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Old 2006-06-08, 05:29   #1
Citrix
 
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Default Fibonacci numbers

For the fibonacci series consider the following

Code:
x--> ratio of f(x)/f(x-1)
2--> 1
3--> 2
4--> 1.5
5--> 1.66
6--> 1.60
7--> 1.625
8--> 1.615
9--> 1.619
10-->1.617
1) if you plot this on a graph, you will not get a smooth curve, the ratio increases on odd values and decreases on even values. Can someone explain why?
2) Does this continue for ever? Can there be a decrease in ratio over 2 values and then an increase in ratio over the third, must the ratio follow the above pattern.

Thanks for the answers
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Old 2006-06-08, 06:38   #2
drew
 
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Quote:
Originally Posted by Citrix
For the fibonacci series consider the following

Code:
x--> ratio of f(x)/f(x-1)
2--> 1
3--> 2
4--> 1.5
5--> 1.66
6--> 1.60
7--> 1.625
8--> 1.615
9--> 1.619
10-->1.617
1) if you plot this on a graph, you will not get a smooth curve, the ratio increases on odd values and decreases on even values. Can someone explain why?
2) Does this continue for ever? Can there be a decrease in ratio over 2 values and then an increase in ratio over the third, must the ratio follow the above pattern.

Thanks for the answers
If you look at the sequence of ratios:

Rn+1 = 1+1/Rn

If you vary Rn, the function has a negative slope and Rn+1 == Rn at Phi. Therefore, values higher than Phi yield values lower than Phi, and vice-versa. The series oscillates and converges indefinitely.

Drew

Last fiddled with by drew on 2006-06-08 at 06:40
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Old 2006-06-08, 08:15   #3
Greenbank
 
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Default

http://mathforum.org/library/drmath/view/52681.html

and more from this link:

http://mathforum.org/dr.math/faq/faq.golden.ratio.html
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Old 2006-06-08, 16:04   #4
mfgoode
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Thumbs up Oscillating and convergent.

Quote:
Originally Posted by drew
If you look at the sequence of ratios:

Rn+1 = 1+1/Rn

If you vary Rn, the function has a negative slope and Rn+1 == Rn at Phi. Therefore, values higher than Phi yield values lower than Phi, and vice-versa. The series oscillates and converges indefinitely.

Drew

Drew: I've got to give it to you boy!

A masterful analysis in a nut shell, concise and elegant.

Here's some more for you Greenbank which elaborates on Drew's explanantion.

http://www.people.bath.ac.uk/rdmg20/...0-%20Fib1.html

Mally

P.S. BTW this is my 1000th post and Im happy its on math.

Last fiddled with by mfgoode on 2006-06-08 at 16:06
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Old 2006-06-10, 16:35   #5
mfgoode
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Question Slope

Quote:
Originally Posted by drew
If you look at the sequence of ratios:

Rn+1 = 1+1/Rn

If you vary Rn, the function has a negative slope and Rn+1 == Rn at Phi. Therefore, values higher than Phi yield values lower than Phi, and vice-versa. The series oscillates and converges indefinitely.

Drew

Drew: could you please explain why is it a negative slope?
Mally
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Old 2006-06-10, 17:55   #6
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Quote:
Originally Posted by mfgoode

Drew: could you please explain why is it a negative slope?
Mally
Because 1/R decreases as R increases. The graph is a hyperbola shifted upward by 1 unit.

Drew
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Old 2006-06-11, 08:40   #7
mfgoode
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Cool Slope tangent.

Quote:
Originally Posted by drew
If you look at the sequence of ratios:

Rn+1 = 1+1/Rn

If you vary Rn, the function has a negative slope and Rn+1 == Rn at Phi. Therefore, values higher than Phi yield values lower than Phi, and vice-versa. The series oscillates and converges indefinitely.

Drew


From your own equation I get phi^2-phi-1 =0 which is well known for the Fib series.

Put this equal to y,

Therefore y=p^2 -p -1 which I liberally use as the asymptotic function

To get the general slope we differentiate.

Hence dy/dx =2p -1 but p>1 therefore the slope is +ve

Please correct me if I am wrong. Thanks.

Mally
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Old 2006-06-11, 14:05   #8
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Mally, drew is right.

Let Rn+1 = 1 + 1/Rn

If Rn > phi, then 1/Rn < 1/phi so:

Rn+1 = 1 + 1/Rn < 1 + 1/phi

But 1 + 1/phi = phi, so:

Rn > phi implies Rn+1 < phi.

If Rn < phi, then 1/Rn > 1/phi so:

Rn+1 = 1 + 1/Rn > 1 + 1/phi

But 1 + 1/phi = phi, so:

Rn < phi implies Rn+1 > phi.
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Old 2006-06-11, 19:41   #9
drew
 
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Quote:
Originally Posted by mfgoode


From your own equation I get phi^2-phi-1 =0 which is well known for the Fib series.

Put this equal to y,

Therefore y=p^2 -p -1 which I liberally use as the asymptotic function

To get the general slope we differentiate.

Hence dy/dx =2p -1 but p>1 therefore the slope is +ve

Please correct me if I am wrong. Thanks.

Mally
What are you doing? You're taking an arbitrary expression for phi...one which I did not claim to be decreasing, and finding the slope of that. It's not even a function that represents anything.

We can play that game and get whatever we want:

0=p2-p -1

The negation of that is true as well:

0 = -p2+p+1

Set that equal to y:

y = -p2+p+1

differentiate:

dy/dp = -2p+1

Now it's negative! Although I'm unsure what y is supposed to represent.

The expression I gave was for the seqence of ratios. Rn+1 = 1 + 1/Rn. *That* function increases with Rn. You thought it was simple and elegant the other day. What happened?

By the way, you carelessly typed dy/dx when it was clearly a function of p as expressed. Poor attention to detail such as this suggest you have an attention deficit disorder.

Last fiddled with by drew on 2006-06-11 at 19:52
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Old 2006-06-12, 10:55   #10
mfgoode
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Thumbs down Drew's Brew!

Quote:
Originally Posted by drew
What are you doing?
#~~~
By the way, you carelessly typed dy/dx when it was clearly a function of p as expressed. Poor attention to detail such as this suggest you have an attention deficit disorder.

Drew:
Your post deserves the yellow card, but being more mature than you, we will carry on the game, regardless of the foul, as it’s to my advantage.

You have actually secreted venom, but, like the poisonous tree frog called the kambo’, a native of Brazil, it may be put to good use in medicine (NYT.) we can make use of ‘Drew’s Brew’ as helpful if not in Pharmacy, then in Math.

Well you are good Drew ( now don’t get a swollen head!) and we need you for your expert comments and solutions, also for cross checking primes and wiki entries. I’m still reeling over your explanation in Wacky’s coin-canoe problem !

To get back on track the eqn. x^2-x-1 =0 is not as arbitrary as you remark.
It can be derived from your eqn. R (n+1) = 1+1/R (n) thus
Multiplying across we get Rn+1*Rn =Rn +1.
In the limit Rn+1 =Rn =phi (p for short)
Therefore p^2-p-1 = 0 To introduce a variable ratio we denote the eqn. by x^2-x-1
Now dy/dx =2x-1 which is the eqn. of the tangent at x whatever its value. It stabilises when phi (p) is put instead. As x increases so does the value of the tangent until at the asymptotes its value is infinity and this occurs on both sides of the axis ot the parabola

I’m sorry that you have been confused in stating that the eqn is arbitrary.

Maybe you need to understand what slope really means.

You have used the term LOOSELY by stating a negative slope AND it is NOT decreasing.

To any competent analyst this is jarring to the ears as well as repulsive to the eyes and understanding. Every slope which is a tangent is either up or down depending in which direction you take – up or down.

Its the rate of change of value that counts.

The second derivative D2y/Dx2 = 2 in this case 2 (+ve) is the rate of change of the tangent whether the value is increasing or decreasing and is clearly increasing. As the value of x increases the value of the tangent increases until at the asymptotes it becomes infinite (tan pi/2)

The other point you have made is that the function is a parabola moved 1 unit up.
This is not true, as the vertex is at (1/2, -5/4) i.e. the vertex lies below the x axis.

Mally
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Old 2006-06-12, 12:16   #11
mfgoode
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Smile

Quote:
Originally Posted by alpertron
Mally, drew is right.
~~~~~

Rn < phi implies Rn+1 > phi.

Alpertron:
I am not disputing the fact made by Drew that the Fib. Series converges by oscillating above and below the value of Phi.

This is more than aptly derived by you in your last post.

Further the astute observation, by sheer brute force, made by Citrix is exemplary that the series oscillates – lower than phi in the even terms and higher in the odd terms.
As a matter of fact it oscillates between the values 1 (R2) and 2 (R3) using Drew’s terminology.

The Fib. Series is 1, 1, 2, 3, 5, 8…
…….
Here R1 =1 and R2 =2 and for the other ratios in between these two extremes. The odds converge upwards from 1 and the evens downwards from 2 tending to phi in the limit..

If you would be kind enough to plot the graph of the parabola y=x^2-x-1 with vertex at (1/2,-5/4) and points at (0,-1) ;( 1,-1); (-1, 1) ;( 2,-1) ;(-2, 1) (1, ½, -1/4)
(Phi, 0); (-phi, 0) [hope there is no typo] I’m sure we will all get a better perspective of the problem.

You may also draw tangents at (phi, 0) and (–phi, 0). These angles should work out as 65.905 or 66* appx [incidentally another 666 number!}

Mally
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