Cassini and Catalan identities explained

__notoc__Cassini's identity (sometimes called Simson's identity) and Catalan's identity are mathematical identities for the Fibonacci numbers. Cassini's identity, a special case of Catalan's identity, states that for the nth Fibonacci number,

Fn-1Fn+1-

2
F
n

=(-1)n.

Note here

F0

is taken to be 0, and

F1

is taken to be 1.

Catalan's identity generalizes this:

2
F
n

-Fn-rFn+r=(-1)n-r

2.
F
r

Vajda's identity generalizes this:

Fn+iFn+j-FnFn+i+j=

nF
(-1)
i

Fj.

History

Cassini's formula was discovered in 1680 by Giovanni Domenico Cassini, then director of the Paris Observatory, and independently proven by Robert Simson (1753). However Johannes Kepler presumably knew the identity already in 1608.[1]

Catalan's identity is named after Eugène Catalan (1814–1894). It can be found in one of his private research notes, entitled "Sur la série de Lamé" and dated October 1879. However, the identity did not appear in print until December 1886 as part of his collected works . This explains why some give 1879 and others 1886 as the date for Catalan's identity .

The Hungarian-British mathematician Steven Vajda (1901–95) published a book on Fibonacci numbers (Fibonacci and Lucas Numbers, and the Golden Section: Theory and Applications, 1989) which contains the identity carrying his name.[2] [3] However, the identity had been published earlier in 1960 by Dustan Everman as problem 1396 in The American Mathematical Monthly,[4] and in 1901 by Alberto Tagiuri in Periodico di Matematica.[5]

Proof of Cassini identity

Proof by matrix theory

A quick proof of Cassini's identity may be given by recognising the left side of the equation as a determinant of a 2×2 matrix of Fibonacci numbers. The result is almost immediate when the matrix is seen to be the th power of a matrix with determinant -1:

Fn-1Fn+1-

2 =\det\left[\begin{matrix}F
F
n+1

&Fn\\Fn&Fn-1\end{matrix}\right] =\det\left[\begin{matrix}1&1\\1&0\end{matrix}\right]n =\left(\det\left[\begin{matrix}1&1\\1&0\end{matrix}\right]\right)n =(-1)n.

Proof by induction

Consider the induction statement:

Fn-1Fn+1-

2
F
n

=(-1)n

The base case

n=1

is true.

Assume the statement is true for

n

. Then:

Fn-1Fn+1-

2
F
n

+FnFn+1-FnFn+1=(-1)n

Fn-1Fn+1+FnFn+1-

2
F
n

-FnFn+1=(-1)n

Fn+1(Fn-1+Fn)-Fn(Fn+Fn+1)=(-1)n

2
F
n+1

-FnFn+2=(-1)n

FnFn+2-

2
F
n+1

=(-1)n+1

so the statement is true for all integers

n>0

.

Proof of Catalan identity

We use Binet's formula, that

F
n=\phin-\psin
\sqrt5
, where
\phi=1+\sqrt5
2
and
\psi=1-\sqrt5
2
.

Hence,

\phi+\psi=1

and

\phi\psi=-1

.

So,

2
5(F
n

-Fn-rFn+r)

=(\phin-\psin)2-(\phin-r-\psin-r)(\phin+r-\psin+r)

=(\phi2n-2\phin\psin+\psi2n)-(\phi2n-\phin\psin(\phi-r\psir+\phir\psi-r)+\psi2n)

=-2\phin\psin+\phin\psin(\phi-r\psir+\phir\psi-r)

Using

\phi\psi=-1

,

=-(-1)n2+(-1)n(\phi-r\psir+\phir\psi-r)

and again as

\phi=-1
\psi
,

=-(-1)n2+(-1)n-r(\psi2r+\phi2r)

Ln

is defined as
n+\psi
L
n=\phi

n

, so

=-(-1)n2+(-1)n-rL2r

Because

L2n=5

2
F
n

+2(-1)n

=-(-1)n2+(-1)n-r(5

2
F
r

+2(-1)r)

=-(-1)n2+(-1)n-r2(-1)r+(-1)n-r5

2
F
r

=-(-1)n2+(-1)n2+(-1)n-r5

2
F
r

=(-1)n-r5

2
F
r

Cancelling the

5

's gives the result.

Notes

  1. Miodrag Petkovic: Famous Puzzles of Great Mathematicians. AMS, 2009,, S. 30-31
  2. Douglas B. West: Combinatorial Mathematics. Cambridge University Press, 2020, p. 61
  3. Steven Vadja: Fibonacci and Lucas Numbers, and the Golden Section: Theory and Applications. Dover, 2008,, p. 28 (original publication 1989 at Ellis Horwood)
  4. Thomas Koshy: Fibonacci and Lucas Numbers with Applications. Wiley, 2001,, pp. 74-75, 83, 88
  5. Alberto Tagiuri: Equation (3) in Di alcune successioni ricorrenti a termini interi e positivi, Periodico di Matematica 16 (1901), pp. 1–12.

References

External links