## The Golden Ratio

For the discussion of math. Duh.

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phi1.618
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### The Golden Ratio

hey i just wanted to start a post on my favorite subject..... The Golden Ratio!!!!! all i can say is that 1.618 is a good looking number and if you count the spirals on a sunflower going in one direction and divide them by the number going in the opposite direction then you get around 1.618!!!!!! its like magic. Anyways post some things you think are interesting about The Divine Proportion, The Golden Section, or The Golden Ratio.
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Blatm
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### Re: The Golden Ratio

I've heard of many claims of phi appearing in nature (the distance from head to toe divided by the distance from belly-button to toe and similar relations in the human body), but I've never heard an explanation of WHY this is true, and as a result, I get the impression that much of this is simply coincidence. I'd love it if someone could justify these claims?

And to contribute: Phi and -1/Phi are the only two numbers that are one less than their squares (duh).

heyitsguay
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### Re: The Golden Ratio

[imath]\varphi+1.53256[/imath] is approximately [imath]\pi[/imath].

Proof that God exists!

t0rajir0u
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### Re: The Golden Ratio

Consult the following resources:

http://www.amazon.com/Golden-Ratio-Worl ... 0767908163
http://en.wikipedia.org/wiki/Golden_ratio
http://mathworld.wolfram.com/GoldenRatio.html

Incidentally, that book happens to discuss the appearance of the golden ratio in nature in depth and concludes that as far as aesthetics go the actual "golden ratio" is probably just 1.6. It's good to be careful about these things.

NathanielJ
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### Re: The Golden Ratio

Blatm wrote:I've heard of many claims of phi appearing in nature (the distance from head to toe divided by the distance from belly-button to toe and similar relations in the human body), but I've never heard an explanation of WHY this is true, and as a result, I get the impression that much of this is simply coincidence. I'd love it if someone could justify these claims?

I've heard many things like that about the body too (such as ratio of arm length to forearm length), but that's all nonsense. It's not phi, it's just some ration in the 1.5 - 2.0 regio, as there's absolutely no reason for body proportions to have anything to do with phi.

For things like sunflowers, on the other hand, it's quite true and it's simply because the ratio of successive terms in the Fibonacci sequence approaches phi. That wiki page that I just linked to even has the sunflower example near the bottom.
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qinwamascot
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### Re: The Golden Ratio

Phi has a continued fractional approximation of [1,1,1...] which means it's rational approximations converge slowest of any number. There are some reasons to believe phi might occasionally show up in nature because of this property. But it's certainly being overblown right now.
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gmalivuk
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### Re: The Golden Ratio

The Fibonacci sequence does show up in natural spirals because of the properties of efficiently packing things (like sunflower seeds) in that arrangement. And then, yeah, phi shows up due to its connection with that sequence. But it's far, far more interesting mathematically than naturally.
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Xanthir
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### Re: The Golden Ratio

gmalivuk wrote:The Fibonacci sequence does show up in natural spirals because of the properties of efficiently packing things (like sunflower seeds) in that arrangement. And then, yeah, phi shows up due to its connection with that sequence. But it's far, far more interesting mathematically than naturally.

Gmalivuk got it. The Fibonacci sequence actually *is* useful in nature, and the ratio of successive terms of the Fibonacci sequence (in general, successive terms of Lucas sequences) converges to phi.

As noted by NathanielJ, all the nonsense about the Golden Ratio having some connection with human body and such is just that - nonsense. Phi is approximately 1.6, and a lot of things in the body have approximately a ratio of 1.6 (*very* approximately in many of the supposed Golden Ratio connections).

It *is* really cool mathematically, though. In addition to having a continued fraction of [1;1,1,1,1,1...] it has an infinite nested radical of sqrt(1+sqrt(1+sqrt(1+...))).
(defun fibs (n &optional (a 1) (b 1)) (take n (unfold '+ a b)))

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### Re: The Golden Ratio

I like [imath]\phi=\frac{1+\sqrt{5}}{2}[/imath] because it keeps cropping up in the geometry of the pentagon (and by extension, the pentagram and the dodecahedron).

Fun trigonometry practice: prove that [imath]\cos(\pi/5)[/imath] is equal to half the golden ratio.

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Lóng the Dragon
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### Re: The Golden Ratio

About the golden ratio appearing in nature, I might have a viable explanation:

The golden ratio is, by the mind, perceived as the "ideal", most aesthetic ratio*. Now if through evolution, animals would pick their partners based on looks, and the golden ratio is the ideal look, animals would acquire more and more golden ratios over time, because they are the ones that would reproduce the most (natural selection).

