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 with only two solutions? $(f(2x))^3=f(4x)((f(x))^2+xf(x))$

Recently I started studying functional equations. Now I'm trying to find all solutions to the following functional equation: $$(f(2x))^3=f(4x)((f(x))^2+xf(x)).$$ Unfortunately, I was able to show only few things about this equation, moreover, I…
Peter95
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Functional Equation: When $f(x+y)=f(x)+f(y)-(xy-1)^2$

How does one solve the following functional equation when $f:\mathbb{R}\rightarrow\mathbb{R}$ $$f(x+y)=f(x)+f(y)-(xy-1)^2$$ When I assumed it was a polynomial equation, it can be seen through induction that $$f(nx)=nf(x)-\sum _{ i=1 }^{ n-1 }{…
Chad Shin
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d'Alembert functional equation: $f(x+y)+f(x-y)=2f(x)f(y)$

The d'Alembert functioal equation is: $$f(x+y)+f(x-y)=2f(x)f(y)$$ This equation plays a central role in determining the sum of two vectors in Euclidean and non-Euclidean geometries. Is there a good characterization of the solutions of this equation?
Mohsen Shahriari
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How to prove a function is periodic from a given functional equation?

Given that $f: \mathbb R \to \mathbb R$, and that for some $a \in \mathbb R$, $f(x+a)={1\over 2}+\sqrt{f(x)-f^3(x)}$; prove $f(x)$ is a periodic fucntion. I know that to prove a function is periodic one needs to prove $f(x+bK)=f(x)$ for all $b \in…
Rescy_
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Functional equation $f(x+y)=f(x)+2xy+f(y)$

I am interested in classifying solutions $f\,:\,\mathbb R\longrightarrow \mathbb R$ to the functional equation \begin{equation} f(x+y)=f(x)+f(y)+2xy\qquad\qquad(\dagger) \end{equation} and in particular, how to minimize the underlying assumptions of…
Jason
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Transformation of the functional equation $f(x+y)=f(x+1)f(y)$

Is there a way to reduce the following functional equation $$ f(x+y)=f(x+1)f(y),\qquad x,y>0, $$ to the equation $$ f(x+y)=f(x)f(y),\qquad x,y>0, $$ whose solutions are known? Thanks in advance.
axl
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Finding a function $h$ that satisfies $h \left ( \frac{x}{x^2+h(x)} \right )=1$

Someone gave me a random maths problem to solve: Given that $h \left ( \dfrac{x}{x^2+h(x)} \right )=1$, what is $h(x)$ The restrictions given were: $h(x) \neq constant$ $\exists \frac{dh}{dx}$ $\exists h^{-1}(x)$ $\exists…
Mike 'Pomax' Kamermans
  • 542
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Functional equation $f(f(x))=2x$ on $\mathbb{Z}_{>0}$

I have this functional equation: $$f(f(x))=2x$$ with $f: \mathbb{Z}_{>0}\rightarrow \mathbb{Z}_{>0}$. And I want to know if it is possible to list all solutions. I already know that $f(2x)=2f(x)$, further, if $f(1)=c$, then $f(2^kc)=2^{k+1}$ by…
wythagoras
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Integer functional equation $f(f(f(n)))=f(n+1)+1$

Can you find all functions $f:\mathbb N\rightarrow\mathbb N$ satisfying the functional equation $$ f(f(f(n)))=f(n+1)+1 $$
Vincent P
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Is there a function $f:\mathbb{R}\to\mathbb{R}$ such that $f\circ f\circ f=Id.$, but $f\neq Id.$?

Is there any function $f:\mathbb{R}\to\mathbb{R}$, other than the identity, such that $$f\circ f\circ f=Id.?$$ That is a pretty simple question, but surprisingly I am not able to say anything about it. If we require only $$f\circ f=Id.,$$ then,…
user225615
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Find $f(x)$ satisfy $f(2x)=2f(x)+x$

I would appreciate if somebody could help me with the following problem: Find $f(x)$, given that: $f \colon \mathbb{R} \rightarrow \mathbb{R}$, $f$ is continuous at $x=0$, and $f(2x)=2f(x)+x$ I tried but couldn’t get it that way.
Young
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Find all real functions $f:\mathbb {R} \to \mathbb {R}$ satisfying the equation $f(x^2+y.f(x))=x.f(x+y)$

Find all real functions $f:\mathbb {R} \to \mathbb {R}$ satisfying the equation $f(x^2+y.f(x))=x.f(x+y)$ My attempt - Clearly $f(0)=0$ Putting $x^2=x,y.f(x)=1$, we have $f(x+1)=x.f(x+y)$. Now putting $x=x-1$,we have $f(x)=(x-1)f(x-1+y)$ Putting…
Snehil Sinha
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Functional Equation: Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $(x+y)(f(x)-f(y))=(x-y)f(x+y)$

Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $$(x+y)(f(x)-f(y))=(x-y)f(x+y)$$ My attempt: If $x=-y \not = 0$ then $0= 2x f(0)$ so $f(0)=0$. Suppose for the sake of contradiction that $f(x)=f(x+\epsilon)$ for some $x$ and…
John Marty
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How find this function such $f(2010f(n)+1389)=1+1389+1389^2+1389^3+\cdots+1389^{2010}+n$

Question: Find all function: $f:N\to N$, such that $$f(2010f(n)+1389)=1+1389+1389^2+1389^3+\cdots+1389^{2010}+n,\forall n\in N$$ Maybe this is 2010 Mathematical olympiad problem.But I can't find…
math110
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Solve $f(f(n))=n!$

What am I doing wrong here: ( n!=factorial ) Find $f(n)$ such that $f(f(n))=n!$ $$f(f(f(n)))=f(n)!=f(n!).$$ So $f(n)=n!$ is a solution, but it does not satisfy the original equation except for $n=1$, why? How to solve $f(f(n))=n!$?
TROLLHUNTER
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