golden ratio proof

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monroetransfer
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golden ratio proof

Postby monroetransfer » Sun Feb 01, 2009 1:35 am UTC

can anybody explain why
[math]\varphi^{n+1} = \varphi^n + \varphi^{n-1}\,[/math]
?

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Re: golden ratio proof

Postby Token » Sun Feb 01, 2009 1:42 am UTC

Because [imath]\varphi^{2} = \varphi + 1[/imath]. Multiply both sides by [imath]\varphi^{n-1}[/imath].
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Re: golden ratio proof

Postby monroetransfer » Sun Feb 01, 2009 1:44 am UTC

oh yeah! thanks. :D

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Re: golden ratio proof

Postby PM 2Ring » Mon Feb 02, 2009 5:29 pm UTC

You may enjoy playing with the "Phibonacci" sequence:

[math]\varphi^0 = 1[/math][math]\varphi^1 = \varphi[/math][math]\varphi^2 = \varphi + 1[/math][math]\varphi^3 = 2\varphi + 1[/math][math]\varphi^4 = 3\varphi + 2[/math][math]\varphi^5 = 5\varphi + 3[/math]
etc

(Sorry about the poor formatting. I'd better learn some LaTex, I guess.:))

EDIT:

This pic features phi rather heavily: http://i2.photobucket.com/albums/y43/PM2Ring/DodecF8wS.jpg. Raytraced using POV-Ray.
Last edited by PM 2Ring on Mon Feb 02, 2009 9:51 pm UTC, edited 1 time in total.

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Macbi
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Re: golden ratio proof

Postby Macbi » Mon Feb 02, 2009 9:00 pm UTC

While were on the topic, does anyone know a slick proof of the fact that the diagonal of a pentagon is phi times its side length? It's quite a useful fact to know, but whenever I use it I find it hard to justify why it is true.
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Re: golden ratio proof

Postby Cleverbeans » Mon Feb 02, 2009 9:27 pm UTC

Macbi wrote:While were on the topic, does anyone know a slick proof of the fact that the diagonal of a pentagon is phi times its side length? It's quite a useful fact to know, but whenever I use it I find it hard to justify why it is true.


Some googling seems to have turned it up. Check out addendum #1

http://www.jimloy.com/geometry/pentagon.htm
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Re: golden ratio proof

Postby Rhubarb » Tue Feb 03, 2009 12:04 am UTC

PM 2Ring wrote:You may enjoy playing with the "Phibonacci" sequence


Whoa... there's a part of me that wishes I did maths instead of physics... that is a very cool sequence. Now I'm trying to prove that the ratio of two adjacent Fibonacci numbers F(n)/F(n-1) approaches phi for n->infinity, but I am not sure how... I tried dividing adjacent terms in that sequence but I'm not sure how to proceed from there- hmm well I have some idea but I'm not 100% on it. (No this isn't homework, I'm just curious.)

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Re: golden ratio proof

Postby Talith » Tue Feb 03, 2009 1:51 am UTC

Rhubarb wrote:
PM 2Ring wrote:You may enjoy playing with the "Phibonacci" sequence


Whoa... there's a part of me that wishes I did maths instead of physics... that is a very cool sequence. Now I'm trying to prove that the ratio of two adjacent Fibonacci numbers F(n)/F(n-1) approaches phi for n->infinity, but I am not sure how... I tried dividing adjacent terms in that sequence but I'm not sure how to proceed from there- hmm well I have some idea but I'm not 100% on it. (No this isn't homework, I'm just curious.)


It might help you to know Binet's formula which says that for the nth fibbonachi number

[math]F(n)=\frac{\phi^n-(1-\phi)^n}{\sqrt5}[/math]

and infact you'll find it also holds for any two starting values of the reccurance (the standard is 1,1) as long as it isn't 0,0.

