9
votes
Prove that $f(\mathbb R) = \mathbb R$. Partially solved.
[Note: There have been some important technical revisions since the first version of this answer, which was flawed.]
I realize this is quite old and everyone has likely moved on, but since it got ...
- 7,122
5
votes
Prove that $g(z)$ is identically zero
Denote by $D(R)$ the disk centered in zero with radius $R$.
Let $M(R)$ be the maximum modulus of $g$ on the closed disk $D(R)$.
Let $K$ be $M(10)$.
Then for all $w\in D(81)$ we find a $z$ in $D(9)$ ...
- 31.5k
4
votes
Accepted
How solve the problem $f(x+2)=f(x)+4x+4$ for any $x$
Given, $f(x+2)=f(x)+4x+4$
$$1. \quad{f(x+2+2)=f(x+2)+4 \cdot (x+2) +4\\
2. \quad f(x+2+4){=\color{green}{f(x+4)}+4 \cdot (x+4) +4\\=\color{green}{f(x+2)+4 \cdot (x+2)+4}+4 \cdot (x+4) +4\\=f(x+2)+4\...
- 1,484
4
votes
Accepted
Is $f'(x)=f(1/x)$ solvable?
Everything you're doing is fine, although I would apply the known initial(ish) condition $f'(1) = f(1)$ before expanding everything out.
Doing this, we get
$$\begin{eqnarray} f(x) & = & c_1 x^{...
- 22.4k
4
votes
Accepted
Find $f(x)$ such that $f(x^2)=f(x)+1$
Claim: all continuous solutions $f:(1,\infty)\to\mathbb R$ take the following form: $$f(x) = g(\log_2 \log_2 x) + \log_2\log_2 x$$ where $g:\mathbb R\to\mathbb R$ is any continuous function of period $...
- 39.4k
3
votes
Find $f(x)$ such that $f(x^2)=f(x)+1$
This solution is valid for the previous question where the domain was all of $\mathbb{R}$
There are no solutions at all. Set $x=0$ you get
$$f(0) = f(0)+1$$ which is a contradiction.
- 13.5k
3
votes
Accepted
Proving the uniqueness of solution of a functional equation.
This equation has a unique smooth solution, just by the continuity assumption of $f$.
Starting from your last step, defining the function $g(x)=f(x)-x^2/2$ and substituting in the functional equation ...
- 10.6k
3
votes
Accepted
Thailand MO $f(x)f(y)f(x-y) = x^2f(y) - y^2f(x)$
$$(1) \ \ \ \ \ f(0)=0$$
You can trivially show $f(x) = 0$ is a solution.
Otherwise,
$$(2) \ \ \ \ \ f(-t) = -f(t)$$
$$(3) \ \ \ \ f(2x)\cdot f(x) = 2x^2$$
$$(4) \ \ \ \ \ 2\cdot f(x)=f(2x)$$
Hence, $...
- 1,085
3
votes
Does there exist an $f: \mathbb{R} \to \mathbb{R}$ such that $f(f(x)) = x^3 + 1$?
This spells out an argument proposed by user tomasz. Each orbit of the action of $g(x)=x^3+1$ on $\mathbb R$ is a copy of $\mathbb N$, with $g$ acting on each orbit by $n\mapsto n+1$. Let $X$ be the ...
- 40.1k
3
votes
Accepted
find all the polynomial f that satisfies $f(x^2)+f(x)f(x+1)=0$.
Zero-set
You have correctly identified two transformations which send roots of $f$ to roots of $f$. Explicitly, setting $u(z)=z^2$ and $v(z)=(z-1)^2$, we have $u(S)\subset S$ and $v(S)\subset S$. Let'...
- 3,356
3
votes
How solve the problem $f(x+2)=f(x)+4x+4$ for any $x$
Given that $f(x+2)-f(x)=4x+4$
here you can make a telescopic sum
$$\require{cancel}\cancel{f(x+2)}- f(x)=4x+4$$
$$\require{cancel}\cancel{f(x+4)}-\require{cancel}\cancel{f(x+2)}=4x+12$$
$$\cancel{f(x+...
