47 votes

Function satisfying $f^{-1} =f'$

Seems like we have here yet another vindication of Léo Sauvé's famed dictum... I am going to try to sketch below the awe-inspiring solution to this problem with which A. C. Hindmarsh came up and ...
José Hdz. Stgo.'s user avatar
42 votes
Accepted

For a continuous function $f:\mathbb{R}^{+}\to\mathbb{R}^{+}$ does $(f(x)-f(y)) (f(\frac{x+y}{2}) - f(\sqrt{xy}))=0$ imply that $f$ is constant?

Yes. If $f$ were not constant, then (since ${\bf R}^+$ is connected) it could not be locally constant, thus there exists $x_0 \in {\bf R}^+$ such that $f$ is not constant in any neighbourhood of $x_0$...
Terry Tao's user avatar
  • 108k
36 votes
Accepted

How do i solve this : $\displaystyle \ f'=e^{{f}^{-1}}$?

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 $x\to -\infty$,...
Christian Remling's user avatar
30 votes

Is there a general solution for the differential equation $f''(x) = f(f(x))$?

Remark: I had a little time to write a draft of my notes on the proofs of the claims I make below and have posted it on my home webpage here. (It would have made a very long post on MO, so I decided ...
Robert Bryant's user avatar
23 votes

Is there a general solution for the differential equation $f''(x) = f(f(x))$?

The equation has solutions with powers, $f(x) = ax^b$. Inserting this ansatz, one has $$ a b (b-1) x^{b-2} = a (a x^b)^b = a^{b+1} x^{b^2} \ , $$ so the requirements on $a$ and $b$ are $$ b-2 = b^2 \ \...
Michael Engelhardt's user avatar
18 votes
Accepted

How may I find all continuous and bounded functions g with the following property?

$\newcommand\de\delta$Considering $g$ a distribution (in the generalized-function sense), let $\hat g$ be the Fourier transform of $g$. Then your functional equation yields $$4\hat g(t)=e^{it}\hat g(t)...
Iosif Pinelis's user avatar
17 votes
Accepted

$f'=e^{f^{-1}}$, again

There is no analytic local solution at $0$ to $f'=e^{f^{-1}}$, $f(0)=0$, that is, the formal power series solution is diverging. Together with the solution given in comments by fedja, this means the ...
Pietro Majer's user avatar
  • 56.3k
14 votes

Finding f such that f(f(x))=g(x) given g

Assuming that $g(x) > x$ rather than $g(x) \geq x$ for all $x$, and also that $g$ is strictly increasing (no critical points), one can obtain a solution $f$ to $g = f \circ f$ which piecewise has ...
Terry Tao's user avatar
  • 108k
14 votes
Accepted

A conjecture about certain values of the Fabius function

I have posted in arXiv:1702.05442 the English translation of a paper about Fabius function that I published in Spanish in 1982 (we will refer to it as (A)). With the Theorems in this paper the ...
juan's user avatar
  • 6,966
14 votes
Accepted

Equality in $\mathbb F_q\left(\left(\frac1T\right)\right)$

This is a field automorphism of order $p$, since $T+p=T$, so it fixes an index $p$ subfield. $\mathbb F_q (( \frac{1}{T^p - T } ))$ is an index $p$ subfield and is fixed, hence is the fixed field. ...
Will Sawin's user avatar
  • 135k
13 votes
Accepted

A variant of Cauchy-type functional equation conjecture

counterexample $$f(x)=1-e^{\Re(x) i}$$
math110's user avatar
  • 4,220
12 votes
Accepted

On the consistency of the definition of the conductor for automorphic forms

These definitions are consistent, though it's not immediate. The conductor quantifies the extent to which $\pi$ is ramified. As an aside, I prefer to write $c(\pi)$ for the conductor exponent of $\pi$,...
Peter Humphries's user avatar
11 votes
Accepted

Is there a recurrence for the coefficients of the Laurent series expansion of $\frac{1}{1-e^{e^x - 1}}$?

$e^{e^x-1}$ is the exponential generating function for Bell numbers ${\cal B}_n$: $$e^{e^x-1} = \sum_{n\geq 0} {\cal B}_n \frac{x^n}{n!}.$$ Then $$g(x) := \frac{e^{e^x-1}-1}{x} = \sum_{n\geq 0} {\cal ...
Max Alekseyev's user avatar
10 votes

On the consistency of the definition of the conductor for automorphic forms

Yes this is true, but I don't think it's completely obvious. The following argument is taken from Roberts--Schmidt ``Local newforms for $GSp_4$'', which uses the same argument as the $GL_2$ case. ...
Martin Dickson's user avatar
10 votes
Accepted

Does it follow that $F^{(1)}(z)=F^{(2)}(z)$ for all $z \in \mathbb H$?