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Re: golden ratio proof

Postby PM 2Ring » Tue Feb 03, 2009 2:11 am UTC

Rhubarb wrote:
PM 2Ring wrote:You may enjoy playing with the "Phibonacci" sequence


Whoa... there's a part of me that wishes I did maths instead of physics... that is a very cool sequence. Now I'm trying to prove that the ratio of two adjacent Fibonacci numbers F(n)/F(n-1) approaches phi for n->infinity, but I am not sure how... I tried dividing adjacent terms in that sequence but I'm not sure how to proceed from there- hmm well I have some idea but I'm not 100% on it. (No this isn't homework, I'm just curious.)


Glad you like it. :) The "Phibonacci" sequence can be used to derive Binet's formula. The trick is to note that a similar series can be found for barphi = 1 - phi.

Also, the Lucas numbers can be found using

L(n) = [imath]\phi^n + (1-\phi)^n[/imath]

But to answer your question,

Let xn+1 = Fn+1 / Fn
= (Fn + Fn-1 ) / Fn
= Fn / Fn + Fn-1 / Fn
= 1 + Fn-1 / Fn
xn+1 = 1 + 1 / xn

Let phi = limit of xn as n approaches infinity
Then phi = 1 + 1 / phi and
phi2 = phi + 1

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Re: golden ratio proof

Postby masher » Tue Feb 03, 2009 2:55 am UTC

Macbi wrote:While were on the topic, does anyone know a slick proof of the fact that the diagonal of a pentagon is phi times its side length? It's quite a useful fact to know, but whenever I use it I find it hard to justify why it is true.



I had a quick look around, and a bit of a fiddle...

It all comes down to the fact that 2cos(pi/5)* = phi

Proving that is where I got stuck. I tried looking at series expansions, but I got lost.

*== 36º

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Re: golden ratio proof

Postby Nitrodon » Tue Feb 03, 2009 3:31 am UTC

Macbi wrote:While were on the topic, does anyone know a slick proof of the fact that the diagonal of a pentagon is phi times its side length? It's quite a useful fact to know, but whenever I use it I find it hard to justify why it is true.

A proof lies in the two similar 36-72-72 triangles in this hastily-drawn diagram.
Image

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Re: golden ratio proof

Postby gmalivuk » Fri Feb 06, 2009 11:46 pm UTC

PM 2Ring wrote:This pic features phi rather heavily: http://i2.photobucket.com/albums/y43/PM2Ring/DodecF8wS.jpg. Raytraced using POV-Ray.

Are the colors in there due to colored lights, or has POV-Ray gotten sophisticated enough to model prisms?
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Re: golden ratio proof

Postby PM 2Ring » Sat Feb 07, 2009 10:58 am UTC

gmalivuk wrote:Are the colors in there due to colored lights, or has POV-Ray gotten sophisticated enough to model prisms?

POV-Ray has been able to model dispersion for a while now, but it does slow down rendering a bit, as it has to trace more rays, so I rarely use this feature, and when I do I don't use many dispersion samples, hence the visible banding in that image. Here's a link to the relevant POV-Ray docs: Dispersion

It also has a forward ray-tracing feature known as photon mapping, to do realistic focusing effects, but I don't think I used it in that scene. Photon mapping

I've done a few pics of diamonds using photons and dispersion, but they take a long time to render, on this old steam-powered computer. :) FWIW, I've been using POV-Ray for ages, and started with it's predecessor, DKBtrace, on the Amiga.

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Re: golden ratio proof

Postby gmalivuk » Sat Feb 07, 2009 3:28 pm UTC

PM 2Ring wrote:
gmalivuk wrote:Are the colors in there due to colored lights, or has POV-Ray gotten sophisticated enough to model prisms?

POV-Ray has been able to model dispersion for a while now, but it does slow down rendering a bit, as it has to trace more rays, so I rarely use this feature, and when I do I don't use many dispersion samples, hence the visible banding in that image. Here's a link to the relevant POV-Ray docs: Dispersion

Huh. For some reason I'd never come across that back when I was fiddling more with it.

I probably just assumed it wouldn't do it properly because white is just treated as rgb instead of a spectrum, but yeah, obviously it could still stimulate spectrum effects when it "knows" you want white light...
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Re: golden ratio proof

Postby PM 2Ring » Sun Feb 08, 2009 6:58 am UTC

gmalivuk wrote:Huh. For some reason I'd never come across that back when I was fiddling more with it.

