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What is the appropriate way to simplify such an expression. i am unsure of how to use the series i know to apply to this situation


do i modify such a series as power series, or is there a more efficient series to use here?

thank you very much!!

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If $M$ were $\infty$, then there is a power-series-based method to simplify the sum. Would it help to have that explained? Or do you need $M$ to be finite? – alex.jordan Jun 19 '12 at 18:36
M is a number that isnt infiniti. thank you for reply, but how to solve for M not infiniti? – nanme Jun 19 '12 at 18:37
out of being curious suppose $\sum_{L=0}^{M}s^{L}L$ would the same method follow – nanme Jun 19 '12 at 18:54
The method I've used in my answer can apply to a sum where the terms are $s^LL^k$, where $k$ is an integer. Even if $k$ is negative, then you just have to make the inside look like an integral rather than a derivative. (However, executing the integral at the end will be harder than executing the derivative.) So yes, this method works with $k=1$ too. – alex.jordan Jun 19 '12 at 18:58
@ alex.jordan : An algebraic explanation to your method is one basis chang on $\mathbb R[X]$. We try express the canonical basis $(1,X,X^2,...)$ with the basis : $(1,X,X(X-1),X(X-1)(X-2), ...à$ and this expression gives a simple way to calculate generalized expression ... – Mohamed Jun 19 '12 at 19:04
up vote 3 down vote accepted

Try to make the inner expression look like a derivative: $$ \begin{align} \sum_{L=0}^M\left(Ls^{L-1}\right)sL & =s\sum_{L=0}^M\left(\partial_ss^L\right)L\\ & =s\partial_s\sum_{L=0}^Ms^LL\\ & =s\partial_s\sum_{L=0}^M\left(Ls^{L-1}\right)s\\ & =s\partial_s\left(s\sum_{L=0}^M\left(Ls^{L-1}\right)\right)\\ & =s\partial_s\left(s\sum_{L=0}^M\partial_ss^L\right)\\ & =s\partial_s\left(s\partial_s\sum_{L=0}^Ms^L\right)\\ & =s\partial_s\left(s\partial_s\frac{s^{M+1}-1}{s-1}\right)\\ \end{align}$$ Now just take it from here, simplifying from the inside out.

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Write : $L^2=L(L-1)+L$ and use derivative. For $ L \geq 2$ : $$L^2 s^L = s^2L(L-1)s^{L-2}+ s Ls^{L-1}= s^2(s^L)'' + s (s^L)'$$ We get : $$\sum_{L=0}^M L^2s^L=0^2+1^2 s + s^2 \left(\sum_{L=2}^{M} s^L \right)''+s \left(\sum_{L=2}^{M} s^L \right)'$$

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$L$ is the index in your sum, so I think you must means something else for the first two terms of your left-hand side. – alex.jordan Jun 19 '12 at 20:00
Oops I meant right-hand side :) – alex.jordan Jun 19 '12 at 20:22
@ alex.jordan Thank you very much.I mean :$0^2 s^0 + 1^2 s^1$ – Mohamed Jul 13 '12 at 1:30

Rewrite $L^2 = L(L-1)+L. $ Then,

$$\begin{align} \sum_{L=0}^{M} { L^2 s^L } &= \sum_{L=0}^{M} {L(L-1)s^L} + \sum_{L=0}^{M} {Ls^L}\\ &=s^2 \cdot \frac{\partial^2}{\partial s^2} \sum_{L=0}^{M}{s^L} + s \cdot \frac{\partial}{\partial s}\sum_{L=0}^{M} {s^L}\\ \end{align}$$

You can find formulas for the summations with detailed descriptions of their derivations by searching for "Geometric Progression."

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