Can (x+y)^(1/2) be expanded using the binomial series?

[gentsagree]

Is it possible to do a binomial expansion of (x+y)^{1/2}? I tried to compute it with the factorial expression for the binomial coefficients, but the second term already has n=1/2 and k=1, which makes the calculation for the binomial coefficient (n 1) weird, I think.

Any advice?

 

[jedishrfu]

The binomial expansion is for n=1,2,3,... only.

However Newton generalized the expansion see section on Newton's Generalized Binomial Theorem:

http://en.wikipedia.org/wiki/Binomial_expansion

and for more info:

http://en.wikipedia.org/wiki/Binomial_series

 

[R136a1]

Here you go: http://en.wikipedia.org/wiki/Binomial_series

Edit: seems like jedi power has beaten me!

 

[jedishrfu]

Here you go: http://en.wikipedia.org/wiki/Binomial_series

Edit: seems like jedi power has beaten me!

You don't need to answer this post and these droids aren't the ones you're looking for, move along, move along.

 

[TheoMcCloskey]

You may want to look at something like
http://en.wikipedia.org/wiki/Binomial_series

Assuming neither x or y are zero (and both are positive), I would recommend factoring out the larger of x or y and let your task reduce to that of finding (1+z)^{1/2} with z<1.

For example, assume y < x, then your expression would be

f = \sqrt{x}\,(1+z)^{1/2}

Expand (1+z)^{1/2} using the binomial series. The expansion will be an infinite series due to the non-integer exponent.