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Consider this problem;

Show that the function $\tan{x}-x$ can be inverted and find the corresponding inverse for which applies

$$ f^{-1} (y) = (f^{-1}(y) + y)^{-2}$$

So to prove that a function can be inverted it has to be bijective, which means it has to be surjective and injective

Now to prove for injective I derived the function $\tan{x}$ and got that it is constantly growing so it is injective. To prove that it is surjective we are already given a solution. Now what I did is derived the function $\tan{x}-x$ and got that its simply $\tan^2{x}$ since the $1$ and $-1$ cancel out. Then out of the starting equation I found a way to explicitly write $\tan{x}$;

$$\tan{x}=f(x)+x$$ and with that we can also determine this;

$$\tan^2{x}=(f(x)+x)^2$$

Now I've used the formula $$ (f^{-1})'(y) = \frac {1}{f'(x)} $$ and this is where I am stuck, the only progress I could make is if instead of $f'$ I insert $\tan^2{x}$ but that is it.

Looking forward to your answers !

EDIT: The domain of the function was also given (my bad)

$$\left(\mbox{-}\frac{\pi}{2}, \frac{\pi}{2}\right)$$

codeisfun
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    Actually $\tan x -x$ can not be inverted. For example, $$\tan x - x=0$$ has infinitely many solutions. – Crostul Jan 25 '21 at 16:05
  • Okay but since we are given solutions I'd asumme there has to be something to it because why would it be given to us in the first place? – codeisfun Jan 25 '21 at 16:10
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    My previous comment had a purpose: in order to invert $\tan x -x$ we need to fix a domain (for example $(- \pi /2; \pi /2)$. Then we can think about looking for the inverse function. However, I doubt we can find such a function in terms of elementary function. – Crostul Jan 25 '21 at 16:18
  • Uh my bad actually the domain was given as well,I'll correct that real quick – codeisfun Jan 25 '21 at 16:24
  • Specifying a domain is a necessary part of defining a function -- over which domain is your function defined? This is crucial to the question of inversion. – Allawonder Jan 25 '21 at 16:31
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    Yes,I know,it happens we are all human – codeisfun Jan 25 '21 at 16:37
  • Wolfram Alpha is definitely unhappy about this (ignoring domain of definition issues which are easily rectified by only looking at two branches): https://www.wolframalpha.com/input/?i=inverse+function+calculator&assumption=%7B%22F%22%2C+%22InverseFunction%22%2C+%22invfunction%22%7D+-%3E%22tan%28x%29-x%22 – Cameron L. Williams Jan 25 '21 at 16:55
  • I am pretty sure that inverse function theorem states $$(f^{-1})'(y) = \frac{1}{f'(x)}$$ not $f^{-1}(y) = \frac {1}{f'(x)}$. – macton Jan 25 '21 at 17:05
  • I actually thought I put that in, guess not.I'll correct it – codeisfun Jan 25 '21 at 17:07
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    I would assume an error in the left hand side of the expression to be proved. It should be $(f^{-1})'(y)$. – user Jan 25 '21 at 17:33
  • @user, I think you are quite right. Maybe you should flesh your comment out into a full fledged answer. – Barry Cipra Jan 25 '21 at 18:14
  • It simplifies your typing a lot if you choose to use the symbol $g$ as inverse of $f$. Then $g'(y) =1/f'(x)$ where $f(x) =y$. – Paramanand Singh Jan 26 '21 at 01:52
  • Your function is surjective because the mapping rule (the function expression) assigns an $x$ to every $f(x)$. The inverse rule of differentiation won't get you anywhere. See what is said about the inverse at https://math.stackexchange.com/questions/442332/express-solutions-of-equation-tan-x-x-in-closed-form. Consider that the question for solving the one-variable equations is a different mathematical problem – IV_ Jul 19 '22 at 13:16

3 Answers3

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Here are 2 closed form solutions, one with a limit, for the inverse of $\tan(x)-x$ using the Bessel J Zero and Inverse Beta Regularized functions: $$\tan(y)-y=x\implies x=\tan^{-1}\left(\text j_{\frac32,n+\frac x\pi}\right)+n\pi$$

which works and $n$ gives the $n$th root in terms of increasing size.

