# Difference between revisions of "Binet's Formula"

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==Proof== | ==Proof== | ||

+ | *Written by Mathlete2017* | ||

To derive a general formula for the Fibonacci numbers, we can look at the interesting quadratic<cmath>x^2-x-1=0.</cmath>Begin by noting that the roots of this quadratic are <math>\frac{1\pm\sqrt{5}}{2}</math> according to the quadratic formula. This quadratic can also be written as<cmath>x^2=x+1.</cmath> From this, we can write expressions for all <math>x^n</math>: | To derive a general formula for the Fibonacci numbers, we can look at the interesting quadratic<cmath>x^2-x-1=0.</cmath>Begin by noting that the roots of this quadratic are <math>\frac{1\pm\sqrt{5}}{2}</math> according to the quadratic formula. This quadratic can also be written as<cmath>x^2=x+1.</cmath> From this, we can write expressions for all <math>x^n</math>: | ||

− | \begin{align} | + | <cmath>\begin{align*} |

x&= x\\ | x&= x\\ | ||

x^2 &= x+1\\ | x^2 &= x+1\\ | ||

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x^6 &= 8x+5\\ | x^6 &= 8x+5\\ | ||

&\dots | &\dots | ||

− | \end{align} | + | \end{align*}</cmath> |

We note that<cmath>x^n=f_nx+f_{n-1}.</cmath>Let the roots of our original quadratic be <math>\sigma=\frac{1+\sqrt 5}{2}</math> and <math>\tau=\frac{1-\sqrt 5}{2}.</math> Since both <math>\sigma</math> and <math>\tau</math> are roots of the quadratic, they must both satisfy <math>x^n=f_nx+f_{n-1}.</math> So<cmath>\sigma^n=f_n\sigma+f_{n-1}</cmath>and<cmath>\tau^n=f_n\tau+f_{n-1}.</cmath>Subtracting the second equation from the first equation yields\begin{align*}\sigma^n-\tau^n=f_n(\sigma-\tau)+f_{n-1}-f_{n-1} \\ \left(\frac{1+\sqrt 5}{2}\right)^n - \left(\frac{1-\sqrt 5}{2}\right)^n = f_n \left(\frac{1+\sqrt 5}{2} - \frac{1-\sqrt 5}{2}\right)\end{align*} | We note that<cmath>x^n=f_nx+f_{n-1}.</cmath>Let the roots of our original quadratic be <math>\sigma=\frac{1+\sqrt 5}{2}</math> and <math>\tau=\frac{1-\sqrt 5}{2}.</math> Since both <math>\sigma</math> and <math>\tau</math> are roots of the quadratic, they must both satisfy <math>x^n=f_nx+f_{n-1}.</math> So<cmath>\sigma^n=f_n\sigma+f_{n-1}</cmath>and<cmath>\tau^n=f_n\tau+f_{n-1}.</cmath>Subtracting the second equation from the first equation yields\begin{align*}\sigma^n-\tau^n=f_n(\sigma-\tau)+f_{n-1}-f_{n-1} \\ \left(\frac{1+\sqrt 5}{2}\right)^n - \left(\frac{1-\sqrt 5}{2}\right)^n = f_n \left(\frac{1+\sqrt 5}{2} - \frac{1-\sqrt 5}{2}\right)\end{align*} | ||

This yields the general form for the n[sup]th[/sup] Fibonacci number:<cmath>\boxed{f_n = \frac{\left(\frac{1+\sqrt 5}{2}\right)^n - \left(\frac{1-\sqrt 5}{2}\right)^n}{\sqrt 5}}</cmath> | This yields the general form for the n[sup]th[/sup] Fibonacci number:<cmath>\boxed{f_n = \frac{\left(\frac{1+\sqrt 5}{2}\right)^n - \left(\frac{1-\sqrt 5}{2}\right)^n}{\sqrt 5}}</cmath> |

## Revision as of 18:05, 6 November 2018

**Binet's formula** is an explicit formula used to find the th term of the Fibonacci sequence.
It is so named because it was derived by mathematician Jacques Philippe Marie Binet, though it was already known by Abraham de Moivre.

## Formula

If is the th Fibonacci number, then .

## Proof

- Written by Mathlete2017*

To derive a general formula for the Fibonacci numbers, we can look at the interesting quadraticBegin by noting that the roots of this quadratic are according to the quadratic formula. This quadratic can also be written as From this, we can write expressions for all : We note thatLet the roots of our original quadratic be and Since both and are roots of the quadratic, they must both satisfy SoandSubtracting the second equation from the first equation yields\begin{align*}\sigma^n-\tau^n=f_n(\sigma-\tau)+f_{n-1}-f_{n-1} \\ \left(\frac{1+\sqrt 5}{2}\right)^n - \left(\frac{1-\sqrt 5}{2}\right)^n = f_n \left(\frac{1+\sqrt 5}{2} - \frac{1-\sqrt 5}{2}\right)\end{align*} This yields the general form for the n[sup]th[/sup] Fibonacci number: