Fibonacci numbers

[Citrix]

For the fibonacci series consider the following

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

[drew]

Originally Posted by Citrix

For the fibonacci series consider the following

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:

R_n+1 = 1+1/R_n

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

Drew

[Greenbank]

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

and more from this link:

http://mathforum.org/dr.math/faq/faq.golden.ratio.html

[mfgoode]

Originally Posted by drew

If you look at the sequence of ratios:

R_n+1 = 1+1/R_n

If you vary R_n, the function has a negative slope and R_n+1 == R_n 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.

[mfgoode]

Originally Posted by drew

If you look at the sequence of ratios:

R_n+1 = 1+1/R_n

If you vary R_n, the function has a negative slope and R_n+1 == R_n 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

[drew]

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

[mfgoode]

Originally Posted by drew

If you look at the sequence of ratios:

R_n+1 = 1+1/R_n

If you vary R_n, the function has a negative slope and R_n+1 == R_n 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

[alpertron]

Mally, drew is right.

Let R_n+1 = 1 + 1/R_n

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

R_n+1 = 1 + 1/R_n < 1 + 1/phi

But 1 + 1/phi = phi, so:

R_n > phi implies R_n+1 < phi.

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

R_n+1 = 1 + 1/R_n > 1 + 1/phi

But 1 + 1/phi = phi, so:

R_n < phi implies R_n+1 > phi.

[drew]

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. R_n+1 = 1 + 1/R_n. *That* function increases with R_n. 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.

[mfgoode]

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

[mfgoode]

Originally Posted by alpertron

Mally, drew is right.
~~~~~

R_n < phi implies R_n+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