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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Find $f: \Bbb{R} \to \Bbb{R}$ satisfy this function

Find function $f:\Bbb{R} \to \Bbb{R}$ such that $$\{f(x)\}^2 +2yf(x)+f(y)=f(y+f(x))$$where $\{ f(x) \}$ is the fractional part of $f(x)$ This is an upgrade from JMO Finals 2006. The original problem was $f(x)^2+2y\cdot f(x)+f(y)=f(y+f(x))$ for $x,y…
user1052563
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functional equation (conti-function $f(x)$)

I would appreciate if somebody could help me with the following problem Q: Find conti-function $f(x)=?$ $$4(1-x)^{2} f \left({1-x\over 2} \right)+16f \left({1+x\over 2} \right)=16(1-x)-(1-x)^{4}$$
Young
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How find the $f(x+f(y))+f(y+f(z))+f(z+f(x))=0,$

:$f:R\longrightarrow R$ ,and is continuous such that $$f(x+f(y))+f(y+f(z))+f(z+f(x))=0,$$ find all $f$
math110
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Solving Functional Equation with an essence of Number Theory

PROBLEM STATEMENT Find all functions $f: \mathbb N \rightarrow \mathbb N$ which satisfy: (a) $ f $ is a surjective function; (b) $m|n$ if and only if $f(m)|f(n)$, for any two natural numbers $m, n$. $ \ \ \ \ $ $(1)$ MY…
AbVk1718
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$(x-y)(f(f(x)^2)-f(f(y)^2))=(f(x)-f(y))(f(x)^2-f(y)^2$

Find $f: \mathbb{R} \to \mathbb{R}$ which satisfies: $f\small(0)\normalsize=0, f\small(1)\normalsize=2015.…
RDK
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Is there any non-constant function $f(x)$ satisfying $f(x) f(y) = f(x) + f(y)$?

I am interested in the following functional equation: $\begin{equation} f \left(x \right) f \left(y \right) = f \left(x \right) + f \left(y \right) \end{equation}$ In particular, I would like to know if there are any non-constant solutions for $f…
NRavoisin
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Find all the functions $f:R\to R$ which satisfy the identity $f(x+y)+f(xy)=f(x)\cdot f(y)+1$ for all $x,y\in R$

Find all the functions $f:R\to R$ which satisfy the identity $$f(x+y)+f(xy)=f(x)\cdot f(y)+1$$ for all $x,y\in R$ My solution Taking $x=y=0$, we get $f(0)=1$. Taking $y=-x$, we get $$f(-x^2)=f(x)\cdot f(-x)$$ which is true for all $x\in R$. Let's…
An Alien
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Finding $f:\mathbb R\to\mathbb R$ satisfying $f\bigl(yf(x+y)+f(x)\bigr)=4x+2yf(x+y)$

Find $f:\mathbb R\to\mathbb R$ satisfying $f\bigl(yf(x+y)+f(x)\bigr)=4x+2yf(x+y)$. This is my attempt: \begin{align} & \text{Let } f(0)=t \text. \\ &P(0, 0): f(t)=0 \text. \\ &P(0, t): t = 0 \text. \implies f(0)=0 \text. \\ \\ &\text{Assume) }…
RDK
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Finding all pairs of functions $f, g: \mathbb{R} \to \mathbb{R} $ which satisfy $f(x+y) = g\left(\frac{1}{x}+\frac{1}{y}\right)(xy)^{2008}$

Find two functions $f, g: \mathbb{R} \to \mathbb{R} $ which satisfy the condition: $$ f(x+y) = g\left(\frac{1}{x}+\frac{1}{y}\right)(xy)^{2008} $$ The first one I thought of was this: $$ f(x)=x^{2008}, g(x)=x^{2008} $$ or $$ f \equiv 0, g \equiv…
RDK
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Finding all $f: \mathbb{R} \to \mathbb{R}$ such that $f\bigl(xf(y)+y\bigr)+f\bigl(-f(x)\bigr)=f\bigl(yf(x)-y\bigr)+y $

Find all $f: \mathbb{R} \to \mathbb{R}$ such that $$f\bigl(xf(y)+y\bigr)+f\bigl(-f(x)\bigr)=f\bigl(yf(x)-y\bigr)+y $$ for all $x,y \in \mathbb{R}$. Help me solving this. My expectation of the answer is $f(x) = x+1$. My try: $$ P(x, y):…
RDK
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Solve the functional equation $f(y)f(xf(y))=f(x+y)$

Find all functions f defined over the positive reals. $f(y)f(xf(y))=f(x+y)$ I proved that $01$ exists, By setting $x$ as $\frac{t}{f(t)-1}$, and $y$ as $t$, We get $f(t)=1$, which is a…
HJS
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Find all odd functions $f:\mathbb R\to\mathbb R$ that satisfy $f(x+1)=f(x)+1$ and $f\left(\frac{1}{x}\right)=\frac{f(x)}{x^2}$

Find all functions: $f:\mathbb R\to\mathbb R$ that satisfy all the following three conditions: $f(-x)=-f(x)$ $f(x+1)=f(x)+1$ $f\left(\frac{1}{x}\right)=\frac{f(x)}{x^2}$ I assume $f(x)=x$ satisfies the conditions. I could prove $f(0)=0 $ and…
Beginner
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Function on integers satisfying two functional equation with values at a point + non-value at a point

Let $ f : \mathbb{Z} \to \mathbb{Z}$ be a function satisfying $f(0) \neq 0 =f(1)$. Assume also that $f$ satisfies equations $(A)$ and $(B)$ below: $$ f(xy) = f(x) + f(y) - f(x)f(y) \tag{1}$$ $$ f(x-y)f(x)f(y) =f(0)f(x)f(y) \tag{2}$$ For all…
tryst with freedom
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Trying to prove surjectivity

We are given $f$ : $R$ $\rightarrow$ $R$ such that $f(f(x)f(y)) = f(x) + f(y)$ for all reals $x$ and $y$ To show that the function is surjective, we plug in $y = 0$ and let $z = f(x)$ and $c = f(0)$, then we get $f(cz) = z + c$. Does this imply that…
Kritesh Dhakal
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The Functional Equation $ f \left( x ^ 2 \right) = f ( x ) ^ 2 $

Consider the following functional equation: $$ f \left( x ^ 2 \right) = f ( x ) ^ 2 \text , $$ where $ f : \mathbb Z \to \mathbb Z $. The only two solutions I could find so far are $ f ( x ) = 1 $ and $ f ( x ) = 0 $. Are there any other solutions?
theworldiscooll
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