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knightfischer
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Start of Fibonacci series
« on: Mar 1st, 2008, 5:02am » |
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Some authors start the series at Fo=1 and some at Fo=0. I'm studying for a teacher's exam, and may need to know the answer to "the fourth term of the series", which may be 2 or may be 3, depending where you start. Can anyone help me wit this?
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Sir Col
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Re: Start of Fibonacci series
« Reply #1 on: Mar 1st, 2008, 6:34am » |
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The Fibonacci is defined by the second order recurrence relation Fn+2 = Fn + Fn+1.
It is my understanding that convention defines F0 = 0 and F1 = F2 = 1; we would never write F0 = 1. But clearly it is only necessary to provide two of these terms to properly define the sequence.
So it all comes down to where you start your sequence, and depending on the context it makes sense to start with F0 or F1. Hence you will sometimes see people write the sequence 0, 1, 1, 2, 3, ... and other times 1, 1, 2, 3, 5, ... .
But in answer to your question, the 0th term, F0 = 0; the 1st term, F1 = 1; the 2nd term, F2 = 1; the third term, F3 = 2; and the fourth term, F4 = 3; and so on.
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knightfischer
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Re: Start of Fibonacci series
« Reply #2 on: Mar 1st, 2008, 9:27am » |
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Thanks for your help.
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Benny
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Re: Start of Fibonacci series
« Reply #3 on: Mar 4th, 2008, 1:22pm » |
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pi(N) means the number of primes between 1 and N (we include N if N is prime.)
Fibo numbers : 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765,...
pi(2) = 1
pi(3) = 2
pi(5) = 3
pi( = 4 (there are 4 primes between 1 and 8, namely 2, 3, 5 and 7)
pi(11) = 5.
So, the number of primes is a Fibonacci number.
Can this happen with two larger Fibonacci numbers?
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Christine
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Re: Start of Fibonacci series
« Reply #4 on: Mar 11th, 2008, 12:45pm » |
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We see the Fibonacci numbers and the golden mean in nature (plants, genes...) But do the Fibonacci numbers and the golden mean exist on the grand scale, star systems, i mean on a cosmic level?
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Michael Dagg
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Interesting question. I didn't know the answer until now.
There is another reference beside the PDF attached regarding the
golden ratio that my subscriptions won't let me access:
Searching for the golden ratio
Mario Livio
Astronomy (ISSN 0091-6358), Vol. 31, No. 4, p. 52 - 57 (2003)
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Regards,
Michael Dagg
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Benny
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Re: Start of Fibonacci series
« Reply #6 on: Mar 15th, 2008, 2:39pm » |
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Michael_Dagg, thanks for the info.
I've found that The Golden Ratio also describes the ever-expanding nature of what is termed a logarithmic spiral, we see the logarithmic spiral in a familiar seashell belonging to a creature called the chambered nautilus. The logarithmic spiral is a key shape for anything that grows, because with growth the ratio does not change. But logarithmic spirals appear in totally unrelated phenomena. They also appear, interestingly enough, when a falcon dives toward its prey, the flight pattern allows the bird to maintain a constant angle. Head cocked, its eyes never waver. It allows the falcon to keep its prey continuously in sight.
What about the spinning storms and hurricanes and spiral arms of galaxies ?
We see the presence of the Golden ratio.
Hurricanes and galaxies share few physical traits. Gravity and angular momentum, or spin, play a role in both cases.
Galaxies, by contrast, rotate either direction depending on your point of view.
Hurricanes and galaxies do look similar, but they are not twins.
The physics and scales are so different. Hurricanes are structures in the gravitational field of the Earth, while galaxies are self-gravitating objects in space.
That makes the Golden ratio all the more remarkable.
