ODE discrete version

If $f$ is a function that is defined over natural numbers ($f:N \to R$), let $f^*$ denote the function $f^*(n) = f(n) - f(n-1)$. Find all functions such that $$n(f^*)^*(n) + (n-1)f^*(n) = f(n)$$ for all $n > 2$

Solution We can check that $f(n) = A(n-1) + B/2^n$ for any constant $A,B$ satisfies the given equation. Now we show that there's nothing else.

Define $g = f^*+f$. The equation is identical to: $$n(f^*+f)^*(n) =(f^*+f)(n)$$ $$ n(g(n) - g(n-1)) = ng^*(n) = g(n)$$ $$\frac{g(n)}{n} = \frac{g(n-1)}{n-1}$$ for all $n>2$, therefore $g(n)/n$ must be a constant. Remember that $g(1)$ is not defined because $f$ is only defined for $n \geq 1$ thus $g$ is only defined for $n \geq 2$. Furthermore, the value $g(2)$ depends on $f(1)$ and $f(2)$.

Case 1: $g(2) = 0$ In this case then $g(n) = 0 \forall n$. $$(f^*+f)(n) = 0$$ $$f(n) = f(n-1)/2$$ so we easily get $f(n) = k/2^n$ for some constant $k$.

Case 1: $g(2) \neq 0$ Note that if $f$ is a solution to the problem then $kf$ is also a solution to the problem, for all $k$ real numbers. If we replace $f$ by $kf$, then $g$ is also replaced by $kg$. So WLOG, we may assume that $g(2) = 2$

Because $g(n) / n$ is a constant then $$g(n) / n = g(2) / 2 = 1$$ so $g(n) = n \forall n >2$ $$f(n) = \frac{f(n-1) + n}{2}$$ for all $n > 2$.

Because $g(2) = 2 = 2f(2) - f(1)$ then $f(2) = 1+ f(1)/2$ Now notice this pattern: $$f(3) = \frac{f(2) + 3}{2} = 2 + f(1)/4$$ $$f(4) = \frac{f(3) + 4}{2} = 3 + f(1)/8$$ Easy to prove by induction that $f(n) = n-1 + f(1)2^{1-n}$.

From the two cases above, it's clear that any solution of the equation must be in the form of $f(n) = A(n-1) + B/2^n$ for any constant $A,B$

Function from N to N

Find all functions $ f:\mathbb{N}^{*}\rightarrow\mathbb{N}^{*} $
such that

$f(n)+f(n+1)+f(f(n))=3n+1 (\forall)n\in\mathbb{N}^{*} $

Functional Equation

Find all functions $f:N \to N$ such that:

$$f(f(m+n)) = f(m) + f(n) \forall m,n \in N$$

Function Composition

Are there real-valued continuous functions $f,g$ defined on real numbers such that
$$f(g(x)) = x^{2011}$$
and
$$g(f(x)) = 2011x$$

for all $x$?

Solution

Applying $g$ to both sides, we have:
$$g(f(g(x))) = g(x^{2011})$$
$$2011g(x) = g(x^{2011})$$

Let $x = 1$, we have $g(1) = 0$

Let $P(y) = g(e^y)$ defined on all $y$ real numbers.
So $P(2011y) = g(e^{2011y}) = g((e^y)^{2011}) = 2011g(e^y) = 2011P(y)$
We conjecture that $P(y) = ay$ for some $a$. Setting $x = e^y$, we have:
$$g(x) = g(e^y) = P(y) = ay = a \log x$$

Now plugging back in, our second equation becomes:
$$a \log f(x) = 2011 x$$
$$f(x) = e^{2011x / a}$$

Testing for compatibility with first equation:

$$f(g(x)) = e^{2011 a \log x / a} = e^{2011 \log x} = x^{2011}$$

Symmetric Function in 3 Variables

A function $f$ is defined from a triplet of positive reals to a positive real number $f:R_+^3 \to R_+$ and satisfies the following:

1. $f$ is symmetric in all 3 variables. That is $f(a,b,c) = f(b,a,c) = f(a,c,b) = ...$
2. For any positive real $t$, $f(ta,tb,tc) = tf(a,b,c)$
3. $f(1/a, 1/b, 1/c) = 2011^2/f(a,b,c)$
4. $f(a,b,c) = f(\sqrt{ab}, \sqrt{ab}, c)$

Determine how many triplet of positive integers $(x,y,z)$ are there such that $f(1/x,1/y,1/z) = 1$

Solution

From rule 3, we have $f(1,1,1) = 2011^2/f(1,1,1)$, so $f(1,1,1) = 2011$.

Now, for any $u,v > 0$,
$f(u,v,1/(uv)) = f(\sqrt{uv}, \sqrt{uv}, 1/(uv))$ using rule 4
and
$f(uv,1,1/(uv)) = f(\sqrt{uv}, \sqrt{uv}, 1/(uv))$ using rule 4
So
$f(u,v,1/(uv)) = f(uv,1,1/(uv)) = f(uv, 1/(uv), 1) = f(1,1,1) = 2011$

Now, for any $t,u,v$, let $k = \sqrt[3]{tuv}$, then $t = k^3/uv$
$f(t,u,v) = f(k^3/uv, u, v) = kf(k^2/uv, u/k, v/k) = kf(u/k, v/k, 1/((u/k)(v/k))) = 2011k = 2011\sqrt[3]{tuv}$

Thus $f(t,u,v) = 2011\sqrt[3]{tuv}$ for all $t,u,v$.

Now we need to determine the number of positive integer triples $(x,y,z)$ such that
$$f(1/x,1/y,1/z) = \frac{2011}{\sqrt[3]{xyz}} = 1$$
$$xyz = 2011^3$$
Since 2011 is a prime, there are only the following possibilities:
1. $(1,1,2011^3)$ and all its permutations, there are three triplets.
2. $(1,2011, 2011^2)$ and all its permutations, there are six triplets.
3. $(2011,2011,2011)$, there is one triplet.

So in total, there are ten triplets that satisfy the condition.

The above conditions could be tailored to fit any symmetric functions. For example, for $f = a^2 + b^2 + c^2 + k$ one can have the following conditions:

1. $f$ is symmetric
2. $f(\sqrt{a^2+t},b,c) = f(a,b,c) + t$
3. Any condition that would determine $f(0,0,0)$
4. $f(a,b,c) = f(\sqrt{(a^2+b^2)/2},\sqrt{(a^2+b^2)/2},c)$

Likewise, for $f = ab+bc+ca$ one can have the following conditions (really hard):
1. $f$ is symmetric
2. $f(a+t,b,c) = f(a,b,c) + t(b+c)$
3. Any condition that would determine $f(0,0,0)$
4. $f(a,b,c) = f(k,k,c)$ where $k = \sqrt{(a+c)(b+c)}-c$