0
$\begingroup$

How to ask Mathematica to give $y$ in terms of $x$ in this function? Is there a way to obtain a closed form expression for $y$ in terms of $x$?

$x=\frac{\left(y^4-1\right) \left(\coth (2 y)-\frac{1}{\sinh (2 y)}\right)}{y^7-3 y^4}$

($x$ and $y$ are greater than zero)

Solve and Reduce do not work in this case.

$\endgroup$
  • $\begingroup$ You have a complicated transcendental equation, you should not expect to be able to find a symbolic inverse. The best you can do is to invert the equation numerically. Over what range are you hoping to work with? $\endgroup$ – Carl Woll Nov 5 '19 at 23:40
3
$\begingroup$
Clear["Global`*"]

f[y_] = (y^4 - 1) (Coth[2 y] - 1/Sinh[2 y])/(y^7 - 3 y^4);

Graphically, the inverse is

ParametricPlot[{f[y], y}, {y, -10, 10},
 Frame -> True,
 FrameLabel -> (Style[#, 14, Bold] & /@ {x, y}),
 AspectRatio -> 1/GoldenRatio]

enter image description here

InverseFunction will provide one branch

g[x_] := InverseFunction[f][x]

Plot[g[x], {x, -0.1, 0.15},
 Frame -> True,
 FrameLabel -> (Style[#, 14, Bold] & /@ {x, y}),
 PlotRange -> {-10, 10},
 ImageSize -> 500,
 Epilog -> Inset[
   Plot[g[x], {x, -0.1, 0.15},
    Frame -> True],
   {0.08, -5}]]

enter image description here

Numerically finding the inverse

g2[x_?NumericQ] := y /. NSolve[x == f[y], y, Reals]

ListPlot[
 Flatten[Table[{x, #} & /@ g2[x],
   {x, -0.1, 0.15, 0.005}], 1],
 Frame -> True,
 FrameLabel -> (Style[#, 14, Bold] & /@ {x, y})]

enter image description here

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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