# Show that the $n$-th Fibonacci number is given by $\frac{\cosh na}{\cosh a}$ or $\frac{\sinh na}{\cosh a}$, where $\sinh a=1/2$

This question is taken from book: Advanced Calculus: An Introduction to Classical Analysis, by Louis Brand. The book is concerned with introductory real analysis.

I request to help find the solution.

Show that the general term of Fibonacci sequence $$1,1,2,3,5,\cdots$$, is given by :
$$f_n = \frac{\cosh\, n\alpha}{\cosh\, \alpha}$$ ($$n$$ odd),
$$f_n = \frac{\sinh\, n\alpha}{\cosh\, \alpha}$$ ($$n$$ even),
where $$\sinh \alpha= \frac 12$$, and that $$\lim \frac{f_{n+1}}{f_n} = e^\alpha.$$

As given here:
The hyperbolic functions $$\sinh$$, $$\cosh$$ are given by: $$\sinh\,z = \frac{e^z - e^{-z}}{2}, \cosh\,z = \frac{e^z + e^{-z}}{2}$$ where $$\,z= x+iy\,\,$$ is a complex variable.
These functions satisfy the identities:
$$\cosh^2\,z-\sinh^2\,z = 1,\,\, \cosh\,iy = \cosh\,y,\,\, \sinh\,iy = i\,\sinh\,y$$

If $$\sinh\,z = \frac{e^z - e^{-z}}{2} = \frac 12$$, then $$\cosh^2\,z-\sinh^2\,z = 1\,\,$$ gives $$\cosh^2\,z=\frac 54$$

This yields nothing, so consider opposite approach of taking help by another means of generating Fibonacci sequence terms.

Say, the polynomial $$f(x) = x+1$$ yields terms for successive values for $$x \in \mathbb{N}$$ as $$2,3,\dots$$.

So, took help from paper here that concerns with fibonacci polynomials, with text portion from page 1 copied below:

The Fibonacci polynomials $$\{F_n (x)\}$$ are defined by

(1.1) $$F_1(x) = 1, F_2(x) = x$$, and $$F_{n+1}(x) = xF_n(x) + F_{n-1}x$$.

Notice that, when $$x = 1, F_n(1) = F_n$$ , the $$n$$ Fibonacci number. It is easy to verify that the relation
(1.2) $$F_{-n}(x) = (-1)^{n+1} F_n(x)$$ extends the definition of Fibonacci polynomials to all integral subscripts.

But, this also doesn't help.

• Did you try a proof by induction? May 26, 2019 at 7:59
• @MartinR Never thought that way. May 26, 2019 at 8:00
• Incidentally, there is a general formula for the Fibonacci numbers given here: math.stackexchange.com/questions/65011/…. May 26, 2019 at 8:00
• @MartinR Request elaboration as am confused that for given $\sinh \alpha= \frac 12$, how to apply induction. Better post an answer. May 26, 2019 at 8:09
• @OscarLanzi Thanks. But, an answer towards finding the limit would be too much helpful. May 26, 2019 at 8:31

Hint for induction. By the addition formula for $$\cosh$$ (see wiki), $$\cosh((n+1)\alpha)=\cosh(n\alpha)\cosh(\alpha)+\sinh(n\alpha)\sinh(\alpha)$$ and $$\cosh((n-1)\alpha)=\cosh(n\alpha)\cosh(\alpha)-\sinh(n\alpha)\sinh(\alpha).$$ Hence $$\cosh((n+1)\alpha)-\cosh((n-1)\alpha)=2\sinh(n\alpha)\sinh(\alpha)$$ and, after dividing by $$\cos(\alpha)$$, if $$n$$ is even we get $$f_{n+1}-f_{n-1}=2f_{n}\sinh(\alpha)=f_n\implies f_{n+1}=f_n+f_{n-1}.$$ In a similar way, by using the addition formula for $$\sinh$$, we verify that the same recurrence holds when $$n$$ is odd.
As regards the limit you may use the unified formula $$f_n=\frac{e^{n\alpha} - (-1)^ne^{-n\alpha}}{e^{\alpha}+e^{-\alpha}}$$ Since $$\alpha>0$$, it follows that, as $$n\to \infty$$, $$\frac{f_{n+1}}{f_n}=\frac{e^{(n+1)\alpha} - (-1)^{n+1}e^{-(n+1)\alpha}}{e^{n\alpha} - (-1)^ne^{-n\alpha}}\to e^{\alpha}.$$
P.S. Note that $$e^{\alpha}=\frac{1+\sqrt{5}}{2}=\varphi$$ is the Golden Ratio and therefore the above unified formula can be written as $$f_n=\frac{\varphi^n-(-1)^n\varphi^{-n}}{\varphi+\varphi^{-1}}=\frac{\varphi^n-(-\varphi)^{-n}}{\sqrt{5}}$$ which is the usual closed-form expression for the Fibonacci numbers.