When I plug the below sum (from here) into Mathematica 12.1.1, why do I get something different from the original function ($1 + x^2$)?
S = HoldForm[Sum[2/(n*Pi)^3*((2 L^2 - n^2*Pi^2*(1 + L^2))*(-1)^n - 2*L^2 + n^2*Pi^2)*Sin[n*Pi*x/L], {n, 1, Infinity}]]
FullSimplify[ReleaseHold[S], Assumptions -> 0 <= x <= L]
According to Find the function $f(x)$ by using its fourier expansion, it looks like it is usually not possible to get $f(x)$ from the infinite series.
PowerExpand[ FunctionExpand[-(1/\[Pi]^3) I (-\[Pi]^2 Log[1 - E^(I \[Pi] x)] + \[Pi]^2 Log[ E^(-I \[Pi] x) (-1 + E^(I \[Pi] x))] + 2 \[Pi]^2 Log[1 + E^(I \[Pi] x)] - 2 \[Pi]^2 Log[E^(-I \[Pi] x) (1 + E^(I \[Pi] x))] - 2 PolyLog[3, -E^(-I \[Pi] x)] + 2 PolyLog[3, E^(-I \[Pi] x)] + 2 PolyLog[3, -E^(I \[Pi] x)] - 2 PolyLog[3, E^(I \[Pi] x)])]] does not work
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Commented
Jul 19, 2020 at 16:17
L = 5; S = HoldForm[Sum[ 2/(n*Pi)^3*((2 L^2 - n^2*Pi^2*(1 + L^2))*(-1)^n - 2*L^2 + n^2*Pi^2)*Sin[n*Pi*x/L], {n, 1, Infinity}]]; FullSimplify[ReleaseHold[S], Assumptions -> 0 <= x <= L];Plot[% - x^2 - 1, {x, 0, L}]works well in version12.0. The result ofFullSimplifydiffers from yours. $\endgroup$L=5;s = Sum[2/(n*Pi)^3*((2 L^2 - n^2*Pi^2*(1 + L^2))*(-1)^n - 2*L^2 + n^2*Pi^2)*Sin[n*Pi*x/L], {n, 1, Infinity}];FullSimplify[D[s , {x, 3}], Assumptions -> x >= 0 && x <= L]which performs0. $\endgroup$