Difference between revisions of "Binet's Formula"

Revision as of 17:55, 19 May 2023

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: $\boxed{F_n = \frac{\left(\frac{1+\sqrt 5}{2}\right)^n - \left(\frac{1-\sqrt 5}{2}\right)^n}{\sqrt 5}}$

Proof using Recursion

The Fibonacci recursive relation is $F_n = F_{n-1} + F_{n-2}.$ This is a constant coefficient linear homogenous recurrence relation. We also know that $F_0 = 0$ and $F_1 = 1.$ Thus, it’s characteristic equation is $x^2-x-1=0$ which has solutions $\frac{1\pm\sqrt{5}}{2}.$ Let $v= \frac{1+\sqrt{5}}{2}$ and $p= \frac{1-\sqrt{5}}{2}.$ We get that $F_n = \lambda_1 v^n + \lambda_2 p^n.$ Plugging in our initial conditions, we get

$0 = \lambda_1 + \lambda_2$ $1= \lambda_1 v + \lambda_2 p$

Since $0 = \lambda_1 + \lambda_2 \implies 0 = \lambda_1 v+ \lambda_2 v,$ subtracting $0 = \lambda_1 v+ \lambda_2 v$ from $1 = \lambda_1 v + \lambda_2 p,$ we get $1 = \lambda_2(p-v) \implies \lambda_2 = \frac{1}{p-v} = -\frac{1}{\sqrt{5}}.$ Since $\lambda_2 = -\frac{1}{\sqrt{5}}$ and $0 = \lambda_1 + \lambda_2,$ $\lambda_1 = \frac{1}{\sqrt{5}}.$ Therefore, $F_n = \lambda_1 v^n + \lambda_2 p^n \implies F_n = \left (\frac{1}{\sqrt{5}} \right ) v^n + \left (-\frac{1}{\sqrt{5}} \right ) p^n.$ Therefore, the general form of the $n$ th Fibonacci number is $\boxed{F_n = \frac{\left(\frac{1+\sqrt 5}{2}\right)^n - \left(\frac{1-\sqrt 5}{2}\right)^n}{\sqrt 5}}$ ~peelybonehead

@@ Line 31: / Line 31: @@
 ==Proof using Recursion==
-The Fibonacci recursive relation is <math>F_n = F_{n-1} + F_{n-2}.</math> This is a constant coefficient linear homogenous recurrence relation. We also know that <math>F_0 = 0</math> and <math>F_1 = 1.</math> Thus, it’s characteristic equation is <math>x^2-x-1=0</math> which has solutions <cmath>\frac{1\pm\sqrt{5}}{2}.</cmath>Let <math>v= \frac{1+\sqrt{5}}{2}</math> and <math>p= \frac{1-\sqrt{5}}{2}.</math> We get that <cmath>F_n = \lambda_1 v^n + \lambda_2 p^n.</cmath>Plugging in our initial conditions, we get \begin{align*} 0 &= \lambda_1 + \lambda_2 \\ 1 &= \lambda_1 v + \lambda_2 p \end{align*}Since <math>0 = \lambda_1 + \lambda_2 \implies 0 = \lambda_1 v+ \lambda_2 v,</math> subtracting <math>0 = \lambda_1 v+ \lambda_2 v</math> from <math>1 = \lambda_1 v + \lambda_2 p,</math> we get <math>1 = \lambda_2(p-v) \implies \lambda_2 = \frac{1}{p-v} = -\frac{1}{\sqrt{5}}.</math> Since <math>\lambda_2 = -\frac{1}{\sqrt{5}}</math> and <math>0 = \lambda_1 + \lambda_2,</math> <math>\lambda_1 = \frac{1}{\sqrt{5}}.</math> Therefore, <cmath>F_n = \lambda_1 v^n + \lambda_2 p^n \implies F_n = \left (\frac{1}{\sqrt{5}} \right ) v^n + \left (-\frac{1}{\sqrt{5}} \right ) p^n.</cmath>Therefore, the general form of the <math>n</math>th Fibonacci number is <cmath>\boxed{F_n = \frac{\left(\frac{1+\sqrt 5}{2}\right)^n - \left(\frac{1-\sqrt 5}{2}\right)^n}{\sqrt 5}}</cmath>
+The Fibonacci recursive relation is <math>F_n = F_{n-1} + F_{n-2}.</math> This is a constant coefficient linear homogenous recurrence relation. We also know that <math>F_0 = 0</math> and <math>F_1 = 1.</math> Thus, it’s characteristic equation is <math>x^2-x-1=0</math> which has solutions <cmath>\frac{1\pm\sqrt{5}}{2}.</cmath>Let <math>v= \frac{1+\sqrt{5}}{2}</math> and <math>p= \frac{1-\sqrt{5}}{2}.</math> We get that <cmath>F_n = \lambda_1 v^n + \lambda_2 p^n.</cmath>Plugging in our initial conditions, we get
+<cmath>0 = \lambda_1 + \lambda_2 </cmath> <cmath> 1= \lambda_1 v + \lambda_2 p </cmath>
+Since <math>0 = \lambda_1 + \lambda_2 \implies 0 = \lambda_1 v+ \lambda_2 v,</math> subtracting <math>0 = \lambda_1 v+ \lambda_2 v</math> from <math>1 = \lambda_1 v + \lambda_2 p,</math> we get <math>1 = \lambda_2(p-v) \implies \lambda_2 = \frac{1}{p-v} = -\frac{1}{\sqrt{5}}.</math> Since <math>\lambda_2 = -\frac{1}{\sqrt{5}}</math> and <math>0 = \lambda_1 + \lambda_2,</math> <math>\lambda_1 = \frac{1}{\sqrt{5}}.</math> Therefore, <cmath>F_n = \lambda_1 v^n + \lambda_2 p^n \implies F_n = \left (\frac{1}{\sqrt{5}} \right ) v^n + \left (-\frac{1}{\sqrt{5}} \right ) p^n.</cmath>Therefore, the general form of the <math>n</math>th Fibonacci number is <cmath>\boxed{F_n = \frac{\left(\frac{1+\sqrt 5}{2}\right)^n - \left(\frac{1-\sqrt 5}{2}\right)^n}{\sqrt 5}}</cmath>
 ~peelybonehead

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Difference between revisions of "Binet's Formula"

Revision as of 17:55, 19 May 2023

Contents

Formula

Proof

Proof using Recursion

See Also