* If this is taught, or completely untrue, never mind.
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qinwamascot
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### Re: The Golden Ratio

That doesn't really make sense at all. Phi being "aesthetically appealing" to humans would make little difference in the sexual selection of any other organism, and it's unlikely that it would even affect humans because mathematical ability is something that is learned, not innate. Plus, this would not explain why phi would appear in plants or fungi, both of which don't undergo the same process of sexual selection that animals do.
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Xanthir
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### Re: The Golden Ratio

Lóng the Dragon wrote:About the golden ratio appearing in nature, I might have a viable explanation:

The golden ratio is, by the mind, perceived as the "ideal", most aesthetic ratio*. Now if through evolution, animals would pick their partners based on looks, and the golden ratio is the ideal look, animals would acquire more and more golden ratios over time, because they are the ones that would reproduce the most (natural selection).

* If this is taught, or completely untrue, never mind.

Incorrect, as qinwamascto pointed out, for several reasons. Phi only appears in nature because of the efficiency of using the Fibonacci sequence in some cases, and because it's extremely easy to find things that have a ratio of approximately 1.6.
(defun fibs (n &optional (a 1) (b 1)) (take n (unfold '+ a b)))

greycloud
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### Re: The Golden Ratio

Lóng the Dragon wrote:About the golden ratio appearing in nature, I might have a viable explanation:

The golden ratio is, by the mind, perceived as the "ideal", most aesthetic ratio*. Now if through evolution, animals would pick their partners based on looks, and the golden ratio is the ideal look, animals would acquire more and more golden ratios over time, because they are the ones that would reproduce the most (natural selection).

* If this is taught, or completely untrue, never mind.

I think this makes sense with some animals, though its probably just a coincidence that this number is similar to the golden ratio. People who look 'proportional', even today, are considered good looking. This is not neccessarily the case always, or with all species.

scikidus
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### Re: The Golden Ratio

Heh, I've done some research on (i.e. doodled incessantly on various sem-important pieces of paper about) the Fibonacci sequence.

Int he end I only came up with one "discovery," and I'm not even sure if I'm the first person to come up with this:

Fn refers to the nth Fibonacci number.

For any two numbers m and n:

Fn = ( Fm * Fn-m+1 ) + ( Fm-1 * Fn-m )

Example:

m = 5, n = 12

F12 = ( F5 * F12-5+1 ) + ( F5-1 * F12-5 )
F12 = ( F5 * F8 ) + ( F4 * F7 )
F12 = ( 5 * 21 ) + ( 3 * 13 )
F12 = ( 105 ) + ( 39 )
F12 = 144, and indeed it is.

I'm pretty darn sure that it's true, but I haven't rigorously proved it. (That is, I may have proved it at the time, but I no longer have that proof.)

The formula is interesting, however, because it allows you to quickly find Fibonacci numbers, and also to extend the sequence when n<0.

I haven't tried imaginary Fibonacci numbers. Anyone?
Happy hollandaise!

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Blatm
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### Re: The Golden Ratio

Spoiler:
By induction:

[imath]F_n = F_m F_{n-m+1} + F_{m-1} F_{n-m}[/imath] is true for m = 2.

If it's true for m, it's true for m+1, because the m+1 case reduces to the m case:

$F_{m+1}F_{n-m} + F_m F_{n-m-1} = F_{n-m}(F_m + F_{m-1}) + F_m F_{n-m-1} = F_m F_{n-m} + F_{m-1} F_{n-m} + F_m F_{n-m-1} = F_m(F_{n-m} + F_{n-m-1}) + F_{m-1} F_{n-m} = F_m F_{n-m+1} + F_{m-1} F_{n-m}$

Phi
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### Re: The Golden Ratio

I'm honestly a fan of: [imath]{1 \over \phi} = \phi - 1\,[/imath] and how simply badass that is.
I also like using the letter: I try to use it in place of theta for angles where I can, and then I get to use it for flux and I get happier.

scikidus
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### Re: The Golden Ratio

Blatm wrote:
Spoiler:
By induction:

[imath]F_n = F_m F_{n-m+1} + F_{m-1} F_{n-m}[/imath] is true for m = 2.

If it's true for m, it's true for m+1, because the m+1 case reduces to the m case:

$F_{m+1}F_{n-m} + F_m F_{n-m-1} = F_{n-m}(F_m + F_{m-1}) + F_m F_{n-m-1} = F_m F_{n-m} + F_{m-1} F_{n-m} + F_m F_{n-m-1} = F_m(F_{n-m} + F_{n-m-1}) + F_{m-1} F_{n-m} = F_m F_{n-m+1} + F_{m-1} F_{n-m}$

Thank you.

Do you know if this is well-known?
Happy hollandaise!