I probably just assumed it wouldn't do it properly because white is just treated as rgb instead of a spectrum, but yeah, obviously it could still stimulate spectrum effects when it "knows" you want white light...


POV-Ray has so many features, it's understandable that you may have missed one or two. Dispersion also works with non-white light, but of course the effects aren't quite as dramatic.

IMHO, POV-Ray's a great way to make pretty mathematical (or non-mathematical :) pictures. My most recent POV images involve various patterns on the Riemann sphere.

http://i2.photobucket.com/albums/y43/PM2Ring/RiemannGA6S_85.jpg
http://i2.photobucket.com/albums/y43/PM2Ring/RiemannB0S.png


In a vague attempt to return to the topic, here's another phi-related image created with POV, using two Penrose tilings combined.

http://i2.photobucket.com/albums/y43/PM2Ring/PenroseS4.jpg

And a Penrose tiling on the Riemann sphere

http://i2.photobucket.com/albums/y43/PM2Ring/RiemannD6S95.jpg

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Re: golden ratio proof

Postby PM 2Ring » Sun Feb 15, 2009 9:52 am UTC

Here's a small article I wrote a few years ago, that I thought might be appropriate for this thread. ( And it gives me an opportunity to practice my newly-acquired LaTeX skills. :))

Fibonacci numbers, the Golden ratio phi, and Prèvost's Constant

Introduction
The Fibonacci sequence is generated as follows:

[imath]F_0 = F_1 = 1[/imath]
[imath]F_{i+1} = F_i + F_{i-1}[/imath]

Hence [imath]F_{i+1} / F_i = 1 + F_{i-1} / F_i[/imath]
Let [imath]\phi_i = F_i / F_{i-1}[/imath]
Thus [imath]\phi_{i+1} = 1 + 1/\phi_i[/imath]
([imath]\phi[/imath] is the Greek letter, "phi")

Let [imath]\phi = \lim_{i\to \infty} \phi_i[/imath]
Then [imath]\phi = 1 + 1/\phi[/imath]
or [imath]\phi^2 = \phi + 1[/imath]
and [imath]1 = \phi^{-1} + \phi^{-2}[/imath]
In general, [imath]\phi^{n+1} = \phi^n + \phi^{n-1}[/imath]

Now [imath]\phi^2 = \phi + 1[/imath] is a quadratic equation;
the other solution is [imath]\bar\phi[/imath] (phibar),
where [imath]\phi + \bar\phi = 1[/imath] and [imath]\phi \bar\phi = -1[/imath]
[imath]\phi = (1 + \sqrt 5) / 2 \approx 1.618033988749895[/imath]
[imath]\bar\phi = (1 - \sqrt 5) / 2 \approx -.618033988749895[/imath]
Notice that [imath]\phi - \bar\phi = \sqrt 5[/imath]
Both [imath]\phi[/imath] and [imath]-\bar\phi[/imath] have been known as the Golden ratio.

Some important relations

[imath]\phi^i = \phi F_i + F_{i-1} = F_{i+1} - \bar\phi F_i[/imath]
[imath]\bar\phi^i = \bar\phi F_i + F_{i-1} = F_{i+1} - \phi F_i[/imath]

The table below shows how these arise.

Combining these two equations yields:
[imath]F_i = (\phi^i - \bar\phi^i) / \sqrt 5[/imath]

These formulas can be used to derive many interesting relations, eg
[imath]F_{i+j} = \phi^iF_j + \bar\phi^jF_i[/imath]

The "Phibonacci" sequence: the Fibonacci sequence of [imath]\phi[/imath] and its powers
[imath]\begin{array}{|r||r|r|}

n & F_n & \phi^n \\

0 & 0 & 1 \\
1 & 1 & \phi + 0 \\
2 & 1 & \phi + 1 \\
3 & 2 & 2\phi + 1 \\
4 & 3 & 3\phi + 2 \\
5 & 5 & 5\phi + 3 \\
6 & 8 & 8\phi + 5 \\
7 & 13 & 13\phi + 8 \\
8 & 21 & 21\phi + 13 \\
9 & 34 & 34\phi + 21 \\
10 & 55 & 55\phi + 34 \\
11 & 89 & 89\phi + 55 \\
12 & 144 & 144\phi + 89 \\

