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Fibonacci number

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Image:FibonacciBlocks.svg
A tiling with squares whose sides are successive Fibonacci numbers in length
Image:Fibonacci spiral.svg
A Fibonacci spiral, created by drawing arcs connecting the opposite corners of squares in the Fibonacci tiling shown above; see Golden spiral
Image:Fibonacci Sequence Plot.PNG
A plot of the Fibonacci sequence from 0 to 1597


In mathematics, the Fibonacci numbers are a sequence of numbers named after Leonardo of Pisa, known as Fibonacci, whose Liber Abaci published in 1202 introduced the sequence to Western European mathematics.

The sequence is defined by the following recurrence relation:

Failed to parse (Missing texvc executable; please see math/README to configure.): F(n):= \begin{cases} 0 & \mbox{if } n = 0; \\ 1 & \mbox{if } n = 1; \\ F(n-1)+F(n-2) & \mbox{if } n > 1. \\ \end{cases}


That is, after two starting values, each number is the sum of the two preceding numbers. The first Fibonacci numbers (sequence A000045 in OEIS), also denoted as Fn, for n = 0, 1, 2, … , are:

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368, 75025, 121393, ...[1]

Each third number of the series is an even number.

The sequence named after Fibonacci was first described in Indian mathematics.[2][3]

The sequence extended to negative index n satisfies Fn = Fn−1 + Fn−2 for all integers n, and F-n = (−1)n+1Fn:

.., -8, 5, -3, 2, -1, 1, followed by the sequence above.

Contents

Origins

The Fibonacci numbers first appeared, under the name mātrāmeru (mountain of cadence), in the work of the Sanskrit grammarian Pingala (Chandah-shāstra, the Art of Prosody, 450 or 200 BC). Prosody was important in ancient Indian ritual because of an emphasis on the purity of utterance. The Indian mathematician Virahanka (6th century AD) showed how the Fibonacci sequence arose in the analysis of metres with long and short syllables. Subsequently, the Jain philosopher Hemachandra (c.1150) composed a well-known text on these. A commentary on Virahanka's work by Gopāla in the 12th century also revisits the problem in some detail.

Sanskrit vowel sounds can be long (L) or short (S), and Virahanka's analysis, which came to be known as mātrā-vṛtta, wishes to compute how many metres (mātrās) of a given overall length can be composed of these syllables. If the long syllable is twice as long as the short, the solutions are:

1 mora: S (1 pattern)
2 morae: SS; L (2)
3 morae: SSS, SL; LS (3)
4 morae: SSSS, SSL, SLS; LSS, LL (5)
5 morae: SSSSS, SSSL, SSLS, SLSS, SLL; LSSS, LSL, LLS (8)
6 morae: SSSSSS, SSSSL, SSSLS, SSLSS, SLSSS, LSSSS, SSLL, SLSL, SLLS, LSSL, LSLS, LLSS, LLL (13)
7 morae: SSSSSSS, SSSSSL, SSSSLS, SSSLSS, SSLSSS, SLSSSS, LSSSSS, SSSLL, SSLSL, SLSSL, LSSSL, SSLLS, SLSLS, LSSLS, SLLSS, LSLSS, LLSSS, SLLL, LSLL, LLSL, LLLS (21)

A pattern of length n can be formed by adding S to a pattern of length n−1, or L to a pattern of length n−2; and the prosodicists showed that the number of patterns of length n is the sum of the two previous numbers in the series. Donald Knuth reviews this work in The Art of Computer Programming as equivalent formulations of the bin packing problem for items of lengths 1 and 2.

In the West, the sequence was first studied by Leonardo of Pisa, known as Fibonacci, in his Liber Abaci (1202)[4]. He considers the growth of an idealised (biologically unrealistic) rabbit population, assuming that:

  • in the first month there is just one newly-born pair,
  • new-born pairs become fertile from after their second month
  • each month every fertile pair begets a new pair, and
  • the rabbits never die

Let the population at month n be F(n). At this time, only rabbits who were alive at month n−2 are fertile and produce offspring, so F(n−2) pairs are added to the current population of F(n−1). Thus the total is F(n) = F(n−1) + F(n−2).[5]

Relation to the golden ratio

Closed form expression

Like every sequence defined by linear recurrence, the Fibonacci numbers have a closed-form solution. It has become known as Binet's formula, even though it was already known by Abraham de Moivre:

Failed to parse (Missing texvc executable; please see math/README to configure.): F\left(n\right) = {{\varphi^n-(1-\varphi)^n} \over {\sqrt 5}}={{\varphi^n-(-\varphi)^{-n}} \over {\sqrt 5}}\, ,
where Failed to parse (Missing texvc executable; please see math/README to configure.): \varphi
is the golden ratio (note, that Failed to parse (Missing texvc executable; please see math/README to configure.): 1-\varphi=-1/\varphi

, as can be seen from the defining equation below). The Fibonacci recursion

Failed to parse (Missing texvc executable; please see math/README to configure.): F(n+2)-F(n+1)-F(n)=0\,


is similar to the defining equation of the golden ratio in the form

Failed to parse (Missing texvc executable; please see math/README to configure.): x^2-x-1=0,\,


which is also known as the generating polynomial of the recursion.

