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Difference between revisions of "Fibonacci sequence"
 
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As with many linear recursions, we can run the Fibonacci sequence backwards by solving its recursion relation for the term of smallest index, in this case <math>F_{n - 2} = F_{n} - F_{n - 1}</math>.  This allows us to compute, for example, that <math>F_0 = F_2 - F_1 = 0</math>, <math>F_{-1} = 1</math>, <math>F_{-2} = -2</math>, and so on.  Because these preceding terms are uniquely defined by the recursion, one frequently sees the definition of the Fibonacci sequence given in the form <math>F_0 = 0</math>, <math>F_1 = 1</math> and <math>F_n = F_{n - 1} + F_{n - 2}</math> for <math>n \geq 2</math>.  In general, one can show that <math>F_n = (-1)^{n+1}F_{-n}</math>.
 
As with many linear recursions, we can run the Fibonacci sequence backwards by solving its recursion relation for the term of smallest index, in this case <math>F_{n - 2} = F_{n} - F_{n - 1}</math>.  This allows us to compute, for example, that <math>F_0 = F_2 - F_1 = 0</math>, <math>F_{-1} = 1</math>, <math>F_{-2} = -2</math>, and so on.  Because these preceding terms are uniquely defined by the recursion, one frequently sees the definition of the Fibonacci sequence given in the form <math>F_0 = 0</math>, <math>F_1 = 1</math> and <math>F_n = F_{n - 1} + F_{n - 2}</math> for <math>n \geq 2</math>.  In general, one can show that <math>F_n = (-1)^{n+1}F_{-n}</math>.
  
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== <math>\varphi</math> and Binet's Formula==
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== <math>\varphi</math>, <math>\psi</math>, and Binet's Formula==
 
{{main|Binet's formula}}
 
{{main|Binet's formula}}
  
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The ratios <math>\frac{1}{1}</math>, <math>\frac{2}{1}</math>, <math>\frac{3}{2}</math>, <math>\frac{5}{3}</math>, <math>\frac{8}{5}</math>, ..., between successive terms of the sequence tend towards the limit <math>\frac{1 + \sqrt{5}}{2}</math>, a constant often denoted <math>\varphi</math> (the Greek letter [[phi]], also written <math>\phi</math>).  <math>\varphi</math> is a solution of the quadratic equation <math>x^2-x-1=0</math>.  The other root is <math>-\varphi^{-1} = \frac{1-\sqrt{5}}{2}</math>. One possible explanation for this fact is that the Fibonacci numbers are given explicitly by Binet's formula. It is  
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The ratios <math>\frac{1}{1}</math>, <math>\frac{2}{1}</math>, <math>\frac{3}{2}</math>, <math>\frac{5}{3}</math>, <math>\frac{8}{5}</math>, ..., between successive terms of the sequence tend towards the limit <math>\frac{1 + \sqrt{5}}{2}</math>, a constant often denoted <math>\varphi</math> (the Greek letter [[phi]], also written <math>\phi</math>).  <math>\varphi</math> is a solution of the quadratic equation <math>x^2-x-1=0</math>.  The other root is <math>\psi = \frac{1-\sqrt{5}}{2}</math> (the Greek letter [[psi]]). One possible explanation for this fact is that the Fibonacci numbers are given explicitly by Binet's formula. It is <math>F_n = \frac{\varphi^n - \psi^n}{\sqrt{5}}</math>. (Note that this formula is valid for all integers <math>n</math>.) It is so named because it was derived by mathematician Jacques Philippe Marie Binet, though it was already known by Abraham de Moivre and Euler before him. Another explanation is that you can continue the fibonacci numbers backward, too. Then you get something like <math>\dots -8,5,-3,2,-1,1,0,1,1,2,3,5,8\dots</math> . The ratios between successive terms has you continue backwards is <math>\psi</math>.
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<math>F_n = \frac{\varphi^n - (-\varphi)^{-n}}{\sqrt{5}}</math>.
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(Note that this formula is valid for all integers <math>n</math>.)
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==When to use them==
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It is so named because it was derived by mathematician Jacques Philippe Marie Binet, though it was already known by Abraham de Moivre.
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Usually, fibonacci-like recursions show up when organisms like rabbits produce babies but babies need to grow up which takes time. Two other places are when you make domino tilings in a <math>2\times n</math> rectangle or when you make a string of a's and b's of a fixed length but you can't have 2 a's together. Many things can be turned into either the fibonacci recursion <math>F_n=F_{n-1}+F_{n-2}</math> or another recursion that also produces the fibonacci numbers: <math>a_n=b_{n-1}</math> and <math>b_n=b_{n-1}+a_{n-1}</math>. There are other recursions, but these two are the most frequent.
  
 
== Identities ==
 
== Identities ==
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*<math>F_{n-1}\cdot F_{n+1} = F_{n}^2 + (-1)^n</math>
 
*<math>F_{n-1}\cdot F_{n+1} = F_{n}^2 + (-1)^n</math>
 
*<math>F_{m+n+1} = F_{m+1} \cdot F_{n+1} + F_{m} \cdot F_n</math>
 
*<math>F_{m+n+1} = F_{m+1} \cdot F_{n+1} + F_{m} \cdot F_n</math>
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==Problems==
 
==Problems==

Latest revision as of 13:57, 24 June 2024

The Fibonacci sequence is a sequence of integers in which the first and second terms are both equal to 1 and each subsequent term is the sum of the two preceding it. The first few terms are $1, 1, 2, 3, 5, 8, 13, 21, 34, 55,...$.

Recursion

The Fibonacci sequence can be written recursively as $F_1 = F_2 = 1$ and $F_n=F_{n-1}+F_{n-2}$ for $n \geq 3$. This is the simplest nontrivial example of a linear recursion with constant coefficients. There is also an explicit formula below.

