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William Cleghorn

Answered question

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}

Answer & Explanation

aquariump9

Beginner2022-01-07Added 40 answers

Write :

L^{2} = L (L - 1) + L

and use derivative. For

L \geq 2

L^{2} s^{L} = s^{2} L (L - 1) s^{L - 2} + s L s^{L - 1} = s^{2} (s^{L})^{″} + s {(s^{L})}^{'}

We get:

\sum_{L = 0}^{M} L^{2} s^{L} = 0^{2} + 1^{2} s + s^{2} (\sum_{L = 2}^{M} s^{L})^{″} + s {(\sum_{L = 2}^{M} s^{L})}^{'}

Neunassauk8

Beginner2022-01-08Added 30 answers

Try to make the inner expression look like a derivative:

\sum_{L = 0}^{M} (L s^{L - 1}) s L = s \sum_{L = 0}^{M} (d_{s} s^{L}) L

= s d_{s} \sum_{L = 0}^{M} s^{L} L

= s d_{s} \sum_{L = 0}^{M} (L s^{L - 1}) s

= s d_{s} (s \sum_{L = 0}^{M} (L s^{L - 1}))

= s d_{s} (s \sum_{L = 0}^{M} d_{s} s^{L})

= s d_{s} (s d_{s} \sum_{L = 0}^{M} s^{L})

= s d_{s} (s d_{s} \frac{s^{M + 1} - 1}{s - 1})

Now just take it from here, simplifying from the inside out.

karton

Expert2022-01-11Added 613 answers

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

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