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Let $f(x),h(x)$ be two differentiate on $\mathbb{R}$ functions, $f(0)=h(0)=1$. Solve the functional equation $$ q \, \frac{f(x+1)}{f(x)}=\frac{h(x+1)}{h(x)}, $$ here $q$ is a constant.

For $q>0$ it is easy to find a solution: $f(x)=e^{ax}, h(x)=e^{bx}$ for some suitable $a,b.$

Questions. Are there another solutions for $q>0$? What about the case $q < 0?$

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  • $\begingroup$ Hint: You have also $q f(1)=h(1)$ so $q=\frac{h(1)}{f(1)}$ if $f(1)$ is non zero. $\endgroup$ – Khadija Mbarki Jul 20 '15 at 16:24
  • $\begingroup$ For any function $g(x)$ you can find infinitely many functions such that $f(x+1)=g(x)f(x)$ $\endgroup$ – Michael Galuza Jul 20 '15 at 16:25
  • $\begingroup$ Yes, @MichaelGaluza, I gave a full derivation of the expression. Probably not exactly what the OP was looking for, but I think it's necessary to mention, for the sake of completeness $\endgroup$ – frogeyedpeas Jul 20 '15 at 17:06
  • $\begingroup$ @frogeyedpeas, My idea was: you can define $f(x)$ in any way on $[0, 1)$, for example, and continue $f(x)$ to $[1, 2)$ using $f(x+1)=f(x)g(x)$ $\endgroup$ – Michael Galuza Jul 20 '15 at 17:10
  • $\begingroup$ That's correct and it's basically all I am doing as well, but for an exact symbolic closed form expression of that procedure one can use the techniques below (there might be a more elegant formulation out there but not that I personally am aware of). $\endgroup$ – frogeyedpeas Jul 20 '15 at 17:12
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This can be transformed into a standard problem in linear finite differences. First we multiply both sides by $f(x)$ and divide by $q$ to arrive at

$$ f(x+1) = \frac{1}{q} \frac{h(x+1)}{h(x)} f(x) $$

Then subtract $f(x)$ to yield

$$ f(x+1) - f(x) = \left( \frac{h(x+1) - qh(x)}{qh(x)} \right)f(x) $$

Which can be written as

$$ D_{1,x} f+ \left( \frac{qh(x) - h(x+1)}{qh(x)} \right)f(x) = 0 $$

Whereas $$D_{h,x}f = \frac{f(x+h) - f(x)}{h}$$ the use of $h=1$ above naturally gives us $f(x+1) - f(x)$ as desired.

So given any $h(x)$, I mean any! we can find a find an f that satisfies above. The full proof requires the use of the theory of finite differences. First observe

$$ D_{1,x}[2^x] = 2^{x+1} - 2^x = 2^x$$

Furthermore we can generalize that result to,

$$ D_{1,x}[2^{g(x)}] = 2^{g(x+1)} - 2^{g(x)} = 2^{g(x)}(2^{g(x+1)-g(x)} - 1) = 2^{g(x)} (2^{D_{1,x}[g(x)]} - 1) $$

Furthermore, we create a product rule of sorts,

$$ D_{1,x}[f(x)g(x)] = D_{1,x}[f(x)] g(x) + f(x)D_{1,x}[g(x)] + D_{1,x}[f(x)]D_{1,x}[g(x)] $$

To verify that last product rule, just expand each $$D_{1,x}$$ term into the defintion from above, and just algebraically simplify. One final tool to remark on is the idea of

$$D_{h,x}^{-1}[f] $$

Which is simply the function $g(x)$ such that $D_{h,x}[g] = f$$ If you want a more rigorous treatment of how to define it, please mention in comments, I can give more intuition into it. For the remainder of this we will assume that it is a well defined operator, that can be taken of its argument.

From here we wish to solve for $f(x)$

$$ D_{1,x}[f(x)]+ \left( \frac{qh(x) - h(x+1)}{qh(x)} \right)f(x) = 0 $$

We will utilize a pair of integration factors. Let us attempt to find two functions $u_1(x), u_2(x)$ such that

$$ D_{1,x}[u_1(x)f(x)] = u_2(x) \left( D_{1,x}[f(x)]+ \left( \frac{qh(x) - h(x+1)}{qh(x)} \right)f(x) \right) $$

After distributing we arrive at

$$ D_{1,x}[u_1(x)f(x)] = D_{1,x}[u_1(x)]f(x) + u_1(x)D_{1,x}[f(x)] + D_{1,x}[u_1(x)]D_{1,x}[f(x)] $$

$$ u_2(x) \left( D_{1,x}[f(x)]+ \left( \frac{qh(x) - h(x+1)}{qh(x)} \right)f(x) \right) = $$ $$ u_2(x) D_{1,x}[f(x)] + u_2(x) \left( \frac{qh(x) - h(x+1)}{qh(x)} \right)f(x) $$

So we can equate terms

$$ u_1(x)D_{1,x}[f(x)] + D_{1,x}[u_1(x)]D_{1,x}[f(x)] = u_2(x) D_{1,x}[f(x)] $$ $$ D_{1,x}[u_1(x)]f(x) = u_2(x) \left( \frac{qh(x) - h(x+1)}{qh(x)} \right)f(x)$$

