20060608, 05:29  #1 
Jun 2003
3·5·107 Posts 
Fibonacci numbers
For the fibonacci series consider the following
Code:
x> ratio of f(x)/f(x1) 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 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 
20060608, 06:38  #2  
Jun 2005
2·191 Posts 
Quote:
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 viceversa. The series oscillates and converges indefinitely. Drew Last fiddled with by drew on 20060608 at 06:40 

20060608, 08:15  #3 
Jul 2005
2·193 Posts 

20060608, 16:04  #4  
Bronze Medalist
Jan 2004
Mumbai,India
2052_{10} Posts 
Oscillating and convergent.
Quote:
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 20060608 at 16:06 

20060610, 16:35  #5  
Bronze Medalist
Jan 2004
Mumbai,India
2^{2}×3^{3}×19 Posts 
Slope
Quote:
Drew: could you please explain why is it a negative slope? Mally 

20060610, 17:55  #6  
Jun 2005
2·191 Posts 
Quote:
Drew 

20060611, 08:40  #7  
Bronze Medalist
Jan 2004
Mumbai,India
2052_{10} Posts 
Slope tangent.
Quote:
From your own equation I get phi^2phi1 =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 

20060611, 14:05  #8 
Aug 2002
Buenos Aires, Argentina
10111010110_{2} Posts 
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. 
20060611, 19:41  #9  
Jun 2005
101111110_{2} Posts 
Quote:
We can play that game and get whatever we want: 0=p^{2}p 1 The negation of that is true as well: 0 = p^{2}+p+1 Set that equal to y: y = p^{2}+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. Last fiddled with by drew on 20060611 at 19:52 

20060612, 10:55  #10  
Bronze Medalist
Jan 2004
Mumbai,India
2^{2}×3^{3}×19 Posts 
Drew's Brew!
Quote:
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 coincanoe problem ! To get back on track the eqn. x^2x1 =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^2p1 = 0 To introduce a variable ratio we denote the eqn. by x^2x1 Now dy/dx =2x1 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 

20060612, 12:16  #11  
Bronze Medalist
Jan 2004
Mumbai,India
4004_{8} Posts 
Quote:
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^2x1 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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