The Fibonacci Sequence is by far the most famous and probably the oldest recursive sequence orginating from a problem published 1202 by Leonardo Pisano Fibonacci in his book Liber abacci.
It also represents the first and thus simplified attempt to describe dynamic structures within mathematics:
A man puts a young pair of rabbits into a closed garden. After two months they start to produce offspring: two little rabbits every month. How many pairs of rabbits live in that garden after a year's time?
Counting the rabbits for, let's say seven months, we get the following table (k denotes a juvenile, k a one month old, K an adult rabbit):
Obviously the number of pairs coincides with the first seven Fibonacci numbers. From that we can solve the rabbit problem with the twelfth Fibonacci number: 144.
Besides this modest start of population dynamics the Fibonacci numbers unfold a surprising richness of mathematical coherence. There is for example a relation to the golden section i The golden section is the solution of a partition task: A given line of length s>0 MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrVepeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4Caiabg6da+iaaicdaaaa@38A3@ is to be devided into two parts, so that the ratio 'whole line to bigger segment' is exactly the same as 'bigger segment to smaller one'. Let the given line be the interval [0,s] MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrVepeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaai4waiaaicdacaGGSaGaam4Caiaac2faaaa@3A0B@ . Then we have to find a real number x from ]0,s[ MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrVepeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaiyxaiaaicdacaGGSaGaam4CaiaacUfaaaa@3A0B@ satisfying s x = x s−x . MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrVepeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacaWGZbaabaGaamiEaaaacqGH9aqpdaWcaaqaaiaadIhaaeaacaWGZbGaeyOeI0IaamiEaaaaaaa@3CE3@ Thus our problem is solved as soon as we could solve the quadratic equation s 2 −sx= x 2 MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrVepeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4CamaaCaaaleqabaGaaGOmaaaakiabgkHiTiaadohacaWG4bGaeyypa0JaamiEamaaCaaaleqabaGaaGOmaaaaaaa@3DA2@ for x. With x∈]0,s[ MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrVepeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabgIGiolaac2facaaIWaGaaiilaiaadohacaGGBbaaaa@3C8C@ however we get from the p/q-formula: x 2 +sx− s 2 =0 ⇔ x=− s 2 + s 2 4 + s 2 ⇔ x=− s 2 + s 2 5 ⇔ x=s⋅ −1+ 5 2 ⇔ x=s⋅ϕ. MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrVepeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@6CCA@ So we get the value for x simply by multiplying the length s with the golden section number ϕ MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrVepeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqy1dOgaaa@37B5@ . which we will use to find a non-recursive representation for the Fibonacci sequence. To that end we need the golden section numbers
i
The golden section is the solution of a partition task: A given line of length s>0 MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrVepeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4Caiabg6da+iaaicdaaaa@38A3@ is to be devided into two parts, so that the ratio 'whole line to bigger segment' is exactly the same as 'bigger segment to smaller one'.
Let the given line be the interval [0,s] MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrVepeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaai4waiaaicdacaGGSaGaam4Caiaac2faaaa@3A0B@ . Then we have to find a real number x from ]0,s[ MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrVepeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaiyxaiaaicdacaGGSaGaam4CaiaacUfaaaa@3A0B@ satisfying
Thus our problem is solved as soon as we could solve the quadratic equation s 2 −sx= x 2 MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrVepeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4CamaaCaaaleqabaGaaGOmaaaakiabgkHiTiaadohacaWG4bGaeyypa0JaamiEamaaCaaaleqabaGaaGOmaaaaaaa@3DA2@ for x. With x∈]0,s[ MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrVepeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabgIGiolaac2facaaIWaGaaiilaiaadohacaGGBbaaaa@3C8C@ however we get from the p/q-formula:
So we get the value for x simply by multiplying the length s with the golden section number ϕ MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrVepeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqy1dOgaaa@37B5@ .
Consider:
There are some interesting properties of the section numbers. We need the following ones for the remainder.
