Difference between revisions of "Binet's Formula"
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==Proof== | ==Proof== | ||
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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>: |
Revision as of 18:07, 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: