Difference between revisions of "Binet's Formula"

m (Proof)
m (Proof)
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==Proof==
 
==Proof==
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*}
+
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}
 
x&= x\
 
x&= x\
 
x^2 &= x+1\
 
x^2 &= x+1\
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x^6 &= 8x+5\
 
x^6 &= 8x+5\
 
&\dots
 
&\dots
\end{align*}
+
\end{align}
 
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 yieldsσnτn=fn(στ)+fn1fn1(1+52)n(152)n=fn(1+52152)  
 
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 yieldsσnτn=fn(στ)+fn1fn1(1+52)n(152)n=fn(1+52152)  
 
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>
 
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>

Revision as of 18:04, 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 $x^n$: x=xx2=x+1x3=xx2=x(x+1)=x2+x=(x+1)+x=2x+1x4=xx3=x(2x+1)=2x2+x=2(x+1)+x=3x+2x5=5x+3x6=8x+5 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 yieldsσnτn=fn(στ)+fn1fn1(1+52)n(152)n=fn(1+52152) 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