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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Complicated real to real functional equation

Find all $f:\mathbb{R} \rightarrow \mathbb{R}$ satisfying $$f(f(y)+x^2+1)+2x=y+(f(x)+1)^2$$ for all $x,y \in \mathbb{R}.$ So far I have proved that $f$ is bijective. How should I continue?
mathguy56
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Solve the following functional equation $f(xf(y))+f(yf(x))=2xy$

Find all function $f:\mathbb{R}\rightarrow \mathbb{R}$ so that $f(xf(y))+f(yf(x)=2xy$. By putting $x=y=0$ we get $f(0)=0$ and by putting $x=y=1$ we get $f(f(1))=1$. Let $y=f(1)\Rightarrow f(x)+f(f(x)f(1))=2x$, which tells us that $f$ is an injective…
CryoDrakon
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Solving a functional relation $f\left( x \cdot f(y)\right)=x^2 \cdot y^a$

I have this functional relation - $$f\left( x \cdot f(y)\right)=x^2 \cdot y^a$$ which I am trying to solve. I put $x=1$, then I put $f(y)=\dfrac{1}{x}$. I also tried out $y=f^{-1}(1)$, but it doesn't seem to work out. Please help me out. Thank you.
user1001001
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Additive involutions: $ f ( x + y ) = f ( x ) + f ( y ) $ and $ f \big( f ( x ) \big) = x $

Find all functions $ f : \mathbb R \to \mathbb R $ satisfying $$ f ( x + y ) = f ( x ) + f ( y ) $$ and $$ f \big( f ( x ) \big) = x $$ for all $ x , y \in \mathbb R $ This is a problem involving Cauchy's additive functional equation, but I don't…
primitiveroot
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Finding all real valued functions that satisfy $f(f(y) + xf(x)) = y + (f(x))^2$

I would like some help with finding all real valued functions that satisfy this equation: $f(f(y) + xf(x)) = y + (f(x))^2$ I tried the usual substitutions like $x = y = 0$, but my experience with this kind of problem is very limited. EDIT: I'm an…
Lucas Virgili
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$f:\Bbb Z\to\Bbb Z$ such that $f(f(n)-2n)=2f(n)+n$ for all $n\in\Bbb Z$

Does there exists a function $f:\Bbb Z\to\Bbb Z$ such that $$f(f(n)-2n)=2f(n)+n$$ for all $n\in\Bbb Z$
boh02
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$\forall x\in\mathbb R$, $|x|\neq 1$ it is known that $f\left(\frac{x-3}{x+1}\right)+f\left(\frac{3+x}{1-x}\right)=x$. Find $f(x)$.

$\forall x\in\mathbb R$, $|x|\neq 1$ $$f\left(\frac{x-3}{x+1}\right)+f\left(\frac{3+x}{1-x}\right)=x$$Find $f(x)$. Now what I'm actually looking for is an explanation of a solution to this problem. I haven't really ever had any experience with…
user26486
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Find all functions $f:R_{+}\rightarrow R_{+}$ such that for all $x,y\in R_{+}, f(x)=f(f(f(x))+y)+f(xf(y))f(x+y).$

Let $\mathbb{R}_{+}$ be the set of all positive real numbers. Find all functions $f{:}~ \mathbb{R}_{+} \mapsto \mathbb{R}_{+}$ such that for all $x,y\in R_{+}$, we have $$ f(x)=f(f(f(x))+y)+f(xf(y))f(x+y). $$ This is the third question from USAMO…
Arachnephob1a
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Any chance for a solution of $f[f(x)]=x^2+1$?

Not sure if this is a closed or open question. But the question is suppose $f[f(x)]=x^2+1$ then what is $f(x)$? Though the question does not refer to the domain let us suppose it is defined on $\Bbb R$. Thanks.
user85356
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Continuous functions satisfying $f(f(x))=x$, for all $x \in \mathbb{R}$, and $\int_{-x}^{0} f(t)dt - \int_{0}^{x^2}f(t)dt=x^3$ for $x>0$

Find all continuous functions $f: \mathbb{R} \to \mathbb{R}$ satisfying: I. For all $x\in \mathbb{R}$, $f(f(x))=x$ II. For all $x>0$, $\displaystyle\int_{-x}^{0} f(t)dt - \displaystyle\int_{0}^{x^2}f(t)dt=x^3$ So far I got that $f$ is strictly…
Hypernova
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Help in solving a simple functional equation: $3f(2x+1)=f(x) + 5x$

I need to find all continuous functions satisfying: $$3f(2x+1)=f(x) + 5x$$ The functional equation looks simple but I am unable to solve it. I tried to convert it into a Cauchy type equation but I wasn't able to do so.
saisanjeev
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Existence of two functions $f$ and $g$ for which $f\circ g (x)=x^2 , g\circ f (x)=x^3$

Do there exist two functions $f$ and $g$ from the reals to itself satisfying $f\circ g (x)=x^2 , g\circ f (x)=x^3$ for any $x\in\mathbb{R}$? From the given equations I could get the following information: $f$ is injective. $g$ is surjective and…
Fermat
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Proving a function is constant when $f(x)f(y) + f(\frac{a}{x})f(\frac{a}{y}) = 2f(xy)$

I've been working on the following homework problem: Consider a function $f : (0,∞) → \mathbb{R}$ and a real number $a > 0$ such that $f(a) = 1$. Prove that if $f(x)f(y) + f(\frac{a}{x})f(\frac{a}{y}) = 2f(xy)$ for all $x, y ∈ (0,∞)$, then $f$…
SSumner
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Functions satisfying $g(1 - g(x)) = x$

I need to find all functions with $g(1 - g(x)) = x$ and $x\in \mathbb{R}$. I must also provide a specific example of such function. My first attempt was to invert the expression yielding $g(x) = 1 - g^{-1}(x)$ and inverting again $x = g^{-1}(1 -…
Steve_Steve
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How to find a such function $f:3\mathbb{N}+1\to 4\mathbb{N}+1$

How to find a bijective function $f: 3\mathbb{N}+1\to 4\mathbb{N}+1$ such that $$f(xy)=f(x)f(y),\forall x,y\in 3N+1$$ If i let $x,y\in 3\mathbb{N}+1$ then there exists $n,m\in \mathbb{N}$ such that $x=3n+1,y=3m+1$ but I have no idea how I can find…
Vrouvrou
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