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Grimbal
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Re: Start of Fibonacci series
« Reply #7 on: Mar 16th, 2008, 4:41am » |
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Sorry to highjack this thread and speak about the Fibonacci sequence 
on Mar 1st, 2008, 6:34am, Sir Col wrote:| It is my understanding that convention defines F0 = 0 and F1 = F2 = 1; we would never write F0 = 1. But clearly it is only necessary to provide two of these terms to properly define the sequence. |
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It is a matter of convention where you start, but starting with F(0) = 0 allows for the remarkable relation
n|m => F(n)|F(m)
where a|b is "a divides b", i.e. b is a multiple of a.
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Benny
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Re: Start of Fibonacci series
« Reply #8 on: May 8th, 2008, 9:23am » |
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I noticed that for the 38 fibo numbers
http://www.research.att.com/~njas/sequences/A000045
Every 3rd Fibonacci number is divisible by 2, every 4th Fibonacci number is divisible by 3, every 5th Fibonacci number is divisible by 5, every 6th Fibonacci number is divisible by 8, every 7th Fibonacci number is divisible by 13, and every 8th Fibonacci number is divisible by 21.
And, that the divisors listed are successive Fibonacci numbers. Is there a way to prove that this is true for every 3-th, 4-th, 5-th, 6-th, 7-th, 8-th, etc. number in the sequence?
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Grimbal
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Re: Start of Fibonacci series
« Reply #9 on: May 8th, 2008, 10:05am » |
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Uh ... that's exactly what I just said. But I don't know how to prove it. I have faith in the Fibonacci numbers.
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Benny
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Re: Start of Fibonacci series
« Reply #10 on: May 8th, 2008, 2:12pm » |
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on May 8th, 2008, 10:05am, Grimbal wrote:| Uh ... that's exactly what I just said. But I don't know how to prove it. I have faith in the Fibonacci numbers. |
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Sorry, Grimbal. I was distracted. I'm trying with modulos.
fibonacci numbers are 1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987...
which is 1,1,2,0,2,2,1,0,1,1,2,0,2,2,1,0... (mod 3)
1,1,2,0,2,2,1,0 is repeating.
take mod 4
1,1,2,3,1,0,1,1,...
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Benny
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Re: Start of Fibonacci series
« Reply #11 on: May 8th, 2008, 5:52pm » |
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If n is divisible by m, then fn is divisible by fm.
Proof by Induction:
Let n be divisible by m, i.e., n = m * k where k is some integer.
Assume that fm * fk is divisible by fm .
Consider fm * fk + 1.
fm (k + 1) = fmk + m.
and fmk + m = fmk -1fm + fmkfm + 1
Since fmk -1fm is divisible by fm, fmk1fm + 1 is also divisible by fm
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towr
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Re: Start of Fibonacci series
« Reply #12 on: May 9th, 2008, 12:24am » |
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on May 8th, 2008, 5:52pm, BenVitale wrote:| Assume that fm * fk is divisible by fm . |
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Why assume? The first factor is fm, so it obviously is divisible by fm.
Quote:Consider fm * fk + 1.
fm (k + 1) = fmk + m. |
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Err. What are you trying to do here?
fm (k+1) = k*fm + fm
Also k*fm is not fmk
And let's not even mention the rest.
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Wikipedia, Google, Mathworld, Integer sequence DB
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pex
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Re: Start of Fibonacci series
« Reply #13 on: May 9th, 2008, 12:40am » |
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on May 9th, 2008, 12:24am, towr wrote:| Err. What are you trying to do here? |
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Probably fm(k+1) = fmk+m. Nothing shocking... but true.
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Benny
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Re: Start of Fibonacci series
« Reply #14 on: May 9th, 2008, 12:19pm » |
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I'm sorry guys. I still cannot figure out how to use the typesetting. I cannot spare much time on that.
I will on the weekend, though.
My demonstration is not readable enough. Could anyone please make it more readable by using the typesetting?
Thanks
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towr
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Re: Start of Fibonacci series
« Reply #15 on: May 10th, 2008, 1:23am » |
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on May 9th, 2008, 12:19pm, BenVitale wrote:| I'm sorry guys. I still cannot figure out how to use the typesetting. I cannot spare much time on that. |
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Then try using parentheses. You can write Fm(k+1) or F(m(k+1)) without causing confusion.
