Question: Let f(x) be a piecewise continuous function on -L,L, and for N0, let SN(x)=12a0n=1Nancos(nxL)bnsin(nxL) be the N-th partial sum of FSf(x). Let PN(x)=12c0n=1Ncncos(nxL)dnsin(nxL) be an

Let f(x) be a piecewise continuous function on -L,L, and for N0, let SN(x)=12a0n=1Nancos(nxL)bnsin(nxL) be the N-th partial sum of FSf(x). Let PN(x)=12c0n=1Ncncos(nxL)dnsin(nxL) be an arbitrary trigonometric polynomial of degree N .(a) Show that -LL[f(x)-PN(x)]2dx=-LL[f(x)]2dx-2-LLPN(x)SN(x)dx-LLPN(x)2dx=-LL[f(x)]2dx--LLSN(x)2dx-LL[PN(x)-SN(x)]2dx.(b) Why does part (a) show that the mean-square error -LL[f(x)-PN(x)]2dx is least when PN(x) is chosen to be SN(x)?

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