I have the following Mathematica code:

f[U_] = 1 - (Un/U)^3; 

Un = 1;

L[Us_] := 3 NIntegrate[
   Us^(5/2)/U^(3/2) Sqrt[f[Us]/f[U]] 1/
    Sqrt[U^5 f[U] - Us^5 f[Us]], {U, Us, Infinity}]

UsvsL = Table[{Us, L[Us]}, {Us, 1.01, 2.01, 0.01}]

I am trying to find Us as a function of L or Us[L], for which I tried InverseSeries:

l[Us_] = 
  Series[Us^(5/2)/U^(3/2) Sqrt[f[Us]/f[U]] 1/
    Sqrt[U^5 f[U] - Us^5 f[Us]], {U, 0, 4}], U]

Us1 = InverseSeries[l[Us], l] // PowerExpand // Simplify 

but, to no avail. I need a table of LvsUs where I can vary the L values to obtain the corresponding Us values. Any help in this regard would be truly beneficial!


1 Answer 1


Look at your data:

ListLinePlot[UsvsL, AxesLabel -> {"U", "L"}]

![enter image description here

Note that for some Ls there two Us. Therefore the inverse function is not single valued. We therefore create two inverse functions, One with the data from U= 1.01 .. 1.21 and a second one with data from U= 1.21 .. 2.01.

dat1 = Select[UsvsL, #[[1]] <= 1.21 &];
dat2 = Select[UsvsL, #[[1]] >= 1.21 &];

Now, how to create the inverse funtions? We already have a table of L[Us]. If we reverse the table entries, we have a table Us[L], but maybe not for the requested L. Toward this aim, we can calculate an interpolating function.

intpol1 = Interpolation[Reverse /@ dat1]
intpol2 = Interpolation[Reverse /@ dat2]

enter image description here

With this two functions you may get the Us for given Ls:

Plot[intpol1[l], {l, Min[dat1[[All, 2]]], Max[dat1[[All, 2]]]}, 
 AxesLabel -> {"L", "U"}]
Plot[intpol2[l], {l, Min[dat2[[All, 2]]], Max[dat2[[All, 2]]]}, 
 AxesLabel -> {"L", "U"}]

enter image description here

  • $\begingroup$ Thank you for the answer! I just have one small doubt. How can I create the two tables from the above interpolating functions? $\endgroup$
    – codebpr
    Sep 26 at 13:31
  • 1
    $\begingroup$ That is easy. If you want the U value to a certain L value you say e.g. : interpol1[L]. For a whole table: Table[interpol1[x],{x,xmin,xmax, xdelta}]. Note to take care that xmin and xmax are not outside the definition range, otherwise you may get arbitrary results. $\endgroup$ Sep 26 at 14:56

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