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My last three posts:

- The nth term of a Fibonacci series.
- More on that cool Fibonacci formula.
- Guessing the nth Fibonacci number formula.

were all about a cool formula for the n-th term of the Fibonacci series. Here’s the final chapter of the story of that play. A recap:

## Definition 1.1: Fibonacci series.

\begin{equation*}

F_n = F_{n-2} + F_{n-1}.

\end{equation*}

## Theorem 1.1: Nth term of the Fibonacci series.

F_n = \frac{ \lr{ 1 + \sqrt{5} }^n – \lr{ 1 – \sqrt{5} }^n }{ 2^n \sqrt{5} }.

\end{equation*}

## Deriving the nth Fibonacci formula.

There was a particularly unsatisfactory aspect of the guess that we made in the last post. In particular, we didn’t have any reason to guess the form of that solution, except for the fact that we already knew the answer. Now we will attempt to attack this in a more systematic fashion, so that each step along the way seems logical. First, we need to put a couple goodies in our toolbox.

## Definition 1.1: Discrete sum.

\begin{equation}\label{eqn:fibonacciblog:1540}

F_n = \sum_{k = a}^n G_k + C,

\end{equation}

where \( C \) is an arbitrary boundary value constant.

## Definition 1.2: Difference operators.

\begin{equation}\label{eqn:fibonacciblog:1560}

\Delta X_n = X_n – X_{n-1}.

\end{equation}

The difference operator is a discrete analogue of a differential operator. It is also possible to define a (forward) difference operator as \( \Delta X_n = X_{n+1} – X_{n} \), but the choice is arbitary, and we can find the same results either way.

## Lemma 1.1: Antidifference of discrete sum.

\begin{equation}\label{eqn:fibonacciblog:1580}

\Delta F_n = G_n.

\end{equation}

### Start proof:

\begin{equation}\label{eqn:fibonacci:960}

\Delta F_n =

\sum_{k = a}^n G_k + C

– \lr{ \sum_{k = a}^{n-1} G_k + C }

= G_n.

\end{equation}

### End proof.

Computing differences is pretty easy. What we want to do is the inverse operation (analogous to integration), where we find a closed form representation of \( F_n \) given a difference equation \( \Delta F_n = G_n \). Just as we can compute antiderivatives for \( x^n \), we may do the same for \( n^k \) antidifferences, but the results are messier. The first few such antidifferences are

## Theorem 1.1: Antidifferences for powers of \(n\).

\begin{aligned}

1 &= \Delta n \\

n &= \Delta \lr{ \frac{n}{2}\lr{ n + 1} } \\

n^2 &= \Delta \lr{ \frac{n}{6}\lr{2 n + 1}\lr{n + 1} } \\

n^3 &= \Delta \lr{ \frac{n^2}{4}\lr{n + 1}^2 }.

\end{aligned}

\end{equation}

### Start proof:

The \( \Delta n \) identity is easily verified

\begin{equation}\label{eqn:fibonacci:1000}

\Delta n = n – (n-1) = 1.

\end{equation}

For higher orders it is a bit tedious to verify directly, but we can iteratively build up those results by evaluating the difference operator on each of the powers of \( n \).

\begin{equation}\label{eqn:fibonacci:660}

\begin{aligned}

\Delta n^2

&= n^2 – (n-1)^2 \\

&= n^2 – (n^2 – 2 n + 1) \\

&= 2 n – 1, \\

&= 2 n – \Delta n.

\end{aligned}

\end{equation}

Because the difference operator is linear, we can rearrange to find

\begin{equation}\label{eqn:fibonacci:1020}

\Delta \lr{ n^2 + n } = 2 n.

\end{equation}

Dividing through by \( 2 \) and factoring out an \( n \), recovers the desired result.