Proof by induction

Any root of the equation above satisfies Failed to parse (Missing texvc executable; please see math/README to configure.): \begin{matrix}x^2=x+1,\end{matrix}\,

and multiplying by Failed to parse (Missing texvc executable; please see math/README to configure.): x^{n-1}\,
shows:
Failed to parse (Missing texvc executable; please see math/README to configure.): x^{n+1} = x^n + x^{n-1}\,


By definition Failed to parse (Missing texvc executable; please see math/README to configure.): \varphi

is a root of the equation, and the other root is Failed to parse (Missing texvc executable; please see math/README to configure.): 1-\varphi=-1/\varphi\, .

. Therefore:

Failed to parse (Missing texvc executable; please see math/README to configure.): \varphi^{n+1} = \varphi^n + \varphi^{n-1}\,


and

Failed to parse (Missing texvc executable; please see math/README to configure.): (1-\varphi)^{n+1} = (1-\varphi)^n + (1-\varphi)^{n-1}\, .


Both Failed to parse (Missing texvc executable; please see math/README to configure.): \varphi^{n}

and Failed to parse (Missing texvc executable; please see math/README to configure.): (1-\varphi)^{n}=(-1/\varphi)^{n}

are geometric series (for n = 1, 2, 3, ...) that satisfy the Fibonacci recursion. The first series grows exponentially; the second exponentially tends to zero, with alternating signs. Because the Fibonacci recursion is linear, any linear combination of these two series will also satisfy the recursion. These linear combinations form a two-dimensional linear vector space; the original Fibonacci sequence can be found in this space.

Linear combinations of series Failed to parse (Missing texvc executable; please see math/README to configure.): \varphi^{n}

and Failed to parse (Missing texvc executable; please see math/README to configure.): (1-\varphi)^{n}

, with coefficients a and b, can be defined by

Failed to parse (Missing texvc executable; please see math/README to configure.): F_{a,b}(n) = a\varphi^n+b(1-\varphi)^n
for any real Failed to parse (Missing texvc executable; please see math/README to configure.): a,b\, .


All thus-defined series satisfy the Fibonacci recursion

Failed to parse (Missing texvc executable; please see math/README to configure.): \begin{align} F_{a,b}(n+1) &= a\varphi^{n+1}+b(1-\varphi)^{n+1} \\ &=a(\varphi^{n}+\varphi^{n-1})+b((1-\varphi)^{n}+(1-\varphi)^{n-1}) \\ &=a{\varphi^{n}+b(1-\varphi)^{n}}+a{\varphi^{n-1}+b(1-\varphi)^{n-1}} \\ &=F_{a,b}(n)+F_{a,b}(n-1)\,. \end{align}

Requiring that Failed to parse (Missing texvc executable; please see math/README to configure.): F_{a,b}(0)=0

and Failed to parse (Missing texvc executable; please see math/README to configure.): F_{a,b}(1)=1
yields Failed to parse (Missing texvc executable; please see math/README to configure.): a=1/\sqrt 5
and Failed to parse (Missing texvc executable; please see math/README to configure.): b=-1/\sqrt 5

, resulting in the formula of Binet we started with. It has been shown that this formula satisfies the Fibonacci recursion. Furthermore, an explicit check can be made:

Failed to parse (Missing texvc executable; please see math/README to configure.): F_{a,b}(0)=\frac{1}{\sqrt 5}-\frac{1}{\sqrt 5}=0\,\!


and

Failed to parse (Missing texvc executable; please see math/README to configure.): F_{a,b}(1)=\frac{\varphi}{\sqrt 5}-\frac{(1-\varphi)}{\sqrt 5}=\frac{-1+2\varphi}{\sqrt 5}=\frac{-1+(1+\sqrt 5)}{\sqrt 5}=1,


establishing the base cases of the induction, proving that

Failed to parse (Missing texvc executable; please see math/README to configure.): F(n)={{\varphi^n-(1-\varphi)^n} \over {\sqrt 5}}
for all Failed to parse (Missing texvc executable; please see math/README to configure.):  n\, .


Therefore, for any two starting values, a combination Failed to parse (Missing texvc executable; please see math/README to configure.): a,b

can be found such that the function Failed to parse (Missing texvc executable; please see math/README to configure.): F_{a,b}(n)\,
is the exact closed formula for the series.

