Appendix

Proofs and notes

March 3, 2025

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Proof: F(n)⋅F(−n−1)=F(−1)F(n)\cdot F(-n - 1) = F(-1)

Where

F(n)=1+(p−q)⋅pn−qnpn+qnF(n) = 1+(p-q) \cdot \frac{p^n - q^n}{p^n + q^n}

and 0<p<10 < p < 1 and p+q=1p + q = 1 we will show that

F(n)⋅F(−n−1)=F(−1)F(n)\cdot F(-n - 1) = F(-1)

We will denote

g(n)=pn−qnpn+qnf(n)=(p−q)⋅g(n)=(2p−1)⋅g(n)\begin{align} g(n) &= \frac{p^n - q^n}{p^n + q^n} \notag \\ f(n) &= (p-q) \cdot g(n) \notag \\ &= (2p - 1) \cdot g(n) \notag \\ \end{align}

so

F(n)=1+f(n)F(n) = 1+f(n)

We also know that

g(n)=g(−n)g(n) = g(-n)

since

g(−n)=p−n−q−np−n+q−n=1pn−1qn1pn+1qn=qnpnqn−pnpnqnqnpnqn+pnpnqn=qn−pnpn+qn=−pn−qnpn+qng(−n)=−g(n)\begin{align} g(-n) &= \frac{p^{-n} - q^{-n}}{p^{-n} + q^{-n}} \notag \\ &= \frac{\frac{1}{p^n} - \frac{1}{q^n}}{\frac{1}{p^n} + \frac{1}{q^n}} \notag \\ &= \frac{\frac{q^n}{p^nq^n} - \frac{p^n}{p^nq^n}}{\frac{q^n}{p^nq^n} + \frac{p^n}{p^nq^n}} \notag \\ &= \frac{q^n - p^n}{p^n + q^n} \notag \\ &= -\frac{p^n - q^n}{p^n + q^n} \notag \\ g(-n) &= -g(n) \notag \end{align}

When p=0.5p = 0.5 this is trivial since F(n)=0∀n∈ZF(n) = 0 \forall n \in \Z.

In the case p≠0.5p \neq 0.5 we begin by showing

F(−1)=1+(p−q)⋅p−1−q−1p−1+q−1=1+(p−(1−p))⋅p−1−(1−p)−1p−1+(1−p)−1=1+(2p−1)⋅1p−11−p1p+11−p=1+(2p−1)⋅1−pp(1−p)−pp(1−p)1−pp(1−p)+pp(1−p)F(−1)=1+(2p−1)⋅(1−2p)\begin{align} F(-1) &= 1+(p-q) \cdot \frac{p^{-1} - q^{-1}}{p^{-1} + q^{-1}} \notag \\ &= 1+(p-(1-p)) \cdot \frac{p^{-1} - (1-p)^{-1}}{p^{-1} + (1-p)^{-1}} \notag \\ &= 1+(2p-1) \cdot \frac{\frac{1}{p} - \frac{1}{1-p}}{\frac{1}{p} + \frac{1}{1-p}} \notag \\ &= 1+(2p-1) \cdot \frac{\frac{1-p}{p(1-p)} - \frac{p}{p(1-p)}}{\frac{1-p}{p(1-p)} + \frac{p}{p(1-p)}} \notag \\ F(-1) &= 1+(2p-1) \cdot (1-2p) \end{align}

Now consider

F(n)⋅F(−n−1)=(1+f(n))⋅(1+f(−n−1))=1+f(n)+f(−n−1)+f(n)⋅f(−n−1)=1+(2p−1)g(n)+(2p−1)g(−n−1)+(2p−1)2g(n)⋅g(−n−1)=1+(2p−1)(g(n)−g(n+1)−(2p−1)g(n)⋅g(n+1))\begin{align} &F(n)\cdot F(-n - 1) = (1+f(n))\cdot (1+f(-n-1)) \notag \\ &= 1+f(n)+f(-n-1)+f(n)\cdot f(-n-1) \notag \\ &= 1+(2p-1)g(n)+(2p-1)g(-n-1)+(2p-1)^2g(n)\cdot g(-n-1) \notag \\ &= 1+(2p-1)(g(n)-g(n+1)-(2p-1)g(n)\cdot g(n+1)) \\ \end{align}

Comparing equations (1)(1) and (2)(2) we see that the equations have equality if we show that

g(n)−g(n+1)−(2p−1)g(n)⋅g(n+1)=1−2pg(n)-g(n+1)-(2p-1)g(n)\cdot g(n+1) = 1-2p

or, negating both sides,

(2p−1)g(n)⋅g(n+1)−g(n)+g(n+1)=2p−1(2p-1)g(n)\cdot g(n+1)-g(n)+g(n+1) = 2p-1

Moving terms we obtain

(2p−1)g(n)⋅g(n+1)−(2p−1)=g(n)−g(n+1)(2p-1)g(n)\cdot g(n+1) - (2p-1) = g(n) - g(n+1)

Where we can factor to obtain

(2p−1)(g(n)⋅g(n+1)−1)=g(n)−g(n+1)(2p-1) (g(n)\cdot g(n+1) - 1) = g(n) - g(n+1)

Then after dividing and moving terms, which we can do because p≠0.5p \neq 0.5, we get

g(n)⋅g(n+1)=1+g(n)−g(n+1)2p−1g(n)\cdot g(n+1) = 1 + \frac{g(n) - g(n+1)}{2p-1}

So we will show that this equality holds.

