# Show $f(n)= \frac{ (1-\alpha) a^{n+1} e^{-a}+\alpha b^{n+1} e^{-b} }{ (1-\alpha) a^{n} e^{-a}+\alpha b^{n} e^{-b} }$ is unique for $(\alpha,a,b)$

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.

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 $$n=0,1,2,3,4, \ldots$$

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 $$f(n)=\frac{F(n+1)}{F(n)}$$.

• 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? – Boby Feb 14 at 18:33
• All right, but you should also exclude the other case I mentioned, because if $a=b$, then $f$ does not depend on $\alpha$. – Wolfgang Kais Feb 14 at 19:16
• Done. I made the changes. Please let me know how else I can improve the question. – Boby Feb 14 at 19:30
• 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. – Wolfgang Kais Feb 15 at 0:43
• 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? – Boby Feb 15 at 13:48