Mathematics Stack Exchange
Mathematics Stack Exchange

Questions tagged [functional-equations]

The term "functional equation" is used for problems where the goal is to find all functions satisfying the given equation and possibly other conditions. Solving the equation means finding all functions satisfying the equation. For basic questions about functions use more suitable tags like (functions) or (elementary-set-theory).

The term "functional equation" is used for problems where the goal is to find all functions satisfying the given equation(s) and possibly other conditions; e.g., the goal can be to find all continuous solutions. Solving the equation means finding all functions satisfying the given equation(s) and any additional conditions.This is different from the more common use of the word "equation", where the solutions are numbers. It is also different from the more common use of the word "functional", referring to a mapping from a space into the reals or complexes. For basic questions about functions use more suitable tags like or .

A common technique used in solving functional equations is finding some properties of satisfying functions by substituting variables for certain values in the equation. Proving properties of satisfying functions is also helpful - finding that a function is injective, surjective, involutive, and so on, is often a key step in finding all possible solutions. Other techniques such as exploiting symmetry, considering fixed points, and even using certain properties of domains (e.g. well-ordering) sometimes help.

Some well-known functional equations are:

More information can be found at Wikipedia.

3976 questions
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Find the period of a function $\varphi:\mathbb{R}\backslash\{3\}\to \mathbb{R}$

Let $\varphi:\mathbb{R}\backslash\{3\}\to \mathbb{R}$ a periodic function so that forall $x\in \mathbb{R}$ $$\varphi(x+4)=\frac{\varphi(x)-5}{\varphi(x)-3}$$ Find the period the $\varphi$.
Roiner Segura Cubero
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Find all real functions $f$ such that $f(xf(y))=(1-y)f(xy)+x^2y^2f(y)$

Find all functions $f:\mathbb{R}\rightarrow \mathbb{R}$ such that for all real numbers $x,y$ we have: $f(xf(y))=(1-y)f(xy)+x^2y^2f(y)$ I have solved this problem but the solution is "bruteforce", so I wanted to ask if there is a more elegant way…
CryoDrakon
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Understanding of solution for the functional equation $(x-y)f(x+y) - (x+y)f(x-y) = 4xy\left(x^2 -y^2\right)$

Problem For all $x,y \in \mathbb{R}$ which is $x^2 \not = y^2$, a function $f$ satisfies the following. $$(x-y)f(x+y) - (x+y)f(x-y) = 4xy\left(x^2 -y^2\right)$$ Find the function $f$. Solution Divide both side of the given formula by…
cokecokecoke
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Extension of the additive Cauchy functional equation

Let $f\colon (0,\alpha)\to \def\R{\mathbf R}\R$ satisfy $f(x + y)=f(x)+f(y)$ for all $x,y,x + y \in (0,\alpha)$, where $\alpha$ is a positive real number. Show that there exists an additive function $A \colon \R \to \R$ such that $A(x) = f(x)$ for…
M'smary
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A functional equation: $4f(x)^3 +f(3x)=3f(x)$

Find all functions $f:\mathbb{R}\to\mathbb{R}$ such that $$4f(x)^{3}+f(3x)=3f(x)$$ I know of 2 functions that satisfy the equation but I do not know how to prove that they are the only ones. Thanks Andrew
Andrew
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Problem in solving functional equation $f(x^2 + yf(x)) = xf(x+y)$

To find all functions $f$ which is a real function from $\Bbb R \to \Bbb R$ satisfying the relation $$f(x^2 + yf(x)) = xf(x+y)$$ It can be easily seen that the identity function $i.e.$ $f(x)=x$ and $f(x)=0$ (verified just now) satisfies the above…
User8976
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Solve the following functional equation: $[f(x)+f(y)][f(x+2y)+f(y)]=[f(x+y)]^2+f(2y)f(y)$

Find all functions $f:\mathbb{R} \rightarrow \mathbb{R}$ so that: a) $[f(x)+f(y)][f(x+2y)+f(y)]=[f(x+y)]^2+f(2y)f(y)$ b) for every real $a>b\ge 0$ we have $f(a)>f(b)$ As much as I know: putting $x=y=0$ we get $f(0)=0$. let $x+y=0$, the we get…
CryoDrakon
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Solve the functional equation $f(xf(y)+x)=xy+f(x)$

Find all $f:\mathbb{R}\rightarrow\mathbb{R}$ so that $f(xf(y)+x)=xy+f(x)$. If you put $x=1$ it's easy to prove that f is injective. Now putting $y=0$ you can get that $f(0)=0$. $y=\frac{-f(x)}{x}$ gives us $f(\frac{-f(x)}{x})=-1$
CryoDrakon
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Problem in Putnum competition?

Suppose $f:\mathbb{R}\longrightarrow \mathbb{R}$ is a continuous function and $f(2x^2 -1)=2xf(x)$ for all $x\in \mathbb{R}$. Prove $f(x)=0\,\,\text{for all} \, x\in [-1,1].$
Ramand
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Find a function such that $f(\log(x)) = x \cdot f(x) $

I recently read an article in which the author describes how to find some functions that obey to certain recursion relationships. If we want to find a function that satisfies, for example, $f(x^a) = x \cdot f(x)$, then the author explains we can…
Max Muller
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Is there a continuous {permutation,duplicate,translation,stretch}-invariant function on ordered sets of vectors that returns a vector?

Is there any example of such a function $f$, preferably one defined on all $V^n$ and all positive integers $n$ where $V$ is some vector space? It must satisfy the following: $f(T) = f(\sigma(T))$ for any $T \in V^n$ and permutation…
user21820
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Solutions of the functional equation $f\big(x+g(y)\big)=f(x)+f\big(g(y)\big)$

I know the solutions of the well-known Cauchy functional equation $$f(x+y)=f(x)+f(y)\text.$$ But what does it change if I have the following form $$f\big(x+g(y)\big)=f(x)+f\big(g(y)\big)\text?$$ What can I say about $g$?
William
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$ f(x)=f(x+1) $ and $ f(-1/x)=f(x)$

Is there any function $f$ which would satisfy $f(x)=f(x+1)$ and $f(-1/x)=f(x)$ for every $x$ or at least positive $x$? For the widest possible domains of $x$? If I could turn this functional equation into differential equations, I could use some…
Jose Garcia
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Is there such a function $f:R\rightarrow R$?

Is there such a function $f:\mathbb R\rightarrow\mathbb R$, that for any real $x$ and $y$, we have the equality: $$ \frac{f(x)+f(y)}{2}=f\left({\frac{x+y}{2}}\right)+|x+y|\;\;\;? $$
nanolab
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$f(x+1)=f(x)+1$ and $f(g(x))=g(f(x))$

Let $g_1(x)=x+1$ and $g_2(x)=x^2$ be two real functions. Then it is known that whenever $f$ commutes with $g_1$ and $g_2$, $f$ is the identity function. But in this example we choosed 2 particular functions $g_1$ and $g_2$, and we had the conclusion…
Leyv Zahav
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