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Talith
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### Re: The Golden Ratio

I'm surprised noone's mentioned this awsome formula which can be found by solving the linear difference equation for the fibonacci sequence.

Spoiler:
$F(n) =\:\:\: \frac{\phi^n-\left(1-\phi\right)^n}{\sqrt5}$

Buttons
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### Re: The Golden Ratio

scikidus wrote:Fn refers to the nth Fibonacci number.

For any two numbers m and n:

Fn = ( Fm * Fn-m+1 ) + ( Fm-1 * Fn-m )

Though Blatm's proof is fine, everyone knows there's only one good way to prove Fibonacci identities, and that's with domino tilings. Recall that [imath]F_n[/imath] is the number of ways to tile a [imath]1\times (n-1)[/imath] board using squares and dominos. (Why? See the second spoiler.) So:
Spoiler:
Question: How many ways can you tile a [imath]1\times (n-1)[/imath] board using squares and dominos?

Answer 2: Consider any [imath]m < n-1[/imath], and look at the fault line between the [imath]m-1[/imath]th spaces and the [imath]m[/imath] space. Is this line spanned by a domino? If not, there are [imath]F_m[/imath] ways to tile the first [imath]m-1[/imath] spaces, and [imath]F_{n-m+1}[/imath] ways to tile the other [imath]m-1[/imath] spaces. If so, there are [imath]F_{m-1}[/imath] ways to tile the first [imath]m-2[/imath] spaces before the domino, and then there are [imath]F_{n-m}[/imath] ways to tile the [imath]n-m-1[/imath] spaces after the domino. In total, there are [imath]F_m\cdot F_{n-m+1} + F_{m-1} \cdot F_{n-m}[/imath] ways to do it.

Since the two answers count the same thing, they are equal.
Elaboration:
Spoiler:
There's one way to "tile" the empty board, of course, and there's one way to tile a [imath]1\times1[/imath] board. How many ways can you tile a [imath]1\times(n-1)[/imath] board? Well, each tiling ends with a square or a domino, so if you remove it you're left with a tiling of a [imath]1\times(n-2)[/imath] board or a tiling of a [imath]1\times(n-3)[/imath] board. So the number of tilings satisfies the same initial conditions and recurrence relation as the Fibonacci numbers.
Oh, and is this well-known? To those who know it, sure. I'm almost positive it's in Proofs that Count, though I don't have a copy on hand.

Xanthir
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### Re: The Golden Ratio

Buttons wrote:Though Blatm's proof is fine, everyone knows there's only one good way to prove Fibonacci identities, and that's with domino tilings. Recall that [imath]F_n[/imath] is the number of ways to tile a [imath]1\times (n-1)[/imath] board using squares and dominos. (Why? See the second spoiler.)

What I think even more interesting is that, if Wikipedia is to be believed, this was actually discovered 1400 years ago.
(defun fibs (n &optional (a 1) (b 1)) (take n (unfold '+ a b)))

skeptical scientist
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### Re: The Golden Ratio

Buttons wrote:Recall that [imath]F_n[/imath] is the number of ways to tile a [imath]1\times (n-1)[/imath] board using squares and dominos.

And here all along I thought it was the number of ways to tile a [imath]2\times (n-1)[/imath] board using dominos.
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nazlfrag
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### Re: The Golden Ratio

How about logarithmic spirals? They can be approximated by a sequence of golden rectangles with the square of the shortest side removed, and appear quite often in nature. I love phi, it's the second most delicious number.

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### Re: The Golden Ratio

nazlfrag wrote:How about logarithmic spirals? They can be approximated by a sequence of golden rectangles with the square of the shortest side removed, and appear quite often in nature. I love phi, it's the second most delicious number.

That's really only true when the logarithmic spiral is a golden spiral, whose growth rate is already related to phi. A general logarithmic spiral really has nothing to do with the golden ratio, and so this isn't an occurrence of the golden ratio in nature.
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jjono
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### Re: The Golden Ratio

This is to do with the Fibonacci sequence, rather than the golden ratio, but I like how if [imath]n[/imath] divides [imath]m[/imath], then [imath]F_n[/imath] divides [imath]F_m[/imath].

Admittedly, this is true for a large class of sequences, but I still like it =)

t0rajir0u
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### Re: The Golden Ratio

The Mathworld and Wikipedia articles contain some nice gems. I blog about the Fibonacci numbers and phi a fair bit: see

- Problem 3 here,
- a nice problem involving phi here,
- the Fibonacci numbers and continued fractions,
- the Fibonacci numbers and matrices (and the subsequent follow-up about the Fibonacci numbers and generating functions).

chapel
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### Re: The Golden Ratio

scikidus wrote:Heh, I've done some research on (i.e. doodled incessantly on various sem-important pieces of paper about) the Fibonacci sequence.

