2
$\begingroup$

I'm trying to make a make a plot of a table of expressions, which should be pre-evaluated (since they are infinite series with analytic sums). I'm able to make the plot normally with Evaluate @ Table:

Plot[Evaluate@Table[Sum[t^k/k! Exp[-t], {k, 0, Infinity, n}], {n, 1, 5}], {t, 0, 5}]

But when I surround it with a Manipulate it no longer works (there are errors that indicate Mathematica is evaluating the summation with a fixed value of t, which doesn't work at t = 0, k = 0).

Manipulate[
  Plot[Evaluate @
    Table[Sum[ ((a t)^k)/k! Exp[-a t], {k, 0, Infinity, n}], {n, 1, 5}], {t, 0, 5}],
  {a, 0, 3}]

I know I can just evaluate the table and copy-paste, but I was hoping for a more elegant solution.

$\endgroup$

3 Answers 3

3
$\begingroup$

It can work with a=0 too, the problem is that a is applied before closed form of the sum is calculated. We can force this:

DynamicModule[{t, k, tab, a}, 
 Column[{
   Dynamic@Plot[tab[a, t], {t, 0, 5}, ImageSize -> 400],
   Slider[Dynamic[a], {0, 11, 1}]
   }],
 Initialization :> {tab[a_, t_] = Table[Sum[((a t)^k)/k! Exp[-a t], {k, 0, Infinity, n}],
                                        {n, 1, 5}];
 }]

enter image description here

$\endgroup$
3
$\begingroup$

With a few bells and whistles:

Manipulate[
 Module[{plts},

  plts[a_, t_] =
   Table[Tooltip[
     Sum[((a t)^k)/k! Exp[-a t],{k, 0, Infinity, n}] // Simplify,
     StringForm["n = ``", n]], {n, 5}];

  Plot[Evaluate[plts[a, t]], {t, 0, 5},PlotRange -> {0, 1}]],

 {{a, 1.5}, 0, 3, 0.05, Appearance -> "Labeled"}]

Bob Hanlon

$\endgroup$
2
$\begingroup$

The problem is how to evaluate the sums without before a is set to 0. That can be done with With. If you put Dynamic around Plot, then only the Plot will be updated when the slider for a is moved.

Manipulate[
 With[{plots = Table[Sum[((a0 t)^k)/k! Exp[-a0 t], {k, 0, Infinity, n}], {n, 1, 5}]},
  Dynamic @ Plot[Evaluate[plots /. a0 -> a], {t, 0, 5}, PlotRange -> {0, 1}]],
 {a, 0, 3}]

Manipulate output

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.