How to calculate binomial (n choose k) coefficients when exponent is negative?

[tommymato]

I'm using Pascal's (n choose k) method for calculating the coefficients of the terms of a binomial expansion. However, if the exponent is a negative integer, how can one use this method, seeing as factorials for negative integers are undefined.

For example, how could one determine the coefficients of (a + b) ^ -2

 

[chisigma]

I'm using Pascal's (n choose k) method for calculating the coefficients of the terms of a binomial expansion. However, if the exponent is a negative integer, how can one use this method, seeing as factorials for negative integers are undefined.

For example, how could one determine the coefficients of (a + b) ^ -2

In the general case the binomial series is that which has here...

Binomial Series -- from Wolfram MathWorld

... in detail...

$\displaystyle (1 + x)^{r} = 1 + r\ x + \frac{1}{2}\ r\ (r-1) \ x^{2} + \frac{1}{6}\ r\ (r-1)\ (r-2)\ x^{3} + ... \ (1)$

$\displaystyle (1 + x)^{- r} = 1 + r\ x + \frac{1}{2}\ r\ (r+1) \ x^{2} + \frac{1}{6}\ r\ (r+1)\ (r+2)\ x^{3} + ... \ (2)$

Kind regards

$\chi$ $\sigma$

 

[chisigma]

In the general case the binomial series is that which has here...

Binomial Series -- from Wolfram MathWorld

... in detail...

$\displaystyle (1 + x)^{r} = 1 + r\ x + \frac{r}{2}\ r\ (r-1) \ x^{2} + \frac{r}{6}\ r\ (r-1)\ (r-2)\ x^{3} + ... \ (1)$

$\displaystyle (1 + x)^{- r} = 1 + r\ x + \frac{r}{2}\ r\ (r+1) \ x^{2} + \frac{r}{6}\ r\ (r+1)\ (r+2)\ x^{3} + ... \ (2)$

Kind regards

$\chi$ $\sigma$

... of course for x = -1 and r = 0 using the (1) or (2) is $\displaystyle (1-1)^{0} = 0^{0} = 1$... an happy 2015 and many more years of happiness to those who still believe that $0^{0}$ is an 'indeterminate form '(Happy)...

Kind regards

$\chi$ $\sigma$