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Homederivation of Binet formula

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# derivation of Binet formula

The characteristic polynomial for the Fibonacci recurrence $f_{n}=f_{{n-1}}+f_{{n-2}}$ is

$x^{2}=x+1.$ |

The solutions of the characteristic equation $x^{2}-x-1=0$ are

$\phi=\frac{1+\sqrt{5}}{2},\qquad\psi=\frac{1-\sqrt{5}}{2}$ |

so the closed formula for the Fibonacci sequence must be of the form

$f_{n}=u\phi^{n}+v\psi^{n}$ |

for some real numbers $u,v$. Now we use the boundary conditions of the recurrence, that is, $f_{0}=0,f_{1}=1$, which means we have to solve the system

$0=u\phi^{0}+v\psi^{0},\qquad 1=u\phi^{1}+v\psi^{1}$ |

The first equation simplifies to $u=-v$ and substituting into the second one gives:

$1=u\left(\frac{1+\sqrt{5}}{2}\right)-u\left(\frac{1-\sqrt{5}}{2}\right)=u\left% (\frac{2\sqrt{5}}{2}\right)=u\sqrt{5}.$ |

Therefore

$u=\frac{1}{\sqrt{5}},\qquad v=\frac{-1}{\sqrt{5}}$ |

and so

$f_{n}=\frac{\phi^{n}}{\sqrt{5}}-\frac{\psi^{n}}{\sqrt{5}}=\frac{\phi^{n}-\psi^% {n}}{\sqrt{5}}.$ |

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