I want to extend this function into a new function according to its period
Plot[-((E^x + E^(-x))/2), {x, -1, 1}, AxesOrigin -> {0, 0}] and then draw this function as follows
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0 asked Jan 17, 2020 at 3:07
A little mouse on the pampas
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4 Answers
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Here is a somewhat more general approach. It allows the basic period of the periodic extension to be any interval in its source function's domain.
Clear[f, xf] f[x_] := -((E^x + E^(-x))/2) f[x_, lo_, hi_] /; lo ≤ x < hi := f[x] xf[x_, lo_, hi_] := With[{span = hi - lo}, Piecewise[{ {f[x + span Quotient[hi - x, span], lo, hi], x < lo}, {f[-x + span Quotient[x - lo, span], -hi, -lo], x > hi}}, f[x, lo, hi]]] The plot you request is then:
Plot[xf[x, -1, 1], {x, -5, 5}, AxesOrigin -> {0, -1.54}] 
But a plot making the period of xf the asymmetric interval {-1, .5] is just as easy.
Plot[xf[x, -1, .5], {x, -4, 5}, AxesOrigin -> {0, -1.54}] 
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One possible way
T = 2; (*period*) f[x_] := -((E^x + E^(-x))/2); fExtended[x_] := Piecewise[{{f[x], -T/2 < x < T/2}, {fExtended[x - T], x > T/2}, {fExtended[x + T], x < T/2}}]; Plot[fExtended[x], {x, -1, 1}, AxesOrigin -> {0, 0}, AxesOrigin -> {0, 0}] 
Plot[fExtended[x], {x, -2 T, 2 T}, AxesOrigin -> {0, 0}, AxesOrigin -> {0, 0}] 
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Let's replace x with Mod[x, 2, -1]
Plot[-((E^Mod[x, 2, -1] + E^(-Mod[x, 2, -1]))/2), {x, 0, 10}] We can get the result we want.
answered Jan 17, 2020 at 3:30
A little mouse on the pampas
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This answer adds to Go with the wind's answer.
T = 4; (* period *) x0 = -2; (* start of the function sampling*) f[x_] := -((E^x + E^(-x))/2); xExt[x_] := Mod[x - x0, T] + x0; fExt = f@*xExt; Plot[fExt[x], {x, -5, 5}] 
fExt now repeats f in the interval $[x_0,x_0+T]$. The '@*' means function composition. If you define h=f@*g; then h[x]==f[g[x]].
Edit: made a function out of it for ease of use. This gives the same output as above.
MakePeriodic[f_, T_, x0_: 0] := Module[{xExt}, xExt = Function[x, Mod[x - x0, T] + x0]; f@*xExt ] fExt = MakePeriodic[f,4,-2]; Plot[fExt[x], {x,-5, 5}] 