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Inversing my function doesn't work and maybe simplification oddity

Jérémy Bézairie
Posted 1 year ago

Hi

The function : InverseFunction[(a^2 Integrate[(1/(1 + b Cos[x]))^2, x])/c]

https://www.wolframalpha.com/input?i2d=true&i=InverseFunction+Divide%5Ba%C2%B2*antiderivative+%5C%2840%29Divide%5B1%2C1%2Bb*cos%5C%2840%29x%5C%2841%29%5D%5C%2841%29%C2%B2%2Cc%5D

https://www.wolframalpha.com/input?i2d=true&i=InverseFunction%5C%2891%29Divide%5B%5C%2840%29Power%5Ba%2C2%5D+%5C%2840%29Divide%5B%5C%2840%29-2+ArcTanh%5BDivide%5B%5C%2840%29%5C%2840%29-1+%2B+b%5C%2841%29+Tan%5BDivide%5Bx%2C2%5D%5D%5C%2841%29%2CSqrt%5B-1+%2B+Power%5Bb%2C2%5D%5D%5D%5D%5C%2841%29%2CPower%5B%5C%2840%29-1+%2B+Power%5Bb%2C2%5D%5C%2841%29%2C%5C%2840%29Divide%5B3%2C2%5D%5C%2841%29%5D%5D+%2B+Divide%5B%5C%2840%29b+Sin%5Bx%5D%5C%2841%29%2C%5C%2840%29%5C%2840%29-1+%2B+Power%5Bb%2C2%5D%5C%2841%29+%5C%2840%291+%2B+b+Cos%5Bx%5D%5C%2841%29%5C%2841%29%5D%5C%2841%29%5C%2841%29%2Cc%5D%5C%2893%29

: no solution .

I also tried with Mathematica :

InverseFunction[(a^2 Integrate[(1/(1 + b Cos[x]))^2, x])/c]

gives 

InverseFunction[(a^2 ((b sin(x))/((b^2-1) (b cos(x)+1))-(2 tanh^-1(((b-1) tan(x/2))/Sqrt[b^2-1]))/(b^2-1)^(3/2)+Subscript[c, 1]))/c]

InverseFunction[(a^2 ((b sin(x))/((b^2-1) (b cos(x)+1))-(2 tanh^-1(((b-1) tan(x/2))/Sqrt[b^2-1]))/(b^2-1)^(3/2)+d))/c]

gives 

InverseFunction[(a^2 ((b sin(x))/((b^2-1) (b cos(x)+1))-(2 (b-1) tan(x/2) coth(x))/(b^2-1)^2+d))/c]

it's simplified here ? Why not above ? So the same :
InverseFunction[(a^2 ((b sin(x))/((b^2-1) (b cos(x)+1))-(2 (b-1) tan(x/2) coth(x))/(b^2-1)^2+d))/c] gives
same .

Putting this in WolframAlpha :

https://www.wolframalpha.com/input?i2d=true&i=InverseFunction%5C%2891%29Divide%5B%5C%2840%29Power%5Ba%2C2%5D+%5C%2840%29Divide%5B%5C%2840%29b+sin%5C%2840%29x%5C%2841%29%5C%2841%29%2C%5C%2840%29%5C%2840%29Power%5Bb%2C2%5D-1%5C%2841%29+%5C%2840%29b+cos%5C%2840%29x%5C%2841%29%2B1%5C%2841%29%5C%2841%29%5D-Divide%5B%5C%2840%292+%5C%2840%29b-1%5C%2841%29+tan%5C%2840%29Divide%5Bx%2C2%5D%5C%2841%29+coth%5C%2840%29x%5C%2841%29%5C%2841%29%2CPower%5B%5C%2840%29Power%5Bb%2C2%5D-1%5C%2841%29%2C2%5D%5D%2Bd%5C%2841%29%5C%2841%29%2Cc%5D%5C%2893%29

still no solution .

Integrate[(1/(1 + a Cos[x]))^2, x] in Mathematica and WolframAlpha

https://www.wolframalpha.com/input?i2d=true&i=antiderivative+%5C%2840%29Divide%5B1%2C1%2Ba*cos%5C%2840%29x%5C%2841%29%5D%5C%2841%29%C2%B2
( " Standard computation time exceeded... " )

don't give the maybe simplification .

I haven't Wolfram|Alpha Pro .

You can see https://www.physicsforums.com/threads/solving-keplers-1st-law-as-a-function-of-time.1047141/ to understand where the function come from : the goal is to have the Kepler's 1st law in function of time .