The answer is positive. Indeed, $f=F^{(1)}-F^{(2)}$ is a bounded analytic function in the right half-plane (this follows from your conditions $|a_k|\leq k^{-2}$ and $\lambda_k\to\infty$). But a ...
Alexandre Eremenko's user avatar
10 votes
Accepted

Are there any functions $f$ beyond trivial examples where $\int f(x +f(x + f(x +\dotsb)))\,dx$ $= F(x+F(x+F(x +\dotsb))) + C$ for some function $F$?

Based on others and my comments, one can construct arbitrary solutions to the problem using a Taylor series ansatz. Use the functional equations from the comments \begin{align} \phi(x)&=f(x+\phi(x)...
Fred Hucht's user avatar
  • 2,695
9 votes
Accepted

Solving functional equation $f(xy)=f(x+y)$ and Diophantine equations

Too long for a comment. My guess is that for $x\geq 7$ we that that $f(x)=c$ by strong induction. We need to check the cases $7\leq x\leq 20$ by hand . Let us now suppose we have $y\geq 20$ and ...
Vlad Matei's user avatar
9 votes

Solving functional equation $f(xy)=f(x+y)$ and Diophantine equations

$7 \sim 12$ via $3, 4$ $12 \sim 35$ via $5, 7$ $35 \sim 264$ via $11, 24$ $264 \sim 41$ via $8, 33$ $41 \sim 420$ via $20, 21$ $420 \sim 43$ via $15, 28$ $43 \sim 156$ via $4, 39$ $156 \sim 25$ via $...
Peter Taylor's user avatar
  • 6,471
8 votes
Accepted

A functional equation in real analysis

A theorem of Sergei Bernstein says that if $u$ is continuous, then the sequence of functions on the left-hand side converges uniformly to $u$ on $[0,1]$. The polynomials on the left hand side are ...
Liviu Nicolaescu's user avatar
8 votes
Accepted

Does there exist another form of the derivative for polynomials?

It's not hard to see that $H$ must be of the form $\alpha ux + \beta uz + \gamma yx + \delta yz$ by $\mathbb{R}$-linearity of $F$ (see Jan-Cristoph Schlage-Puchta's answer for a fuller explanation). ...
user44191's user avatar
  • 4,961
8 votes

Existence of function satisfying $f(f'(x))=x$ almost everywhere

Looking for a solution of the form $f(x)=ax^b$, $x>0$, one finds $$ a = \phi^{-\phi/(\phi+1)}, ~~~ b=\phi $$ where $\phi=\frac{\sqrt{5}+1}{2}$ is the Golden ratio.
Marc Chamberland's user avatar
8 votes
Accepted

A functional equation in two complex variables

$Hello$, Tomasz! (for some reason the MO prohibits saying "Hi" or "Hello" in the normal text mode). Nice to see you back. Apparently you are still asking the same question whether ...
fedja's user avatar
  • 59.5k
8 votes
Accepted

Proving the simple form of a function from statistical mechanics

We can indeed prove this for reasonable functions, $\log f_0\in C^2$, say. Let me write $F=\log f_0$. By replacing $F$ by $F(v)-C-d\cdot v$, we can also assume that $F(0),\nabla F(0)=0$. If $a,v$ are ...
Christian Remling's user avatar
7 votes
Accepted

Intuition behind the Riemann $\zeta$ functional equation

Your intuition breaks down because $\zeta(1-2k)$ has a closed form in terms of Bernoulli numbers, but no powers of $\pi$ at all. This was known to Euler (via Abel summation as the series is of course ...
Stopple's user avatar
  • 10.8k
7 votes

Polynomial satisfying a functional equation

Nice exercise, though I doubt that it can be considered as a research question. Put $y=x^2$, and write $q(x)= a(y)+xb(y)$. I claim that the polynomials $a$ and $b$ are coprime: if a non-constant ...
abx's user avatar
  • 37.1k
7 votes
Accepted

The functional equation $f(xy) - 2 f(\frac{x+y}{2}) + f(x+y- x\cdot y) = 0$

Adding a constant to f does not change the property, so we may assume f(0)=0. Under this assumption, I claim that the property holds if and only if f is additive. The if part is obvious. For the only ...
Michael Renardy's user avatar
7 votes

On the functional equation $f(xf(y))=\frac{f(f(x))}y$ on arbitrary groups

For not necessarily abelian groups, we can interpret the division in (1) as the right or left division. Let $G$ be any group. The answer to the question is given by Theorem: The following three ...
Iosif Pinelis's user avatar
7 votes

Progress in robustifying mathematics - i.e. making mathematical theorems robust to small changes in hypotheses

There is a direction of research in statistics called robust statistics. There also is a direction of research in probability, initiated by Zolotarev, concerned with stability problems in probability ...
Iosif Pinelis's user avatar
6 votes
Accepted

Linearizing a power series by conjugation

You may find useful information in a recent article by D. Sauzin and al. "Explicit linearization of one-dimensional germs through tree-expansions" here, where they use "mould calculus" (introduced by ...
Loïc Teyssier's user avatar

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