# Expanding a series

1. Oct 11, 2006

### Poop-Loops

I have the series (cosx)^-1/2 (so it's square-rooted and on the bottom)

And I need to get the first few terms. My idea has been to do this: [1 + (cosx - 1)]^-1/2

But I have no idea how to take the binomial expansion of that. Do I:

Take the BE first, substituting "cosx - 1" as just X or something, then after I get the first few terms of that, substitute in the series for "cosx - 1" back into X and adding everything up? The only reason I haven't tried this yet is because it seems like this would take an infinite amount of work... :yuck:

2. Oct 11, 2006

### quasar987

if you only need the first few terms, why no just differentiate (cosx)^-1/2 a few times?

$$\frac{1}{\sqrt{\cos x}} = 1+ \frac{1}{2}\left[ \frac{d}{dx}\frac{1}{\sqrt{\cos x}}\right]_{x=0}x+...$$

3. Oct 11, 2006

### Poop-Loops

Because your 1st derivative is missing a sinx (Unless I don't get your mathematical notation...), and the next derivatives only get more gruesome. This teacher keeps spouting "we're smart, but lazy", so I'm guessing there is a trick I am missing here.

4. Oct 11, 2006

### quasar987

it's not messy at all after you evaluate the derivative at x=0. All the cos become 1 and all the sines become 0.

5. Oct 11, 2006

quasar, you are doing a MacLaurin series right?

6. Oct 11, 2006

### quasar987

Are you going to say that after a few derivatives, nasty sines come poping in the denominator?

Just to be on the same side, make that a Taylor series centered on $\pi/4$ where both cos and sine are $1/\sqrt{2}$.

Last edited: Oct 11, 2006
7. Oct 11, 2006

### Poop-Loops

Errr... when can I evaluate at x=0? Don't I need to keep it a variable until I am completely done deriving?

8. Oct 11, 2006

### quasar987

Yes. It's true you are lazy, lol.

9. Oct 11, 2006

### Poop-Loops

I tried this once already. I got to through 2 derivations before it became really ugly. And my f(0)'s were something like 1, 0, and 1/2. That only makes it 2 terms, really, unless I want to cheat and say it's 0x. But I probably can't. So if all the odd n terms are supposed to be 0, I have a lot of deriving to go. I said I'm lazy, but smart. Smart enough to know that there has to be a different approach. This isn't a calc class where we are learning this the first time and have to go through this just to know how it's done. This is math physics, so it's mainly just a review + more critical thinking.

10. Oct 11, 2006

### quasar987

Alright, so let's get back to your first idea.

$$(\cos x)^{-1/2} = (1+(\cos x - 1))^{-1/2} \equiv (1-X)^{-1/2}$$

Expand this in a binomial series, then resubstitute X--> cos(x)-1. You will have what you want: a series expansion for (cosx)^-1/2. It is not a power series, but it is a series expansion.

It is frequent that calc classes do not cover the binomial series for non-positive integer exponents, but any physics class will assume that you know it. See

http://en.wikipedia.org/wiki/Binomial_series

The series you find will converge for all values of x such that |X(x)|<1. Find what that corresponds to in terms of x.

Last edited: Oct 11, 2006
11. Oct 11, 2006

### Poop-Loops

I have to get a power series expansion (or whatever it's called, where it's 1 + x/2! etc). So would I then expand cosx - 1 and plug those results into each individual binomial expansion? That sounds like just as much work. But I guess I have to try something.

12. Oct 11, 2006

### quasar987

yeah, do that then. Since X = cos(x) - 1, and cos(x)'s power series's first term is 1, the 1's cancel and then the term of lowest degree in $X^n$ is $x^n$. So if you decide to keep only the terms up to degree 2, you're left with the first two terms (in x) of X and the first term (in x) of X².

Last edited: Oct 11, 2006