Then divide through by common terms to find:

$$ u_1(x) + D_{1,x}[u_1(x)] = u_2(x) $$ $$ D_{1,x}[u_1(x)] = u_2(x) \left( \frac{qh(x) - h(x+1)}{qh(x)} \right)$$

Subtract the second equation from the first to find

$$ u_1(x) = \left(1 - \frac{qh(x) - h(x+1)}{qh(x)} \right) u_2(x) $$

Now look at the second equation in the system of 2. We assume that $u_2(x) = 2^{E(x)} $ then it follows that

$$ D_{1,x}[u_1(x)] = D_{1,x}\left[1 - \frac{qh(x) - h(x+1)}{qh(x)} \right]u_2(x) + \left(1 - \frac{qh(x) - h(x+1)}{qh(x)} \right)D_{1,x}[u_2(x)] = u_2(x) \left( \frac{qh(x) - h(x+1)}{qh(x)} \right) $$

Which means

$$ D_{1,x}\left[1 - \frac{qh(x) - h(x+1)}{qh(x)} \right] + \left(1 - \frac{qh(x) - h(x+1)}{qh(x)} \right)2^{D_{1,x}[E(x)]} = \left( \frac{qh(x) - h(x+1)}{qh(x)} \right) $$

Thus it follows,

$$ 2^{D_{1,x}[E(x)]} = \frac{\left( \frac{qh(x) - h(x+1)}{qh(x)} \right) - D_{1,x}\left[1 - \frac{qh(x) - h(x+1)}{qh(x)} \right]}{\left(1 - \frac{qh(x) - h(x+1)}{qh(x)} \right)} $$

Giving us

$$ E(x) = D_{1,x}^{-1} \left[ \log_2 \left(\frac{\left( \frac{qh(x) - h(x+1)}{qh(x)} \right) - D_{1,x}\left[1 - \frac{qh(x) - h(x+1)}{qh(x)} \right]}{\left(1 - \frac{qh(x) - h(x+1)}{qh(x)} \right)} \right) \right] $$

Meaning

$$ u_2(x) = 2^{D_{1,x}^{-1} \left[ \log_2 \left(\frac{\left( \frac{qh(x) - h(x+1)}{qh(x)} \right) - D_{1,x}\left[1 - \frac{qh(x) - h(x+1)}{qh(x)} \right]}{\left(1 - \frac{qh(x) - h(x+1)}{qh(x)} \right)} \right) \right]}$$

$$ u_1(x) = \left(1 - \frac{qh(x) - h(x+1)}{qh(x)} \right)2^{D_{1,x}^{-1} \left[ \log_2 \left(\frac{\left( \frac{qh(x) - h(x+1)}{qh(x)} \right) - D_{1,x}\left[1 - \frac{qh(x) - h(x+1)}{qh(x)} \right]}{\left(1 - \frac{qh(x) - h(x+1)}{qh(x)} \right)} \right) \right]} $$

And since

$$ D_{1,x}[u_1(x)f(x)] = u_2(x) \left( D_{1,x}[f(x)]+ \left( \frac{qh(x) - h(x+1)}{qh(x)} \right)f(x) \right) = 0$$

It follows that

$$ u_1(x)f(x) = C$$

(C is the finite difference integration constant) and from here

$$ f(x) = \frac{C}{u_1(x)} $$

Giving us

$$ f(x) = \frac{C}{1 - \frac{qh(x) - h(x+1)}{qh(x)})}2^{-D_{1,x}^{-1} \left[ \log_2 \left(\frac{\left( \frac{qh(x) - h(x+1)}{qh(x)} \right) - D_{1,x}\left[1 - \frac{qh(x) - h(x+1)}{qh(x)} \right]}{\left(1 - \frac{qh(x) - h(x+1)}{qh(x)} \right)} \right) \right]} $$

So in short, name your h(x), ANY h(x), and that freakish clusterf--- of an expression will give you the f(x) that satisfies your problem.

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Let $g(x)=e^{ax}\frac{f(x)}{h(x)}$.

Then the functional equation Yields: $$\frac{q}{e^{a}}g(x+1)=g(x)$$

Pick $a$ so that $e^a=|q|$. Then

$$g(x+1)= sgn(q) g(x)$$

Now, lets work backwards.

If $q>0$ Pick any function $g$ which is nonvanishing, differentiable and periodic with period one. There are many such functions.

Let $f(x)$ be any differentiable non-vanishing function.

Then $$h(x)=e^{x \ln q} \frac{f(x)}{g(x)} =q^x \frac{f(x)}{g(x)}$$ satisfies the given relation, and we seen above that any solution is of this form.

If $q<0$ Pick any function $g$ which is nonvanishing, differentiable and $g(x+1)=-g(x)$. There are many such functions.

Let $f(x)$ be any differentiable non-vanishing function.

Then $$h(x)=e^{x \ln q} \frac{f(x)}{g(x)} =q^x \frac{f(x)}{g(x)}$$ satisfies the given relation, and we seen above that any solution is of this form.

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  • $\begingroup$ what is the general solution of the equation $g(x+1)=g(x)?$ $\endgroup$ – Leox Jul 20 '15 at 16:34
  • $\begingroup$ @Leox The differentiable solutions to this equation are exactly the same functions as the set of differentiable functions on the circle $\mathbb R/\mathbb Z$. This is a simple characterization, but also pretty vague, I doubt there is any better description. $\endgroup$ – N. S. Jul 20 '15 at 16:38

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