Difference between revisions of "Binet's Formula"

Revision as of 23:36, 14 November 2022

Binet's formula is an explicit formula used to find the $n$ 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 $F_n$ is the $n$ th Fibonacci number, then $F_n=\frac{1}{\sqrt{5}}\left(\left(\frac{1+\sqrt{5}}{2}\right)^n-\left(\frac{1-\sqrt{5}}{2}\right)^n\right)$ .

Proof

To derive a general formula for the Fibonacci numbers, we can look at the interesting quadratic $x^2-x-1=0.$ Begin by noting that the roots of this quadratic are $\frac{1\pm\sqrt{5}}{2}$ according to the quadratic formula. This quadratic can also be written as $x^2=x+1.$ From this, we can write expressions for all $x^n$ : $\begin{align*} x&= x\\ x^2 &= x+1\\ x^3 &= x\cdot x^2\\ &= x\cdot (x+1)\\ &= x^2+x\\ &= (x+1) + x\\ &= 2x+1\\ x^4 &= x \cdot x^3\\ &= x\cdot (2x+1)\\ &= 2x^2+x\\ &=2(x+1)+x\\ &=3x+2\\ x^5 &= 5x+3\\ x^6 &= 8x+5\\ &\dots \end{align*}$ We note that $x^n=F_nx+F_{n-1}.$ Let the roots of our original quadratic be $\sigma=\frac{1+\sqrt 5}{2}$ and $\tau=\frac{1-\sqrt 5}{2}.$ Since both $\sigma$ and $\tau$ are roots of the quadratic, they must both satisfy $x^n=F_nx+F_{n-1}.$ So $\sigma^n=F_n\sigma+F_{n-1}$ and $\tau^n=F_n\tau+F_{n-1}.$ 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 nth Fibonacci number: $\boxed{F_n = \frac{\left(\frac{1+\sqrt 5}{2}\right)^n - \left(\frac{1-\sqrt 5}{2}\right)^n}{\sqrt 5}}$

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 &\dots
 \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<cmath>\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*}</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<cmath>\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*}</cmath>
-This yields the general form for the nth 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 nth 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>
 ==See Also==
 *[[Fibonacci number]]
 [[Category:Theorems]]

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Difference between revisions of "Binet's Formula"

Revision as of 23:36, 14 November 2022

Formula

Proof

See Also