# How Would Martin Gardner Prove It?

Someone recently asked me where I get enough ideas for blog posts that I can keep up such a ‘prolific’ schedule. (Two posts a week? Prolific? If you say so.) The answer is straightforward:

One reliable source of interesting stuff is @WWMGT - What Would Martin Gardner Tweet? in real life. For example, here’s a recent tweet that had me scratching my head for some time:

arccot(1) = arccot(2) + arccot(5) + arccot(13) + arccot(34) + ..... using every other term of the Fibanacci series (D. H. Lehmer)

— Martin Gardner tweet (@WWMGT) April 18, 2017

I don’t normally have much time for Fibonacci stuff - it’s wildly overplayed in most cases - but this one caught me. You might like to have a go at proving it yourself before reading on.

I want to show that:

$\arccot(1) = \arccot(2) + \arccot(5) + \arccot(13) + … \arccot(F_{2n+1}) + …$,

where $F_n$ is the $n$th Fibonacci number (numbered such that $F_1 = F_2 = 1$).

To do this, I’ll need to use two identities, one trigonometric, and one Fibonaccic:

- The arccotangent identity: $\arccot(A)-\arccot(B) = \arccot\left(\frac{AB+1}{B-A}\right)$;
- Cassini’s identity: $F_{2n-1}F_{2n+1}=F_{2n}^2 + 1$

I’m going to use a recursive argument here ((It feels a little cowboy, and I’m sure I’ll be pulled up on the details)), and show that the arccotangent of an even-numbered Fibonacci term is the same as the sum of the arccotangents of the following pair - the second of which is even-numbered and can be split up in the same way.

That is to say, if I can show that $\arccot(F_{2n})=\arccot(F_{2n+1})+\arccot(F_{2n+2})$ - or equivalently, $\arccot(F_{2n+2})=\arccot(F_{2n})-\arccot(F_{2n+1})$ for $n=1,2,3,…$, I’m away.

The base case, for $n=1$, is straightforward - I’m not sure we need it for the proof, but it’s instructive all the same: I want to show that $\arccot({F_4})=\arccot({F_2})-\arccot({F_3})$, or that $\arccot(3)=\arccot(1)-\arccot(2)$.

I mean, I *know* that’s true, but let’s do it properly using the identity: $\arccot(1)-\arccot(2)=\arccot\left(\frac{(1)(2)+1}{2-1}\right) = \arccot(3)$. That’s reassuring.

Let’s look at the general case, now: does $\arccot(F_{2n})-\arccot(F_{2n+1})$ work out to be $\arccot(F_{2n+2})$ for all $n$?

We can use the arccotangent identity as before: we get $\arccot(F_{2n})-\arccot(F_{2n+1}) = \arccot\left(\frac{F_{2n}F_{2n+1}+1}{F_{2n+1}-F_{2n}}\right)$.

That doesn’t look promising to begin with, but it turns out ok. The bottom can be easily simplified to $F_{2n-1}$, using the definition of the Fibonacci sequence. Similarly, $F_{2n+1}=F_{2n-1}F_{2n}$, so we have: $\arccot\left(\frac{F_{2n}\left(F_{2n-1}+F_{2n}\right)+1}{F_{2n-1}}\right)$

Let’s just look at the top of the inner fraction: $F_{2n}\left(F_{2n-1}+F_{2n}\right)+1$, which expands to $F_{2n}F_{2n-1} + F_{2n}^2 + 1$. Those last two terms ring a bell, don’t they? They’re the right hand side of Cassini’s identity, so we can replace them with $F_{2n-1}F_{2n+1}$, which makes the top $F_{2n}F_{2n-1} + F_{2n-1}F_{2n+1}$, or $F_{2n-1}(F_{2n}+F_{2n+1})$.

So now we have: $\arccot\left(\frac{F_{2n-1}\left(F_{2n+1}+F_{2n}\right)}{F_{2n-1}}\right)$ - and there’s a factor of $F_{2n-1}$ top and bottom to remove, leaving: $\arccot\left(F_{2n+1}+F_{2n}\right)$.

Aha! Those are Fibonacci terms, which sum to $F_{2n+2}$ as required!

So, we know that:

- $\arccot(F_2)=\arccot(F_3)+\arccot(F_4)$; and
- $\arccot(F{2n})=\arccot(F_{2n+1})+\arccot(F_{2n+2})$ for $n=1,2,3,…$

In particular, $\arccot(F_4)=\arccot(F_5)+\arccot(F_6)$, so $\arccot(F_2)=\arccot(F_3)+\arccot(F_5)+\arccot(F_6)$; similarly, $\arccot(F_6)$ can be replaced with $\arccot(F_7)+\arccot(F_8)$ - and so on *ad infinitum*, leaving only the odd Fibonacci terms.

It’s a lovely result, and a proof I’m sure can be tightened up! Suggestions, as always, are more than welcome.