Articles of functional equations

Functions such that $f(f(n))=n+2015$

This question already has an answer here: IMO 1987 – function such that $f(f(n))=n+1987$ 6 answers

Variant of Cauchy Functional Equation

Consider the equation $$f(kx-f(x))=x=kf(x)-f(f(x))$$ for montonic $f$. What can we say about the solutions to this equation. Comparing with Cauchy equation $f(x+y)=f(x)+f(y)$, I think the solution must be somewhat close to being linear. Any hints. Thanks beforehand.

Functional equation $P(X)=P(1-X)$ for polynomials

I have encountered the following problem : Find all polynomials $P$ such as $P(X)=P(1-X)$ on $\mathbb{C}$ and then $\mathbb{R}$. I have found that on $\mathbb{C}$ such polynomials have an even degree. Because for each $a$ root, $1-a$ must be a root too. I struggle to find whether we can do something with polynomials of degree […]

How to solve $f'(x)=f'(\frac{x}{2})$

How do we solve this given $f'(0)=-1$. It does not look separable. I can integrate both sides but end up with a functional equation with is not helpful.

Find all real real functions that satisfy the following eqation $f(x^2)+f(2y^2)=$

Find all real functions $f:\Bbb R\rightarrow\Bbb R$ so that $f(x^2)+f(2y^2)=[f(x+y)+f(y)][f(x-y)+f(y)]$, for all real numbers $x$ and $y$. $f(x)=x^2$ is the only solution I think. So far I have got: $f(x^2)=[f(x)]^2+2f(x)f(0)+[f(0)]^2-f(0)$ $f(2y^2)=4[f(y)^2]-f(0)$ And by putting $x=y$ and $x=-y$ you get that $[f(x)-f(-x)][f(2x)-f(0)]=0$. If $f(2x)=f(0)$ then $f$ is a constant function of value $0$ or $1/2$, both […]

Differentiability of $f(x+y) = f(x)f(y)$

This question already has an answer here: Prove that $f'$ exists for all $x$ in $R$ if $f(x+y)=f(x)f(y)$ and $f'(0)$ exists 2 answers

Solving functional equation gives incorrect function

Let $f:\mathbb{R} \to \mathbb{R}$ be a function which satisfies $e^xf(y)+e^yf(x)=2e^{x+y}-e^{x-y}$ for all real x and y. If I place $x=y$, I get $f(x)=e^x-\frac{1}{2}e^{-x}$ which does not satisfy the original equation. Now instead, if I set$x=y=0$, I get $f(0)=\frac{1}{2}$. Now setting $y=0$ in the original equation, I get $f(x)=\frac{e^x}{2}$ My question is that why do we […]

Minimum period of function such that $f\left(x+\frac{13}{42}\right)+f(x)=f\left(x+\frac{1}{6}\right)+f\left(x+\frac{1}{7}\right) $

Let $ f$ be a function from the set of real numbers $ \mathbb{R}$ into itself such for all $ x \in \mathbb{R},$ we have $ |f(x)| \leq 1,f(x)\neq constant $ and $$f\left(x+\dfrac{13}{42}\right)+f(x)=f\left(x+\dfrac{1}{6}\right)+f\left(x+\dfrac{1}{7}\right)$$ Prove that $ f$ is a periodic function (that is, there exists a non-zero real number $ c$ such $ f(x+c)=f(x)(c>0)$ for […]

Suppose a function $f : \mathbb{R}\rightarrow \mathbb{R}$ satisfies $f(f(f(x)))=x$ for all $x$ belonging to $\mathbb{R}$.

Suppose a function $f : \mathbb{R}\rightarrow\mathbb{R}$ satisfies $f(f(f(x)))=x$ for all $x$ belonging to $\mathbb{R}$. Show that (a) $f$ is one-to-one. (b) $f$ cannot be strictly decreasing, and (c) if $f$ is strictly increasing, then $f(x)=x$ for all $x$ belonging to $\mathbb{R}$. Now, I’ve done the first part(the simplest that is) and I need someone to […]

Solving functional equation $f(x)f(y) = f(x+y)$

I’m having some trouble solving the following equation for $f: A \rightarrow B$ where $A \subseteq \mathbb{R}$ and $B \subseteq \mathbb{C}$ such as: $$f(x)f(y) = f(x+y) \quad \forall x,y \in A$$ The solution is supposed to be the form $f(x) = e^{kx}$ where k can be complex but I don’t see how to show it. […]