Similarly from:

Inverse of $x - \tanh(x)$:

$$x\in\text{Imaginary}\ \Bbb I:\tan(y)-y= x\implies y=\pm i\lim_{a\to0}\tanh^{-1}\left(\sqrt{\text I^{-1}_{\pm aix}\left(\frac32,\frac a2\right)}\right)\sim 1-ix$$

which works. Note the $2$ symmetrical “branches” of the inverse where both “$+$”s or “$-$”s are taken based on the sign of $x$. Please correct me and give me feedback!

This inverse is probably the only one that can be done with a Bessel Zero and a quantile function in Mathematica.

Тyma Gaidash
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COMMENT.-There is no difficulty in verifying that the function $f (x) = \tan (x) -x$ is a bijection of $(-\dfrac{\pi}{2},\dfrac{\pi}{2})$ on $\mathbb R$. Furthermore, as many points of the inverse function can be constructed in the Cartesian plane as desired. Since if $(x, f (x))$ is a point on the initial curve then the symmetric with respect to the first diagonal, $(f (x), x)$ is a point of $f^{-1}$. In the attached figure, the red points of the inverse function have been drawn as symmetrical to the green points on the curve $f (x) = \tan (x) -x$ and the curve corresponding to $f^{-1}$ has been drawn with imperfect drawing in blue color.

I don't think there is a simple expression for the inverse function. But you can find, as I repeat, as many points of the graph of $f^{-1}$ as you want.

enter image description here

Ataulfo
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  • Okay I've been stuck at this for ages and I've simply looked in the solutions,thing is I am more confused now than I was before.I'll post the solution and maybe someone can enlighten me. – codeisfun Jan 25 '21 at 21:25
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Okay so as I've just mentioned in the comments I've been unable to solve this so I looked in the solutions.The solution is very unclear to me,so I'd like if someone could try to explain or give me a step in the right direction.This is what the solution in its full detail looks.

$$ f(x) = tan x - x => tan(x) = f(x) + x$$ $$ f'(x) = 1 + tan^2(x) - 1 = tan^2(x) $$ $$ tan^2(x) = (f(x) + x)^2 $$

$$ (f^-1)'(x) = \frac{1}{f'(f^-1)} = \frac{1}{tan^2(f^-1)} = \frac{1}{(f(f^-1)+f^-1)^2} =\frac{1}{(x+f^-1(x))^2}$$ $$ (f^-1)'(y) = \frac{1}{f'(f^-1(y))}$$

Now the only thing more standing in the solutions is that $$(f(f^-1) = id(x) = x $$ Could anyone give me insight on what happened here?

Thank you.

codeisfun
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  • The inverse function $f^{-1}$ transforms $f$ back to the identity, well it's the definition of an inverse function. – macton Jan 26 '21 at 08:59
  • Okay that is something new for me.If I look at it that way that makes more sence for sure but still not 100% clear.We substitute f' for tan^2 that is clear with me. But how do we get to the next step, where does the f(f^-1) + (f-1)^2 come from? What was used to get that, I'm looking at the upper equations but I just dont see it. – codeisfun Jan 26 '21 at 11:36
  • $$\tan^2(x) = (f(x) + x)^2$$ that is the equation you derived. Apply $x = f^{-1}(y)$. – macton Jan 26 '21 at 11:42
  • Take the derivative of $ f^{-1}(f(x))=x$ and eventually also consider $fof^{-1})=f^{-1}of$ – Ataulfo Jan 26 '21 at 13:25
  • Okay so if i take $$ x = f{-1} (y) $$ I am able to get to the last part,actually the part where the $$(f(f^{-1}) = id(x) = x $$ but now the only question is how does one know that, and also how did you know that we can express x as $$ x= f^{-1}(y) $$ – codeisfun Jan 26 '21 at 14:31
  • If you mean how can one ensure we can find $y$ such that $x = f^{-1}(y)$, it is because the derivative of $f$ is $$\sec^2(x) - 1 = \tan^2(x)$$ which is positive for all $x$, so in fact $f$ is monotone increasing, and thus is bijective (continuity). – macton Jan 27 '21 at 08:03
  • Ah okay thanks! – codeisfun Jan 27 '21 at 12:27