I have to find the sine series of x3 using the the cosine series of x22 x22=l26+2l2π2[∑(−1)nn2cos⁡(nπxl)]

Question

I have to find the sine series of x3 using the the cosine series of x22  x22=l26+2l2π2[∑(−1)nn2cos⁡(nπxl)]

Louis Page · Accepted Answer

Obviously when you integrate you get  x22=l26+2l2π2[∑(−1)nn2cos⁡(nπxl)]  You now need to represent x in terms of a Fourier sine series to finish this off. That is, express  x=∑n=1∞Bnsin⁡(nπxl)  where  Bn=1l∫0ldxxsin⁡(nπxl)=(−1)n+1πnl

Debbie Moore · Accepted Answer

Hint: a sine series for x can be obtained from differentiating the series for x22 term-by-term. That can be used to get rid of the l26x that you get from integrating. You don&#039;t need to worry about a constant of integration, because a sine series has no constant term.

user_27qwe · Accepted Answer

Starting with the series x22=L26+2Lπ2[∑(−1)nn2cos⁡(πnxl)] then upon integration from zero to x it is seen that ∫0x[∑n=1∞(−1)nn2cos⁡(nπuL)]du=π22L2∫0x(u22−L26)du which becomes ∑n=1∞(−1)nn3sin⁡(nπxL)=π312L2(x3−L2x) By performing closed limit integration there is no constant of integration to attend to. It is with minimal effort to show the Fourier Sine series of x is x=−2Lπ∑n=1∞(−1)nnsin⁡(nπxL) which then makes the prior formula to be in the form x3=2L3π3∑n=1∞(−1)n−1n3(π2n2−6)sin⁡(nπxL).

I have to find the sine series of x^3 using

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