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 all possible choices of real-valued functions $f(x)$ and $g(y)$ such that $f (x) + g(y) = log (1 + x + x y + y)$ for all positive $x$ and $y$.

Question: Find all possible choices of real-valued functions $f(x)$ and $g(y)$ such that $$f(x) + g(y) = \log (1 + x + x y + y)$$ for all positive $x$ and $y$. Attempt: Note that $$f(x)+g(y)=\log(1+x)+\log(1+y)\tag{1}$$ Plug in $x=0$, to…
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When are there solutions to $f(x+y)-f(x-y) = axy+bx+cy+d $ and what are they?

This is a generalization of Solving the functional equation When are there solutions to $f(x+y)-f(x-y) = axy+bx+cy+d $ (i.e., what restrictions are there on $a, b, c, d$) and what are they? I have a solution, but want to see what others come up…
marty cohen
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Find function $f(ax+b)+c \le x \le f(x+c)+b$

Find all functions $f: \Bbb R \rightarrow \Bbb R$ such that $f(ax+b)+c \le x \le f(x+c)+b$ for all $x \in \Bbb R$ and $a,b,c$ real constant. I've done some trivial idea as $x=0$, $x=-c$, $x=-b/a$ but I got nothing more.
Arnaldo
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Find all functions $f:\mathbb R \to \mathbb R$ that satisfy $f(x) + 3 f\left( \frac {x-1}{x} \right) = 7x$ for all nonzero $x$.

Find all functions $f:\mathbb R \to \mathbb R$ that satisfy $$ f(x) + 3 f\left( \frac {x-1}{x} \right) = 7x $$ for all nonzero $x$. I don't know what's going on today, my brain is not moving! I'm stuck on this problem. Solutions are greatly…
Yuna Kun
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Functional Equation Analysis

Given: $$F(F(n)) = n$$ $$F(F(n + 2) + 2) = n$$ $$F(0) = 1$$ where n is a non-negative integer. $$F(129) = ?$$ How can we solve such kind of functional equations? Is there any simpler approach other than doing iteration over the values manually?
user6374
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Functional equation high school problem $f(x) = f(-x) - 2x $

Let \begin{cases} f: \mathbb R \to \mathbb R\\ f(x) = f(-x) - 2x \end{cases} I can see that one solution is: $f(x) = -x$ , but I have no idea how to prove that this is the only solution. Thank you for your help
Kelly
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How to prove that the following two sets are equal?

Let $M=\{x|f(x)=x\},~N=\{x|f(f(x))=x\}$, and $f(x)=x^2+ax+b$, where $a,b\in R$ are such that $4b=(a-1)^2$. Show that $M=N$.
math110
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Functional Equation: $f \left(x+\cos(2017y) \right)=f(x)+2017\cos\left(f(y)\right)$

Find all functions $f: \mathbb R \rightarrow \mathbb R$, such that for all $x,y \in \mathbb R$ satisfies the equation: $$f \left(x+\cos(2017y) \right)=f(x)+2017\cos\left(f(y)\right)$$ My work so far: Let $f(0)=c$. 1) $x=0 \Rightarrow…
Roman83
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Determining a function through three equations

I have the following assignment question, and I'm having trouble even getting started: Consider the set of functions $\mathcal{F}=\{f,g\}$, with $f:\mathbb{R}^2\to\mathbb{R}$, and $g:\mathbb{R}\to\mathbb{R}$, defined by $f(x,y)=xy$ and $g(x)=x+1$.…
FPP
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How to solve this functional equation: $2f(x) = f(x-1)+f(x+1)$?

After some calculations, I came up with this functional equation: $f(x-1)+f(x+1)=2f(x)$. I found linear function is one possible answer, but don't know how to derive it. I don't know much about the techniques to solve this kind of equation. Can…
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How do I determine periodicity of a function through a system of functional equations?

I was given these equations, $f(k+x) = f(k-x)$, $f(2k+x)= -f(2k-x)$ . k is assumed a constant. I was asked to comment whether $f(x)$ is even or odd. By solving I came to the equation, $f(-x)=-f(x)$,which states that $f(x)$ is an odd function. But…
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The functional equation and differentiability

Find all functions $f: \mathbb R\rightarrow \mathbb R$, at the same time satisfying the following two conditions: a) $f (x + yf (x)) = f (x) f (y)$ b) the function $f$ can be represented in the form $f (x) = (\varphi (x)) ^ 2, x \in \mathbb R,$…
Roman83
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Functional equation $g(2x )= 1/2 g(x)$

I am trying to solve functional equation for $g: \ (0, \infty) \mapsto ( 0, \infty)$ $$ g( 2x ) =\frac 12 g(x)$$ Wolfram claims, and it is intuitive, that the function is $g(x) = C \frac{1}{x}$. But how to prove it?
wroobell
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Prove that there are no two functions $f$ and $g$ defined over real numbers that satisfy either of the following functional equations

Prove that there are no two functions $f$ and $g$ defined over real numbers that satisfy either of the following functional equations: $f(x)g(y) = x+y$ and $f(x) + g(y) = xy$. I think we should prove this using casework. That is, show it is…
user19405892
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Can I prove that $f(x,y)$ can be written as $g(x+y)$ under certain conditions.

I have $f(x,y):R^2\rightarrow R$. I know $f(x,y)=f(y,x)$ and $f(x+d,y)=f(x,y+d)$. Can I prove that I can express $f(x,y)$ as $g(x+y)$. This is where I got: $f(x+d,y)=f(x,y+d)$, I plug in $x=0$ Gives me $f(d,y)=f(0,y+d)$, I plug in $d=x$, and use the…