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Difference between revisions of "Binet's Formula"

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==Proof==
 
==Proof==
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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 1.61904761905, \frac{55}{34} \approx 1.61764705882, \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>.
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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>:\begin{align*}
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x&= x\\
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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.
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x^2 &= x+1\\
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x^3 &= x\cdot x^2\\
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&= x\cdot (x+1)\\
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&= x^2+x\\
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&= (x+1) + x\\
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&= 2x+1\\
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x^4 &= x \cdot x^3\\
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&= x\cdot (2x+1)\\
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&= 2x^2+x\\
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&=2(x+1)+x\\
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&=3x+2\\
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x^5 &= 5x+3\\
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x^6 &= 8x+5\\
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&\dots
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\end{align*}
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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*}
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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>
  
 
==See Also==
 
==See Also==
 
*[[Fibonacci number]]
 
*[[Fibonacci number]]
 
[[Category:Theorems]]
 
[[Category:Theorems]]

Revision as of 18:03, 6 November 2018

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 n[sup]th[/sup] Fibonacci number:\[\boxed{f_n = \frac{\left(\frac{1+\sqrt 5}{2}\right)^n - \left(\frac{1-\sqrt 5}{2}\right)^n}{\sqrt 5}}\]

See Also

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