Binomial expansion with n=1/2

  • Thread starter gentsagree
  • Start date
  • #1
96
1
Is it possible to do a binomial expansion of [itex](x+y)^{1/2}[/itex]? 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?
 

Answers and Replies

  • #5
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 [itex](1+z)^{1/2}[/itex] with [itex]z<1[/itex].

For example, assume [itex]y < x[/itex], then your expression would be

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

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

Related Threads on Binomial expansion with n=1/2

  • Last Post
Replies
1
Views
2K
Replies
25
Views
3K
  • Last Post
Replies
7
Views
4K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
7
Views
5K
Replies
2
Views
646
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
4
Views
1K
Replies
6
Views
2K
Top