Readers should be wary: some authors give the Fibonacci sequence with the initial conditions $F_0 = F_1 = 1$ (or equivalently $F_1 = 1, F_2 = 2$). This change in indexing does not affect the actual numbers in the sequence, but it does change which member of the sequence is referred to by the symbol $F_n$ and so also changes the appearance of certain identities involving the Fibonacci numbers.

Running Backwards

As with many linear recursions, we can run the Fibonacci sequence backwards by solving its recursion relation for the term of smallest index, in this case $F_{n - 2} = F_{n} - F_{n - 1}$. This allows us to compute, for example, that $F_0 = F_2 - F_1 = 0$, $F_{-1} = 1$, $F_{-2} = -2$, and so on. Because these preceding terms are uniquely defined by the recursion, one frequently sees the definition of the Fibonacci sequence given in the form $F_0 = 0$, $F_1 = 1$ and $F_n = F_{n - 1} + F_{n - 2}$ for $n \geq 2$. In general, one can show that $F_n = (-1)^{n+1}F_{-n}$.

$\varphi$, $\psi$, and Binet's Formula

Main article: Binet's formula

The ratios $\frac{1}{1}$, $\frac{2}{1}$, $\frac{3}{2}$, $\frac{5}{3}$, $\frac{8}{5}$, ..., between successive terms of the sequence tend towards the limit $\frac{1 + \sqrt{5}}{2}$, a constant often denoted $\varphi$ (the Greek letter phi, also written $\phi$). $\varphi$ is a solution of the quadratic equation $x^2-x-1=0$. The other root is $\psi = \frac{1-\sqrt{5}}{2}$ (the Greek letter psi). One possible explanation for this fact is that the Fibonacci numbers are given explicitly by Binet's formula. It is $F_n = \frac{\varphi^n - \psi^n}{\sqrt{5}}$. (Note that this formula is valid for all integers $n$.) It is so named because it was derived by mathematician Jacques Philippe Marie Binet, though it was already known by Abraham de Moivre and Euler before him. Another explanation is that you can continue the fibonacci numbers backward, too. Then you get something like $\dots -8,5,-3,2,-1,1,0,1,1,2,3,5,8\dots$ . The ratios between successive terms has you continue backwards is $\psi$.

When to use them

Usually, fibonacci-like recursions show up when organisms like rabbits produce babies but babies need to grow up which takes time. Two other places are when you make domino tilings in a $2\times n$ rectangle or when you make a string of a's and b's of a fixed length but you can't have 2 a's together. Many things can be turned into either the fibonacci recursion $F_n=F_{n-1}+F_{n-2}$ or another recursion that also produces the fibonacci numbers: $a_n=b_{n-1}$ and $b_n=b_{n-1}+a_{n-1}$. There are other recursions, but these two are the most frequent.

Identities

The most important identity regarding the Fibonacci sequence is its recursive definition, $F_{n+1} = F_n + F_{n-1}$. The following identities involving the Fibonacci numbers can be proved by induction.

  • $F_0 + F_1 + \cdots + F_{n} = F_{n+2} - 1$
  • $F_0 - F_1 + F_2 - \cdots - F_{2n-1} + F_{2n} = F_{2n-1} - 1$
  • $F_0^2 + F_1^2 + F_2^2 + \cdots + F_n^2 = F_n \cdot F_{n+1}$
  • $F_{n-1}\cdot F_{n+1} = F_{n}^2 + (-1)^n$
  • $F_{m+n+1} = F_{m+1} \cdot F_{n+1} + F_{m} \cdot F_n$


Problems

Introductory

  1. The Fibonacci sequence $1,1,2,3,5,8,13,21,\ldots$ starts with two 1s, and each term afterwards is the sum of its two predecessors. Which one of the ten digits is the last to appear in the units position of a number in the Fibonacci sequence?

    $\mathrm{(A) \ 0 } \qquad \mathrm{(B) \ 4 } \qquad \mathrm{(C) \ 6 } \qquad \mathrm{(D) \ 7 } \qquad \mathrm{(E) \ 9 }$
  2. A colony has $1$ rabbit. A rabbit produces one offspring every month. An offspring rabbit takes one month to grow up. Find a formula for the number of rabbits (including offspring) in the $n$th month.
    1. How about if the colony starts with $a$ rabbits and $b$ offspring?
    2. Use this result to prove the identity $F_{m+n+1} = F_{m+1} \cdot F_{n+1} + F_{m} \cdot F_n$.
  3. Find $\gcd(F_n,F_{n+1})$.
  4. Prove the above identites.

Intermediate

  1. Seven line segments, with lengths no greater than 10 inches, and no shorter than 1 inch, are given. Show that one can choose three of them to represent the sides of a triangle.
    (Manhattan Mathematical Olympiad 2004)
  2. Except for the first two terms, each term of the sequence $1000, x, 1000 - x,\ldots$ is obtained by subtracting the preceding term from the one before that. The last term of the sequence is the first negative term encounted. What positive integer $x$ produces a sequence of maximum length?
  3. A fair coin is to be tossed $10_{}^{}$ times. Let $\frac ij^{}_{}$, in lowest terms, be the probability that heads never occur on consecutive tosses. Find $i+j_{}^{}$.
  4. Find $a$ if $a$ and $b$ are integers such that $x^2 - x - 1$ is a factor of $ax^{17} + bx^{16} + 1$.

Olympiad

  1. Determine the maximum value of $m^2 + n^2$, where $m$ and $n$ are integers satisfying $m, n \in \{ 1,2, \ldots , 1981 \}$ and $( n^2 - mn - m^2 )^2 = 1$.

See also

External Links

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