# Annoying Mclaurin Series

Suppose that we’re given the function , find the ordinary generating function associated with it in the form of . Furthermore, find/compute .

We can employ Taylor’s theorem, but to my knowledge, there’s no easy way to construct the derivatives of [that’s left to the more ambitious reader ;)] Instead, we focus on classical algebraic techniques to carry us through the day.

Notice that

recall that all terms of the form , so this becomes

Okay, let’s suppose that , then the above tells us that

If we compute the first few out, we will see that it looks like

Why, this is the Fibonacci sequence! It seems that for the first few terms we looked at, obeys the fibonacci recurrence

#### Lemma (Fibonacci Sequence)

For ,

Proof: Substituting the definition of in, we have

Finally, recall that , so

so we would expect

If we want a closed form, we can find that, letting

then

Finally, we can compute

##### Posted By Lee

Woosh

• flood

heh well that’s an interesting… who would have expected to get fibonacci from that

and there is a (sorry to say this again) trivially easy way to get the derivatives. it’s just partial fractions and I believe you recover binet’s formula if you add up the taylor expansions of the two fractions.

• flood

by the way, try this
http://mathbin.net/441855

• I’ll look at this after lunch :P

You should start a blog/facebook, this trend seems to be heavily one-sided. On the other hand it gives me an (and the only) incentive to post

• Haha, I feel like you of all people are probably one of the very few that would be allowed to say trivial

• on the partial function thing, factoring gives B(z) = a/(z + phi) + b/(z + psi) with a + b = 1, aphi+bpsi = 1, and the solution would just be the addition of the [z^n] of a/(z + phi) and b/(z + psi) which is just – (aphi^{1-n} + bpsi^{1-n})