For the next power, we have

\begin{equation}\label{eqn:fibonacci:680}

\begin{aligned}

\Delta n^3

&= n^3 – (n-1)^3 \\

&= n^3 – (n^3 – 3 n^2 + 3 n – 1) \\

&= 3 n^2 – 3 n + 1 \\

&= 3 n^2 – 3 \Delta \frac{n}{2}\lr{ n + 1 } + \Delta n,

\end{aligned}

\end{equation}

or

\begin{equation}\label{eqn:fibonacci:1040}

\begin{aligned}

3 n^2

&=

\Delta \lr{n} \lr{ n^2 + \frac{3}{2}\lr{ n + 1} – 1 } \\

&=

\Delta \frac{n}{2} \lr{ 2 n^2 + 3\lr{ n + 1} – 2} \\

&=

\Delta \frac{n}{2} \lr{ 2 n^2 + 3 n + 1 } \\

&=

\Delta \frac{n}{2} \lr{ 2 n + 1}\lr{ n + 1 }

\end{aligned}

\end{equation}

Dividing through by \( 3 \) recovers the desired result.

The final result is left to the reader. It can be derived or verified easily with a couple lines of Mathematica code.

### End proof.

## Problem: Sum some series.

Find the sums \( \sum_{k = 1}^n k^m \), for \( m = 1, 2, 3 \).

## Answer

- \( m = 1 \). This is the (probably apocryphal) sum of Gauss’s grade school classroom:

\begin{equation}\label{eqn:fibonacci:1060}

F_n = \sum_{k = 1}^n k = 1 + 2 + \cdots n,

\end{equation}

satisfying

\begin{equation}\label{eqn:fibonacci:1080}

\begin{aligned}

\Delta F_n

&= F_n – F_{n-1} \\

&=

(n + (n-1) + \cdots + 1)

–

((n-1) + \cdots + 1) \\

&= n \\

&= \Delta \frac{n}{2}(n + 1).

\end{aligned}

\end{equation}

We must have

\begin{equation}\label{eqn:fibonacci:1100}

F_n = \frac{n}{2}\lr{ n + 1} + C.

\end{equation}

To fix \( C \) consider \( F_1 \)

\begin{equation}\label{eqn:fibonacci:1180}

F_1 = \inv{2}(1 + 1) + C = 1,

\end{equation}

so \( C = 0 \), so we find Gauss’s summation formula

\begin{equation}\label{eqn:fibonacci:1200}

\sum_{k = 1}^n k = \frac{n}{2}\lr{ n + 1},

\end{equation}

as expected. - \( m = 2 \). Now let’s do the sum of squares

\begin{equation}\label{eqn:fibonacci:1120}

F_n = \sum_{k = 1}^n k^2,

\end{equation}

for which we have

\begin{equation}\label{eqn:fibonacci:1140}

\Delta F_n = n^2 = \Delta \frac{n}{6}( 2 n + 1 )(n+1),

\end{equation}

so

\begin{equation}\label{eqn:fibonacci:1160}

F_n = \frac{n}{6}( 2 n + 1 )(n+1) + C.

\end{equation}

Clearly \( C = 0 \) satisfies the boundary condition, leaving

\begin{equation}\label{eqn:fibonacci:1220}

\sum_{k = 1}^n k^2 =

\frac{n}{6}( 2 n + 1 )(n+1).

\end{equation} - \( m = 3 \). We see the pattern, so for the sum of cubes, we can just write down the answer

\begin{equation}\label{eqn:fibonacci:1240}

\sum_{k = 1}^n k^3 =

\frac{n^2}{4}\lr{n + 1}^2

.

\end{equation}

Now that we have some basic comfort with the ideas of difference equations, and their solutions,

let’s get back to the Fibonacci problem. In that case, we have

\begin{equation}\label{eqn:fibonacci:1260}

F_n = F_{n-1} + F_{n-2}.

\end{equation}

Stated as a difference equation, this is

\begin{equation}\label{eqn:fibonacci:1280}

\Delta F_n = F_{n-2}.

\end{equation}

Before tackling the Fibonacci problem, let’s try one that slightly simpler.

## Problem: A simpler problem.

Solve \( \Delta F_n = F_{n-1} \), where \( F_0 = 0, F_1 = 1 \).

## Answer

The problem to solve is just

\begin{equation}\label{eqn:fibonacci:1300}

F_n = 2 F_{n-1}.