\end{array}[/imath]

The sum of the Fibonacci reciprocals

[imath]1/F_n = \sqrt 5 / (\phi^n - \bar\phi^n)[/imath]
[imath]= \sqrt 5\phi^{-n} / (1 - \phi^{-n}\bar\phi^n)[/imath],
which is the sum of a geometric progression with first term [imath]\sqrt 5\phi^{-n}[/imath]
and ratio [imath]\phi^{-n}\bar\phi^n[/imath].

Hence,
[imath]1/F_n = \sqrt 5(\phi^{-n} + \bar\phi^{3n} +
\phi^{-5n} + \bar\phi^{7n} + \ldots)[/imath], and similarly

[imath]1/F_{n+1} = \sqrt 5(\phi^{-n-1} + \bar\phi^{3n+3} +
\phi^{-5n-5} + \bar\phi^{7n+7} + \ldots)[/imath]

[imath]1/F_{n+2} = \sqrt 5(\phi^{-n-2} + \bar\phi^{3n+6} +
\phi^{-5n-10} + \bar\phi^{7n+14} + \ldots)[/imath]

[imath]1/F_{n+k} = \sqrt 5(\phi^{-n-k} + \bar\phi^{3n+3k} +
\phi^{-5n-5k} + \bar\phi^{7n+7k} + \ldots)[/imath]

Therefore,
[imath]\Sigma_{i=n}^{\infty} 1/F_i = \sqrt 5\left(
(\phi^{-n} + \phi^{-n-1} +
\phi^{-n-2} + \ldots + \phi^{-n-k} + \ldots ) +

(\bar\phi^{3n} + \bar\phi^{3n+3} +
\bar\phi^{3n+6} + \ldots + \bar\phi^{3n+3k} + \ldots ) + \right.[/imath]
[imath]\left. (\phi^{-5n} + \phi^{-5n-5} +
\phi^{-5n-10} + \ldots + \phi^{-5n-5k} + \ldots ) +

(\bar\phi^{7n} + \bar\phi^{7n+7} +
\bar\phi^{7n+14} + \ldots + \bar\phi^{7n+7k} + \ldots ) + \ldots \right)[/imath]

Summing these geometric progressions,

[imath]\Sigma_{i=n}^{\infty} 1/F_i =
\sqrt 5 \left( \phi^{-n}/(1-\phi^{-1}) +
\bar\phi^{3n}/(1-\bar\phi^3) +
\phi^{-5n}/(1-\phi^{-5}) +
\bar\phi^{7n}/(1-\bar\phi^7) + \ldots \right)[/imath]

The sum of the Fibonacci reciprocals was studied by Marc Prèvost,
hence [imath]\Sigma_{i=1}^{\infty} 1/F_i[/imath] is sometimes referred to as
Prèvost's Constant.

Also see http://mathworld.wolfram.com/ReciprocalFibonacciConstant.html

Here is Prèvost's constant, calculated to 500 digits using the above
series in bc.

3.
35988 56662 43177 55317 20113 02918 92717 96889 05133 73196
84864 95553 81532 51303 18996 68338 36154 16216 45679 00872
97045 34292 88539 13304 13678 90171 00883 67959 13517 33077
11907 85803 33550 33250 77531 87599 85048 71797 77897 00603
95645 09215 37589 27752 65673 35402 40331 69441 79929 39346

10992 62625 79646 47651 86865 94497 10216 55898 43608 81472
69324 95910 79473 87367 33785 23326 87749 97627 27757 94685
36769 18541 98146 76687 42998 76738 20969 13901 21772 20244
05208 15109 42649 34951 37454 16672 78955 34447 07777 75847
80259 63407 69074 84741 55579 10420 06750 15203 41070 53352


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