# Can I solve a certain equation?

Neither Mathematica nor Wolfram|Alpha can resolve the following. How else can I reach a solution?

 Reduce[(1 + x)^(2/m) - (1 - x)^(2/m) == (1 - x^2)^(1/m), x]

• FindInstance[(1 + x)^(2/m) - (1 - x)^(2/m) == (1 - x^2)^(1/m), {x, m}] will give you a solution... restricting that to reals gives another... – ciao Jul 18 '15 at 0:12
• @ciao Also with FindInstance[..., Reals] :) – Dr. belisarius Jul 18 '15 at 0:16
• @belisarius: You must have been typing as I edited my comment ;-) – ciao Jul 18 '15 at 0:19

Manipulate[
Show[
ContourPlot[
(1 + x)^(2/m) - (1 - x)^(2/m) - (1 - x^2)^(1/m),
{x, -.1, 1.1}, {m, .1, 5}, Contours -> {0}],
pt = {x /. FindRoot[
(1 + x)^(2/m) - (1 - x)^(2/m) - (1 - x^2)^(1/m),
{x, m/5}],
m};
Graphics[{
Text[ToString[Round[pt, .001]], pt, {-1.25, 0}],
Red, AbsolutePointSize, Point[pt]}]],
{{m, 1}, .1, 5, .01, Appearance -> "Labeled"}] • Nice use of FindRoot, +1 – ciao Jul 18 '15 at 0:39

This problem is solvable analytically, although we use Mathematica for some of the algebra. To begin, write the initial expression as

(1 + x)^(2/m) - (1 - x)^(2/m) - (1 - x^2)^(1/m);


and divide through by the third term.

Expand[-%/%[]] /. (1 - x^2)^(-1/m) -> (1 - x)^(-1/m) (1 + x)^(-1/m)
(* -1 - (1 - x)^(1/m) (1 + x)^(-1/m) + (1 - x)^(-1/m) (1 + x)^(1/m) *)


Replacing ((1 + x)/(1 - x))^(1/m) by z casts the expression into the form

z - 1 - 1/z


which can be solved by

zsol = Solve[z - 1 - 1/z == 0, z]
(* {{z -> 1/2 (1 - Sqrt)}, {z -> 1/2 (1 + Sqrt)}} *)


Choose the second solution to obtain real x.

zsol[[2, 1]] /. z -> ((1 + x)/(1 - x))^(1/m) /. Rule -> Equal;
Solve[%, x][[1, 1]]
(* x -> (-1 + (1/2 + Sqrt/2)^m)/(1 + (1/2 + Sqrt/2)^m) *)


For comparison with the numerical solution by Bob Hanlon,

%[] /. m -> 1 // N
(* 0.236068 *)

• Very good. Note that (x -> (-1 + (1/2 + Sqrt/2)^m)/(1 + (1/2 + Sqrt/2)^ m)) // FullSimplify reduces to x -> Tanh[1/2 m ArcCsch] – Bob Hanlon Jul 18 '15 at 14:05