Talk:Multinomial Theorem

Induction type
The text indicates that the proof uses strong induction, but I see only weak induction and implicit dependence on the binomial case. I will probably change that if no one objects or beats me to it. Dfeuer (talk) 16:43, 29 November 2012 (UTC)


 * The second to third line of the induction step uses the induction hypothesis on $n - j$ for $j = 0 \ldots n$, so strong induction is used. --Lord_Farin (talk) 17:20, 29 November 2012 (UTC)


 * Hm, let me revoke that. I will explicate that the induction is on $m$, not on $n$. This was likely the cause of confusion. --Lord_Farin (talk) 18:14, 29 November 2012 (UTC)


 * I too suspect that was the source of confusion, as I briefly made the same mistake when I first wrote essentially the same proof (before I saw it here). I'm new to ProofWiki, so I don't know if the proof given on Wikipedia should be imported here. I find the Wikipedia proof uglier myself, but others may disagree—that proof combines the last two terms to apply the multinomial theorem inductively before applying the binomial theorem.Dfeuer (talk) 18:27, 29 November 2012 (UTC)


 * I hope that the current version is to your liking; thanks for pointing this out. --Lord_Farin (talk) 18:30, 29 November 2012 (UTC)

The basic formula making up the body of the theorem is repeated too many times, as a result of a rather formal description of the induction. If such formality is really necessary, it may be good to name the propositions. This should also make it easier to clarify the sense of the conclusion of the inductive step: as written, it says that P_{m+1} for all m, while what it should say is that P_m\implies P_{m+1} for all m.Dfeuer (talk) 23:18, 29 November 2012 (UTC)


 * Especially because the previous proof was shown to be prone to misunderstanding, it is really necessary to spell out the induction, so as to eliminate as much grounds for confusion as possible. I have added a few words to explicate that we used the IH. --Lord_Farin (talk) 23:24, 29 November 2012 (UTC)


 * I switched it to a different style, similar to what I've seen in many texts. Does that look okay to you? Dfeuer (talk) 00:58, 1 December 2012 (UTC)


 * It does to me; I've changed some minor stylistic and formatting things to accommodate for house style. --Lord_Farin (talk) 08:44, 1 December 2012 (UTC)

Additional proof(s)

 * There is value in adding the proof on WP since its use of $K$ makes it easier to understand what happens with the summations. Feel free to add it on Multinomial Theorem/Proof 2 (simply copy the theorem statement, then write your proof. I'll try to drop by to help you get it through the house style check. Lastly, welcome! :) --Lord_Farin (talk) 18:34, 29 November 2012 (UTC)
 * Isn't Wikipedia's licensing arrangement compatible with GFDL? Couldn't I just copy their proof directly? Dfeuer (talk) 18:52, 29 November 2012 (UTC)


 * Some stuff on here has indeed been coped from Wikipedia (no worries), but it invariably needs to be tidied up as our house style is far more strict (and, in general, completely incompatible with that of WP) so feel free to import anything you like. Expect the source code to be hacked about, though.

Need to add generalized form(s) for exponents that aren't positive integers
I saw a very long and somewhat incoherent discussion of such a form somewhere or other, but I believe the basic idea is to adapt Newton's binomial series to multinomials. Dfeuer (talk) 18:49, 29 November 2012 (UTC)

What can the x_i be?
I made a first stab at it. I know (based on the page on the binomial theorem), that they can be from certain rings, not only fields, but the notation in that article gets kind of intimidating. Also, I imposed the restriction that at least one x_i must not be 0. This isn't strictly necessary, but if all the x_i are 0, n cannot be. Dfeuer (talk) 01:08, 1 December 2012 (UTC)


 * Even if all $x_i$ and $n$ are $0$, then the clause "Whenever $0^0$ may appear, we give it the value $1$" makes the theorem true - even in that degenerate case. Also, I have brought the proof up to house style again. Most notably, consider Help:Editing to familiarise yourself with the intricacies of it; e.g. it is adopted on this site that in displayed formulae, accuracy in punctuation is sacrificed to eliminate possible causes of confusion. That is, a displayed formula should not end with a comma or a semicolon, even if grammar requires so. --Lord_Farin (talk) 08:37, 1 December 2012 (UTC)