Given the plot of $(v, f(v))$ how to scale horizontal axis with $v \to 87\sqrt{-2\log(1 - v)})$, while keeping the vertical axis unchanged?

Question

In my previous post @herbertfederer provided a script to plot $\frac{\partial C}{\partial u}\Big|_{u=0.5}$ where $C(u,v)= \min\left[ u^{0.0325}v, uv^{-0.1018} \right]$,

c[u_, v_] := Min[u^0.0325 v, u v^-0.1018]
Plot[D[c[u, v], u] /. u -> .5 // Evaluate, {v, 0, 1}]

Here I modified the range of $v$ to $(0,1)$. Now it plots a function say $f(v)$ for the horizontal $v$ values. I need to assign $87\sqrt{-2\log(1 - v)}$ to $v$ and plot the function based on this new horizontal axis. So I tried,

c[u_, v_] := Min[u^0.0325 v, u v^-0.1018]
Plot[D[c[u, v], u] /. u -> 0.5 /. v -> (87*Sqrt[-2*Log[1 - v]]) // Evaluate, {v, 0, 0.9999}]

I choose "{v, 0, 0.9999}" in order to $\log(1-v)$ be defined. But after running the script I got errors. Can you help me to fix the issue?

Answer 1

$Version

(* "14.0.0 for Mac OS X x86 (64-bit) (December 13, 2023)" *)

c[u_, v_] := Min[u^0.0325  v, u  v^-0.1018]

Using ParametricPlot as suggested by Ulrich Neumann

ParametricPlot[{(87*Sqrt[-2*Log[1 - v]]), Derivative[1, 0][c][0.5, v]}, 
  {v, 0, 1},
 AspectRatio -> 1/GoldenRatio,
 ColorFunction -> (ColorData["Rainbow"][#3] &),
 PlotLegends -> BarLegend[{"Rainbow", {0, 1}},
   LegendLabel -> Style[v, 14],
   LegendMarkerSize -> 220],
 Frame -> True,
 FrameLabel -> (Style[#, 14] & /@ {(87*Sqrt[-2*Log[1 - v]]),
     HoldForm[Derivative[1, 0][c][0.5, v]]})]

Answer 2

The new v looks like:

Plot[-87*Sqrt[-2*Log[1 - v]], {v, 0, 0.9999}]

Now rise this to the power of : -0.1018:

Table[(-87*Sqrt[-2*Log[1 - v]])^-0.1018, {v, 0, 0.9, 0.1}]

{ComplexInfinity, 0.6522 - 0.215998 I, 0.627758 - 0.207903 I, 
 0.612949 - 0.202998 I, 0.601844 - 0.199321 I, 0.592567 - 0.196248 I, 
 0.584208 - 0.19348 I, 0.576145 - 0.19081 I, 0.567696 - 0.188011 I, 
 0.55744 - 0.184615 I}

You see that the new v is complex.

Answer 3

From your comment, I take it that you are looking for: f'( Inverse(f)(x) ).

Lets set u=0.1. Then the derivative is is defined for x>0. However, the inverse function creates a problem. Look at:

Plot[c[0.1, x], {x, 0, 1}, PlotRange -> All]

The inverse function c takes the values between 0.145 and 0.1 twice. Therefore, it it is not unique, but double valued. Therefore, you have to restrict the definition range to make the function single valued..