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 all $f : \mathbb{N} \to \mathbb{N}$ such that $f(a) + f(b) \mid a+b, \ \forall \ a, b \in \mathbb{N}$

Find all $f : \mathbb{N} \to \mathbb{N}$ such that $$f(a) + f(b) \mid a+b, \ \forall \ a, b \in \mathbb{N}$$ All I can find is the following: If we put $a=b=n$ we get $f(n)\mid n$, so $f(n)\leq n$ for all $n \in \mathbb{N}$. So $f(1)=1$ and we have…
nonuser
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There does not exist $f(f(x))=x^2-2016$

Prove that there does not exist any function $f:\mathbb{R}\to\mathbb{R}$ such that $$f(f(x))=x^2-2016$$ for all real number $x$. My attempt : Substitute $x=f(x)$ in $f(f(x))=x^2-2016$, we have $f(f(f(x)))=f(x)^2-2016$ so $f(x^2-2016) =…
user403160
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A functional equation $\varphi(x) = \epsilon(x)\epsilon(x+x_1)\varphi(x+x_2)$

First, let $\epsilon:\mathbb{R}\rightarrow \mathbb{R}$ defined by $1$ over $[-1,1)$ and extended over $\mathbb{R}$ by $2$-antiperiodicity (for example, $\forall x\in [-2,-1)$, $\varepsilon(x)=-1$). I would like to find all $2$-antiperiodic…
anderstood
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$(f(x))^2=f(2x)+2f(x)-2$ Functional Equation

$f$ is a function on real numbers: $$f(x)^2=f(2x)+2f(x)-2$$ and $$f(1)=3$$ What is the value of $f(6)$? I find a solution $f(x)=2^x+1$. But, I don't know is there more solutions?
scarface
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How to solve $\frac{f^{-1}(x)f(x)}{x}=\frac{f^{-1}(x)+f(x)}{2}$?

I've come across the following functional equation: Determine all surjective functions $f:\mathbb{R_{>0}}\to\mathbb{R_{>0}}$ which satisfy for all $x\in\mathbb{R_{>0}}$: $$ 2xf(f(x))=f(x)\left(x+f(f(x))\right) $$ My approach so far: The equation is…
Redundant Aunt
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solving functional equation $f(x+y) +f(x)f(y)=f(x)+f(y)+f(xy)$ for all real numbers

The functional equation to be solved is $$ f ( x + y ) + f ( x ) f( y ) = f ( x ) +f ( y ) + f ( x y ) $$ for $ f : \mathbb R \to \mathbb R $. I found about four possible solutions to the equation but ran into a fundamental problem with all of them.…
Manan
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Solve the functional equation $2f(x)=f(ax)$ for some $a$.

I am trying to solve the following functional equation, and could use some help.$$ 2f(x)=f(ax)$$ For some $a\in\mathbb{R}$. By repeated adding $2f(x)$ together we notice that $$2nf(x)=f(a^nx).$$ Also $$2mf(a^{-m}x)=f(x)\Rightarrow…
kjhvzd
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Solutions to functional equation $f(f(x))=x$

Is there any more solutions to this functional equation $f(f(x))=x$? I have found: $f(x)=C-x$ and $f(x)=\frac{C}{x}$.
Somnium
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functions satisfying $f(x - y) = f(x) f(y) - f(a - x) f(a + y)$ and $f(0)=1$

A real valued function $f$ satisfies the functional equation $$f(x - y) = f(x) f(y) - f(a - x) f(a + y) \tag 1 \label 1$$ Where $a$ is a given constant and $f(0) = 1$. Prove that $f(2a - x) = -f(x)$, and find all functions which satisfy the given…
juantheron
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Functional equation $f(x+f(x+y))=f(x-y)+f(x)^2 \quad \forall x,y\in \mathbb R$

We have to find all functions $f\colon \mathbb R\to\mathbb R$ such that $f$: $$\forall x,y\in \mathbb R \quad f(x+f(x+y))=f(x-y)+f(x)^2.$$ Could somebody help me solve this problem? Thank you.
Steve
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A functional equation over a circle: $f(r \cos \phi)+f(r\sin \phi)=f(r)$

I am interested in the functional equation $$f(r \cos \phi)+f(r\sin \phi)=f(r),\qquad r\geq 0,\ \ \phi\in[0,\pi/2]\text.$$ Let's assume that $f:[0,\infty)\to\mathbb R$ is monotone. Clearly, $f(x)=ax^2$ is a solution. Are there any other solutions?
UserNew
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Proving the uniqueness of solution of a functional equation.

Problem Determine all functions $f$ such that $$f(2x)-2f(x+y)+f(2y)=(x-y)^2$$ where $f$ is continuous and defined for all $x,y\in{\mathbb{R}}$. I've found a particular solution: $$f(x)= \frac{x^2}{2} + bx + c $$ where $b,c\in{\mathbb{R}}$. Are there…
hejhoppbong
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$f(x+f(y))=f(x)+f(f(y))+3xy^6+3x^2f(y)$

Find all functions $f : \mathbb{R} \mapsto \mathbb{R}$ that check $f(x+f(y))=f(x)+f(f(y))+3xy^6+3x^2f(y)$ for all $x,y \in \mathbb{R}$ It's a very difficult functional equation, I tried many substitution Can someone help me, please
user1144884
6
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1 answer

trouble understanding a solution to symmetric equations between three functions

Find all triples $(f,g,h)$ of injective functions from the set of real numbers to itself satisfying \begin{align*} f(x+f(y)) &= g(x) + h(y) \\ g(x+g(y)) &= h(x) + f(y) \\ h(x+h(y)) &= f(x) + g(y) \end{align*} for all real numbers $x$ and $y$. I…
math_learner
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Solving the equation $\exp{y\left(x\right)} + y\left(x\right) + 1 = 0$ to find the function $y$

I am trying to solve the following equation and find a real-valued function $y\left(x\right)$ which satisfies $$ \exp{y\left(x\right)} + y\left(x\right) + 1 = 0, \hspace{0.2cm}\forall x>0 $$ The trivial solution $y\left(x\right) = x_0$ where $x_0$…
George
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