Talk:Binomial

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binomial distribution is great!


[edit] a similar problem

we know (a+b)^n=.... what about (a+b)^c=??? with c>0

Jackzhp 21:44, 6 November 2006 (UTC)

Not quite sure what you mean here. (a+b)^c = \sum_{j=0}^{\infty} { c \choose j } a^jb^{c-j}. This works even if c is not a natural number. For example:

\sqrt{1+x} = (1+x)^{\frac12} = 1 + \frac12x + \frac{\frac12 (\frac12 - 1)}{2!}x^2 + \frac{\frac12 (\frac12 - 1)(\frac12 - 2)}{3!}x^3...

\sqrt(1+x) = 1 + \frac{x}2 - \frac18x^2 + \frac1{16}x^3...

This can be used to compute square root provided x is small enough. However, for computing square root there are usually better methods.

salte 16:20, 19 December 2006 (UTC)

[edit] Citations????

Are something this article appears to lack completely, though it's a rather nicely written article. I suggest that this article be tagged for needing some citations. A math-wiki (talk) 05:51, 28 March 2008 (UTC)