# Thread: binomial expansion

1. ## binomial expansion

how would i get sqrt [(1+x)/(1-x)] into an appropriate format for binomial expansion?

i know that i can turn it into sqrt (1+x)[(1-x)^-1], but then what do i do next i.e. what do i do with the sqrt?

2. $\displaystyle \sqrt{ \frac{1+x}{1-x} } = \frac{ \sqrt{1+x}}{ \sqrt{1-x}} = (1+x)^{\frac{1}{2}} (1-x)^{- \frac{1}{2}}$

3. Division? Partial Fractions?

$\displaystyle \frac{1+x}{1-x}\;=\;\frac{2}{x+1}-1$

Basic Principles?

$\displaystyle \frac{1+x}{1-x}\;=\;\frac{1}{1-x}+\frac{x}{1-x}$

Other choices?

4. Division? Partial Fractions?

$\displaystyle \frac{1+x}{1-x}\;=\;\frac{2}{x+1}-1$
There is a typo here. I believe TKHunny meant $\displaystyle \frac{1+x}{1-x}\;=\;\frac{2}{1-x}-1$

5. Originally Posted by nmanik90
how would i get sqrt [(1+x)/(1-x)] into an appropriate format for binomial expansion?

i know that i can turn it into sqrt (1+x)[(1-x)^-1], but then what do i do next i.e. what do i do with the sqrt?
$\displaystyle \sqrt{\frac{1+x}{1-x}}=\sqrt{\frac{(1+x)^2}{(1-x)(1+x)}}=|1+x| (1-x^2)^{-1/2}$

6. Can someone do the binomial expansion on their answer? I am confused by how this would work, and would appreciate an example.

7. Originally Posted by CaptainBlack
$\displaystyle \sqrt{\frac{1+x}{1-x}}=\sqrt{\frac{(1+x)^2}{(1-x)(1+x)}}=|1+x| (1-x^2)^{-1/2}$
When $\displaystyle |x|<1$,

$\displaystyle |1+x| (1-x^2)^{-1/2}=(1+x) (1-x^2)^{-1/2}$

........... $\displaystyle =(1+x)\left[ 1+(-1/2)(-x^2)+\frac{(-1/2)(-3/2)}{2!}(-x^2)^2 + ...\right]$

RonL

8. Originally Posted by CaptainBlack
When $\displaystyle |x|<1$,

$\displaystyle |1+x| (1-x^2)^{-1/2}=(1+x) (1-x^2)^{-1/2}$

........... $\displaystyle =(1+x)\left[ 1+(-1/2)(-x^2)+\frac{(-1/2)(-3/2)}{2!}(-x^2)^2 + ...\right]$

RonL
Thank you. I am still a little confused, my understanding is that the Binomial Expansion is this:

$\displaystyle (a+b)^n = \sum_{i=0}^{n} {n\choose i}a^{n-i}b^n$

So in this case, n=-1/2 and it is unclear to me how you can do a summation or a choose in this situation.

9. Originally Posted by angel.white
Thank you. I am still a little confused, my understanding is that the Binomial Expansion is this:

$\displaystyle (a+b)^n = \sum_{i=0}^{n} {n\choose i}a^{n-i}b^n$

So in this case, n=-1/2 and it is unclear to me how you can do a summation or a choose in this situation.
The good Captain is using the generalised binomial theorem ..... Another link

10. Originally Posted by angel.white
Thank you. I am still a little confused, my understanding is that the Binomial Expansion is this:

$\displaystyle (a+b)^n = \sum_{i=0}^{n} {n\choose i}a^{n-i}b^n$

So in this case, n=-1/2 and it is unclear to me how you can do a summation or a choose in this situation.
I just want to add a little bit more. What you have is the finite binomial theorem when the exponent is an integer.
It has the form,
$\displaystyle (1 + x)^n = 1 + kx + \frac{n(n-1)}{2!}x^2+\frac{n(n-1)(n-2)}{3!}x^3+...$
It must be finite because if $\displaystyle n$ is a positive integer then eventually among $\displaystyle n,n-1,n-2,n-3,...$ we hit zero, and the series shall terminate.

But if you think of $\displaystyle n$ as a non-integer then it will go on forever. And it will be the binomial (infinite) expansion. Which will converge to the function.

11. Originally Posted by angel.white
Thank you. I am still a little confused, my understanding is that the Binomial Expansion is this:

$\displaystyle (a+b)^n = \sum_{i=0}^{n} {n\choose i}a^{n-i}b^n$

So in this case, n=-1/2 and it is unclear to me how you can do a summation or a choose in this situation.
That is only for positive powers (after you sort out the powers to give the actual expansion), but the binomial expansion works for any power.

The general form used here is:

$\displaystyle (1+x)^n= \sum_{r=0}^{\infty} \frac{(n)\times (n-1)\times .. \times(n-(r-1))}{r!} x^r$

Note if $\displaystyle n$ is an integer the coefficients are all zero for $\displaystyle r>n$.

You also have to worry about convergence for non-integer powers.

RonL