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I have the equation

$$ x(\phi)=\frac{1}{2\sqrt2}\left(\frac{-2}{\phi}+\log\left(\frac{1+\phi}{1-\phi}\right)\right) $$ I need to find the dependence $\phi(x)$, so I use NSolve:

B = 20;
NSteps = 2*B*100; 
H = N[2*B/NSteps];
X = Range[-B, B, H];
F[i_] := NSolve[x[ϕ] == i && (-1.0 <= ϕ <= 0.0), ϕ];
Monitor[PhiTable = Table[ϕ /. F[i], {i, -B, B, H}], i]

And I get this output:

Output

NSolve can't find the solution and it gives an empty result at the values from about -15 and lower. I suppose this bug is caused by exponential asymptotic of the function $\phi(x)$.

But when I write 

F[i_] := NSolve[x[ϕ] == i, ϕ, Reals];
Monitor[PhiTable = Table[ϕ /. F[i][[1]], {i, -B, B, H}], i]

It finds roots at the begining but freezes at the point 16.44 and higher.

Is there a way to fix it? Or shall I use another function, not NSolve?

Added:

x[ϕ_] := 1/(2 Sqrt[2]) (-2/ϕ + Log[(1 + ϕ)/(1 - ϕ)]);
Michael E2
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Chertan
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     –  Aug 11 '16 at 18:35
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    Could you give Mathematica code for your definition of x[\[Phi]]? Could be similar to this question. – Chris K Aug 11 '16 at 18:55
  • It looks like your equation $x(\phi) = x_0$ has two solutions for all values of $x_0$, one with positive $\phi$ and one with negative $\phi$. Do you have a preference as to which one is returned? – Michael Seifert Aug 11 '16 at 19:03
  • @ChrisK definition: x[[Phi]_] := 1/(2 Sqrt[2]) (-2/[Phi] + Log[(1 + [Phi])/(1 - [Phi])]); I also added it to the post as screenshot – Chertan Aug 11 '16 at 19:04
  • @MichaelSeifert yes, I need negative one. Therefore I used the restrictions on phi in the first example – Chertan Aug 11 '16 at 19:07
  • Sorry, I can't add the screenshot of the function x[[Phi]] due to my reputation. But it has the same view as the equation at the begining of the question. – Chertan Aug 11 '16 at 19:16
  • A side issue, unrelated to the problem: You should avoid starting your symbol names with a capital letter, especially single character variables. Quite a few are reserved symbols used by Mathematica. – Michael E2 Aug 11 '16 at 19:50

1 Answers1

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The root tracker TrackRoot I wrote here can be applied to this problem. First, run TrackRoot from that link.

enter image description here

Then:

x[ϕ_] := 1/(2 Sqrt[2]) (-2/ϕ + Log[(1 + ϕ)/(1 - ϕ)]);
tr = TrackRoot[{x[ϕ] - xval}, {ϕ}, {xval, -20, 20}, 0, {-0.5}];
Plot[ϕ[x] /. tr, {x, -20, 20}]

enter image description here

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Chris K
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  • Thanks a lot, it works! Is it possible to correct this function and not to use interpolation? I'd like to get only values of phi. Sorry if it is a very foolish and simple question, I have never worked with manual functions before. – Chertan Aug 12 '16 at 14:22
  • @Chertan I'm not sure if this is exactly what you mean by correcting this function, but you can extract particular values of \[Phi] like this: \[Phi][1] /. tr gives -0.506903. – Chris K Aug 13 '16 at 01:39
  • yes, that's exactly what I meant. Thank you! – Chertan Aug 13 '16 at 13:42