# Q: Is there a formula to find the Nth term in the Fibonacci sequence?

**Physicist**. Hells yes! It’s

Everything after this is a detailed, math-heavy explanation of where this formula comes from.

The “Fibonacci sequence” is defined as a sequence of numbers

Explicitly, the Fibonacci sequence is: 1, 1, 2, 3, 5, 8, 13, 21, … That is, the recursion says that every term is the sum of the previous two.

You can also talk about “generalized Fibonacci sequences”, where these restrictions and/or the recursion are changed. For example:

Every now and again it’s useful to encode a string of numbers in a “generating function “. For obscure (and unimportant to this post) reasons, you can write many functions as infinitely long polynomials. For example:

You can take the recursion and use it to find a relationship between these three slightly different functions. Here’s a good first guess:

You can write this out, group by powers of x, and then use the recursion. However (if you look at the definitions above for g, xg, and x 2 g), each sum starts at a different value of n. That needs to be dealt with first:

ii) pulling out the n=0,1 terms

iii-iv) grouping by powers of x

vi) f_{0} =1 and f_{1} =1

This doesn’t quite line up, so the guess wasn’t perfect. There should have been an extra “+1″ on the right side of the original equation:

Armed

with this latest equation we can actually solve for g:

Here

So far, using what is known about the Fibonacci sequence, we’ve found a nice closed equation for the generating function (g), which “encodes” the sequence. Hopefully, we can use this fancy new equation to figure out what each f_{n} must be. Again, the function (g) itself does nothing. The only reason it’s around is so that we can look at the coefficients when it’s written in the form of a (Taylor ) polynomial.

Now using “partial fractions” you can pull this one kinda-complicated fraction into two no-so-complicated fractions (that’s where the “

It so happens (and this is the point of the entire excercise) that functions of the form “

So, with that in mind:

But g was originally defined as n in front of each power of x must be the same as these weird things above.

Notice: 1)

Were you so inclined you could take any initial conditions (the f_{0} and f_{1} ) and any recursion (of the form f_{n} = Af_{n-1} +Bf_{n-2} ) and, using the method above, find a closed form for it as well. The only problem you may run into is finding yourself with a polynomial that can’t be factored (x 2 +x-1 had factors, but it needn’t have). If that happens, that’s bad…

Don’t know what to tell you.

Source: www.askamathematician.com

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