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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Solve the functional equation $f(xy) = f(x)f(y) - f(x + y) + 1$ and $f(1) = 2$

I'm working on this problem already, the domain and co-domain are the set of Rational Numbers. we see that setting $x = 1$ and $y = n$, we obtain $f(n + 1) = f(n) + 1$ But I'm trying to prove that $f(n) = n + 1$ for all $n$, the case when $n = 1, 2$…
Icosahedron
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How do we know this is the unique solution to this functional equation?

I have the functional equation $$F(x)+F(y)=F(\sqrt {x^2+y^2})$$ I know that one solution to this equation is $F(x)=a\cdot x^2$. However, I am told that this is the unique solution, but I don't know how to show this? How do we know that this is the…
user56834
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Find the sum of all values of $ f ( 2017 ) $ given $ f ^ { f ( a ) } ( b ) f ^ { f ( b ) } ( a ) = f ( a + b ) ^ 2 $.

Let $ f :\mathbb N \to \mathbb N $ be an injective function such that $$ f ^ { f ( a ) } ( b ) f ^ { f ( b ) } ( a ) = f ( a + b ) ^ 2 $$ for all $ a , b \in \mathbb N $. Let $ S $ be the sum of all possible values of $ f ( 2017 ) $. Find $ S \bmod…
Abhash Jha
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Solve the functional Equation:- $f(x+y)=3^xf(y)+9^yf(x)\forall x,y\in \mathbb{R}$

Problem Statement:- Consider a differentiable function $f:\mathbb{R}\rightarrow\mathbb{R}$ for which $f(1)=6$ and $$f(x+y)=3^x\cdot f(y)+9^y\cdot f(x),\;\;\forall x,y\in \mathbb{R}$$ then find $f(x)$. My attempt:- As the function is differentiable…
user350331
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Functional equation: finding all pairs of functions $f,g:\mathbb{R}\to\mathbb{R}$ with $f$ strictly increasing and $ f(xy) = g(y)f(x) + f(y)$

Find all pairs of functions $ f,g : \mathbb{R}\to \mathbb{R}$ such that (a) if $ x < y$, then $ f(x) < f(y)$; (b) for all $ x,y \in \mathbb{R}$, $ f(xy) = g(y)f(x) + f(y)$. My work : Let $ P(x,y) : f(xy) = g(y)f(x) + f(y)$ $x
user403160
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Functional equation $(x + y)f(f(x)y) = x^2 f(f(x) + f(y))$

Find all functions $f:\mathbb{R}^+ \to \mathbb{R}^+$ such that $$(x + y)f(f(x)y) = x^2 f(f(x) + f(y)) \mbox{, for all } x,y\in\mathbb{R}^+.$$ I tried out various substitutions such as $x=y$, $x=1$, $x+y=x^2$, and nothin' works. Thanks in advance for…
Pathbreaker
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Functional equation: $f: \mathbb{R} \rightarrow \mathbb{R}$, $f((x + 1) f(y)) = y (f(x) + 1)$

Determine all $f: \mathbb{R} \rightarrow \mathbb{R}$ such that, for all $x, y \in \mathbb{R}$, $$f((x + 1) f(y)) = y (f(x) + 1).$$ Source: Vuong Lam Huy, on a Facebook group.
Triskele
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A Non-Standard Functional Equation: $\big(1+f(x^{-1})\big)\big(f(x)-f(x)^{-1}\big)=\frac{(x-a)(1-ax)}{x}$

Suppose that $a\in(0,1)$ and that $f:\mathbb R \to \mathbb R$. I am trying to solve for $f$ such that $$ \big(1+f(x^{-1})\big)\big(f(x)-f(x)^{-1}\big)=\frac{(x-a)(1-ax)}{x} .$$ I really don't know where to start from though. I've tried the method of…
mzp
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What solutions are there to $f(f(x))=x^4$?

I know $f(x)=x^2$ is a solution, but I can't seem to find any others and I have no idea how to approach this.
Miles Johnson
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Find all continuous functions from positive reals to positive reals such that $f(x)^2=f(x^2)$

Find all continuous functions $f:\mathbb R^{+}\to\mathbb R^{+}$ such that $f(x)^2=f(x^2)$ for all positive reals $x$. If $f$ is a constant function, then $f(x)=1$. If $f$ is non-constant, I'm suspecting the only solutions are of the form $f(x)=x^k$,…
John Smith
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$f(1-f(x))=f(x)$

Find all continuous $f:[0,1] \rightarrow [0,1]$ such that $f(1-f(x))=f(x)$.
Benji
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If $f$ is a continuous function from $R \rightarrow R$ and $f(x)=f(x+f(x))$ then prove that $f$ is constant.

If $f$ is a continuous function from $R \rightarrow R$ and $f(x)=f(x+f(x))$ then prove that $f$ is constant. I could prove that $f(x)=f(x+f(x))=..=f(x+nf(x))$ after $n$ iterations.Then , how will I proceed?
user321656
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Find all functoins $f: \mathbb R \rightarrow \mathbb R$ such that $\forall x,y \in \mathbb R f(xy+f(x))=xf(y)+f(x)$

Find all functoins $f: \mathbb R \rightarrow \mathbb R$ such that $\forall x,y \in \mathbb R$ the equality:: $$f(xy+f(x))=xf(y)+f(x)$$ My work so far: 1) $f(0)=0$; Let $f(a)=f(b)\not=0$ $x=a, y=b \Rightarrow f(ab+f(a))=af(b)+f(a)$; $x=b, y=a…
Roman83
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Find all functions satisfying the condition $f(x+y)+f(x)f(y)=f(xy)+f(x)+f(y)$

Find all functions $f:\mathbb{R}\to{\mathbb{R}}$ such that $f(x+y)+f(x)f(y)=f(xy)+f(x)+f(y)$ for all $x,y\in{\mathbb{R}}$ first I put $x=y=0$ so I got $f(0)=0$ or $f(0)=2$. For the case $f(0)=2$, putting $y=0$, I got $f(x)=2$ for all real $x$. For…
Satvik Mashkaria
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How to solve the functional equation $f(x+a)=f(x)+a$

Are there any other solutions of the functional equation $f(x+a)=f(x)+a$ ($a=\mathrm{const}$, $a\in\mathbb{R}\setminus\left\{0\right\}$) apart from $f(x)=x+C$ ($C=\mathrm{const}$)? Edit: $a$ is a fixed number here.
Constructor
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