Computation by rounding

Since Failed to parse (Missing texvc executable; please see math/README to configure.): \begin{matrix}|1-\varphi|^n/\sqrt 5 < 1/2\end{matrix}

for all Failed to parse (Missing texvc executable; please see math/README to configure.): n\geq 0

, the number Failed to parse (Missing texvc executable; please see math/README to configure.): F(n)

is the closest integer to Failed to parse (Missing texvc executable; please see math/README to configure.): \varphi^n/\sqrt 5\, .
Therefore it can be found by rounding, or in terms of the floor function:
Failed to parse (Missing texvc executable; please see math/README to configure.): F(n)=\bigg\lfloor\frac{\varphi^n}{\sqrt 5} + \frac{1}{2}\bigg\rfloor.


Limit of consecutive quotients

Johannes Kepler observed that the ratio of consecutive Fibonacci numbers converges. He wrote that "as 5 is to 8 so is 8 to 13, practically, and as 8 is to 13, so is 13 to 21 almost”, and concluded that the limit approaches the golden ratio Failed to parse (Missing texvc executable; please see math/README to configure.): \varphi .[6]

Failed to parse (Missing texvc executable; please see math/README to configure.): \lim_{n\to\infty}\frac{F(n+1)}{F(n)}=\varphi,

This convergence does not depend on the starting values chosen, excluding 0, 0.

Proof:

It follows from the explicit formula that for any real Failed to parse (Missing texvc executable; please see math/README to configure.): a \ne 0, \, b \ne 0 \,

Failed to parse (Missing texvc executable; please see math/README to configure.): \begin{align} \lim_{n\to\infty}\frac{F_{a,b}(n+1)}{F_{a,b}(n)} &= \lim_{n\to\infty}\frac{a\varphi^{n+1}-b(1-\varphi)^{n+1}}{a\varphi^n-b(1-\varphi)^n} \\ &= \lim_{n\to\infty}\frac{a\varphi-b(1-\varphi)(\frac{1-\varphi}{\varphi})^n}{a-b(\frac{1-\varphi}{\varphi})^n} \\ &= \varphi \end{align}

because Failed to parse (Missing texvc executable; please see math/README to configure.): \bigl|{\tfrac{1-\varphi}{\varphi}}\bigr| < 1

and thus Failed to parse (Missing texvc executable; please see math/README to configure.): \lim_{n\to\infty}\left(\tfrac{1-\varphi}{\varphi}\right)^n=0 .


Decomposition of powers of the golden ratio

Since the golden ratio satisfies the equation

Failed to parse (Missing texvc executable; please see math/README to configure.): \varphi^2=\varphi+1,\,

this expression can be used to decompose higher powers Failed to parse (Missing texvc executable; please see math/README to configure.): \varphi^n

as a linear function of lower powers, which in turn can be decomposed all the way down to a linear combination of Failed to parse (Missing texvc executable; please see math/README to configure.): \varphi
and 1. The resulting recurrence relationships yield Fibonacci numbers as the linear coefficients, thus closing the loop:
Failed to parse (Missing texvc executable; please see math/README to configure.): \varphi^n=F(n)\varphi+F(n-1).

This expression is also true for Failed to parse (Missing texvc executable; please see math/README to configure.): n \, <\, 1 \,

if the Fibonacci sequence Failed to parse (Missing texvc executable; please see math/README to configure.): F(n) \,
is  extended to negative integers using the Fibonacci rule Failed to parse (Missing texvc executable; please see math/README to configure.): F(n) = F(n-1) + F(n-2) . \,


Matrix form

A 2-dimensional system of linear difference equations that describes the Fibonacci sequence is

Failed to parse (Missing texvc executable; please see math/README to configure.): {F_{k+2} \choose F_{k+1}} = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix} {F_{k+1} \choose F_{k}}


or

Failed to parse (Missing texvc executable; please see math/README to configure.): \vec F_{k+1} = A \vec F_{k}.\,


The eigenvalues of the matrix A are Failed to parse (Missing texvc executable; please see math/README to configure.): \varphi\,\!

and Failed to parse (Missing texvc executable; please see math/README to configure.): (1-\varphi)\,\!

, and the elements of the eigenvectors of A, Failed to parse (Missing texvc executable; please see math/README to configure.): {\varphi \choose 1}

and Failed to parse (Missing texvc executable; please see math/README to configure.): {1 \choose -\varphi}

, are in the ratios Failed to parse (Missing texvc executable; please see math/README to configure.): \varphi\,\!

and Failed to parse (Missing texvc executable; please see math/README to configure.): (1-\varphi\,\!)

.

This matrix has a determinant of −1, and thus it is a 2×2 unimodular matrix. This property can be understood in terms of the continued fraction representation for the golden ratio:

Failed to parse (Missing texvc executable; please see math/README to configure.): \varphi =1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{\;\;\ddots\,}}} \;.

The Fibonacci numbers occur as the ratio of successive convergents of the continued fraction for Failed to parse (Missing texvc executable; please see math/README to configure.): \varphi\,\! , and the matrix formed from successive convergents of any continued fraction has a determinant of +1 or −1.

