How Does Binomial Expansion Work for Rational Indices?

[sparsh]

Hi

I wanted to know what is the expansion of (1+x)^n when n is a rational number and |x|<1 ...
Please let me know as soon as possible..

Thanks for your help
Sincerely
Sparsh

 

[Hootenanny]

$$(1+x)^{n} = 1 + nx + \frac{n(n-1)}{1\cdot 2}x^{2} + ... + \frac{n(n-1)...(n-r+1)}{1\cdot 2 ... r}x^{r}$$

Where $|x|<1$ and n is any real number. This can be derived from the general binomial expansion of $(a+b)^n$.

Regards,
~Hoot

 

[HallsofIvy]

I assume you know that, for n a positive integer
$$(1+ x)^n= 1+ nx+ ... + _nC_i x^i+ ...$$
where
$$_nC_i= \frac{n!}{i!(n-i)!}= \frac{n(n-1)...(n-i+1)}{i!}$$
For n a rational number, basically the same formula is true. Only now $_nC_i$ is never 0 so we get an infinite sum.

For example, if n= 1/2 then $_{\frac{1}{2}}C_1= \frac{1}{2}$, $_{\frac{1}{2}}C_2= \frac{\frac{1}{2}(\frac{1}{2}-1)}{2}= -\frac{1}{8}$, $_{\frac{1}{2}}C_3= \frac{\frac{1}{2}(\frac{1}{2}-1)(\frac{1}{2}-2)}{6}= \frac{1}{16}$, etc. so that
$$(1+ x)^{\frac{1}{2}}= 1+ \frac{1}{2}x-\frac{1}{8}x^2+ \frac{1}{16}x^4-...$$
exactly as you would get from the Taylor's series.

 

[sparsh]

Thanks to both .
The post by HallsofIvy was particularly useful .