Int he end I only came up with one "discovery," and I'm not even sure if I'm the first person to come up with this:

Fn refers to the nth Fibonacci number.

For any two numbers m and n:

Fn = ( Fm * Fn-m+1 ) + ( Fm-1 * Fn-m )

I haven't tried imaginary Fibonacci numbers. Anyone?

Or maybe uncountable Fibonacci numbers? If m=2, [imath]F_{\omega_1} = (F_2 * F_{\omega_1 - 2 + 1}) + (F_{2 - 1} + F_{\omega_1 - 2}) = (F_2 * F_{\omega_1 - 1}) + (F_1 * F_{\omega_1 - 2}) = F_{\omega_1 - 1} + F_{\omega_1 - 2}[/imath]

Umm... well that didn't help. At least it is consistent meaning my algebra was right. And was is the ancestor of a limit ordinal?

In more seriousness, I think that using a number other than m=2 and assuming something like the Generalized Continuum Hypothesis might give something a little more useful.

scikidus
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### Re: The Golden Ratio

chapel wrote:Or maybe uncountable Fibonacci numbers? If m=2, [imath]F_{\omega_1} = (F_2 * F_{\omega_1 - 2 + 1}) + (F_{2 - 1} + F_{\omega_1 - 2}) = (F_2 * F_{\omega_1 - 1}) + (F_1 * F_{\omega_1 - 2}) = F_{\omega_1 - 1} + F_{\omega_1 - 2}[/imath]

Umm... well that didn't help. At least it is consistent meaning my algebra was right. :¡This cheese is burning me!: And was is the ancestor of a limit ordinal?

In more seriousness, I think that using a number other than m=2 and assuming something like the Generalized Continuum Hypothesis might give something a little more useful.

Hell, while we're at it, let's use this to find a pattern in the primes.

For any two primes p and q,

Fp = ?

Come on, let's figure it out. There are only, over 2*10^6 dollars at stake.
Happy hollandaise!

"The universe is a figment of its own imagination" -Douglas Adams

chapel
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### Re: The Golden Ratio

scikidus wrote:
chapel wrote:Or maybe uncountable Fibonacci numbers? If m=2, [imath]F_{\omega_1} = (F_2 * F_{\omega_1 - 2 + 1}) + (F_{2 - 1} + F_{\omega_1 - 2}) = (F_2 * F_{\omega_1 - 1}) + (F_1 * F_{\omega_1 - 2}) = F_{\omega_1 - 1} + F_{\omega_1 - 2}[/imath]

Umm... well that didn't help. At least it is consistent meaning my algebra was right. :¡This cheese is burning me!: And was is the ancestor of a limit ordinal?

In more seriousness, I think that using a number other than m=2 and assuming something like the Generalized Continuum Hypothesis might give something a little more useful.

Hell, while we're at it, let's use this to find a pattern in the primes.

For any two primes p and q,

Fp = ?

Come on, let's figure it out. There are only, over 2*10^6 dollars at stake.

Sounds good, anyone want a Field's Medal? All you have to do is plug some prime numbers into this formula.

nazlfrag
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### Re: The Golden Ratio

Fp= q

if p=5 then q=5
and so on.

Therefore, p=q.

Can I have my medal now?

Tibixe
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### Re: The Golden Ratio

A gentleman wearing a black hat would say that [imath]\phi[/imath] and [imath]\frac{1}{\phi}[/imath] are nothing more than the eigenvalues of $\begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}$

oblivimous
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### Re: The Golden Ratio

There was just a problem on the craigslist math/sci forum:

Find a fifth degree polynomial with two equal maximums and two equal minimums.

After a bit of organized trial-and-error I found
Spoiler:
x^5-5x^3+5x

I was suprised to discover that the extrema of this function occur at +/- phi and +/- (phi - 1)

Buttons
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### Re: The Golden Ratio

Tibixe wrote:A gentleman wearing a black hat would say that [imath]\phi[/imath] and [imath]\frac{1}{\phi}[/imath] are nothing more than the eigenvalues of $\begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}$
I hate that guy! He's always inserting sign errors into things.

t0rajir0u
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### Re: The Golden Ratio

oblivimous wrote:After a bit of organized trial-and-error I found
Spoiler:
x^5-5x^3+5x

I was suprised to discover that the extrema of this function occur at +/- phi and +/- (phi - 1)

Oh, but you shouldn't be! This polynomial is the unique fifth-degree polynomial [imath]T_5(x)[/imath] with the property that
$T_5(2 \cos \theta) = 2 \cos 5\theta$
(which is why it has the property you wanted it to have). This also explains the location of the extrema; see Wikipedia for more information.