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From Irrational to Rational

I find the relationship between the Golden Ratio and the Fibonacci numbers amazing.

Here is a derivation:

Consider the following  recursive relationship , where A and B are positive integers:

x_n=A x_{n-1} +Bx_{n-2}

We can represent this in matrix form:

\left (\begin{array}{ll} 0 &1\\B&A\end{array}\right) \left ( \begin{array}{l}x_{n-2}\\x_{n-1}\end{array}\right ) =\left ( \begin{array}{l} x_{n-1}\\x_{n}\end{array}\right )


R =\left (\begin{array}{ll} 0 &1\\B&A\end{array}\right)

The characteristic polynomial of R is

\lambda^2 - A\lambda -B =0

Now by the choice of A and B the characteristic polynomial has distinct and real roots: \alpha and \beta . This follows because the discriminant \forall A, B \in \mathrm{N}, A^2 +4 B >0

R can be represented as:

R =\frac{1}{\beta - \alpha} \left (\begin{array}{ll} 1& 1\\\alpha&\beta\end{array}\right ) \left (\begin{array}{ll} \alpha & 0 \\ 0&\beta\end{array}\right ) \left ( \begin{array}{ll}\beta&-1\\-\alpha&1\end{array}\right )

where the scalar is the determinant of the first matrix, the first matrix holds the eigenvectors of R , the central diagonal matrix the eigenvalues of R and the third matrix (when scalar is included) is the inverse of the first matirx, i.e. given the distinct real eigenvalues:

R\Sigma = \Sigma\Lambda \Longleftrightarrow R=\Sigma\Lambda\Sigma^{-1}


R \vec{x_n} =R^2 \vec{x_{n-1}}= R^n \vec{x_1}=\vec{x_{n+1}}\Longleftrightarrow \Sigma \Lambda^n\Sigma^{-1} \vec{x_1} =\vec{x_{n+1}}

where \vec{x_n}=\left (\begin{array}{l} x_{n-1}\\x_{n}\end{array}\right )

Let x_0=0 and x_1 =1 then

\Sigma^{-1} \vec{x_1} =\frac{1}{\beta-\alpha} \left (\begin{array}{l} -1 \\ 1\end{array}\right )

\Sigma\Lambda^n =\left ( \begin{array}{ll} \alpha^n &\beta^n \\ \alpha^{n+1}&\beta^{n+1}\end{array}\right )


\left ( \begin{array}{l} x_n\\x_{n+1}\end{array}\right ) = \left ( \begin{array}{l} \frac{\beta^n-\alpha^n}{\beta-\alpha}\\\frac{\beta^{n+1}-\alpha^{n+1}}{\beta-\alpha}\end{array}\right )


x_n = \frac{\beta^n-\alpha^n}{\beta-\alpha}

For the Fibonacci series A =B=1 and \beta =\frac{1 +\sqrt{5}}{2}=\phi and \alpha =\frac{1-\sqrt{5}}{2}=1-\phi=-\frac{1}{\phi} and  hence \beta -\alpha =\sqrt{5}


F_n =\frac{\phi^n -(1-\phi)^n}{\sqrt{5}}=\frac{\phi^n -(-\phi^{-1})^n}{\sqrt{5}}

An integer on the left hand side and an expression of irrationals on the right!

Categories: Mathematics

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