# Tagged Questions

The name "functional equation" is used for problems where the goal is to find all functions satisfying the given equation (and maybe some other conditions). So in this case, solving the equation means finding all functions fulfilling the equation. (This is different from the more common use of the ...

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### Functional equation with sqrt solutions [closed]

Let $f:(0,\infty)\to(0,\infty)$ so that for all $x,y\in(0,\infty)$ we have $$f\left(\frac {x} {f(y)}\right)=\frac {x} {f(x\sqrt y)}.$$ Find function $f$.
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### Functional equation: $f(x)=\frac x{\prod_{i|x}^{x-1} f(i)}$

Find an algebraic function $f:\Bbb N\to\Bbb N$ such that $$f(x)=\frac x{\prod_{i|x}^{x-1} f(i)}$$ and $$f(1)=1$$ for all $x\in\Bbb N$ I allready know two things: $f(p^k)=p$ where $p$ is prime ...
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### Find all functions $f: \mathbb{N}\to\mathbb{N}$ such that $f(x)=f(x^y)$

Find all functions $f:\mathbb{N}\to\mathbb{N}$ such that $$f(x)=f(x^y)$$ for all $x,y\in\mathbb{N}$. I'm not intrested in the trivial solution $f(x)=k$, where $k\in\mathbb{N}$.
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### System of linear Volterra integral equation

Consider the Volterra integral equation $$f(t) = g(t) +\int_0^t K(t,s) f(s) ds$$ where $g(t)$ is continuous on $t\in[0,T]$ and $K(t,s)$ is a weakly singular kernel. It is well known that there ...
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### Functional analysis with KKT conditions

I want to solve an optimization problem min $F(x_{ik})$ subject to $x \in X$. $F$ here is a function or functional that I wish to determine. I want my optimal ...
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### $f(x)$ is an analytic function in $\mathbb{R}$ such that $f(-x)f(x)=1$. What else can we find out about $f(x)$?

Well, I know that there are some easy things we can say immediately: $f(0)= \pm 1$, follows immediately $f(x)=\pm 1$ is the obvious solution, so let's look for other solutions. Moreover, let's ...
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### Functional equation that models trigonometric identities

Find, with proof, all continuous functions $f,g : \mathbb{R} \rightarrow \mathbb{R}$ such that $f(x)^2 + g(x)^2 = 1$ and $2f(x)g(x)=f(2x)$. I am aware that the solution pair ...