The matrix representation gives the following closed expression for the Fibonacci numbers:

Failed to parse (Missing texvc executable; please see math/README to configure.): \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}^n = \begin{pmatrix} F_{n+1} & F_n \\ F_n & F_{n-1} \end{pmatrix}.


Taking the determinant of both sides of this equation yields Cassini's identity

Failed to parse (Missing texvc executable; please see math/README to configure.): (-1)^n = F_{n+1}F_{n-1} - F_n^2.\,


Additionally, since Failed to parse (Missing texvc executable; please see math/README to configure.): A^n A^m=A^{m+n}

for any square matrix Failed to parse (Missing texvc executable; please see math/README to configure.): A

, the following identities can be derived:

Failed to parse (Missing texvc executable; please see math/README to configure.): {F_n}^2 + {F_{n-1}}^2 = F_{2n-1},\,
Failed to parse (Missing texvc executable; please see math/README to configure.): F_{n+1}F_{m} + F_n F_{m-1} = F_{m+n}.\,


For the first one of these, there is a related identity:

Failed to parse (Missing texvc executable; please see math/README to configure.): (2F_{n-1}+F_n)F_n = (F_{n-1}+F_{n+1})F_n = F_{2n}.\,

For another way to derive the Failed to parse (Missing texvc executable; please see math/README to configure.): F_{2n+k}

formulas see the "EWD note" by Dijkstra[7].

Recognizing Fibonacci numbers




Occasionally, the question may arise whether a positive integer Failed to parse (Missing texvc executable; please see math/README to configure.): z


is a Fibonacci number. Since Failed to parse (Missing texvc executable; please see math/README to configure.): F(n)
is the closest integer to Failed to parse (Missing texvc executable; please see math/README to configure.): \varphi^n/\sqrt{5}




, the most straightforward, brute-force test is the identity
Failed to parse (Missing texvc executable; please see math/README to configure.): F\bigg(\bigg\lfloor\log_\varphi(\sqrt{5}z)+\frac{1}{2}\bigg\rfloor\bigg)=z,



which is true if and only if Failed to parse (Missing texvc executable; please see math/README to configure.): z


is a Fibonacci number.




Alternatively, an elegant algebraic test states, that a positive integer Failed to parse (Missing texvc executable; please see math/README to configure.): z


is a Fibonacci number if (and only if) Failed to parse (Missing texvc executable; please see math/README to configure.): \bigg(5z^2+4\bigg)
or Failed to parse (Missing texvc executable; please see math/README to configure.): \bigg(5z^2-4\bigg)
is a perfect square.[8] 


A slightly more sophisticated test uses the fact that the convergents of the continued fraction representation of Failed to parse (Missing texvc executable; please see math/README to configure.): \varphi


are ratios of successive Fibonacci numbers, that is the inequality
Failed to parse (Missing texvc executable; please see math/README to configure.): \bigg|\varphi-\frac{p}{q}\bigg|<\frac{1}{q^2}



(with coprime positive integers Failed to parse (Missing texvc executable; please see math/README to configure.): p
, Failed to parse (Missing texvc executable; please see math/README to configure.): q
) is true if and only if Failed to parse (Missing texvc executable; please see math/README to configure.): p


and Failed to parse (Missing texvc executable; please see math/README to configure.): q
are successive Fibonacci numbers. From this one derives the criterion that Failed to parse (Missing texvc executable; please see math/README to configure.): z
is a Fibonacci number if and only if the closed interval
Failed to parse (Missing texvc executable; please see math/README to configure.): \bigg[\varphi z-\frac{1}{z},\varphi z+\frac{1}{z}\bigg]



contains a positive integer.[9]

Identities
  1. F(n + 1) = F(n) + F(n − 1)
  2. F(0) + F(1) + F(2) + … + F(n) = F(n + 2) − 1
  3. F(1) + 2 F(2) + 3 F(3) + … + n F(n) = n F(n + 2) − F(n + 3) + 2
  4. F(0)² + F(1)² + F(2)² + … + F(n)² = F(n) F(n + 1)



These identities can be proven using many different methods.
But, among all, we wish to present an elegant proof for each of them using combinatorial arguments here.
In particular, F(n) can be interpreted as the number of ways summing 1's and 2's to n − 1, with the convention that F(0) = 0, meaning no sum will add up to −1, and that F(1) = 1, meaning the empty sum will "add up" to 0.
Here the order of the summands matters.
For example, 1 + 2 and 2 + 1 are considered two different sums and are counted twice.

Proof of the first identity




Without loss of generality, we may assume n ≥ 1.
Then F(n + 1) counts the number of ways summing 1's and 2's to n.


When the first summand is 1, there are F(n) ways to complete the counting for n − 1; and when the first summand is 2, there are F(n − 1) ways to complete the counting for n − 2.
Thus, in total, there are F(n) + F(n − 1) ways to complete the counting for n.

