Difference between revisions of "Binet's Formula"

Latest revision as of 16:12, 24 February 2024

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, its 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 7: / Line 7: @@
 ==Proof==
-If we experiment with fairly large numbers, we see that the quotient of consecutive terms of the sequence approach <math>1.618\ldots</math>: <math>\frac{1}{1} = 1, \frac{2}{1} = 2, \frac{3}{2} = 1.5, \ldots, \frac{34}{21} \approx 1.617, \frac{55}{34} \approx 1.618, \ldots</math>. Thus we have a sequence resembling that of a [[geometric sequence]]. We let such a geometric sequence have terms <math>G_n = a \cdot r^n</math>. Then, <math>F_{n+1} = F_n + F_{n-1} \Longrightarrow a \cdot r^{n+1} = a \cdot r^{n} + a \cdot r^{n-1} </math> <math>\Longrightarrow r^2 = r + 1</math>. Using the [[quadratic formula]], we find <math>r = \frac{1 \pm \sqrt{5}}{2}</math>.
-We now have two sequences <math>G_n = a \left(\frac{1 + \sqrt{5}}{2}\right)^n</math> and <math>H_n = a \left(\frac{1 - \sqrt{5}}{2}\right)^n</math>, but neither match up with the Fibonacci sequence. In particular, <math>F_0 = 0</math>, but for <math>G_0, H_0</math> to be zero, we need <math>a = 0</math>, but then the sequence just generates a constant <math>0</math>. After a bit of experimenting with these two sequences to find a sequence where the zeroth term being zero, notice that <math>G_{n+1} - H_{n+1} = G_{n} - H_{n} + G_{n-1} - H_{n-1}</math>, so <math>G_{n} - H_{n}</math> also satisfies this recurrence. If we match up the numbers of <math>F_n</math> and <math>G_n - H_n = a\left(\left(\frac{1+\sqrt{5}}{2}\right)^n-\left(\frac{1-\sqrt{5}}{2}\right)^n\right)</math>, we note that <math>F_0 = G_0 - H_0 = 0</math>. However, <math>F_1 = 1 = G_1 - H_1</math>, which implies that <math>a = \frac{1}{\sqrt{5}}</math>. Now, <math>G_n - H_n</math> satisfies the same recurrence as <math>F_n</math> and has the same initial terms, so we are done.
+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>:
+<cmath>\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*}</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>
+==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, its 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
 ==See Also==
 *[[Fibonacci number]]
-{{stub}}
 [[Category:Theorems]]

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Difference between revisions of "Binet's Formula"

Latest revision as of 16:12, 24 February 2024

Contents

Formula

Proof

Proof using Recursion

See Also