First, consider

g(n)⋅g(n+1)=pn−qnpn+qn⋅pn+1−qn+1pn+1+qn+1=p2n+1+q2n+1−pnqn+1−pn+1qnp2n+1+q2n+1+pnqn+1+pn+1qn=1+−2pnqn+1−2pn+1qnp2n+1+q2n+1+pnqn+1+pn+1qn\begin{align} g(n)\cdot g(n+1) &= \frac{p^n - q^n}{p^n + q^n} \cdot \frac{p^{n+1} - q^{n+1}}{p^{n+1} + q^{n+1}} \notag \\ &= \frac{p^{2n+1} + q^{2n+1} - p^nq^{n+1} - p^{n+1}q^n}{p^{2n+1} + q^{2n+1} + p^nq^{n+1} + p^{n+1}q^n} \notag \\ &= 1 + \frac{- 2p^nq^{n+1} - 2p^{n+1}q^n} {p^{2n+1} + q^{2n+1} + p^nq^{n+1} + p^{n+1}q^n} \end{align}

Now consider

1+g(n)−g(n+1)2p−1=1+pn−qnpn+qn−pn+1−qn+1pn+1+qn+12p−1=1+(pn−qn)⋅(pn+1+qn+1)(pn+qn)⋅(pn+1+qn+1)−(pn+1−qn+1)⋅(pn+qn)(pn+qn)⋅(pn+1+qn+1)2p−1=1+p2n+1+pnqn+1−pn+1qn−q2n+1−p2n+1−pn+1qn+pnqn+1+q2n+1p2n+1+q2n+1+pnqn+1+pn+1qn2p−1=1+2pnqn+1−2pn+1qnp2n+1+q2n+1+pnqn+1+pn+1qn2p−1=1+2pnqn+1−2pn+1qn2p−1p2n+1+q2n+1+pnqn+1+pn+1qn\begin{align} &1 + \frac{g(n) - g(n+1)}{2p-1} = 1 + \frac{\frac{p^n - q^n}{p^n + q^n} - \frac{p^{n+1} - q^{n+1}}{p^{n+1} + q^{n+1}}}{2p-1} \notag \\ &= 1 + \frac{\frac{(p^n - q^n)\cdot (p^{n+1} + q^{n+1})}{(p^n + q^n)\cdot (p^{n+1} + q^{n+1})} - \frac{(p^{n+1} - q^{n+1})\cdot (p^n + q^n)}{(p^n + q^n)\cdot (p^{n+1} + q^{n+1})}}{2p-1} \notag \\ &= 1 + \frac{\frac{p^{2n+1} + p^nq^{n+1} - p^{n+1}q^n - q^{2n+1} - p^{2n+1} - p^{n+1}q^n + p^nq^{n+1} + q^{2n+1}}{p^{2n+1} + q^{2n+1} + p^nq^{n+1} + p^{n+1}q^n}}{2p-1} \notag \\ &= 1 + \frac{\frac{2p^nq^{n+1} - 2p^{n+1}q^n}{p^{2n+1} + q^{2n+1} + p^nq^{n+1} + p^{n+1}q^n}}{2p-1} \notag \\ &= 1 + \frac{\frac{2p^nq^{n+1} - 2p^{n+1}q^n}{2p-1}}{p^{2n+1} + q^{2n+1} + p^nq^{n+1} + p^{n+1}q^n} \end{align}

Similar to before, by examining equations (3)(3) and (4)(4) we can determine equality if we can show that

−2pnqn+1−2pn+1qn=2pnqn+1−2pn+1qn2p−1- 2p^nq^{n+1} - 2p^{n+1}q^n = \frac{2p^nq^{n+1} - 2p^{n+1}q^n}{2p-1}

Multiplying by 2p−12p-1 gives

−4pn+1qn+1−4pn+2qn+2pnqn+1+2pn+1qn=2pnqn+1−2pn+1qn- 4p^{n+1}q^{n+1} - 4p^{n+2}q^n + 2p^nq^{n+1} + 2p^{n+1}q^n = 2p^nq^{n+1} - 2p^{n+1}q^n

Canceling and moving terms yields

−4pn+1qn+1−4pn+2qn=−4pn+1qn- 4p^{n+1}q^{n+1} - 4p^{n+2}q^n = -4p^{n+1}q^n

After factoring

−4pn+1qn⋅(q+p)=−4pn+1qn- 4p^{n+1}q^n \cdot (q+p) = -4p^{n+1}q^n

Since q=1−pq = 1-p,

q+p=(1−p)+p=1q + p = (1-p)+p = 1

So

−4pn+1qn=−4pn+1qn- 4p^{n+1}q^n = - 4p^{n+1}q^n

and we have shown equality.

Thus

F(n)⋅F(−n−1)=F(−1)F(n)\cdot F(-n - 1) = F(-1)