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.

3976 questions
5
votes
1 answer

If $f\circ f$ is smooth, is the monotonic function $f$ smooth?

Let $f:\mathbb R\to\mathbb R$ be continuous and monotonic and assume that $f\circ f$ is analytic. Is $f$ necessarily continuously differentiable/smooth/analytic? My question arose from this: Inspired by the thread a continuous function satisfying…
Samuel
  • 5,550
5
votes
3 answers

Functional Equation : $f(x) = f(x + y^2 + f(y))$

This problem is from my textbook: Given : $f:\mathbb R\to\mathbb R$ Solve this functional equation : $f(x) = f(x + y^2 + f(y))$ I think this function is just a simple constant, so I try all my best to prove this, but no result. I'm not sure this way…
hqt
  • 759
  • 4
  • 9
  • 19
5
votes
3 answers

Continuous solutions of $f(x+y+z)=f(x)g(y)+f(y)g(z)+f(z)g(x)$

Consider the following functional equation: $$f(x+y+z)=f(x)g(y)+f(y)g(z)+f(z)g(x)$$ where the equation holds for all $x,y,z \in \mathbb{R}$. One solution is $f(x)=cx$ and $g(x)=1$. What are all the continuous solutions $f, g\colon \mathbb{R} \to…
Malper
  • 1,387
5
votes
2 answers

How find all $f:\mathbb R\to\mathbb R$ such that $f\bigl(x\cdot f(y)\bigr)=y\cdot f(x)+kxy$

Let $k$ be a given real number. Find all the functions $f:\mathbb R\to\mathbb R$ such that $$f\bigl(x\cdot f(y)\bigr)=y\cdot f(x)+kxy\,.$$ My try: Let $x=y=0$, then $$f(0)=0,$$ and $x=y=1$, then $$f\bigl(f(1)\bigr)=f(1)+k.$$ So let $f(1)=a$, then…
user94270
5
votes
1 answer

Does $f(x+f(y))=f(x)+f(y)$ imply $f(x+y)=f(x)+f(y)-f(0)$, given $f(0)\neq0$?

We are given a function $f\colon R\to R$ that satisfies $f(x+f(y))=f(x)+f(y)$ for any real $x,y$, and also $f(0)\neq0$. The equation seems to be entirely dependent on the values of $f$. It seems impossible to have anything to say about $f(x+y)$ just…
noballpointpen
  • 223
5
votes
1 answer

Nontrivial solutions to simple functional equation

I am interested in the functional equation $$ f(x^2) = 2f(x)^2 - 1, $$ where the domain of $f$ is nonzero real numbers. Are there any nontrivial continuous solutions? I know $f(x) = 1$ and $f(x) = \mathrm{sgn}(x)$ are solutions, but I want to know…
jackson
  • 987
5
votes
1 answer

Iterative Functional Equation on $ \mathbb N $: $ f \Big( f \big( f ( n ) \big) \Big) + 6 f ( n ) = 3 f \big( f ( n ) \big) + 4 n + 2001 $

Find all functions $ f : \mathbb N \to \mathbb N $ satisfying $$ f \Big( f \big( f ( n ) \big) \Big) + 6 f ( n ) = 3 f \big( f ( n ) \big) + 4 n + 2001 , \forall n \in \mathbb N \text . $$ After some trial and error, I assumed the solution to be…
Alpha
  • 126
5
votes
2 answers

$f(x+f(y))=f(x)+y^n$ for monotone $f$

Here is the problem: Fix $n\in\mathbb{N}$. Find all monotonic solutions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that $f(x+f(y))=f(x)+y^n$. I've tried to show that $f(0)=0$ and derive some properties from that but have been unable to do so. A…
Ryan
  • 59
5
votes
2 answers

Finding all $f:\mathbb{R} \rightarrow \mathbb{R}$ s.t $f(x^2+f(y))=(x-y)^2f(x+y)$

Find all $f:\mathbb{R} \rightarrow \mathbb{R}$ s.t $$f(x^2+f(y))=(x-y)^2f(x+y)$$ I don't want people to solve this one for me I'd just like to know whether one of my steps is legitimate. So I put $x=y$ and got $f(x^2+f(x))=0$. Putting $y=-x$ I got…
John Marty
  • 3,650
5
votes
1 answer

Solving the functional equation: $f\bigl(f(x)-x\bigr)=2x$

PROBLEM STATEMENT: Find all functions $f :\mathbb{R}^{+}_{0} \to \mathbb{R}^{+}_{0} $ satisfying the functional equation $$f\bigl(f(x)-x\bigr)=2x.$$ MY PROGRESS: This question seemed pretty tough to me. All I could figure out was: $f(x) = 2x$ and…
AbVk1718
  • 501
5
votes
1 answer

Find all functions $f: \mathbb{R} \to\mathbb{R} $ such that if $a+f(b)+f^2(c) = 0$ then $f(a)^3 + bf(b)^2 + c^2f(c) = 3abc$

Find all functions $ f : \mathbb R \to \mathbb R $ such that for all $ a , b , c \in \mathbb R $, if $$ a + f ( b ) + f ^ 2 ( c ) = 0 \tag {Eq1} \label {eqn1} $$ then $$ f ( a ) ^ 3 + b f ( b ) ^ 2 + c ^ 2 f ( c ) = 3 a b c \text . \tag {Eq2}…
texiwi
  • 130
5
votes
2 answers

Find all functions taking rationals to rationals satisfying $f(x+f(y))=f(x)f(y)$

Ok, for this problem, I let $f(y)=a$, and then we have $f(x+a)=f(x)a$. Then, by an easy induction argument, we have $f(x+na)=f(x)a^n$ for some rational numbers $a$ and positive integers $n$. Now, if $f$ can only take $a$, then it follows $a=0$ or…
Maths explorer
  • 677
5
votes
1 answer

Functional equation $f(f(k+1)+3)=k$

Find all functionas $f:\mathbb{Z}\to\mathbb{Z}$ such that $\forall k\in\mathbb{Z}$, $$f(f(k+1)+3)=k$$. Obvious solution is $f(k)=k-2$. I have found following: $k+3=f(f(k+1)+3)+3\implies f(k+3)=f(f(f(k+1)+3)+3)=f(k+1)+2\implies f(k)=\begin{cases}…
Mutse
  • 571
5
votes
0 answers

Find all functions $f \colon \mathbb R \to \mathbb R$ such that for $f(x)f(y) + f(\frac{1}{2} + \sqrt{xy(x + y) + \frac{1}{4}}) = f(xy) + f(x + y)$.

Find all functions $f \colon \mathbb R \to \mathbb R$ such that for $\forall x, y \in \left[-\dfrac{1}{2}, +\infty\right)$, $$f(x)f(y) + f\left(\frac{1}{2} + \sqrt{xy(x + y) + \frac{1}{4}}\right) = f(xy) + f(x + y)$$ Let $P(x, y)$ be the assertion…
Lê Thành Đạt
  • 4,228
5
votes
1 answer

Solutions to the functional equations $f(sx) = tf(x)$ and $f(sx + (1-s)) = tf(x) + (1-t)$ on $[0,1]$

Suppose that $s,t \in (\frac{1}{2},1)$ with $t \ne s$. Does there exists a continuous bijection $f \colon [0,1] \to [0,1]$ which simultaneously satisfies the functional equations $$ f(sx) = tf(x) $$ and $$ f(sx + (1-s)) = tf(x) + (1-t) $$ I…
Zorngo
  • 1,421
1 2 3
…
60 61