Almost everyone has heard of this sequence: . It is named after Leonardo of Pisa who introduced it to the western world in one of the most influential books ever published in mathematics – Liber Abaci. This book introduced Europe to the Hindu numerals 0 through 9, the word zero, the notion of an algorithm and the subject of algebra.

The beauty of the Fibonacci sequence and the golden ratio (which is intimately connected to it) lies in that they are not just another mathematical construct, but occur throughout the nature.

Have you ever taken a look at a pine cone and noticed that the scales of the cone are in spirals? Have you ever counted the spiral in the clockwise and counterclockwise directions? I would be surprised if you have … but the counts turn out to be 5 and 8 (or 8 and 13 for bigger cones).

How about these? Care to count the petals?

… intriguing stuff indeed.

So, once again, I was reading through problems on Yahoo Answers. Most of the questions “reduce” (lol) to plugging numbers into well known equations, or using a calculator such as TI-89, or Mathematics, Maple, or even Google – basically, what engineers do. *Yeah, you “heard” me right!* It takes some effort to found a meaningful question that actually requires some knowledge and skills – *see the difference between a mathematician and an engineer now?* That said, here’s a one I couldn’t resist:

**Problem**: Find the term in the following sequence

There is a more “elegant” way to tackle this problem and I may write about in the future (maybe quite soon actually), but for now I’ll ignore matrices, diagonalization and eigenspaces *(although the reason why the following solution gives the correct result is tightly connected to linear algebra)* and focus, instead, on the recursive nature of the Fibonacci sequence.

**Solution**: To find the term, “all” that is needed is to find the closed form for the n-th terms of the sequence . In general, a Fibonacci sequence is given by

and any particular sequence is fully determined by the initial two terms

and

In our example, we have

and

To find the closed form, let’s assume

Then using our recursive formula we get

and so

Solving the quadratic from gives

Using this result, we can write the closed form formula for the n-th term of our sequence as

where and are constant parameters determined by our initial conditions,

and

We can calculate these values by solving the following system of equations

Plugging in known values gives us

Simplifying turns the above into

Now we need to solve and for and . Although I said I was going to leave matrices alone, this is a perfect situation to use them. I prefer to use the procedure described below for a system of two equations with coefficients such as these. Substitution or elimination method would, of course, work, but they involves messy arithmetic which can be so easily error prone in situations such as this one.

We have the following system:

In general, to solve the equation

we use

which turns into

Hence, to solve we need to find the inverse of our matrix. This is fairly simple for a 2 by 2 matrix … Let

then

and so

Our equation then becomes

which gives us

Finally, by plugging and into we get

We can now “easily” compute which is given by

Google returns which is just an approximation, but now we have a formula for the precise answer.

If you wish to check for yourself that out formula is indeed correct, click on the terms below to see the computation by Google:

On a side note – Raising a number to the power of 386 may seem like a long computation but it really requires only 10 multiplications! Don’t believe me?

Consider this:

To calculate

we use the fact that

…

As you can see, in every step (multiplication) we use the result from the previous one and so it takes only 8 steps (multiplications) to calculate

Finally, in the process of calculating

we get

and

as a bonus, and so to calculate

we now only need to use two more multiplications

This idea can be generalized to any

The (maximal) number of multiplications required to compute

is given by

To put this result in a little bit of perspective, consider

This number is the currently largest known (Mersenne) prime. It has 12,978,189 digits! But and so

It follows that we can calculate

using at most 51 multiplications! Incredible!