rayquaza17
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#1
Report Thread starter 4 years ago
#1
Is this okay to do:

\sum_{n=1}^{\infty }\frac{2al}{n^2\pi ^2}[(-1)^n-1]cos(\frac{n\pi x}{l})=\sum_{m=2n-1}^{\infty }\frac{-4al}{m^2\pi ^2}cos(\frac{m\pi x}{l})
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TeeEm
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#2
Report 4 years ago
#2
(Original post by rayquaza17)
Is this okay to do:

\sum_{n=1}^{\infty }\frac{2al}{n^2\pi ^2}[(-1)^n-1]cos(\frac{n\pi x}{l})=\sum_{m=2n-1}^{\infty }\frac{-4al}{m^2\pi ^2}cos(\frac{m\pi x}{l})
not correct

the summation starts from m=1

and all m inside the summation must me (2m-1)
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rayquaza17
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#3
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(Original post by TeeEm)
not correct

the summation starts from m=1

and all m inside the summation must me (2m-1)
Like this instead?

\sum_{n=1}^{\infty }\frac{2al}{n^2\pi ^2}[(-1)^n-1]cos(\frac{n\pi x}{l})=\sum_{n=1}^{\infty }\frac{-4al}{(2n-1)^2\pi ^2}cos(\frac{(2n-1)\pi x}{l})
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TeeEm
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#4
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(Original post by rayquaza17)
Like this instead?

\sum_{n=1}^{\infty }\frac{2al}{n^2\pi ^2}[(-1)^n-1]cos(\frac{n\pi x}{l})=\sum_{n=1}^{\infty }\frac{-4al}{(2n-1)^2\pi ^2}cos(\frac{(2n-1)\pi x}{l})
that is correct now
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rayquaza17
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#5
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#5
(Original post by TeeEm)
that is correct now
Thanks.

prsom
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