Ask Uncle Colin: Is the Fibonacci series witchcraft?

Ask Uncle Colin is a chance to ask your burning, possibly embarrassing, maths questions -- and to show off your skills at coming up with clever acronyms. Send your questions to colin@flyingcoloursmaths.co.uk and Uncle Colin will do what he can.

Dear Uncle Colin,

I read somewhere that if you work out $\frac{1}{999,999,999,998,999,999,999,999}$, you get the Fibonacci sequence. Is that really true? Is there witchcraft at work?

-- Feeling Inspired By Ordinary Numbers; Arithmetic Calculation Can Intimidate!

Hi, FIBONACCI! First, to put your mind at rest, there's no witchcraft at work here -- just a very nice coincidence!

First up, your sources are correct: $\frac{1}{999,999,999,998,999,999,999,999}$ works out to be:

0. 000 000 000 000 
   000 000 000 001
   000 000 000 001
   000 000 000 002
   000 000 000 003
   000 000 000 005
   000 000 000 008
   000 000 000 013
   000 000 000 021...
   

The bigger question, as you asked, is why? It does seem like witchcraft!

I'm going to start from the other side, and use a bit of notation: let $F_i$ be the $i$th Fibonacci number -- so $F_3 = 1$ and $F_9= 21$. Then, this decimal works out to be:

(1) $x = 10^{-12}F_1 + 10^{-24}F_2 + 10^{-36}F_3 + ... + 10^{-12k} F_k + ...$

Multiply this by $10^{12}$ and you get:

(2) $10^{12} x = F_1 + 10^{-12}F_2 + 10^{-24}F_3 + ... + 10^{-12k}F_{k+1} + ...$

One more time, multiply this by $10^{12}$ and get:

(3) $10^{24}x = 10^{12}F_1 + F_2 + 10^{-12}F_3 + 10^{-24}F_4 + ... + 10^{-12k}F_{k+2} + ...$

Gosh. What now? Well, now you do something clever: work out (3) - (2) - (1).

The left hand side is easy, it's just numbers with an $x$ in there: you get $(10^{24} - 10^{12} - 1) x$.

The right hand side is trickier: you have: $10^{12}F_1 + (F_2 - F_1) + 10^{-12}(F_3 - F_2 - F_1) + 10^{-24}(F_4 - F_3 - F_2) + ... + 10^{-12k} (F_{k+2} - F_{k+1} - F_k) + ...$

How does this help? Well, the Fibonacci sequence is defined as $F_a = F_{a-1} + F_{a-2}$, with initial conditions $F_1 = 0$ and $F_2 = 1$. That immediately allows us to make the first few terms simpler, but it's the later terms -- the ones look like that last one, with all the $k$s in -- that look nastier. At least until you realise that the ones you're taking away add up to the first one, so they vanish:

$F_{k+2} - F_{k+1} - F_k = (F_{k+1} + F_k) - F_{k+1} - F_{k} = 0$

All you're left with is:

$(10^{24} - 10^{12} - 1) x = 1$

That number on the left is 999,999,999,998,999,999,999,999!1

A couple of things to notice: firstly, once the Fibonacci numbers get big enough, carries start getting in the way of the nice pattern; secondly, by adding the same number of nines at the start and the end of this number, you can get as far as you like into the Fibonacci sequence without carries becoming a problem!

-- Uncle Colin

Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.

  1. Exclamation, not factorial. []

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