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I want to solve a nonlinear equation involving the Real part and imaginary part of the Exponentialintegralfunction, i.e. $\mathcal{R}[\text{Ei}[a(1+ib)]]$ and some other complicated terms. As it is not possible to solve my problem analytically, I wanted to use NSolve...but it doesnt work. The thing is: If i look at my plot, the solution is very evident. How can i solve this problem?

The explicit form i want to solve is

 E^(t (0.001998 Im[ExpIntegralEi[0.001 (1 + I t)]] + (
 2 (t Cos[0.001 t] - Sin[0.001 t]))/(1 + t^2)) + 
 2000. (-1 + (Cos[0.001 t] + t Sin[0.001 t])/(1 + t^2)) - 
 999. (-2.01265 - 0.001998 Re[ExpIntegralEi[0.001 (1 + I t)]] + (
 2 (Cos[0.001 t] + t Sin[0.001 t]))/(
 1 + t^2))) (-E^(-t (0.001998 Im[
      ExpIntegralEi[0.001 (1 + I t)]] + (
    2 (t Cos[0.001 t] - Sin[0.001 t]))/(1 + t^2)) - 
 2000. (-1 + (Cos[0.001 t] + t Sin[0.001 t])/(1 + t^2)) + 
 999. (-2.01265 - 0.001998 Re[ExpIntegralEi[0.001 (1 + I t)]] + (
    2 (Cos[0.001 t] + t Sin[0.001 t]))/(1 + t^2))) + 1/(1 + t^2))==0.2

Just an example...of course i want to use some different values. I can plot the r.h.s. and 'see' my result;

edit: Here is the actual command that gives me { } as solution

    NSolve[{E^(t (0.001998 Im[
         ExpIntegralEi[
          0.001 (1 + I t)]] + (2 (t Cos[0.001 t] - 
            Sin[0.001 t]))/(1 + t^2)) + 
    2000. (-1 + (Cos[0.001 t] + t Sin[0.001 t])/(1 + t^2)) - 
    999. (-2.01265 - 
       0.001998 Re[
         ExpIntegralEi[
          0.001 (1 + I t)]] + (2 (Cos[0.001 t] + 
            t Sin[0.001 t]))/(1 + 
          t^2))) (-E^(-t (0.001998 Im[
            ExpIntegralEi[
             0.001 (1 + I t)]] + (2 (t Cos[0.001 t] - 
               Sin[0.001 t]))/(1 + t^2)) - 
       2000. (-1 + (Cos[0.001 t] + t Sin[0.001 t])/(1 + t^2)) + 
       999. (-2.01265 - 
          0.001998 Re[
            ExpIntegralEi[
             0.001 (1 + I t)]] + (2 (Cos[0.001 t] + 
               t Sin[0.001 t]))/(1 + t^2))) + 1/(1 + t^2)) == 
0.2 && 0 < t < 50 }, t]

The Plot suggest that the solution lies at about $t\approx 30$ Thanks already!

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1
  • $\begingroup$ There is no root in-between 0 and 50 (the roots is around 62, see my answer). $\endgroup$
    – anderstood
    Commented Feb 21, 2018 at 15:23

1 Answer 1

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Your equation is not very complicated for a numerical problem. Try:

f[t_?NumericQ] := 
   E^(t (0.001998 Im[ExpIntegralEi[0.001 (1 + I t)]] + (
 2 (t Cos[0.001 t] - Sin[0.001 t]))/(1 + t^2)) + 
 2000. (-1 + (Cos[0.001 t] + t Sin[0.001 t])/(1 + t^2)) - 
 999. (-2.01265 - 0.001998 Re[ExpIntegralEi[0.001 (1 + I t)]] + (
 2 (Cos[0.001 t] + t Sin[0.001 t]))/(
 1 + t^2))) (-E^(-t (0.001998 Im[
      ExpIntegralEi[0.001 (1 + I t)]] + (
    2 (t Cos[0.001 t] - Sin[0.001 t]))/(1 + t^2)) - 
 2000. (-1 + (Cos[0.001 t] + t Sin[0.001 t])/(1 + t^2)) + 
 999. (-2.01265 - 0.001998 Re[ExpIntegralEi[0.001 (1 + I t)]] + (
    2 (Cos[0.001 t] + t Sin[0.001 t]))/(1 + t^2))) + 1/(1 + t^2))-0.2

FindRoot[f[t], {t, 50}] // AbsoluteTiming
(* {0.003991, {t -> 62.2356}} *)

It also has another root at approx. -62.2356, since f is even (I think---haven't proven it).

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  • $\begingroup$ Thanks! Why does this work but not Nsolve? $\endgroup$
    – Martin
    Commented Feb 21, 2018 at 14:53
  • 1
    $\begingroup$ @Martin NSolve and FindRoot are different functions; FindRoot finds one root for an initial point (using e.g. Newton-Raphson) but NSolve try to find all roots, which is much more difficult. But I don't know specifically why NSolve returns an error, I guess it would be because f has complex terms. You could try simplifying f use only real terms and see if it works. $\endgroup$
    – anderstood
    Commented Feb 21, 2018 at 14:57
  • $\begingroup$ Thanks for the answer! Simplifying to only real terms doesn't work, as there is no expression for Im[Ei[a+bi]]. But it thanks, you solved my problem! $\endgroup$
    – Martin
    Commented Feb 21, 2018 at 15:32
  • $\begingroup$ @Martin Well, there is, but it is not elementary :) functions.wolfram.com/GammaBetaErf/ExpIntegralEi/19/02 $\endgroup$
    – anderstood
    Commented Feb 21, 2018 at 15:35

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