\end{equation}

This sequence is \( \setlr{ 1, 2, 4, 8, \cdots } \), so we can solve it by inspection, and the answer is just \( F_n = 2^{n-1} \). We want inspiration for the Fibonacci problem, so let’s pretend that we can’t see the answer, but that we can guess something close, and see if it works. Namely, let’s guess:

\begin{equation}\label{eqn:fibonacci:1320}

F_n = \alpha a^n + C.

\end{equation}

If we plug this trial solution into our difference equation, we get

\begin{equation}\label{eqn:fibonacci:1340}

\begin{aligned}

\alpha a^{n-1} + C

&=

\Delta F_n \\

&= \alpha \lr{ a^n – a^{n-1} } \\

&= \alpha a^{n-1} \lr{ a – 1 }

\end{aligned}

\end{equation}

This can be satisfied by setting \( C = 0 \) and \( a – 1 = 1 \), or \( a = 2 \), as we already knew. To fix the constant \( \alpha \) we utilize our boundary constraints, namely

\begin{equation}\label{eqn:fibonacci:1400}

F_1 = 1 = \alpha 2

\end{equation}

so \(\alpha = 1/2 \).

Compared to just seeing the answer, the procedure above was a lot of work. However, a side effect of this work is discovery of a guessing strategy that is somewhat like using \( f(t) = e^{s t} \) to generate a characteristic equation when solving a differential equation. For a difference equation of this form, it appears we can substitute \( F_n = \alpha a^n + C \) and use the differences to determine the values of \( \alpha, a, C \). Now let’s try this with the Fibonacci difference equation.

## Problem: Find a solution to the Fibonacci difference equation.

Without worrying about boundary constraints, find the solutions to \( \Delta F_n = F_{n-2} \), using a trial solution of \( F_n = \alpha a^n \).

## Answer

Inserting our trial solution, we have

\begin{equation}\label{eqn:fibonacci:1420}

\begin{aligned}

\alpha a^n

&=

F_n \\

&= F_{n-1} + F_{n-2} \\

&= \alpha \lr{ a^{n-1} + a^{n-2} } \\

&= \alpha a^{n-2} \lr{ a + 1 },

\end{aligned}

\end{equation}

so our “characteristic equation” is

\begin{equation}\label{eqn:fibonacci:1440}

a + 1 = a^2.

\end{equation}

Completing the square yields

\begin{equation}\label{eqn:fibonacci:1460}

\lr{ a – \inv{2} }^2 = 1 + \inv{4},

\end{equation}

or

\begin{equation}\label{eqn:fibonacci:1480}

a = \inv{2} \pm \frac{\sqrt{5}}{2}.

\end{equation}

Bamn. There’s our golden ratio, and it’s buddy!

We find that

\begin{equation}\label{eqn:fibonacci:1500}

F_n = \alpha \lr{\frac{1 \pm \sqrt{5}}{2} }^n,

\end{equation}

are solutions to the difference equation \ref{eqn:fibonacci:1280}.

Since we have a second order difference equation, we need a superposition of both solutions to try to satisfy the boundary conditions. In particular, we want to find the constants

\begin{equation}\label{eqn:fibonacci:1520}

F_n =

\alpha_{+} \lr{\frac{1 + \sqrt{5}}{2} }^n

+

\alpha_{-} \lr{\frac{1 – \sqrt{5}}{2} }^n + C.

\end{equation}

However, we already did this when we guessed used \( F_n = \alpha a^n + \beta b^n \) as a trial solution. When we did that, it was just to see if we could find the end result, knowing only the structure of the solution, but none of the specific constants. Now we have justified why that was a reasonable trial solution, since exactly this structure follows naturally from the difference equation itself.

This train of thought, makes me want to dig out my little Dover book on difference equations [1] that I’ve had since I was a kid. I think I only worked through the first chapter of that book. I have a lot of little sad neglected Dover books on mathematics and physics that I bought super cheap at the World’s Biggest Bookstore when I was back in school. It will be interesting to see how to tackle problems such as this, in a still more systematic fashion.

# References

[1] Hyman Levy and Freda Lessman. *Finite difference equations*. Courier Corporation, 1992.