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Infinite Series quick question

Is this okay to do:

∑n=1∞2aln2π2[(−1)n−1]cos(nπxl)=∑m=2n−1∞−4alm2π2cos(mπxl)\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})
Reply 1
Avatar for TeeEm
TeeEm
19
Original post by rayquaza17
Is this okay to do:

∑n=1∞2aln2π2[(−1)n−1]cos(nπxl)=∑m=2n−1∞−4alm2π2cos(mπxl)\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)
Reply 2
Avatar for rayquaza17
rayquaza17
OP
17
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?

∑n=1∞2aln2π2[(−1)n−1]cos(nπxl)=∑n=1∞−4al(2n−1)2π2cos((2n−1)πxl)\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})
Reply 3
Avatar for TeeEm
TeeEm
19
Original post by rayquaza17
Like this instead?

∑n=1∞2aln2π2[(−1)n−1]cos(nπxl)=∑n=1∞−4al(2n−1)2π2cos((2n−1)πxl)\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
Reply 4
Avatar for rayquaza17
rayquaza17
OP
17
Original post by TeeEm
that is correct now


Thanks. :smile:

prsom

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