Binet Revisited

In this section we will prove that Binet's Formula holds true. Binet's Formula states that:

We know that the two numbers inside the smaller sets of parentheses are the roots of the quadratic equation

We have a name for the first value, . We need a name now for the other value, the value in the rightmost set of small parentheses. Let's call this value tau (), and rewrite Binet's Formula in terms of and .

Since we know that is a root of the afore-shown quadratic equation, we can plug it in for y in the equation itself.

We rearrange this equation to obtain:

Now we know what squared is, what if we want to know what cubed is? Or further powers of ?

If we start by multiplying our above equation by , and substituting the above equation at the appropriate time, we can find some values.

We can then find the next power by multiplying this new equation by .

We continue in this fashion for two more powers of .

So we have figured out the following powers of :

What do you notice about the coefficients of these equations? They are all Fibonacci numbers. In fact, they are all consecutive Fibonacci numbers. We can generalize this observation by the equation:

The only reason this works is because is a root of that quadratic equation. We also know that is also a root of that same quadratic equation. Therefore, we can write a similar equation for :

Now subtract the equation in from the equation in to obtain the following:

If we factor out F(a) from this equation, we are left with:

Solving for F(a), we get the following:

Since both and are constants, we can find a numerical value for the denominator of this equation. Looking back at the values for and , it should be evident that the difference between the two is the square root of 5. Therefore:

We can factor the denominator outside the numerator to obtain this familiar equation:

Do you recognize this last equation? It's the rewritten version of Binet's Formula. Therefore, we have proved that Binet's Formula is true.