# Math Help - Expansion of 1/sqrt(1-x)

1. ## Expansion of 1/sqrt(1-x)

How do you find the binomial expansion of 1/sqrt(1-x) in series form?

I know what the term by term expansion is but i'm trying to find the series representation,

The closest i have found involved double factorials and i'm sure there's an easier representation,

I've been trying to use the binomial theorem but i get fractional factorials which just give ∞.

Is there some formula that i haven't been able to find to apply to this?

2. ## Re: Expansion of 1/sqrt(1-x)

Originally Posted by Daniiel
How do you find the binomial expansion of 1/sqrt(1-x) in series form?

I know what the term by term expansion is but i'm trying to find the series representation,

The closest i have found involved double factorials and i'm sure there's an easier representation,

I've been trying to use the binomial theorem but i get fractional factorials which just give ∞.

Is there some formula that i haven't been able to find to apply to this?
\displaystyle \begin{align*} \frac{1}{\sqrt{1 - x}} = (1 - x)^{-\frac{1}{2}} \end{align*}

Now expand using the Binomial Series.

3. ## Re: Expansion of 1/sqrt(1-x)

but alpha = -1/2

using the binomial coefficient formula don't you get

(-1/2)!/k!(-1/2-k)!

the term by term part works, but to find it in terms of the sum is not going so well for me

4. ## Re: Expansion of 1/sqrt(1-x)

Originally Posted by Daniiel
but alpha = -1/2

using the binomial coefficient formula don't you get

(-1/2)!/k!(-1/2-k)!

the term by term part works, but to find it in terms of the sum is not going so well for me

\displaystyle \begin{align*} {\alpha\choose{k}} = \frac{\alpha \cdot (\alpha - 1) \cdot (\alpha - 2) \cdot \dots \cdot (\alpha - k + 1)}{k!} \end{align*}

What will you do in this case where \displaystyle \begin{align*} \alpha = -\frac{1}{2} \end{align*}?

5. ## Re: Expansion of 1/sqrt(1-x)

So it has to be

$\frac{1}{\sqrt{1-x}}=\sum\limits_{k=0}^{\infty }{\frac{\left( -\frac{1}{2} \right)\left( -\frac{3}{2} \right)....\left( \frac{1}{2}-k \right)}{k!}}{{(-x)}^{k}}$

6. ## Re: Expansion of 1/sqrt(1-x)

Not quite, remember that \displaystyle \begin{align*} (1 - x)^{-\frac{1}{2}} = [1 + (-x)]^{-\frac{1}{2}} \end{align*}, so you should actually have \displaystyle \begin{align*} (-x)^k \end{align*} inside your sum.

7. ## Re: Expansion of 1/sqrt(1-x)

ahh right,

ahah I actually had that but my edit was editing the negative out, one of my friends said it shouldn't be negative, guess he was wrong!

haha thanks!

8. ## Re: Expansion of 1/sqrt(1-x)

i still cant understand how did you exactly solve it?

9. ## Re: Expansion of 1/sqrt(1-x)

what do you mean by solve it?

you pretty much put -1/2 = alpha into that forumla,

when I was first trying to do it, I was trying to find a way to do it without having to write it in that form, (a+b)(a+2b)....(a+nb) (just an example of what i was trying to avoid, the ... product), I was hoping to have it in terms of factorials or something,

10. ## Re: Expansion of 1/sqrt(1-x)

Originally Posted by Daniiel
what do you mean by solve it?

you pretty much put -1/2 = alpha into that forumla,

when I was first trying to do it, I was trying to find a way to do it without having to write it in that form, (a+b)(a+2b)....(a+nb) (just an example of what i was trying to avoid, the ... product), I was hoping to have it in terms of factorials or something,
You could always write it in terms of the generalised binomial coefficients...