What is the appropriate way to simplify such an expression.

William Cleghorn 2022-01-06 Answered
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
\(\displaystyle{\sum_{{{L}={0}}}^{{M}}}{s}^{{L}}{L}^{{2}}\)

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Expert Answer

aquariump9
Answered 2022-01-07 Author has 3471 answers
Write : \(\displaystyle{L}^{{2}}={L}{\left({L}-{1}\right)}+{L}\) and use derivative. For \(\displaystyle{L}\geq{2}\):
\(\displaystyle{L}^{{2}}{s}^{{L}}={s}^{{2}}{L}{\left({L}-{1}\right)}{s}^{{{L}-{2}}}+{s}{L}{s}^{{{L}-{1}}}={s}^{{2}}{\left({s}^{{L}}\right)}{''}+{s}{\left({s}^{{L}}\right)}'\)
We get:
\(\displaystyle{\sum_{{{L}={0}}}^{{M}}}{L}^{{2}}{s}^{{L}}={0}^{{2}}+{1}^{{2}}{s}+{s}^{{2}}{\left({\sum_{{{L}={2}}}^{{M}}}{s}^{{L}}\right)}{''}+{s}{\left({\sum_{{{L}={2}}}^{{M}}}{s}^{{L}}\right)}'\)
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Neunassauk8
Answered 2022-01-08 Author has 4276 answers
Try to make the inner expression look like a derivative:
\(\displaystyle{\sum_{{{L}={0}}}^{{M}}}{\left({L}{s}^{{{L}-{1}}}\right)}{s}{L}={s}{\sum_{{{L}={0}}}^{{M}}}{\left({d}_{{s}}{s}^{{L}}\right)}{L}\)
\(\displaystyle={s}{d}_{{s}}{\sum_{{{L}={0}}}^{{M}}}{s}^{{L}}{L}\)
\(\displaystyle={s}{d}_{{s}}{\sum_{{{L}={0}}}^{{M}}}{\left({L}{s}^{{{L}-{1}}}\right)}{s}\)
\(\displaystyle={s}{d}_{{s}}{\left({s}{\sum_{{{L}={0}}}^{{M}}}{\left({L}{s}^{{{L}-{1}}}\right)}\right)}\)
\(\displaystyle={s}{d}_{{s}}{\left({s}{\sum_{{{L}={0}}}^{{M}}}{d}_{{s}}{s}^{{L}}\right)}\)
\(\displaystyle={s}{d}_{{s}}{\left({s}{d}_{{s}}{\sum_{{{L}={0}}}^{{M}}}{s}^{{L}}\right)}\)
\(\displaystyle={s}{d}_{{s}}{\left({s}{d}_{{s}}{\frac{{{s}^{{{M}+{1}}}-{1}}}{{{s}-{1}}}}\right)}\)
Now just take it from here, simplifying from the inside out.
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karton
Answered 2022-01-11 Author has 8659 answers

Rewrite \(L^2=L(L-1)+L\). Then,
\(\sum_{L=0}^M L^2s^L=\sum_{L=0}^M L(L-1)s^L+\sum_{L=0}^M Ls^L\)
\(=s^2\cdot\frac{d^2}{ds^2}\sum_{L=0}^M s^L+s\cdot\frac{d}{ds}\sum_{L=0}^M s^L\)

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