Binomial Expansion

**dd86** · July 10th 2010, 07:15 PM

Hi

Can anyone help me with expanding the following equation?

[1-(1/x)]^1/4

Thank you.

**Also sprach Zarathustra** · July 10th 2010, 07:48 PM

Originally Posted by dd86

Hi

Can anyone help me with expanding the following equation?

[1-{1}/{x}]^1/4

Thank you.

$(1-\frac{1}{x})^{1/4} = \Sigma_{n=-\infty}^0x^n (-1)^n {\frac{1}{4}\choose{-n}}<br />$

For understanding: Binomial Theorem for Rational n

**dd86** · July 10th 2010, 08:00 PM

Sorry, I think one of the brackets have not been closed. I'm a bit confused cos of that

**Prove It** · July 10th 2010, 08:05 PM

The binomial series:

$(1 + x)^n = \sum_{k = 0}^{\infty}{n\choose{k}}x^k$ , where ${n\choose{k}} = \frac{n(n-1)(n-2)\dots (n-k+1)}{k!}$ .

So in your case:

$\left(1 - \frac{1}{x}\right)^{\frac{1}{4}} = \sum_{k = 0}^{\infty}{\frac{1}{4}\choose{k}}\left(-\frac{1}{x}\right)^k$

$= \sum_{k = 0}^{\infty}{\frac{1}{4}\choose{k}}\frac{(-1)^k}{x^k}$

$= 1 - \frac{1}{4x} - \frac{3}{32x^2} - \frac{7}{128x^3} - \frac{77}{2048x^4} + O\left[\left(\frac{1}{x}\right)^5\right]$ .

**chisigma** · July 10th 2010, 08:05 PM

In the formula for the binomial expansion...

$\displaystyle (1+t)^{\alpha} = 1 + \binom {\alpha}{1}\ t + \binom {\alpha}{2}\ t^{2} + \binom {\alpha}{3}\ t^{3} + \dots$ (1)

... You set $\alpha=\frac{1}{4}$ and $t= -x^{-1}$ and obtain...

$\displaystyle (1-\frac{1}{x})^{\frac{1}{4}} = 1 - \frac{1}{4}\ x^{-1} - \frac{3}{2\ 4\ 4}\ x^{-2} - \frac{3\ 7}{6\ 4\ 4\ 4}\ x^{-3} - \dots$ (2)

The expansion (2) converges for $|x|>1$ ...

Kind regards

$\chi$ $\sigma$

**Prove It** · July 10th 2010, 08:19 PM

Originally Posted by chisigma

In the formula for the binomial expansion...

$\displaystyle (1+t)^{\alpha} = 1 + \binom {\alpha}{1}\ t + \binom {\alpha}{2}\ t^{2} + \binom {\alpha}{3}\ t^{3} + \dots$ (1)

... You set $\alpha=\frac{1}{4}$ and $t= -x^{-1}$ and obtain...

$\displaystyle (1-\frac{1}{x})^{- \frac{1}{4}} = 1 + \frac{1}{4}\ x^{-1} + \frac{5}{2\ 4\ 4}\ x^{-2} + \frac{5\ 9}{6\ 4\ 4\ 4}\ x^{-3} + \dots$ (2)

The expansion (2) converges for $|x|>1$ ...

Kind regards

$\chi$ $\sigma$

Sorry, there's something wrong with your calculations.

For one thing, ${\frac{1}{4}\choose{2}} = \frac{\frac{1}{4}\left(-\frac{3}{4}\right)}{2!} = \frac{-\frac{3}{16}}{2} = -\frac{3}{32} \neq \frac{5}{2\,4\,4}$ ...

Your calculation of the $x^{-3}$ term is also incorrect.

Also, all the coefficients except for the $x^0$ term should be negative.

**dd86** · July 10th 2010, 08:36 PM

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?

**chisigma** · July 10th 2010, 08:50 PM

Originally Posted by Prove It

Sorry, there's something wrong with your calculations.

For one thing, ${\frac{1}{4}\choose{2}} = \frac{\frac{1}{4}\left(-\frac{3}{4}\right)}{2!} = \frac{-\frac{3}{16}}{2} = -\frac{3}{32} \neq \frac{5}{2\,4\,4}$ ...

Your calculation of the $x^{-3}$ term is also incorrect.

Also, all the coefficients except for the $x^0$ term should be negative.

The expression...

$\displaystyle (1 - \frac{1}{x})^{-\frac {1}{4}} = 1 + \frac{1}{4}\ x^{-1} + \frac{5}{2\ 4\ 4}\ x^{-2} + \frac{5\ 9}{6\ 4\ 4\ 4}\ x^{-3} + \dots$ (1)

... is perfectly correct... the only problem was that it was requested the expasion of $(1-\frac{1}{x})^{\frac{1}{4}}$ and not of $(1-\frac{1}{x})^{-\frac{1}{4}}$

... fortunately when You have posted that was already been corrected...

Kind regards

$\chi$ $\sigma$

**Prove It** · July 10th 2010, 08:55 PM

Originally Posted by dd86

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 formula is essentially the same, except you need to use an infinite series rather than finite, and you need to use the revised binomial coefficients.

**dd86** · July 10th 2010, 09:13 PM

By revised binomial coefficients, you mean that I have to ensure that I substitute the correct variables in the expansion to get the correct result, right?

**Prove It** · July 10th 2010, 09:31 PM

I mean that for $n$ as a positive integer, then ${n\choose{k}} = \frac{n!}{k!(n-k)!}$ , and you can look it up on Pascal's Triangle.

For $n$ in general, ${n\choose{k}} = \frac{n(n-1)(n-2)\dots(n-k+1)}{k!}$ .

**dd86** · July 10th 2010, 10:00 PM

Thank you very much.

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