Proof of the second identity




We count the number of ways summing 1's and 2's to n + 1 such that at least one of the summands is 2.


As before, there are F(n + 2) ways summing 1's and 2's to n + 1 when n ≥ 0.
Since there is only one sum of n + 1 that does not use any 2, namely 1 + … + 1 (n + 1 terms), we subtract 1 from F(n + 2).


Equivalently, we can consider the first occurrence of 2 as a summand.
If, in a sum, the first summand is 2, then there are F(n) ways to the complete the counting for n − 1.
If the second summand is 2 but the first is 1, then there are F(n − 1) ways to complete the counting for n − 2.
Proceed in this fashion.
Eventually we consider the (n + 1)th summand.
If it is 2 but all of the previous n summands are 1's, then there are F(0) ways to complete the counting for 0.
If a sum contains 2 as a summand, the first occurrence of such summand must take place in between the first and (n + 1)th position.
Thus F(n) + F(n − 1) + … + F(0) gives the desired counting.

Proof of the third identity




This identity can be established in two stages.
First, we count the number of ways summing 1s and 2s to −1, 0, …, or n + 1 such that at least one of the summands is 2.


By our second identity, there are F(n + 2) −  1 ways summing to n + 1; F(n + 1) − 1 ways summing to n; …; and, eventually, F(2) − 1 way summing to 1.
As F(1) − 1 = F(0) = 0, we can add up all n + 1 sums and apply the second identity again to obtain
   [F(n + 2) − 1] + [F(n + 1) − 1] + … + [F(2) − 1]
= [F(n + 2) − 1] + [F(n + 1) − 1] + … + [F(2) − 1] + [F(1) − 1] + F(0)
= F(n + 2) + [F(n + 1) + … + F(1) + F(0)] − (n + 2)
= F(n + 2) + F(n + 3) − (n + 2).



On the other hand, we observe from the second identity that there are
  • F(0) + F(1) + … + F(n − 1) + F(n) ways summing to n + 1;
  • F(0) + F(1) + … + F(n − 1) ways summing to n;



……
  • F(0) way summing to −1.



Adding up all n + 1 sums, we see that there are
  • (n + 1) F(0) + n F(1) + … + F(n) ways summing to −1, 0, …, or n + 1.



Since the two methods of counting refer to the same number, we have
(n + 1) F(0) + n F(1) + … + F(n) = F(n + 2) + F(n + 3) − (n + 2)



Finally, we complete the proof by subtracting the above identity from n + 1 times the second identity.

Identity for doubling n




There is a very simple formula for doubling n :Failed to parse (Missing texvc executable; please see math/README to configure.): F_{2n} = F_{n+1}^2 - F_{n-1}^2 = F_n(F_{n+1}+F_{n-1}) 
. 
[10]


Another identity useful for calculating Fn for large values of n is
Failed to parse (Missing texvc executable; please see math/README to configure.): F_{2n+k} = F_k F_{n+1}^2 + 2 F_{k-1} F_{n+1} F_n + F_{k-2} F_n^2





for all integers n and k. Dijkstra[7] points out that doubling identities of this type can be used to calculate Fn using O(log n) arithmetic operations. Notice that, with the definition of Fibonacci numbers with negative n given in the introduction, this formula reduces to the double n formula when k = 0.


(From practical standpoint it should be noticed that the calculation involves manipulation of numbers with length (number of digits) Failed to parse (Missing texvc executable; please see math/README to configure.): {\rm \Theta}(n)\,
. Thus the actual performance depends mainly upon efficiency of the implemented  long multiplication, and usually is Failed to parse (Missing texvc executable; please see math/README to configure.): {\rm \Theta}(n \,\log n)


or Failed to parse (Missing texvc executable; please see math/README to configure.): {\rm \Theta}(n ^{\log_2 3})




.)

Other identities




Other identities include relationships to the Lucas numbers, which have the same recursive properties but start with L0=2 and L1=1. These properties include
F2n=FnLn.


There are also scaling identities, which take you from Fn and Fn+1 to a variety of things of the form Fan+b; for instance


Failed to parse (Missing texvc executable; please see math/README to configure.): F_{3n} = 2F_n^3 + 3F_n F_{n+1} F_{n-1} = 5F_{n}^3 + 3 (-1)^n F_{n} 


by Cassini's identity.




Failed to parse (Missing texvc executable; please see math/README to configure.): F_{3n+1} = F_{n+1}^3 + 3 F_{n+1}F_n^2 - F_n^3




Failed to parse (Missing texvc executable; please see math/README to configure.): F_{3n+2} = F_{n+1}^3 + 3 F_{n+1}^2F_n + F_n^3




Failed to parse (Missing texvc executable; please see math/README to configure.): F_{4n} = 4F_nF_{n+1}(F_{n+1}^2 + 2F_n^2) - 3F_n^2(F_n^2 + 2F_{n+1}^2)




These can be found experimentally using lattice reduction, and are useful in setting up the special number field sieve to factorize a Fibonacci number. Such relations exist in a very general sense for numbers defined by recurrence relations, see the section on multiplication formulae under Perrin numbers for details.

Power series




The generating function of the Fibonacci sequence is the power series
Failed to parse (Missing texvc executable; please see math/README to configure.): s(x)=\sum_{k=0}^{\infty} F_k x^k.





This series has a simple and interesting closed-form solution for x < 1/Failed to parse (Missing texvc executable; please see math/README to configure.): \varphi
Failed to parse (Missing texvc executable; please see math/README to configure.): s(x)=\frac{x}{1-x-x^2}.





This solution can be proven by using the Fibonacci recurrence to expand each coefficient in the infinite sum defining Failed to parse (Missing texvc executable; please see math/README to configure.): s(x)
Failed to parse (Missing texvc executable; please see math/README to configure.): \begin{align} s(x) &= \sum_{k=0}^{\infty} F_k x^k \\ &= F_0 + F_1x + \sum_{k=2}^{\infty} \left( F_{k-1} + F_{k-2} \right) x^k \\ &= x + \sum_{k=2}^{\infty} F_{k-1} x^k + \sum_{k=2}^{\infty} F_{k-2} x^k \\ &= x + x\sum_{k=0}^{\infty} F_k x^k + x^2\sum_{k=0}^{\infty} F_k x^k \\ &= x + x s(x) + x^2 s(x) \end{align}





Solving the equation Failed to parse (Missing texvc executable; please see math/README to configure.): s(x)=x+xs(x)+x^2s(x)


for Failed to parse (Missing texvc executable; please see math/README to configure.): s(x)
results in the closed form solution.




In particular, math puzzle-books note the curious value Failed to parse (Missing texvc executable; please see math/README to configure.): \frac{s(\frac{1}{10})}{10}=\frac{1}{89}
, or more generally
Failed to parse (Missing texvc executable; please see math/README to configure.): \sum_{n = 1}^{\infty}{\frac {F(n)}{10^{(k + 1)(n + 1)}}} = \frac {1}{10^{2k + 2} - 10^{k + 1} - 1}





for all integers Failed to parse (Missing texvc executable; please see math/README to configure.): k >= 0
.


Conversely,
Failed to parse (Missing texvc executable; please see math/README to configure.): \sum_{n=0}^\infty\,\frac{F_n}{k^{n}}\,=\,\frac{k}{k^{2}-k-1}.

Reciprocal sums




Infinite sums over reciprocal Fibonacci numbers can sometimes be evaluated in terms of theta functions. For example, we can write the sum of every odd-indexed reciprocal Fibonacci number as
Failed to parse (Missing texvc executable; please see math/README to configure.): \sum_{k=0}^\infty \frac{1}{F_{2k+1}} = \frac{\sqrt{5}}{4}\vartheta_2^2 \left(0, \frac{3-\sqrt 5}{2}\right) ,





and the sum of squared reciprocal Fibonacci numbers as
Failed to parse (Missing texvc executable; please see math/README to configure.): \sum_{k=1}^\infty \frac{1}{F_k^2} = \frac{5}{24} \left(\vartheta_2^4\left(0, \frac{3-\sqrt 5}{2}\right) - \vartheta_4^4\left(0, \frac{3-\sqrt 5}{2}\right) + 1 \right).





If we add 1 to each Fibonacci number in the first sum, there is also the closed form
Failed to parse (Missing texvc executable; please see math/README to configure.): \sum_{k=0}^\infty \frac{1}{1+F_{2k+1}} = \frac{\sqrt{5}}{2},





and there is a nice nested sum of squared Fibonacci numbers giving the reciprocal of the golden ratio,
Failed to parse (Missing texvc executable; please see math/README to configure.): \sum_{k=1}^\infty \frac{(-1)^{k+1}}{\sum_{j=1}^k {F_{j}}^2} = \frac{\sqrt{5}-1}{2}.





Results such as these make it plausible that a closed formula for the plain sum of reciprocal Fibonacci numbers could be found, but none is yet known. Despite that, the reciprocal Fibonacci constant
Failed to parse (Missing texvc executable; please see math/README to configure.): \psi = \sum_{k=1}^{\infty} \frac{1}{F_k} = 3.359885666243 \dots





has been proved irrational by Richard André-Jeannin.

Primes and divisibility
Main article: Fibonacci prime



A Fibonacci prime is a Fibonacci number that is prime (sequence A005478 in OEIS). The first few are:
2, 3, 5, 13, 89, 233, 1597, 28657, 514229, …



Fibonacci primes with thousands of digits have been found, but it is not known whether there are infinitely many. They must all have a prime index, except F4 = 3.


Any three consecutive Fibonacci numbers, taken two at a time, are relatively prime: that is,
gcd(Fn,Fn+1) = gcd(Fn,Fn+2) = 1.



