Sum[Cosh[n*Pi*x]/(n*Cosh[n*Pi])*Sin[n*Pi*y], {n, 1, Infinity, 
    2}], {x, -1, 1}, {y, 0, 1}]

On evaluating the plot I am not getting any results nor do I get any error message. How to overcome this problem ?

  • 1
    $\begingroup$ Plot is not the problem here, Mathematica gets stuck trying to evaluate the Sum. $\endgroup$ – Fraccalo Jan 20 at 15:02
  • $\begingroup$ How to overcome it? $\endgroup$ – Gourav Halder Jan 20 at 15:30
  • 1
    $\begingroup$ mathematica doesn't seem to be able to solve it. Probably best thing to do is trying to understand if the series actually converges/can be written as a closed-form function of x,y $\endgroup$ – Fraccalo Jan 20 at 15:35
  • $\begingroup$ it actually converges , this plot is there in literature but i was trying to reproduce it $\endgroup$ – Gourav Halder Jan 20 at 15:38
  • $\begingroup$ "this plot is there in literature" - then how about mentioning the book/paper you saw this in in your question? $\endgroup$ – J. M.'s discontentment Jan 24 at 3:09

Until closed form sum can be found, you could always plot partial sum as the sum increases. This shows it converges. Mathematica not able to find closed forum sum to infinity, but this is close enough if you are only interested in the plot itself.

enter image description here

ClearAll[x, y, n];
  sum = Sum[Cosh[n*Pi*x]/(n*Cosh[n*Pi])*Sin[n*Pi*y], {n, 1, upTo, 2}];
  Plot3D[(4/Pi)*sum, {x, -1, 1}, {y, 0, 1}]
 {{upTo, 10, "number of terms?"}, 2, 200, 1, Appearance -> "Labeled"},
 ContinuousAction -> False,
 TrackedSymbols :> {upTo}
| improve this answer | |
  • 1
    $\begingroup$ This is pretty much what I wanted. Thanks. $\endgroup$ – Gourav Halder Jan 20 at 17:49

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