# Help with function graph

I need to graph the function $$f_s(x)=\frac{1+2x^2-3x^2\Xi(x)}{1-x^2},$$ were $$\Xi(x)= \begin{cases} \frac{1}{\sqrt{1-x^2}}\tanh^{-1}(\sqrt{1-x^2}); & 0\leq x<1\\ \frac{1}{\sqrt{x^2-1}}\tan^{-1}(\sqrt{x^2-1}); & x \geq1 \end{cases}.$$

My effort:

P1 = Plot[(1 + 2 x^2)/(1 - x^2) + (3 x^2 ArcTanh[Sqrt[x^2 - 1]])/\!$$\*SuperscriptBox[\((\*SuperscriptBox[\(x$$, $$2$$] - 1)\), $${\*FractionBox[\(3$$, $$2$$]}\)]\), {x, 1, 14},  PlotTheme -> "Scientific", PlotStyle -> Black,  FrameTicks -> {{{-2, -1.5, -1, 0, 0.5, 1},
None}, {{0, 2, 4, 6, 8, 10, 12, 14}, None}}]

P2 = Plot[(1 + 2 x^2)/(1 - x^2) - (3 x^2 ArcTanh[Sqrt[-x^2 + 1]])/\!$$\*SuperscriptBox[\((\(-\*SuperscriptBox[\(x$$, $$2$$]\) + 1)\), $${\*FractionBox[\(3$$, $$2$$]}\)]\), {x, 0, 1},  PlotTheme -> "Scientific", PlotStyle -> Black,  FrameTicks -> {{{-2, -1.5, -1, 0, 0.5, 1},
None}, {{0, 2, 4, 6, 8, 10, 12, 14}, None}}]

Show[P1, P2]


I want the graph is equal to

or

• Your formula suggests that you should use Tanh, but in your code you use ArcTan. – C. E. Jul 12 '17 at 0:10
• it is true. I'm sorry. – Dinesh Shankar Jul 12 '17 at 0:16

Just use $\Xi(x) = \frac{1}{\sqrt{x^2-1}}\tanh(\sqrt{x^2-1})$ for all $x$ instead. Or use $\frac{1}{\sqrt{1-x^2}}\tan(\sqrt{1-x^2})$ for the $x<1$ range instead.

I meant to make this a comment, but since I answered it instead, here is the plot I think you want:

LogLinearPlot[(1+2x^2-3x^2 Tanh[Sqrt[x^2-1]]/Sqrt[x^2-1])/(1-x^2), {x,0,100}]


Update

Looking at the literature, (e.g., Exact solution of the Thomas-Fermi equation for a trapped Bose-Einstein condensate with dipole-dipole interactions by Claudia Eberlein, Stefano Giovanazzi, and Duncan H. J. O’Dell) it seems that the correct function uses ArcTan, and that the formula for $x<1$ is obtained by analytic continuation. So, I think the plot you want is:

LogLinearPlot[(1+2x^2-3x^2 ArcTan[Sqrt[x^2-1]]/Sqrt[x^2-1])/(1-x^2), {x,0.01,100}]


which seems to perfectly match your expected plot.

One more comment. Note:

FullSimplify[
ComplexExpand[ArcTanh[Sqrt[1-x^2]]/Sqrt[1-x^2]-ArcTan[Sqrt[-1+x^2]]/Sqrt[-1+x^2]],
x > 0
]


Piecewise[{{0, x != 1}}, Indeterminate]

So, there is no need for a Piecewise definition of $\Xi(x)$. Mathematica can analytically continue both formulas into each other.

• My function was wrong. I'm so sorry – Dinesh Shankar Jul 12 '17 at 0:26
• @user3321 Are you sure it's supposed to be ArcTanh? A brief survey of the literature seems to indicate that it's supposed to be ArcTan. – Carl Woll Jul 12 '17 at 0:41
• Now it's right. I'm sorry again – Dinesh Shankar Jul 12 '17 at 0:49

Here is a start:

Ξ[x_] := Piecewise[{{1/Sqrt[1 - x^2]*Tanh[Sqrt[1 - x^2]], 0 <= x < 1}},
1/Sqrt[x^2 - 1]*Tanh[Sqrt[x^2 - 1]]]

f[x_] := (1 + 2*x^2 - 3 x^2*Ξ[x])/(1 - x^2)

Plot[f[x], {x, 0, 16}, PlotRange -> {Automatic, {-2, 2}},
Frame -> True, AxesStyle


• My function was wrong. I'm so sorry. – Dinesh Shankar Jul 12 '17 at 0:21
• @user3321 No problem, just take from my answer how you define a piecewise function and you can use it with your correct formula. – halirutan Jul 12 '17 at 0:31