More generally,
gcd(Fn, Fm) = Fgcd(n,m).[11]

A proof of this striking fact is online at Harvey Mudd College's Fun Math site

Divisibility by prime numbers

If p is a prime number then[12][13]

Failed to parse (Missing texvc executable; please see math/README to configure.): F_{p-\left(\frac{p}{5}\right)} \equiv 0 \pmod p,\;\;\; F_{p} \equiv \left(\frac{p}{5}\right) \pmod p,


where Failed to parse (Missing texvc executable; please see math/README to configure.): \;\left(\tfrac{p}{5}\right)\;

is the Legendre symbol. Failed to parse (Missing texvc executable; please see math/README to configure.): \;\left(\tfrac{p}{5}\right)=-1\;
if p ≡ ±2 (mod 5), Failed to parse (Missing texvc executable; please see math/README to configure.): \;\left(\tfrac{p}{5}\right)=+1\;
if p ≡ ±1 (mod 5), and Failed to parse (Missing texvc executable; please see math/README to configure.): \;\left(\tfrac{5}{5}\right)=0\;

.

For example,

Failed to parse (Missing texvc executable; please see math/README to configure.): (\tfrac{2}{5}) = -1, \,\, F_3 = 2, F_2=1,
Failed to parse (Missing texvc executable; please see math/README to configure.): (\tfrac{3}{5}) = -1, \,\, F_4 = 3,F_3=2,
Failed to parse (Missing texvc executable; please see math/README to configure.): (\tfrac{5}{5}) = \;\;\,0,\,\, F_5 = 5,
Failed to parse (Missing texvc executable; please see math/README to configure.): (\tfrac{7}{5}) = -1, \,\,F_8 = 21,\;\;F_7=13,
Failed to parse (Missing texvc executable; please see math/README to configure.): (\tfrac{11}{5}) = +1, F_{10} = 55, F_{11}=89.

Divisibility by 11

Failed to parse (Missing texvc executable; please see math/README to configure.): \sum_{k=n}^{n+9} F_{k} = 11 F_{n+6}


Right triangles

Starting with 5, every second Fibonacci number is the length of the hypotenuse of a right triangle with integer sides, or in other words, the largest number in a Pythagorean triple. The length of the longer leg of this triangle is equal to the sum of the three sides of the preceding triangle in this series of triangles, and the shorter leg is equal to the difference between the preceding bypassed Fibonacci number and the shorter leg of the preceding triangle.

The first triangle in this series has sides of length 5, 4, and 3. Skipping 8, the next triangle has sides of length 13, 12 (5 + 4 + 3), and 5 (8 − 3). Skipping 21, the next triangle has sides of length 34, 30 (13 + 12 + 5), and 16 (21 − 5). This series continues indefinitely. The triangle sides a, b, c can be calculated directly:

Failed to parse (Missing texvc executable; please see math/README to configure.): \displaystyle a_n = F_{2n-1}
Failed to parse (Missing texvc executable; please see math/README to configure.): \displaystyle b_n = 2 F_n F_{n-1}
Failed to parse (Missing texvc executable; please see math/README to configure.): \displaystyle c_n = {F_n}^2 - {F_{n-1}}^2


These formulas satisfy Failed to parse (Missing texvc executable; please see math/README to configure.): a_n ^2 = b_n ^2 + c_n ^2

for all n, but they only represent triangle sides when Failed to parse (Missing texvc executable; please see math/README to configure.): n > 2

.

Any four consecutive Fibonacci numbers Fn, Fn+1, Fn+2 and Fn+3 can also be used to generate a Pythagorean triple in a different way:

Failed to parse (Missing texvc executable; please see math/README to configure.): a = F_n F_{n+3} \, ; \, b = 2 F_{n+1} F_{n+2} \, ; \, c = F_{n+1}^2 + F_{n+2}^2 \, ; \, a^2 + b^2 = c^2 \,.

Example 1: let the Fibonacci numbers be 1, 2, 3 and 5. Then:

Failed to parse (Missing texvc executable; please see math/README to configure.): \displaystyle a = 1 \times 5 = 5
Failed to parse (Missing texvc executable; please see math/README to configure.): \displaystyle b = 2 \times 2 \times 3 = 12
Failed to parse (Missing texvc executable; please see math/README to configure.): \displaystyle c = 2^2 + 3^2 = 13 \,
Failed to parse (Missing texvc executable; please see math/README to configure.): \displaystyle 5^2 + 12^2 = 13^2 \,.

Example 2: let the Fibonacci numbers be 8, 13, 21 and 34. Then:

Failed to parse (Missing texvc executable; please see math/README to configure.): \displaystyle a = 8 \times 34 = 272
Failed to parse (Missing texvc executable; please see math/README to configure.): \displaystyle b = 2 \times 13 \times 21 = 546
Failed to parse (Missing texvc executable; please see math/README to configure.): \displaystyle c = 13^2 + 21^2 = 610 \,
Failed to parse (Missing texvc executable; please see math/README to configure.): \displaystyle 272^2 + 546^2 = 610^2 \,.


