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 ...
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 $x_0$...
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 $x\to -\infty$,...
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 ...
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
$$
b-2 = b^2 \ \...
18
votes
Accepted
How may I find all continuous and bounded functions g with the following property?
$\newcommand\de\delta$Considering $g$ a distribution (in the generalized-function sense), let $\hat g$ be the Fourier transform of $g$. Then your functional equation yields
$$4\hat g(t)=e^{it}\hat g(t)...
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 ...
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 ...
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 ...
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. ...
13
votes
Accepted
A variant of Cauchy-type functional equation conjecture
counterexample
$$f(x)=1-e^{\Re(x) i}$$
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 $\pi$,...
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 ...
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.
...
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 ...
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)...
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 ...
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 $...
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 ...
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).
...
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.
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 ...
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 ...
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 ...
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 ...
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 ...
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 ...
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 ...
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 ...
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