1
$\begingroup$

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.

enter image description here

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

enter image description here

Notation:

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.

$\endgroup$
2
  • $\begingroup$ 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
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.