- 909
2
votes
Find the solution to $f(x)=-f(1/x)$.
Setting $x=e^t$, your functional equation becomes :
$$f(e^t)=-f(e^{-t})\tag{1}$$
Let $F=f \circ \exp$ ($\circ$ = function's composition).
(therefore $F$ is a function $\mathbb{R} \to \mathbb{R}$).
(1) ...
- 79k
2
votes
How solve the problem $f(x+2)=f(x)+4x+4$ for any $x$
Define $g(2x) = f(x)$, we find the difference sequence of $g$:
$$
0 \quad 12 \quad 32 \quad 60 \quad ... \\
12 \quad 20 \quad 28 \quad ... \\
8 \quad 8 \quad ... \\\
0 \quad ...
$$
Observe that the $4$...
- 1,426
2
votes
Accepted
Tetration and $f(x) = \exp(\int_1^x \ln(f(t)) dt)$
No differentiable function $f$ can satisfy $f(1)=f'(1)=1$ along with $$f(x) = \exp\left(\int^x_1 \ln(f(t))dt\right).$$
Indeed, assuming that $f$ is differentiable, and satisfies the functional ...
- 15.4k
2
votes
Thailand MO $f(x)f(y)f(x-y) = x^2f(y) - y^2f(x)$
Suppose that $f(x) = 0$ for some $x \neq 0$. Then, substituting in any value of $y$ shows that $0 = x^2f(y)$, so $f(y)=0$ for all $y \in \mathbb R$. Meanwhile, letting $y=x=0$ gives $f(0)^3=0$, so $f(...
2
votes
Thailand MO $f(x)f(y)f(x-y) = x^2f(y) - y^2f(x)$
Swapping $x$ and $y$, for any $x$ and $y$ we have:
$f(y)f(x)f(y-x)=y^2f(x)-x^2f(y)=-f(x)f(y)f(x-y) \Rightarrow f(x)=0~or~f(y-x)=-f(x-y) $.
Assuming a nontrivial solution, this means that $f(x)$ is an ...
- 556
1
vote
$f(z) = \sum_{n=0}^{\infty} f(n)^2 z^n$?
I'm not sure if it's correct, but these are my thoughts (I hope they help):
According to the convergence criteria $\lim\limits_{z \to \infty}\left[ f\left( x \right) \right] = \text{constant}$ must ...
- 1,627
1
vote
Functions $f(x)$ for which the set of functions $af(x+b)$ is closed under addition
This will only be a partial answer aiming to give a strong necessary condition in the case where $f$ is assumed to be uniformly continuous over $\mathbb{R}$, making use of the theory of tempered ...
- 4,275
1
vote
Functions $f(x)$ for which the set of functions $af(x+b)$ is closed under addition
So, what we have want is a description of those continuous $f: \mathbb{R} \to \mathbb{R}$ such that the the image of the function $(a,b) \mapsto af(x + b)$, where $a, b \in \mathbb{R}$, is closed ...
- 3,612
1
vote
Proving the uniqueness of solution of a functional equation.
I found a direct solution, (thanks to your previous analysis).
We put
$$\forall x,y \in \mathbb{R}, \qquad f(2x) - 2 f(x+y) + 2f(y) = (x-y)^2 \qquad (E)$$
Of course this makes us think of the ...
- 2,980
1
vote
Accepted
functional equation, continuous functions.
Disclaimer: im 99% sure there's an cleaner solution, but i have been at this for hours now, im just happy to be done. If someone knows an better way PLEASE do let me know
Here we go!
Claim 1: $f$ is ...
- 198
1
vote
Find $f(x)$ such that $f(x^2)=f(x)+1$
With the domain $(1,\infty)$ there are infinitely many solutions, all of which can be found as follows: take any function $g:[2,4) \to \Bbb R$ and define $f$ recursively as follows:
$
f(x)=
\begin{...
- 17.2k
1
vote
Binary with digit 2 allowed
Unless I made a mistake, my brute force implementation seems to indicate the answer to your first question above is $51$.
Here's a list of solutions where $x=2$ that perhaps provides a bit more ...
- 6,948
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