William Cleghorn

2022-01-06

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
$\sum _{L=0}^{M}{s}^{L}{L}^{2}$

aquariump9

Write : ${L}^{2}=L\left(L-1\right)+L$ and use derivative. For $L\ge 2$:
${L}^{2}{s}^{L}={s}^{2}L\left(L-1\right){s}^{L-2}+sL{s}^{L-1}={s}^{2}\left({s}^{L}\right){}^{″}+s{\left({s}^{L}\right)}^{\prime }$
We get:
$\sum _{L=0}^{M}{L}^{2}{s}^{L}={0}^{2}+{1}^{2}s+{s}^{2}\left(\sum _{L=2}^{M}{s}^{L}\right){}^{″}+s{\left(\sum _{L=2}^{M}{s}^{L}\right)}^{\prime }$

Neunassauk8

Try to make the inner expression look like a derivative:
$\sum _{L=0}^{M}\left(L{s}^{L-1}\right)sL=s\sum _{L=0}^{M}\left({d}_{s}{s}^{L}\right)L$
$=s{d}_{s}\sum _{L=0}^{M}{s}^{L}L$
$=s{d}_{s}\sum _{L=0}^{M}\left(L{s}^{L-1}\right)s$
$=s{d}_{s}\left(s\sum _{L=0}^{M}\left(L{s}^{L-1}\right)\right)$
$=s{d}_{s}\left(s\sum _{L=0}^{M}{d}_{s}{s}^{L}\right)$
$=s{d}_{s}\left(s{d}_{s}\sum _{L=0}^{M}{s}^{L}\right)$
$=s{d}_{s}\left(s{d}_{s}\frac{{s}^{M+1}-1}{s-1}\right)$
Now just take it from here, simplifying from the inside out.

karton

Rewrite ${L}^{2}=L\left(L-1\right)+L$. Then,
$\sum _{L=0}^{M}{L}^{2}{s}^{L}=\sum _{L=0}^{M}L\left(L-1\right){s}^{L}+\sum _{L=0}^{M}L{s}^{L}$
$={s}^{2}\cdot \frac{{d}^{2}}{d{s}^{2}}\sum _{L=0}^{M}{s}^{L}+s\cdot \frac{d}{ds}\sum _{L=0}^{M}{s}^{L}$

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