16
$\begingroup$

Let $f$ be a function such that :$f:\mathbb{R}\to \mathbb{R}$ and $f^{-1}$ is a compositional inverse of $f$. I would'd like to know how do I solve this class of differential equation : $$\displaystyle \ f'= e^{\displaystyle {f}^{-1}}?$$

Note 01: $f' =\displaystyle\frac{df}{dx}$.

Edit: ${f}^{-1}$ is the inverse compositional of $f$, for example $\log$ is the inverse application of exp function .

Note 02: I have edited my question to clarify the titled question that related to ${f}^{-1}$

Thank you for any help

Improve this question
$\endgroup$
11
  • 3
    $\begingroup$ Is there any reason to expect a closed form? Or even a unique solution? I can see that there is a unique solution in formal power series around $0$ satisfying $f(0)=0$. You could compute the first few terms and see if they agree with any known closed-form functions. $\endgroup$
    – Will Sawin
    Commented Jan 1, 2017 at 21:04
  • 3
    $\begingroup$ First few coefficients of the unique power series solution are [0, 1, 1/2, 0, 1/24, -1/20, 13/180, -197/1680, 2101/10080, -48203/120960, 2938057/3628800, -23059441/13305600, 74408941/19160064, -9409883317/1037836800] $\endgroup$
    – F. C.
    Commented Jan 1, 2017 at 21:08
  • 8
    $\begingroup$ I calculated the first 100 terms of the formal power series. It is pretty clear that $|a_n|^{-1/n}\to 0$ as $n\to\infty$, so the radius of convergence is zero, so this approach will not give a solution that is an actual function. It is also of some interest that the number $b_n=(-1)^nn!a_n$ appears to always be a positive integer (for $n>1$), but this sequence is not in OEIS. Also $b_n$ it does not factorise in a way that suggests that there could be a simple formula: for example $b_{10}=2938057$, which is prime. $\endgroup$
    – Neil Strickland
    Commented Jan 1, 2017 at 22:55
  • 6
    $\begingroup$ Despite appearances, this is rather different from an ODE since the equation is non-local in the sense that the RHS at $x$ can not be evaluated if one only knows $f$ near $x$. $\endgroup$
    – Christian Remling
    Commented Jan 2, 2017 at 1:34
  • 3
    $\begingroup$ Using the inverse function theorem, this is equivalent to $A'(x)=\exp(-A(A (x)))$ with $A(x)=f^{(-1)}(x)$, which apparently gives oeis.org/A214645, mod signs. $\endgroup$
    – Tom Copeland
    Commented Jan 3, 2017 at 7:05

3 Answers 3

Reset to default
36
$\begingroup$

There is no such function. Since $f$ would have to map $\mathbb R$ onto $\mathbb R$ for the equation to make sense at all $x\in\mathbb R$, it follows that $f^{-1}(x)\to -\infty$ also as $x\to -\infty$, so $f'\to 0$. Thus $f(x)\ge x$, say, for all small enough $x$, hence $f^{-1}(x)\le x$ eventually, but then the equation shows that $f'\le e^x$, which is integrable on $(-\infty, 0)$, so $f$ would approach a limit as $x\to -\infty$ and not be surjective after all.

Improve this answer
$\endgroup$
3
  • $\begingroup$ Oh that integral trick, I never would've thought of that. Nice one. $\endgroup$
    – user78249
    Commented Jan 2, 2017 at 3:36
  • 8
    $\begingroup$ The missing argument (understatement ?): because $f'>0$, $f$ must be increasing. Being onto, it thus satisfies $f(\pm\infty)=\pm\infty$. $\endgroup$
    – Denis Serre
    Commented Jan 2, 2017 at 6:43
  • $\begingroup$ What is true is $e^{-xt}f'(x)dx = exp(-f^{-1}(z)t)dz$ for any inverse pair $(z,x)=(f(x),f^{-1}(z))$, so any "solution" has to be a divergent series to circumvent this analytic fact. $\endgroup$
    – Tom Copeland
    Commented Feb 25, 2018 at 19:14
5
$\begingroup$

A formal Taylor series (e.g.f.) solution about the origin can be obtained a few ways.

Let $f^{(-1)}(x) = e^{b.x}$ with $(b.)^n=b_n \;$ and $ \; b_0=0$.

Then A036040 (Bell polynomials) gives the e.g.f.

$$e^{f^{(-1)}(x)}= e^{e^{b.x}}= 1 + b_1 x + (b_2+b_1^2) \frac{x^2}{2!}+(b_3+3b_1b_2+b_1^3)\frac{x^3}{3!}+\cdots \; ,$$

and the Lagrange inversion / series reversion formula (LIF) A134685 gives

$$f'(x)= \frac{1}{b_1} + \frac{1}{b_1^3} (-b_2) x + \frac{1}{b_1^5}(3b_2^2-b_1b_3)\frac{x^2}{2!}+\cdots \; .$$

Equating the two series and solving recursively gives

$$b_n \rightarrow (0,1,-1,3,-16,126,-1333,...)$$

which is signed A214645. This follows from the application of the inverse function theorem (essentially the LIF again)

$$f'(z) = 1/f^{(-1)}{'}(\omega) \; ,$$

when $(z,\omega)=(f^{(-1)}(\omega),f(z)) $, leading to

$$f^{(-1)}{'}(x) = \exp[-f^{(-1)}(f^{(-1)}(x))],$$

the differential equation defining signed A214645.

