How to Plot an Infinite Series

Question

I want to sketch the graphs of $$u(t,x)=\frac{1}{2}+\sum_{n=1}^\infty \frac{1}{n\pi} ((-1)^n-1)e^{-n^2t}\sin(nx)$$ for $t=0, 0.01, 0.1, 0.5, 1, 10$ on the same axes.

For $t=0$, I input

Plot[{1/2 + Sum[1/(n π) ((-1)^n - 1) Sin [n x]], {n, 1, Infinity}}, {x, -π, π}]

Then Mathematica keeps running...

Could you help me with this? Thanks.

Answer 1

If you define

f[x_, t_, nm_] := 1/2 + Sum[1/(n π) ((-1)^n - 1) Sin[n x] Exp[-t n^2], {n, 1, nm}];

then

Plot[Table[f[x, t, 150], {t, {0, 0.01, 0.1, 0.5, 1, 10}}] // Evaluate, {x, -Pi, Pi}]

produces

and the "Gibbs ringing" i.e. the small oscillations near the sharp edges come from truncation of the sum at 150 instead of $\infty$.

Treating separately the t=0 case, which can be summed to infinity (see Sjoerd's answer), you can get a pretty accurate plot while choosing nm=1000

 Show[{Plot[
 1/2 + Sum[1/(n π) ((-1)^n - 1) Sin[n x], {n, 1, Infinity}] // 
 Evaluate, {x, -π, π},PlotStyle-> Darker[Blue,0.5]],
 Plot[Table[
 1/2 + Sum[1/(n π) Exp[-n^2 t] ((-1)^n - 1) Sin[n x], {n, 1, 
    1000}], {t, {0.01, 0.1, 0.5, 1, 10}}] // 
 Evaluate, {x, -π, π}, PlotPoints -> 50]}]

Answer 2

Your input contains a syntax error. You put the summation range outside the Sum. Another thing that will improve plotting is adding an Evaluate, otherwise Plot will re-calculate the sum for every iteration.

Plot[
  1/2 + Sum[1/(n π) ((-1)^n - 1) Sin[n x], {n, 1, Infinity}] // Evaluate,
  {x, -π, π}
]

Answer 3

This just a slight improvement from previous answers adding more information in the plot:

Plot[{1/2 + Sum[1/(n \[Pi]) ((-1)^n - 1) Sin[n x], {n, 1, Infinity}], 
 Table[1/2 + 
 Sum[1/(n \[Pi]) Exp[-n^2 t] ((-1)^n - 1) Sin[n x], {n, 1, 
   1000}], {t, {0.01, 0.1, 0.5, 1, 10}}]} // 
Evaluate, {x, -\[Pi], \[Pi]}, PlotPoints -> 50, 
PlotStyle -> ColorData[3, "ColorList"][[1 ;; 6]], 
PlotLegends -> Table[t == i, {i, {0, 0.01, 0.1, 0.5, 1, 10}}], 
Exclusions-> True, ExclusionsStyle -> Directive[Black, Dashed, Thickness[0.006]]