Talk:Binomial Theorem

Inductive Hypothesis:

 * $(x+y)^j = \sum_{k=0}^n {j\choose k}x^{j-k}y^k$ for "some" $j\geq 1$

should that be "some", or "any"? User: jehan60188

Some doesn't sound too bad, what do other's think. Oh, and you can sign posts by typing " --~ ". There's also a button on top of the edit box. Second from the right.--Joe 18:18, 2 September 2008 (UTC)

"Some" is good for me, but the upper limit of the summation should be j not n. I changed it in the proof. --prime mover (talk) 18:40, 2 September 2008 (UTC)

Hmm, I guess I was just trying to ascertain a more precise derivation of this proof- does an inductive hypothesis require "some" of the nth steps to be true, or "all" of the steps up to n, or "exactly" the nth step? Or does it matter? --Jehan60188 21:04, 2 September 2008 (UTC)

If for any k, P(k) implies P(k+1), and P(1) is true, it follows that P(n) is true for all (n). --prime mover (talk) 22:05, 2 September 2008 (UTC)

Sorry I don't like "all", that's what we're trying to prove and it makes the argument circular.--prime mover (talk) 05:19, 3 September 2008 (UTC)

Change it if you want, but I can't think of a better phrasing since the inductive hypothesis does have to hold for all j. I don't think it's circular, since as long as you work through one number at a time, you've already proven that it holds for the first one. --cynic 13:03, 3 September 2008 (UTC)

http://en.wikipedia.org/wiki/Mathematical_induction#Axiom_of_induction looks like the axiom wants "any given" not that wikipedia is a end-all-be all authoriy, atleast for mathematics (that's Wolfram, i guess =0P ) --Jehan60188 20:11, 3 September 2008 (UTC)

errr, \geq 1?
So, the theorem had n > 0, but the base case started at 1. i changed the theorem to n \geq 1 and now it's giving a parse error. oddly enough, if i replace n with j, or 1 with 0 it works fine...

yeah, check out the main page talk - there's a problem with LaTeX rendering at the moment.--prime mover (talk) 20:33, 3 September 2008 (UTC)

Well, the point was taken none the less, since the binomial theorem should still apply when n is 0, although the case is trivial. Therefore I fixed the base case so it is n=0. Of course, it's still not rendering properly, but I assume that will fix itself once Joe gets the larger problem fixed. --cynic 20:43, 3 September 2008 (UTC)

This appeared ...
"The proof of the binomial theorem for a non-integer power uses the formula for the derivitive of $x^a = a x^{a-1}$. The usual demonstration of ths formula uses the binomial theorem.  This makes the proof circular unless a proof of the derivitive formula can be given that does not use the binomial formula."

It was unsigned and anonymous, so I've added it here. I'm going to look into its circularity. --prime mover 18:08, 8 December 2010 (UTC)


 * ... no, we're all right, there is a version of the proof of Power Rule for Derivatives for integer $n$ which uses induction, not the binomial theorem (although there is one that does use the BT just for completeness). No circularity from that end of town. --prime mover 18:14, 8 December 2010 (UTC)

Tidied up the talk page while I was about it. --prime mover 18:10, 8 December 2010 (UTC)