0
$\begingroup$

Can anyone help me solve this sum?

 p[x_,a_] := PDF[NormalDistribution[0,a], x];
  a[x_] := Exp[x];
  u[x_] := (a[x]/(a[x] + 1));
  wa[x_, t_] := ((t - 3)*a[x])^(u[x] + .02442);
 Sum[p[x, .5]*wa[x, t], {x, 0, Infinity}, {t, 0, 50}]

Output For me it only gives back a symbolic output. But I would like to have the actual result.

$\endgroup$
6
  • 3
    $\begingroup$ You mean, a closed form? Do you have any justification that it even exists? $\endgroup$
    – corey979
    Jun 24, 2018 at 13:09
  • $\begingroup$ @corey979 Justification? what do you mean? $\endgroup$
    – MarV
    Jun 24, 2018 at 13:10
  • 2
    $\begingroup$ Do you have strong evidence that a closed form (e.g., the sum is convergent) exists (you didn't answer my first question, so I assume you want a closed form), or is it just an expression that you arrived at, threw it in MMA and just expect it to do wonders? $\endgroup$
    – corey979
    Jun 24, 2018 at 13:17
  • 1
    $\begingroup$ I see. Well, first, you can't obtain a result other than numeric because you have inexact numbers in the formulas. E.g., Sum[1./k^2, {k, 1, Infinity}] yields 1.64493, but if you change 1. to 1, you'll get $\pi^2/6$. Second, examine Table[Plot[Evaluate@ReIm[p[x, .5]*wa[x, t]], {x, 0, 5}, PlotRange -> All, Frame -> True, PlotLegends -> {"Re", "Im"}, PlotLabel -> "t = " <> ToString@t], {t, Range[0, 10]~Join~{100}~Join~{1000}}] // FlipView. For every t, the expression decays very quickly with x, so summing up to infinity is an overkill (...) $\endgroup$
    – corey979
    Jun 24, 2018 at 13:30
  • 2
    $\begingroup$ (...) Second (as a side note), the bigger the t, the greater the values of the real part of the function are. Finally, NSum[p[x, .5]*wa[x, t], {x, 0, 3}, {t, 0, 50}] gave me (after several seconds) 303.282 + 4.15345 I - is that a result that satisfies you - the actual result you are after? [Changing {x, 0, 3} to {x, 0, 4} doesn't change the result.] $\endgroup$
    – corey979
    Jun 24, 2018 at 13:32

0

Your Answer

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

Browse other questions tagged or ask your own question.