Is it true that for x∈[0,2π] we have ∑n=1∞cos⁡(nx)n2=x24−πx2+π26

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Is it true that for x∈[0,2π] we have  ∑n=1∞cos⁡(nx)n2=x24−πx2+π26

Gordon Stephens · Accepted Answer

I show that  ∑n=−∞∞sin2ann2=πa  when a∈(0,π). Now, use the fact that  ∑n=1∞1n2=π26  and let S be the sum in question. Then  π26−S=∑n=1∞1−cos⁡nxn2=2∑n=1∞sin2nx2n2  Rewrite the last sum as  2∑n=1∞sin2nx2n2=∑n=−∞∞sin2nx2n2−(x2)2  =πx2−x24  Then  π26−S=πx2−x24→S=x24−πx2+π26

huglaus51 · Accepted Answer

So we have  ∑n=1∞sin⁡nxn=π−x2 (0&amp;lt;x&amp;lt;2π)  Integrate both sides with respect to x  −∑n=1∞cos⁡nxn2=πx2−x24+C  Set x=0⇒C=−ζ(2)=−π26  Giving us  ∑n=1∞x24−πx2+π26

RizerMix · Accepted Answer

Here there is a way using Fourier series. Consider the orthonormal basis of L2([0,2π]), given by the functions en(x)=12πe−inx, n∈Z Since for x∈[0,2π] the indicator function 1[0,x] is in L2([0,2π]), it can be expressed in its development in the basis (en)n. If we denote (f,g) the dot-product of two functions f,g in L2([0,2π]), that is (f,g)=∫[0,2π]f(x)g(x)dx then we have x=(1[0,x],1[0,x])=∑n∈Z(1[0,x],en)(en,1[0,x])=∑n∈Z|(1[0,x],en)|2 The rest is just computing integrals and separating absolutely convergence series. We have (1[0,x],en)=12π∫0xeinydy=(einx−12πni,n≠0),(x2π,n=0): Hence, x=x22π+12π∑n∈Z2−2cos⁡(nx)n2 Using symetric arguments (2−2cos⁡(nx)n2=2−2cos⁡(−nx)(−n)2), we obtain x=x22π+2π∑n=1∞1n2−2π∑n=1∞cos⁡(nx)n2=x22π+π3−2π∑n=1∞cos⁡(nx)n2 From where we finally obtain ∑n=1∞cos⁡(nx)n2=x24−πx2+π26 Outside the interval, just use the periodicity of cos

Is it true that for x\in[0,2\pi] we have \sum_{n=1}^\infty\frac{\cos(nx)}{n^2}=\frac{x^2}{4}-\frac{\pi x}{2}+\frac{\pi^2}{6}

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