Hi
Can anyone help me with expanding the following equation?
[1-(1/x)]^1/4
Thank you.
$\displaystyle (1-\frac{1}{x})^{1/4} = \Sigma_{n=-\infty}^0x^n (-1)^n {\frac{1}{4}\choose{-n}}
$
For understanding: Binomial Theorem for Rational n
The binomial series:
$\displaystyle (1 + x)^n = \sum_{k = 0}^{\infty}{n\choose{k}}x^k$, where $\displaystyle {n\choose{k}} = \frac{n(n-1)(n-2)\dots (n-k+1)}{k!}$.
So in your case:
$\displaystyle \left(1 - \frac{1}{x}\right)^{\frac{1}{4}} = \sum_{k = 0}^{\infty}{\frac{1}{4}\choose{k}}\left(-\frac{1}{x}\right)^k$
$\displaystyle = \sum_{k = 0}^{\infty}{\frac{1}{4}\choose{k}}\frac{(-1)^k}{x^k}$
$\displaystyle = 1 - \frac{1}{4x} - \frac{3}{32x^2} - \frac{7}{128x^3} - \frac{77}{2048x^4} + O\left[\left(\frac{1}{x}\right)^5\right]$.
In the formula for the binomial expansion...
$\displaystyle \displaystyle (1+t)^{\alpha} = 1 + \binom {\alpha}{1}\ t + \binom {\alpha}{2}\ t^{2} + \binom {\alpha}{3}\ t^{3} + \dots $ (1)
... You set $\displaystyle \alpha=\frac{1}{4}$ and $\displaystyle t= -x^{-1}$ and obtain...
$\displaystyle \displaystyle (1-\frac{1}{x})^{\frac{1}{4}} = 1 - \frac{1}{4}\ x^{-1} - \frac{3}{2\ 4\ 4}\ x^{-2} - \frac{3\ 7}{6\ 4\ 4\ 4}\ x^{-3} - \dots $ (2)
The expansion (2) converges for $\displaystyle |x|>1$ ...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
Sorry, there's something wrong with your calculations.
For one thing, $\displaystyle {\frac{1}{4}\choose{2}} = \frac{\frac{1}{4}\left(-\frac{3}{4}\right)}{2!} = \frac{-\frac{3}{16}}{2} = -\frac{3}{32} \neq \frac{5}{2\,4\,4}$...
Your calculation of the $\displaystyle x^{-3}$ term is also incorrect.
Also, all the coefficients except for the $\displaystyle x^0$ term should be negative.
Thank you all for your invaluable help.
I was confused when expanding the above equation as the textbook I am referring to stated different formulae where n is a non-positive integer and when x>1 or x<1.
Can the expansion formula that is used when n is a positive integer be still used when n is a negative integer or when n is not an integer at all?
The expression...
$\displaystyle \displaystyle (1 - \frac{1}{x})^{-\frac {1}{4}} = 1 + \frac{1}{4}\ x^{-1} + \frac{5}{2\ 4\ 4}\ x^{-2} + \frac{5\ 9}{6\ 4\ 4\ 4}\ x^{-3} + \dots$ (1)
... is perfectly correct... the only problem was that it was requested the expasion of $\displaystyle (1-\frac{1}{x})^{\frac{1}{4}}$ and not of $\displaystyle (1-\frac{1}{x})^{-\frac{1}{4}}$ ... fortunately when You have posted that was already been corrected...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
I mean that for $\displaystyle n$ as a positive integer, then $\displaystyle {n\choose{k}} = \frac{n!}{k!(n-k)!}$, and you can look it up on Pascal's Triangle.
For $\displaystyle n$ in general, $\displaystyle {n\choose{k}} = \frac{n(n-1)(n-2)\dots(n-k+1)}{k!}$.