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Benny
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Re: Start of Fibonacci series
« Reply #16 on: May 18th, 2008, 12:52am » |
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Fibonacci numbers :
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, 196418, 317811, 514229, 832040, ....
Define function F to be a function that generates the Fibonacci sequence.
Let's define the Fibonacci number to be a number that fits the following pattern: F(0).F(1)F(2)F(3)... and so on
So, the first few digits of it would be 0.11235813213455...
Is this number irrational? If so, is it transcendental?
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Grimbal
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Re: Start of Fibonacci series
« Reply #17 on: May 18th, 2008, 7:10am » |
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I bet it is. No proof.
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Grimbal
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Re: Start of Fibonacci series
« Reply #18 on: May 18th, 2008, 8:52am » |
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At least it cannot be rational:
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Suppose the number is rational with period N.
Consider the Fibonacci sequence mod M where M = 10N. The sequence mod M depends only on the 2 last numers. Since these are in range 0..M-1 the sequence is bound to repeat after max M2 terms.
Since the sequence can be reconstructed backwards, any 2 numbers repeated in 2 places can be worked back to a repetition of the initial 0 and 1 somewhere along the sequence mod M.
That means that somewhere in the sequence there is a term that is 0 (mod M). But by definition of M, that means that the corresponding Fibonacci number ends with N zeros.
So there must be a sequence of N zeros in the number under investigation.
Since N is the supposed period of that number, that implies all digits past some point are zero. But the construction of the number forbids that. |
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| « Last Edit: May 18th, 2008, 8:57am by Grimbal » |
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Benny
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Re: Start of Fibonacci series
« Reply #19 on: May 18th, 2008, 11:20am » |
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Can't this number be expressed as an infinite summation of fractions?
It can't possibly be rational, otherwise you would have a finite sequence of digits so that all of the Fibonacci numbers concatenated (in base 10) is the same as this sequence concatenated with itself infinitely many times.
Can it be transcendental? if not transcendental, then algebraic.
Do we have reason to believe it can be algebraic?
I say numbers are generally transcendental unless you have some reason to believe otherwise.
Your thoughts, please.
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towr
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Re: Start of Fibonacci series
« Reply #20 on: May 18th, 2008, 11:47am » |
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on May 18th, 2008, 11:20am, BenVitale wrote:| Can't this number be expressed as an infinite summation of fractions? |
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Can't every number?
 di/10i where di is the ith decimal in the decimal expansion.
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Benny
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Re: Start of Fibonacci series
« Reply #21 on: May 18th, 2008, 12:50pm » |
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on May 18th, 2008, 11:47am, towr wrote:
Can't every number?
di/10i where di is the ith decimal in the decimal expansion.
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Yes, you're right. I did not express myself properly. just because a number can be expressed as an infinite summation of fractions, does not make this number rational. I thought it could be rational.
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Obob
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Re: Start of Fibonacci series
« Reply #22 on: May 18th, 2008, 7:54pm » |
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I would certainly expect the number to be transcendental. This is because the number of digits in a Fibonacci number is a very weird thing, that one would not expect to be encoded in some polynomial. However, this doesn't even come close to being a proof. There are many very strange looking infinite sums that converge to algebraic numbers.
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Benny
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Re: Start of Fibonacci series
« Reply #23 on: May 19th, 2008, 10:38am » |
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we know that the golden number is transcendental
f(n+1)/f(n) = phi
could we use this tp prove that
F(0).F(1)F(2)F(3)... and so on
is transcendental?
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ThudnBlunder
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Re: Start of Fibonacci series
« Reply #24 on: May 19th, 2008, 10:45am » |
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on May 19th, 2008, 10:38am, BenVitale wrote:we know that the golden number is transcendental
f(n+1)/f(n) = phi
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Phi = ( 5 + 1)/2
phi = (5 - 1)/2
Neither are transcendental.
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