Thanks .
POSTED BY: Jérémy Bézairie
4 Replies
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y=Integrate[(a/(1 + b Cos[x]))^2/c, x]

So if we have y , a , b , c and the antiderivative's constant how to find x ? Must compute it everytime ?

What about the maybe simplification ?

Also the derivative of antiderivative doesn't give the original function (1/(1 + a Cos[x]))^2 , WolframAlpha and Mathematica gives
(a cos(x))/((a^2-1) (a cos(x)+1))+(a^2 sin^2(x))/((a^2-1) (a cos(x)+1)^2)-((a-1) sec^2(x/2) coth(x))/(a^2-1)^2+(2 (a-1) tan(x/2) csch^2(x))/(a^2-1)^2 .
POSTED BY: Jérémy Bézairie
Updating Name
Updating Name
Posted 1 year ago

y(x)=(a^2 Integrate[(1/(1 + b Cos[x]))^2, x])/c , y(0)=0

So if we have y , how to find x ? Must compute it everytime ?

What about the maybe simplification ?

Also the derivative of antiderivative doesn't give the original function (1/(1 + a Cos[x]))^2 , WolframAlpha and Mathematica gives
(a cos(x))/((a^2-1) (a cos(x)+1))+(a^2 sin^2(x))/((a^2-1) (a cos(x)+1)^2)-((a-1) sec^2(x/2) coth(x))/(a^2-1)^2+(2 (a-1) tan(x/2) csch^2(x))/(a^2-1)^2 .
POSTED BY: Updating Name 
  func = a^2 *(1/(1 + b Cos[x]))^2/c
  int = Integrate[func, x]
  (*(a^2 (-((2 ArcTanh[((-1 + b) Tan[x/2])/Sqrt[-1 + b^2]])/(-1 + b^2)^(
      3/2)) + (b Sin[x])/((-1 + b^2) (1 + b Cos[x]))))/c*)
  D[int, x] // FullSimplify 
  (*a^2/(c (1 + b Cos[x])^2)*)
  func == D[int, x] // FullSimplify
  (*True*)

The derivative of antiderivative gives the original function .

for y[0]= 0 then we have:

 Limit[int, x -> 0](*Then C = 0 *)
 y[x] == int // Expand
 (*y[x] == -((2 ArcTanh[Tan[x/2]/Sqrt[3]])/(3 Sqrt[3])) + (2 Sin[x])/(
   3 (1 + 2 Cos[x]))*)

Solve[y[x] == -((2 ArcTanh[Tan[x/2]/Sqrt[3]])/(3 Sqrt[3])) + (2 Sin[
   x])/(3 (1 + 2 Cos[x])), x](*Solve can' t solve because is a Transcendental equation .*)

We can solve only numerically automatic and view the solution into on the Graph:

  a = 1; b = 2; c = 1; 
   ContourPlot[y == int, {y, -1, 1}, {x, -Pi, Pi}, 
   Frame -> False, Axes -> True, AxesLabel -> Automatic]

More info for Transcendental equation see here.
POSTED BY: Mariusz Iwaniuk

Analytically,it works only for simple functions.See below

    f[x_] := Log[x]; InverseFunction[(f[x] /. x -> #) &][x]
    f[x_] := Sin[x]; InverseFunction[(f[x] /. x -> #) &][x]
    f[x_] := Exp[x] - x; InverseFunction[(f[x] /. x -> #) &][x]
   f[x_] := Exp[x] - x^2; InverseFunction[(f[x] /. x -> #) &][x](*Can't find *)

 InverseFunction[((Integrate[(1/(1 + a Cos[x]))^2, x]) /. x -> #) &][x](*Can't find *)

In Mathematica,from help pages of InverseFunction:

As discussed in "Functions That Do Not Have Unique Values", many mathematical functions do not have unique inverses. In such cases, InverseFunction[f] can represent only one of the possible inverses for f. InverseFunction is generated by Solve when the option InverseFunctions is set to Automatic or True

Numerically with Plot

  Show[{ContourPlot[Sin[x] == y, {y, -1, 1}, {x, -Pi, Pi}], 
    Plot[ArcSin[x], {x, -1, 1}, PlotStyle -> {Dashed, Red}]}](*Example with Sin[x]*)


 Int = Integrate[(1/(1 + a Cos[x]))^2, x] /. a -> 2(*For a=2*)
 ContourPlot[Int == y, {y, -1, 1}, {x, -Pi, Pi}](*Inverse of integral*)
POSTED BY: Mariusz Iwaniuk
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