Applying the LIF to the sequence for $b_n$ gives the e.g.f. $f(x)=e^{a.x}$ equivalent of F.C.'s o.g.f.

$$ a_n \rightarrow (0,1,1,0,1,-6,52,...).$$

As another consistency check, apply the formalism of A133314 for finding the multiplicative inverse of an e.g.f. to find the e.g.f. for $\exp[-A(-x)]=\exp[f^{(-1)}(x)]$ from that for

$$\exp[A(-x)]= 1 - x + 2 \frac{x^2}{2!}-7 \frac{x^3}{3!}+\cdots \; ,$$

which is signed A233335, as noted in A214645. This gives $f'(x)=a. \; e^{a.x}$.

Improve this answer
$\endgroup$
9
  • $\begingroup$ The inverse function theorem here might be more aptly called the inverse formal series theorem. As you can see, the differential equations and inverses here in analytic guise are concise statements of relations among the coefficients of formal series (e.g.f.s or o.g.f.s). $\endgroup$
    – Tom Copeland
    Commented Jan 4, 2017 at 18:29
  • $\begingroup$ When you talk of the Lagrange inversion theorem, do you mean the plain derivation of coefficients of the series for $f^{(-1)}$, or also the residue formula $[x^n] f^{(-1)}={1\over n}\mathrm{Res}(f^{-n})$ ? (which could be of use here to get some explicit formula for the coefficients, although I don't see it) $\endgroup$
    – Pietro Majer
    Commented Jan 16, 2017 at 18:43
  • 2
    $\begingroup$ @PietroMajer, both. See, for example, tcjpn.wordpress.com/2016/11/01/… and the July 2015 formula in oeis.org/A133437. Lagrange a la Lah has different series reps of the Lagrange inversion formula, all available on the OEIS (see cross-references in oeis.org/A145257). $\endgroup$
    – Tom Copeland
    Commented Jan 16, 2017 at 21:33
  • 1
    $\begingroup$ A divergent series doesn't really have any derivatives, so one can't invoke the inverse function theorem in its clearly geometric interpretation but only as a relation beween the coefficients of a formal series and its 'formal inverse' as defined by the series involutions denoted as Lagrange inversion formulas in the OEIS, which apply to convergent analytic series at the origin with f(0)=0 and a nonzero first derivation. These involutions define relations among the coefficients of the pair of series independent of convergence, that are consistent with composition and reciprocation. $\endgroup$
    – Tom Copeland
    Commented Jan 17, 2017 at 1:23
  • 2
    $\begingroup$ @PietroMajer, I pay tribute to Lagrange by calling pretty much any formula an LlF, and there are very many, that give a series expansion equivalent to that of your LIF. Think Lagrange inversion = compostitional inversion via series whether o.g.f.s, e.g.f.s, or other series reps. For non-series inversion, I might use directly $g(g^{-1}(x) )= x$, or a Laplace-like transform with a change of variables--any way to skin the cat analytically--but we don't have a well-defined analytic function, forward or inverse, to begin with here though, so bootstrap methods only come to mind. $\endgroup$
    – Tom Copeland
    Commented Jan 17, 2017 at 22:42
0
$\begingroup$

I have an idea to express solutions of transcend equations. If you replace functions in algebraic equations by operators then this way will suit operator equations. You can find my post in mathoverflow by User 'qian' or 'tag' symbolic computation.

In my post I give the equation $x+e^{x}=0$ an explicit solution:

$x=\Big(I_{1}\{C^{3}_{1,3}[A^{3}_{1,2}(\varphi_{a})]C^{3}_{2}[A^{3}_{2,3}(\varphi_{p})]\}\Big)(0,e)$:

Your equation is $f'=e^{f^{-1}}$, we denote $f'=\alpha(f)$,$f^{-1}=I(f)$,that is say we consider differential and inverse as special unary operators. We denote subtraction and power as $\varphi_{s}$,$\varphi_{p}$ respectively as I used in my post then

$f'=e^{f^{-1}}\Longleftrightarrow \varphi_{s}\{\alpha(f),\varphi_{p}[e,I(f)]\}=0$,

The solution will be:

$f=\Bigg(I_{1}\Big\{C^{3}_{1,3}\Big[\Big(A^{3}_{1,2}\{\varphi_{a}C^{2}_{1}[A^{2}_{1}(\alpha)]\}\Big)C^{3}_{2}\Big(A^{3}_{2,3}\{\varphi_{p}C^{2}_{2}[A^{2}_{2}(I)]\}\Big)\Big]\Big\}\Bigg)(0,e)$

This result is only a reference but I hope it will help you.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Pleas could u explain more , seems interesting $\endgroup$
    – zeraoulia rafik
    Commented Jul 24, 2022 at 21:05

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .