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Carl McTague
mathematician, composer, photographer, fiddler 
A Dynamical Proof That the Ratio of Successive Fibonacci Numbers Approaches the Golden Ratio
In his 1611 essay On the SixCornered Snowflake, Johannes Kepler observed that the ratio of successive Fibonacci numbers $F_n/F_{n1}$ approaches the golden ratio $\tfrac12(1+\sqrt5)$. [Recall that $F_1=F_2=1$ and $F_{n+1}=F_n+F_{n1}$ for $n\ge2$.]
Many proofs have since been found. Here is a quick, dynamical one I thought of which I have been unable to find in the literature.
Proof: Consider the function $f(x)=(1+x)/x$. The sequence of ratios of successive Fibonacci numbers is the orbit of $F_2/F_1=1$ under $f$, i.e. $$ f(F_n/F_{n1})=F_{n+1}/F_n. $$ On the other hand, $f$ has precisely two fixed points, the roots of $x^2x1$, one positive (the golden ratio), the other negative. Since $f'(x)=1/x^2$, $f$ is a decreasing function for $x>0$. And $f(2)=3/2$ while $f(3/2)=5/3$. So $f$ maps the interval $[3/2,2]$ into itself. In fact, $f$ is a contracting mapping on $[3/2,2]$ (with Lipschitz constant $4/9$) since $$ f(x)f(y)=yx/xy\le\tfrac49yx $$ for $x,y\in[3/2,2]$. By the Banach fixed point theorem, then, all orbits in $[3/2,2]$ approach the golden ratio. Although $1$ is not in $[3/2,2]$, $f(1)=2$, so the orbit of $1$ approaches the golden ratio. ∎
This proof can be extended to show that the conclusion holds even if the first two terms $F_1,F_2$ of the Fibonacci sequence are altered, provided their ratio $F_2/F_1$ is not $\tfrac12(1\sqrt5)$ (the negative, unstable fixed point of $f$).
In a sense this explains why the limit is the golden ratio: being a fixed point of $f$ is the defining property of the golden ratio.
Letchworth Lower Falls
How to Recognize Generators for String Bordism
A key result of my paper:
The Cayley Plane and String Bordism, Geometry & Topology 184 (2014), 2045–2078.
unmentioned in the abstract or introduction—and likely of independent interest—is the following characteristicnumbertheoretic criterion for a set to generate the String bordism ring (with 6 inverted).
Theorem 4. A set $S$ generates the ring $\pi_*\mathrm{MO}\langle8\rangle[1/6]$ if:
 For each integer $n>1$, there is an element $M^{4n}$ of $S$ such that for any prime $p>3$: $$\begin{align*} \mathrm{ord}_p \big( \mathrm{s}_n[M^{4n}] \big) = \begin{cases} 1 & \text{if $2n=p^i1$ or $2n=p^i+p^j$ for some integers $0 \le i \le j$} \\ 0 & \text{otherwise} \end{cases} \end{align*}$$ where $\mathrm{s}_n[M]$ is the characteristic number corresponding to the $n$th power sum symmetric polynomial $\sum x_i^n$ in the Pontrjagin roots of the tangent bundle of $M$, and $\mathrm{ord}_p()$ its $p$adic order, i.e. the heighest power of $p$ which divides it.
 For each prime $p>3$ and each pair of integers $0<i<j$, there is an element $N^{2(p^i+p^j)}$ of $S$ such that: $$\begin{align*} \mathrm{s}_{(p^i+p^j)/2}[N^{2(p^i+p^j)}]&=0 \\ \mathrm{s}_{(p^i+1)/2,(p^j1)/2}[N^{2(p^i+p^j)}] &\not\equiv 0 \mod p^2 \end{align*}$$ where $\mathrm{s}_{m,n}[N]$ is the characteristic number corresponding to the symmetric polynomial $\sum_{i\ne j} x_i^mx_j^n$ in the Pontrjagin roots of the tangent bundle of $N$.
This is a consequence of Hovey’s calculation of:
The homotopy of $\mathrm{MString}$ and $\mathrm{MU}\langle6\rangle$ at large primes, Algebr. Geom. Topol. 8 (2008), 2401–2414.
which itself builds on Hopfringtheoretic work of Ravenel and Wilson. It should be compared with the analogous result for the oriented bordism ring (with 2 inverted):
Theorem (Novikov, cf. Stong p. 180). A sequence $\{M^{4n}\}_{n\ge1}$ generates the ring $\pi_*\mathrm{MSO}[1/2]$ if and only if:
 For any integer $n>0$ and any odd prime $p$: $$\begin{align*} \mathrm{ord}_p \big( \mathrm{s}_n[M^{4n}] \big) = \begin{cases} 1 & \text{if $2n=p^i1$ for some integer $i>0$} \\ 0 & \text{otherwise} \end{cases} \end{align*}$$
References

Robert Stong, Notes on Cobordism Theory, Princeton University Press, 1968.

John Milnor, Characteristic Classes, Princeton University Press, 1974.
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