Cassini's Identity

From ProofWiki
Jump to: navigation, search

Theorem

Let $F_k$ be the $k$th Fibonacci number.

Then $F_{n+1}F_{n-1} - F_n^2 = \left({-1}\right)^n$.


This is also sometimes reported (slightly less elegantly) as $F_{n+1}^2 - F_n F_{n+2} = \left({-1}\right)^n$


Proof 1

We see that:

$F_2 F_0 - F_1^2 = 1 \times 0 - 1 = -1 = \left({-1}\right)^1$

so the proposition holds for $n=1$.

We also see that:

$F_3 F_1 - F_2^2 = 2 \times 1 - 1 = \left({-1}\right)^2$

so the proposition holds for $n=2$.

Suppose the proposition is true for $n=k$, that is:

$F_{k+1}F_{k-1} - F_k^2 = \left({-1}\right)^k$

It remains to be shown that it follows from this that the proposition is true for $n=k+1$, that is:

$F_{k+2}F_k - F_{k+1}^2 = \left({-1}\right)^{k+1}$


So:

\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle F_{k+2} F_k - F_{k+1}^2\) \(=\) \(\displaystyle \) \(\displaystyle \left({F_k + F_{k+1} }\right) F_k - F_{k+1}^2\) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \) \(\displaystyle F_k^2 + F_k F_{k+1} - F_{k+1}^2\) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \) \(\displaystyle F_k^2 + F_k F_{k+1} - F_{k+1} \left({F_k + F_{k-1} }\right)\) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \) \(\displaystyle F_k^2 + F_k F_{k+1} - F_k F_{k+1} - F_{k+1} F_{k-1}\) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \) \(\displaystyle F_k^2 - F_{k+1} F_{k-1}\) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \) \(\displaystyle \left({-1}\right) \left({F_{k+1} F_{k-1} - F_k^2}\right)\) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \) \(\displaystyle \left({-1}\right) \left({-1}\right)^k\) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \) \(\displaystyle \left({-1}\right)^{k+1}\) \(\displaystyle \) \(\displaystyle \)                    


By the Principle of Mathematical Induction, the proof is complete.

$\blacksquare$


Note that from the above we have that:

$F_{k+2} F_k - F_{k+1}^2 = \left({-1}\right)^{k+1}$

from which:

$F_{n+1}^2 - F_n F_{n+2} = \left({-1}\right)^n$

follows immediately.


Proof 2

First this identity is proved:

$\begin{bmatrix} F_{n+1} & F_n \\ F_n & F_{n-1} \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^n$

then the determinant of both sides is taken.


Basis for the Induction

$\begin{bmatrix} F_2 & F_1 \\ F_1 & F_0 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^1$


Induction Hypothesis

For $k \in \Z_{>0}$, it is assumed that:

$\begin{bmatrix} F_{k+1} & F_k \\ F_k & F_{k-1} \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^k$

It remains to be shown that:

$\begin{bmatrix} F_{k+2} & F_{k+1} \\ F_{k+1} & F_k \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^{k+1}$


Induction Step

The induction step follows from conventional matrix multiplication:

\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^{k+1}\) \(=\) \(\displaystyle \) \(\displaystyle \begin{bmatrix} F_{k+1} & F_k \\ F_k & F_{k-1} \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}\) \(\displaystyle \) \(\displaystyle \)          by the induction hypothesis          
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \) \(\displaystyle \begin{bmatrix} F_{k+1} + F_k & F_{k+1} \\ F_k + F_{k-1} & F_k \end{bmatrix}\) \(\displaystyle \) \(\displaystyle \)          by conventional matrix multiplication          
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \) \(\displaystyle \begin{bmatrix} F_{k+2} & F_{k+1} \\ F_{k+1} & F_k \end{bmatrix}\) \(\displaystyle \) \(\displaystyle \)                    


So by induction:

$\begin{bmatrix} F_{n+1} & F_n \\ F_n & F_{n-1} \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^n$


Determinant

Now we calculate the determinants:

The LHS follows directly from the order 2 determinant:

$\begin{bmatrix} F_{n+1} & F_n \\ F_n & F_{n-1} \end{bmatrix} = F_{n+1} F_{n-1} - F_n^2$


Now for the RHS:

Basis for the Induction

$\begin{vmatrix} 1 & 1 \\ 1 & 0 \end{vmatrix} = 1 \times 0 - 1 \times 1 = -1 = \left({-1}\right)^1$


Induction Hypothesis

For $k \in \Z_{>0}$, it is assumed that:

$\begin{vmatrix} 1 & 1 \\ 1 & 0 \end{vmatrix}^k = \left({-1}\right)^k$


It remains to be shown that:

$\begin{vmatrix} 1 & 1 \\ 1 & 0 \end{vmatrix}^{k+1} = \left({-1}\right)^{k+1}$


Induction Step

The induction step follows from Determinant of Matrix Product:

$\begin{vmatrix} 1 & 1 \\ 1 & 0 \end{vmatrix}^{k+1} = \begin{vmatrix} 1 & 1 \\ 1 & 0 \end{vmatrix}^k \begin{vmatrix} 1 & 1 \\ 1 & 0 \end{vmatrix} = \left({-1}\right)^k \left({-1}\right) = \left({-1}\right)^{k+1}$


Hence by induction:

$\forall n \in \Z_{>0}: \begin{vmatrix} 1 & 1 \\ 1 & 0 \end{vmatrix}^n = \left({-1}\right)^n$

$\blacksquare$


Source of Name

This entry was named for Giovanni Domenico Cassini.


Sources