1/2, using the above result with

1/[1 + cosx^2(x)] = 1 - cos^2(x) + cos^4(x) -...

and you look at the constant coefficient in the RH, you get 1 - 1 + 1 - 1 +..., so "it must be" that its sum is 1/2.