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.

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Functional equation $f(x+y)\leq yf(x)+f(f(x)).$

Let $f\colon \mathbb{R} \rightarrow \mathbb{R}$ be such that $$f(x+y)\leq yf(x)+f(f(x)).$$ What can be the restriction of $f$ to $\mathbb{R}_{\leq 0}$
user1136275
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Functional equation $f(y)f(x+f(y))=f(xy)+f(y)^2.$

$f: \Bbb{R} \to \Bbb{R}, f(y)f(x+f(y))=f(xy)+f(y)^2.$ \begin{align} P(0, 0): \; & f(0)f(f(0))=f(0)+f(0)^2. \ \\ \text{if } \; & f(0) \neq 0: \\ & f(f(0))=f(0)+1. \\ \ \\ P(-1, f(0)): \; & (f(0)+1)^2=f(-f(0))+(f(0)+1)^2. \\ \therefore \; &…
RDK
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$f:\mathbb{R}\to\mathbb{R},$ such that $f(|f(x)-f(y)|)=f(f(x))-2x^2f(y)+f(y^2)\,\,\forall \, x,y\in\mathbb{R}.$ Find all such $f(x).$

$f:\mathbb{R}\to\mathbb{R},$ such that $f(|f(x)-f(y)|)=f(f(x))-2x^2f(y)+f(y^2)\,\,\forall \, x,y\in\mathbb{R}.$ Find all such $f(x).$ My Working: Let $P(x,y): f(|f(x)-f(y)|)=f(f(x))-2x^2f(y)+f(y^2)\,\,\forall \, x,y\in\mathbb{R}$ $P(0,0):f(f(0))=0;$…
Makar
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Functional equation contradiction

Consider this functional equation: $$f(x+\frac{1}{x})=x, x\ne0$$ We can solve it easily by setting $y = x+\frac{1}{x}$ and solving for $x$. It's just a quadratic equation in $x$: $x=\frac{y\pm\sqrt{y^2-4}}{2}$ so finally we have…
Alma Do
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$f(z_1 z_2) = f(z_1) f(z_2)$ for $z_1,z_2\in \mathbb{C}$ then $f(z) = z^k$ for some $k$

Same as my previous question except domain is complex. I tried assuming that the function was analytic, so for $z_1=z_2=z$ , $f(z^2) = f(z)^2$ $$\sum_{n=0}^\infty a_n z^{2n}=\left(\sum_{n=0}^\infty a_n z^n\right)^2$$ and try to solve it. Ideally,…
kuch nahi
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Could we compute $P(t^2)$?

Let $P$ be an operator such that $P(kx)=kP(x)$, $k \in \mathbb{C}$, $x$ is a variable, $P(xy)=P(xP(y))+P(P(x)y)-P(x)P(y)$, $x, y$ are variables. All variables commute. Let $P(t)=t$. Then $P(t^2)=2P(tP(t))-P(t)P(t)=2P(t^2)-t^2$. Hence $P(t^2)=t^2$.…
LJR
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Given a functional equation with the following data. Prove this problem

The functional equation is $f(2x+y+f(x+y))+f(xy)=yf(x)$. Note that $f$ runs from $\mathbb{R}$ to $\mathbb{R}$. We want to prove that if $f(0)=0$ then $f(x)=0$ for every real number $x$. I have managed to prove for $x$ is an integer, since I can…
Tmt
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if $f(f(x+y))=f(x^2)+f(y^2)$ then $f(x)=?$

if $f(f(x+y))=f(x^2)+f(y^2)$ then $f(x)=?$ for all integers ($f: \mathbb Z \rightarrow \mathbb Z$) I know how to solve the following problem though: if $f(f(x+y))=f(2x)+2f(y)$ then $f(x)=?$ We can easily analyze that $f(x)$ here (in the second…
74H54N3
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Find function $ f $ such that $f(\frac{x-3}{x+1})+f(\frac{x+3}{1-x})=x$

I am looking for functions $ f:\Bbb R \to \Bbb R $ satisfying $$f\Big(\frac{x-3}{x+1}\Big)+f\Big(\frac{x+3}{1-x}\Big)=x$$ I used the substitution $ x=\cos(2t) $ for $ x\in (0,2\pi) $, to use the identities $$1+x=2\cos^2(t) \text{ and }…
hamam_Abdallah
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Does $f(x+1)-f(x)$ constant imply $f$ is linear?

I have a function $f:\mathbb R\to \mathbb R$ and $f$ doesn’t have any restriction, so it might not be continuous and so on. If the condition $$f(x+1)-f(x)=c$$ holds $\forall x\in \mathbb R$, does that mean $f$ is linear? If not why this textbook say…
PNT
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$f(x) = x f(\frac{1}{x})$ for $x>0$

For $x>0$, $f(x)>0$ and continuous. Also $$f(x)=xf\bigg(\frac{1}{x}\bigg).$$ The function $f(x)=x \ \text{for} \ x\in(0,1]$ and $f(x)=1 \ \text{for} \ x>1$ certainly satisfies the above conditions. Is this the unique function satisfying the above…
vnd
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Solving $f(x+f(y)) = f(x) + y$

I found this problem in an old book of mine. The question asks to find all the functions $f: \mathbb Q \to \mathbb R$ which satisfy $f(x+f(y)) = f(x) + y$ for all rational numbers $x,y$. Also I am unsure what $f: \mathbb Q \to \mathbb R$ means…
user687894
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Prove that $f(f(x))=x$ has no roots .... $f$ having a general form

This problem gave me some headache, especially because $f$ have its own general form : let $f(x) = ax^2 + bx + c$. Suppose that $f(x) = x$ has no real roots. Show that equation $f(f(x))=x$ has also no real roots.
Florin M.
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Is there a non-constant function $f: \mathbb{R}_{>0} \to \mathbb{R}$ such that $f(x) = f(x + 1/x)$?

I am looking for a non-constant function $f: \mathbb{R}_{>0} \to \mathbb{R}$ such that $f(x) = f(x + 1/x)$, or a proof that no such function exists. Replacing $x$ by $1/x$ shows we must have $f(x) = f(1/x)$. I am most interested in (non-)existence…
Ricardo Buring
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$f: \mathbb{N}\cup\{0\} \longrightarrow \mathbb{N}\cup\{0\}, \forall x\in\mathbb{N}\cup\{0\}, f(x+1)+1 = f(f(x) + 1)$

I try to solve this question : Find all function $f:\mathbb{N}\cup\{0\}\longrightarrow \mathbb{N}\cup\{0\}$ such that $\forall x\in\mathbb{N}\cup\{0\}, f(x+1)+1 = f(f(x)+1)$. My attempt : By induction, we obtain $$\tag1\forall x,…
ZENG
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