Φ−ϕ=1 MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrVepeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGamaiMfA6agjabgkHiTiabew9aQjabg2da9iaaigdaaaa@3CFD@
Φ⋅ϕ=1 MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrVepeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGamaiMfA6agjabgwSixlabew9aQjabg2da9iaaigdaaaa@3E5A@
(x− Φ)⋅(x+ϕ)= x 2 −( Φ−ϕ)x− Φ⋅ϕ= x 2 −x−1 MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrVepeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaiikaiaadIhacqGHsislcWaGywOPdyKaaiykaiabgwSixlaacIcacaWG4bGaey4kaSIaeqy1dOMaaiykaiabg2da9iaadIhadaahaaWcbeqaaiaaikdaaaGccqGHsislcaGGOaGaeuOPdyKaeyOeI0Iaeqy1dOMaaiykaiaadIhacqGHsislcqqHMoGrcqGHflY1cqaHvpGAcqGH9aqpcaWG4bWaaWbaaSqabeaacaaIYaaaaOGaeyOeI0IaamiEaiabgkHiTiaaigdaaaa@5A85@
From the last equation we see that Φ and −ϕ MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrVepeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeuOPdyKaaeyDaiaab6gacaqGKbGaeyOeI0Iaeqy1dOgaaa@3CEC@ satisfy x 2 −x−1=0 MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrVepeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEamaaCaaaleqabaGaaGOmaaaakiabgkHiTiaadIhacqGHsislcaaIXaGaeyypa0JaaGimaaaa@3D2B@ , that means:
These preparations enable us to present a non-recursive version of the Fibonacci sequence.
Proposition: Let ( a n ) MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrVepeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaiikaiaadggadaWgaaWcbaGaamOBaaqabaGccaGGPaaaaa@3951@ denote the Fibonacci sequence. Then we have:
Proof: We prove by induction. The two stage recursion however now needs a two step basis of induction.
1∈A: Φ−(−ϕ) 5 = Φ+ϕ 5 = 1+ 5 −1+ 5 2 5 =1= a 1 . MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrVepeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaGymaiabgIGiolaadgeacaGG6aWaaSaaaeaacqqHMoGrcqGHsislcaGGOaGaeyOeI0Iaeqy1dOMaaiykaaqaamaakaaabaGaaGynaaWcbeaaaaGccqGH9aqpdaWcaaqaaiabfA6agjabgUcaRiabew9aQbqaamaakaaabaGaaGynaaWcbeaaaaGccqGH9aqpdaWcaaqaaiaaigdacqGHRaWkdaGcaaqaaiaaiwdaaSqabaGccqGHsislcaaIXaGaey4kaSYaaOaaaeaacaaI1aaaleqaaaGcbaGaaGOmamaakaaabaGaaGynaaWcbeaaaaGccqGH9aqpcaaIXaGaeyypa0JaamyyamaaBaaaleaacaaIXaaabeaaaaa@5474@
2∈A: Φ 2 − (−ϕ) 2 5 = Φ+1−(−ϕ+1) 5 = Φ+ϕ 5 =1= a 2 . MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrVepeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaGOmaiabgIGiolaadgeacaGG6aWaaSaaaeaacqqHMoGrdaahaaWcbeqaaiaaikdaaaGccqGHsislcaGGOaGaeyOeI0Iaeqy1dOMaaiykamaaCaaaleqabaGaaGOmaaaaaOqaamaakaaabaGaaGynaaWcbeaaaaGccqGH9aqpdaWcaaqaaiabfA6agjabgUcaRiaaigdacqGHsislcaGGOaGaeyOeI0Iaeqy1dOMaey4kaSIaaGymaiaacMcaaeaadaGcaaqaaiaaiwdaaSqabaaaaOGaeyypa0ZaaSaaaeaacqqHMoGrcqGHRaWkcqaHvpGAaeaadaGcaaqaaiaaiwdaaSqabaaaaOGaeyypa0JaaGymaiabg2da9iaadggadaWgaaWcbaGaaGOmaaqabaaaaa@5964@
n∈A⇒n+1∈A: MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrVepeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOBaiabgIGiolaadgeacqGHshI3caWGUbGaey4kaSIaaGymaiabgIGiolaadgeacaGG6aaaaa@411B@
Φ n+2 − (−ϕ) n+2 5 = Φ n Φ 2 − (−ϕ) n (−ϕ) 2 5 = Φ n ( Φ+1)− (−ϕ) n (−ϕ+1) 5 = Φ n+1 + Φ n − (−ϕ) n+1 − (−ϕ) n 5 = Φ n+1 − (−ϕ) n+1 5 + Φ n − (−ϕ) n 5 = a n+1 + a n = a n+2 . MathType@MTEF@5@5@+=feaafeart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrVepeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@A6C0@
There is much more information on Fibonacci numbers on the net. This site e.g. is very comprehensive.