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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How can I attempt this FE

Prove that there is no function f : ℕ→ℕ such that f (f (n)) = n + 1. Here ℕ is the positive integers {1, 2, 3,...}. I have messed around with the FE but can't seem to produce anything meaningful. I found the solution online which states but I'm…
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If $f: N\rightarrow N $ be such that $f(f(n))+f(n+1)=n+2$ for all $n\in N$ find $f(1)$ and $f(2)$.

Putting $n=1$, we get $f(f(1))+f(2)=3$. Thus there are two possibilities. Either $f(f(1))=1, f(2)=2$ or $f(f(1))=2, f(2)=1$. Also, we observe that $f(f(n))=n+2-f(n+1)\leq n+1$ and $f(n+1)=n+2-f(f(n))\leq n+1$. How do we get the values of $f(1)$ and…
Debashish
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$f(\vec{x_1}+\vec{x_2})=f(\vec{x_1})f(\vec{x_2})$

What is the general solution to $f(\vec{x_1}+\vec{x_2})=f(\vec{x_1})f(\vec{x_2})$ where $\vec{x}$'s are in discrete vector space $x\in \{n_1\vec{e_1}+n_2\vec{e_2}+n_3\vec{e_3},n_1,n_2,n_3 \in Z\}$?
richard
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Solution of the equation $f(ax) = bf(x)$

Given the equation $f(ax) = bf(x)$, with $a, b > 0$, demonstrate that the solution is: $$f(x) = g(\log x)x^{\frac{\log b}{\log a}}$$ where $g(x) = g(x + \log a)$ is an arbitrary periodic function with period $\log(a)$. By the method of induction I…
JaberMac
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Functional equation with three variables

I have a functional equation with three variables. $f(x,y,z)$ is a real function with three variables where y is different from z i.e., $f(x,y,z)$ defined only for $y \neq z$. This function satisfies…
user409680
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Functional equations in one variable..

How do you solve the functional equation involving only one variable...what if and if not given that $f(x)$ is a polynomial... Say for example $f(x)=f(x-1) +2x$
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How to solve a functional equation involving log?

It's given that $$f(xy)=\frac {f (x)}{y}+\frac {f (y)}{x}$$ Also $x,y>0$ and $f(x)$ is differentiable for $x>0$ such that $f(e)=\frac{1}{e}$. By the look of the functional equation I am sure it does involve log at some point . By common…
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Two functional equations

Is there a systematic approach that can be used to solve these two functional equations? $$af(x) = f(bx), \quad\text{where }\ f\colon \mathbb{R}\to\mathbb{R} \tag{1}$$ $$ag(y) + ay = g(ay),\quad\text{where }\…
pizet
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How many polynomial functions exist such that $f(x^2) = (f(x))^2 = f(f(x))$

How many polynomial functions $f$ of degree $\geq1$ satisfy $f(x^2) = (f(x))^2 = f(f(x))$ for all real $x$?
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Solving a complex valued functional equation

I have an equation in the following form: $\frac{1}{r}F(s-j) = F(s) - F^{2}(s)$ Where $r$ is a real valued constant, $j$ is the unit imaginary number, $s$ is a complex variable and $F$ is a complex function. Any idea how to approach this?
Ali
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Functional equation with $f(2x)$

Any other solutions(advice) are welcome. For any $x>0, \;\;\; 2f(\frac{1}{x}+1)+f(2x)=1$ Find all possible f(x). I wish you luck on a good thing in $2019$.
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Find the form of functional

I know the form of $f(x)$ and that of $g(x)$ and I would like to find an expression for a function $H$ such that $H[g(x)] = f(x)$. $f$ is a polynomial and $g$ is more complex and involves some $\exp$ and $\cos$. Is there any procedure to find…
Begbi
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$f(x^2*f(y)^2) = f(x^2) * f(y)$

$$f(x^2 * f(y)^2) = f(x^2) * f(y)$$ $$f:{\mathbb Q}^+ \rightarrow {\mathbb Q}^+$$ $x,y \in {\mathbb Q}^+$ I have been given the following equation and information and am supposed to find all functions which satisfy this equation. I have found…
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$f(y+zf(x))=f(y)+xf(z) $ with $x,y,z\in \mathbb{R}$

Find all $f\colon\mathbb{R} \rightarrow \mathbb{R}$ such that $f(y+zf(x))=f(y)+xf(z)$ with $x,y,z\in \mathbb{R}$ When I study functional equation, I have some difficult in solving above equation. Can anyone help me please?
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Solving functional equation by comparising sides

I have a brain lag. I'm overthinking probably. I have the following equation: $$ f^{-1}(f(x)+f(y))=(g\circ h)^{-1}((g\circ h)(x)+(g\circ h)(y)), $$ where, say, $f,g,h$ are automorphisms on $\mathbb{R}$. Is it enough to compare the relevant parts of…