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Suppose we have the following function \begin{align} f(n)= \frac{ (1-\alpha) a^{n+1} e^{-a}+\alpha b^{n+1} e^{-b} }{ (1-\alpha) a^{n} e^{-a}+\alpha b^{n} e^{-b} }, \text{ where }n=0,1,2,3,4, \ldots \end{align} for some parameters $\alpha \in (0,1)$, $a>0$ and $b>0$ where $a\neq b$.

Can we show the function $f$ is always unique for any choice of $\alpha,a$ and $b$? In other words, if $f'$ corresponds to the triplet $(\alpha',a',b')$, then $f=f'$ if and only if $(\alpha,a,b)=(\alpha',a',b')$.

I tried several examples and think this is true (of course I can be wrong)

What I noticed that think might be helpful is that if we define \begin{align} g(x;a)= e^{-ax} \end{align} then \begin{align} f(n)= - \frac{(1-\alpha) g^{(n+1)}(1;a) + \alpha g^{(n+1)}(1;b)}{(1-\alpha) g^{(n)}(1;a) + \alpha g^{(n)}(1;b)} \end{align} where $g^{(n)}$ is the $n$-th derivative with respect to $x$.

Correction: In the solution, I would like to consider $(\alpha,a,b)$ and $(1-\alpha,b,a)$ to be the same triplet. The reason is that this result in the change of the order of summation the denominator and numerator stay the same.

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I'm afraid no, we can't.

If $f$ is the function for a triplet $(\alpha,a,b)$ with $\alpha \ne \frac12$ or $a \ne b$, then the triplet $(1-\alpha,b,a)$ is a different triplet, but $f'=f$ for the corresponding function $f'$.

This can be seen "almost" immediately: If we swap $\alpha$ with $1-\alpha$ and $a$ with $b$, each two summands in both the numerator and denominator just change places.

Also, if $a=b$ then for all $\alpha \in [0,1]$ we have $f(n)=a$ for all $n=0,1,2,3,4, \ldots$

Added:

According to your definition of $\displaystyle{f=f_{\alpha,a,b}}$ together with the restrictions $a \ne b$ and $a \in ]0,1[$, the following is true:

$\displaystyle{f_{\alpha,a,b}=f_{\alpha',a',b'} \Leftrightarrow (\alpha,a,b) = (\alpha',a',b) \lor (\alpha,a,b) = (1-\alpha',b',a')}$.

Note that if $a=b$, the function does not depend on $\alpha$, and if $\alpha=1$ (or $\alpha=0$), the function does not depend on $a$ (or $b$).

Sketch of proof:

With $p:=a+b$ and $q:=ab$, we can prove (by complete induction) that $f(n+1) = p- \frac{q}{f(n)}$.

$\text{Also (seen via calculation), if }a \ne b\text{, then }\alpha\text{ is uniquely determined by }a\text{, }b\text{ and }f(0)\tag{*}$

Now, if we have $f=f'$ with $\displaystyle{f'=f_{\alpha',a',b'}}$, we define $p':=a'+b'$ and $q'=a'b'$ and we find that for $n \ge 0$:

$$p'+\frac{q'}{f(n)}=p+\frac{q}{f(n)} \Leftrightarrow (p'-p)f(n) = q-q'$$

So, $f$ must be constant or $p=p' \land q=q'$. If the latter is true, $a$ and $b$ are the solutions of $x^2-px+q=0$ and so are $a'$ and $b'$, so $\{a,b\}= \{a',b'\}$ which results in $(\alpha,a,b) = (\alpha',a',b) \lor (\alpha,a,b) = (1-\alpha',b',a')$ (because of (*)).

Otherwise, if $f$ is constant, then $f(n)=c$ for all $n$ and so

$c=p-\frac{q}{c} \Leftrightarrow c^2-pc+q=0 \Leftrightarrow c \in \{a,b\} \Leftrightarrow \alpha \in \{0,1\}$ (again because of (*)).

Remark:

The series $F(n)$ defined by the denominator (and numerator) of $f$, $$F(n):= (1-\alpha)a^n e^{-a}+\alpha b^n e^{-b}$$ is "Fibonacci-like" and obeys to the recursion formula $F(n+1)=p \cdot F(n) - q \cdot F(n-1)$ (with $p=a+b$ and $q=ab$, which does not depend on $\alpha$), and we have $f(n)=\frac{F(n+1)}{F(n)}$.

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  • $\begingroup$ Thank you for your answer. Even though, I didn't mention this. I would like to avoid this case. This is my fault I didn't really formulate the problem in the right? $\endgroup$ – Boby Feb 14 at 18:33
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    $\begingroup$ All right, but you should also exclude the other case I mentioned, because if $a=b$, then $f$ does not depend on $\alpha$. $\endgroup$ – Wolfgang Kais Feb 14 at 19:16
  • $\begingroup$ Done. I made the changes. Please let me know how else I can improve the question. $\endgroup$ – Boby Feb 14 at 19:30
  • $\begingroup$ There's one more thing before the assertion becomes true: $\alpha \notin \{0,1\}$, because otherwise $f$ would be constantly $a$ or $b$ (and $f$ respectively independant of the other). I added a sketch of proof to my answer. $\endgroup$ – Wolfgang Kais Feb 15 at 0:43
  • $\begingroup$ Thanks. I think I am able to follow your proof. The only thing that I can do my self is the recursion. Can please add a bit more details to it? $\endgroup$ – Boby Feb 15 at 13:48

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