Magnitude of Fibonacci numbers

Since Failed to parse (Missing texvc executable; please see math/README to configure.): F_n

is asymptotic to Failed to parse (Missing texvc executable; please see math/README to configure.): \varphi^n/\sqrt5

, the number of digits in the base b representation of Failed to parse (Missing texvc executable; please see math/README to configure.): F_n\,

is asymptotic to Failed to parse (Missing texvc executable; please see math/README to configure.): n\,\log_b\varphi

.

In base 10, for every integer greater than 1 there are 4 or 5 Fibonacci numbers with that number of digits, in most cases 5.

Applications

The Fibonacci numbers are important in the run-time analysis of Euclid's algorithm to determine the greatest common divisor of two integers: the worst case input for this algorithm is a pair of consecutive Fibonacci numbers.

Yuri Matiyasevich was able to show that the Fibonacci numbers can be defined by a Diophantine equation, which led to his original solution of Hilbert's tenth problem.

The Fibonacci numbers occur in the sums of "shallow" diagonals in Pascal's triangle and Lozanić's triangle (see "Binomial coefficient").

Every positive integer can be written in a unique way as the sum of one or more distinct Fibonacci numbers in such a way that the sum does not include any two consecutive Fibonacci numbers. This is known as Zeckendorf's theorem, and a sum of Fibonacci numbers that satisfies these conditions is called a Zeckendorf representation.

Fibonacci numbers are used by some pseudorandom number generators.

Fibonacci numbers are used in a polyphase version of the merge sort algorithm in which an unsorted list is divided into two lists whose lengths correspond to sequential Fibonacci numbers - by dividing the list so that the two parts have lengths in the approximate proportion φ. A tape-drive implementation of the polyphase merge sort was described in The Art of Computer Programming.

Fibonacci numbers arise in the analysis of the Fibonacci heap data structure.

A one-dimensional optimization method, called the Fibonacci search technique, uses Fibonacci numbers.[14]

In music, Fibonacci numbers are sometimes used to determine tunings, and, as in visual art, to determine the length or size of content or formal elements. It is commonly thought that the first movement of Béla Bartók's Music for Strings, Percussion, and Celesta was structured using Fibonacci numbers.

Since the conversion factor 1.609344 for miles to kilometers is close to the golden ratio (denoted φ), the decomposition of distance in miles into a sum of Fibonacci numbers becomes nearly the kilometer sum when the Fibonacci numbers are replaced by their successors. This method amounts to a radix 2 number register in golden ratio base φ being shifted. To convert from kilometers to miles, shift the register down the Fibonacci sequence instead.[15][16][17]

Fibonacci numbers in nature

Image:Helianthus whorl.jpg
Sunflower head displaying florets in spirals of 34 and 55 around the outside

Fibonacci sequences appear in biological settings,[18] such as branching in trees, the fruitlets of a pineapple,[19] an uncurling fern and the arrangement of a pine cone.[20]. In addition, numerous poorly substantiated claims of Fibonacci numbers or golden sections in nature are found in popular sources, e.g. relating to the breeding of rabbits, the spirals of shells, and the curve of waves[citation needed].

Przemyslaw Prusinkiewicz advanced the idea that real instances can be in part understood as the expression of certain algebraic constraints on free groups, specifically as certain Lindenmayer grammars.[21]

A model for the pattern of florets in the head of a sunflower was proposed by H. Vogel in 1979.[22] This has the form

Failed to parse (Missing texvc executable; please see math/README to configure.): \theta = \frac{2\pi}{\phi^2} n

, Failed to parse (Missing texvc executable; please see math/README to configure.): r = c \sqrt{n}

where n is the index number of the floret and c is a constant scaling factor; the florets thus lie on Fermat's spiral. The divergence angle, approximately 137.51°, is the golden angle, dividing the circle in the golden ratio. Because this ratio is irrational, no floret has a neighbor at exactly the same angle from the center, so the florets pack efficiently. Because the rational approximations to the golden ratio are of the form F(j):F(j+1), the nearest neighbors of floret number n are those at n±F(j) for some index j which depends on r, the distance from the center. It is often said that sunflowers and similar arrangements have 55 spirals in one direction and 89 in the other (or some other pair of adjacent Fibonacci numbers), but this is true only of one range of radii, typically the outermost and thus most conspicuous.[23]

Popular culture

Main article: Fibonacci numbers in popular culture

Because the Fibonacci sequence is easy for non-mathematicians to understand, there are many examples of the Fibonacci numbers being used in popular culture.

Generalizations

Main article: Generalizations of Fibonacci numbers

The Fibonacci sequence has been generalized in many ways. These include:

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