This post concerns Chapter 1 section "The Multinomial Theorem" on pages 65-67 of Analysis I by Amann and Escher.

Excerpts from text:

The part that I can't understand is the equation with the summation in this excerpt.

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The multinomial theorem (stated below) was proven immediately prior.

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In case the notation isn't clear, we have the multi-index $\alpha = (\alpha_1, \dots, \alpha_m) \in \mathbb N^m$, and its length is $\lvert \alpha \rvert := \sum_{j = 1}^m \alpha_j$. We have $\alpha ! := \prod_{j = 1}^m (\alpha_j)!$. We also have $a^{\alpha} := \prod_{j = 1}^m (a_j)^{\alpha_j}$.

Questions and comments:

I am assuming that $1 = 1_R$ in the equation I don't understand. I have trouble explaining to myself why the form of the sum in the equation I don't understand is different from the form of the sum (on the right-hand side) in the multinomial theorem (8.4).

The first sentence of the proof is not hard to understand. However, the second sentence doesn't make sense to me. I'm sorry I can't be more specific. I guess I would ask why we need this special form (the equation I don't understand) when in the multinomial theorem, any of the $a_j$ could be equal to $1$ anyway? I can't reconcile the two.

I appreciate any help.

  • $\begingroup$ The second sentence after the proof of Remark 8.6(a) doesn't belong to the proof, it's a separate remark about noncommutative rings (assuming that the given quantities do commute anyway). $\endgroup$
    – Berci
    Aug 23 '20 at 17:35
  • $\begingroup$ @Berci When I wrote "the second sentence" I was referring to the second sentence of the proof. The one that goes "The claim now follows from Theorem 8.5." $\endgroup$
    – Novice
    Aug 23 '20 at 18:17

Exactly as you say, theorem 8.5 can be applied to elements $(a_1,\dots,a_m,a_{m+1})$ where $a_{m+1}=1=1_R$.

And this is what happens in the proof of remark 8.6(a). Let $b$ denote the sequence $(a_1,\dots,a_m,1)$ and note that $b^\beta=a^\alpha$ for the initial segment $\alpha:=(\beta_1,\dots,\beta_m)$ of an exponent sequence $\beta=(\beta_1,\dots,\beta_m,\beta_{m+1})$ because $1^{\beta_{m+1}}=1$.
Note also that the initial segment $\alpha$ uniquely determines the last term, hence the whole sequence $\beta$ if we assume $|\beta|=k$: specifically $\beta_{m+1}$ must be then $k-|\alpha|$, and $\alpha$ can be any sequence of exponents with $\ |\alpha|\,\le \,k$.

Using this correspondence we get $$(1+a_1+\dots+a_m)^k \ =\ \sum_{|\beta|=k} \frac{k!}{\beta!}b^\beta\ =\\ =\ \sum_{\matrix{|\alpha|\,\le\,k \\ \beta:=(\alpha_1,\dots,\alpha_m,k-|\alpha|)}} \frac{k!}{\beta!}b^\beta\ =\ \sum_{|\alpha|\le k} \binom k\alpha a^\alpha\ \ \ .$$


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