golden ratio proof
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 Posts: 26
 Joined: Wed Aug 27, 2008 1:03 am UTC
golden ratio proof
can anybody explain why
[math]\varphi^{n+1} = \varphi^n + \varphi^{n1}\,[/math]
?
[math]\varphi^{n+1} = \varphi^n + \varphi^{n1}\,[/math]
?
Re: golden ratio proof
Because [imath]\varphi^{2} = \varphi + 1[/imath]. Multiply both sides by [imath]\varphi^{n1}[/imath].
All posts are works in progress. If I posted something within the last hour, chances are I'm still editing it.

 Posts: 26
 Joined: Wed Aug 27, 2008 1:03 am UTC
Re: golden ratio proof
oh yeah! thanks.
Re: golden ratio proof
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 POVRay.
[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 POVRay.
Last edited by PM 2Ring on Mon Feb 02, 2009 9:51 pm UTC, edited 1 time in total.
Re: golden ratio proof
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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 Cleverbeans
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Re: golden ratio proof
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
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(n1) 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.)
 Talith
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Re: golden ratio proof
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(n1) 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.
Re: golden ratio proof
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(n1) 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 x_{n+1} = F_{n+1} / F_{n}
= (F_{n} + F_{n1} ) / F_{n}
= F_{n} / F_{n} + F_{n1} / F_{n}
= 1 + F_{n1} / F_{n}
x_{n+1} = 1 + 1 / x_{n}
Let phi = limit of x_{n} as n approaches infinity
Then phi = 1 + 1 / phi and
phi^{2} = phi + 1
Re: golden ratio proof
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
Re: golden ratio proof
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 367272 triangles in this hastilydrawn diagram.
 gmalivuk
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Re: golden ratio proof
PM 2Ring wrote:This pic features phi rather heavily: http://i2.photobucket.com/albums/y43/PM2Ring/DodecF8wS.jpg. Raytraced using POVRay.
Are the colors in there due to colored lights, or has POVRay gotten sophisticated enough to model prisms?
Re: golden ratio proof
gmalivuk wrote:Are the colors in there due to colored lights, or has POVRay gotten sophisticated enough to model prisms?
POVRay 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 POVRay docs: Dispersion
It also has a forward raytracing 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 steampowered computer. FWIW, I've been using POVRay for ages, and started with it's predecessor, DKBtrace, on the Amiga.
 gmalivuk
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Re: golden ratio proof
PM 2Ring wrote:gmalivuk wrote:Are the colors in there due to colored lights, or has POVRay gotten sophisticated enough to model prisms?
POVRay 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 POVRay 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...
Re: golden ratio proof
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...
POVRay has so many features, it's understandable that you may have missed one or two. Dispersion also works with nonwhite light, but of course the effects aren't quite as dramatic.
IMHO, POVRay's a great way to make pretty mathematical (or nonmathematical 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 phirelated 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
Re: golden ratio proof
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 newlyacquired 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_{i1}[/imath]
Hence [imath]F_{i+1} / F_i = 1 + F_{i1} / F_i[/imath]
Let [imath]\phi_i = F_i / F_{i1}[/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^{n1}[/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_{i1} = F_{i+1}  \bar\phi F_i[/imath]
[imath]\bar\phi^i = \bar\phi F_i + F_{i1} = 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}{rrr}
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^{n1} + \bar\phi^{3n+3} +
\phi^{5n5} + \bar\phi^{7n+7} + \ldots)[/imath]
[imath]1/F_{n+2} = \sqrt 5(\phi^{n2} + \bar\phi^{3n+6} +
\phi^{5n10} + \bar\phi^{7n+14} + \ldots)[/imath]
[imath]1/F_{n+k} = \sqrt 5(\phi^{nk} + \bar\phi^{3n+3k} +
\phi^{5n5k} + \bar\phi^{7n+7k} + \ldots)[/imath]
Therefore,
[imath]\Sigma_{i=n}^{\infty} 1/F_i = \sqrt 5\left(
(\phi^{n} + \phi^{n1} +
\phi^{n2} + \ldots + \phi^{nk} + \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^{5n5} +
\phi^{5n10} + \ldots + \phi^{5n5k} + \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
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_{i1}[/imath]
Hence [imath]F_{i+1} / F_i = 1 + F_{i1} / F_i[/imath]
Let [imath]\phi_i = F_i / F_{i1}[/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^{n1}[/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_{i1} = F_{i+1}  \bar\phi F_i[/imath]
[imath]\bar\phi^i = \bar\phi F_i + F_{i1} = 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}{rrr}
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^{n1} + \bar\phi^{3n+3} +
\phi^{5n5} + \bar\phi^{7n+7} + \ldots)[/imath]
[imath]1/F_{n+2} = \sqrt 5(\phi^{n2} + \bar\phi^{3n+6} +
\phi^{5n10} + \bar\phi^{7n+14} + \ldots)[/imath]
[imath]1/F_{n+k} = \sqrt 5(\phi^{nk} + \bar\phi^{3n+3k} +
\phi^{5n5k} + \bar\phi^{7n+7k} + \ldots)[/imath]
Therefore,
[imath]\Sigma_{i=n}^{\infty} 1/F_i = \sqrt 5\left(
(\phi^{n} + \phi^{n1} +
\phi^{n2} + \ldots + \phi^{nk} + \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^{5n5} +
\phi^{5n10} + \ldots + \phi^{5n5k} + \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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