Why does this number work?

MathsIsFun · 2014-03-29 10:50:13

A gentle reader emailed me, after visiting Nature The Golden Ratio and Fibonacci Numbers, saying:

"With the sunflower app we discovered that 0.15656 works really well too. Why is this?"

Is it related to the Golden Ratio? Why does it work?

Agnishom · 2014-03-29 13:44:22

Very Interesting! Could we get more digits?

bobbym · 2014-03-29 13:49:08

It is an empirical result and it is unlikely that we will get more digits.

Agnishom · 2014-03-29 13:52:17

and I am sure that Mr. bumpkin already knows why the number works, yes?

eigenguy · 2014-03-29 13:57:20

It seems to work pretty much the same no matter how many more digits you add. If you assume repeating 56, then it is the rational number 31/198.

(Speaking of which,

But the Golden Ratio (its symbol is the Greek letter Phi, shown at left) is an expert at not being any fraction. It is an Irrational Number (meaning you cannot write it as a simple fraction), but more than that ... it is as far as you can get from being near any fraction.

SAY WHAT??? Technically this is true. It isn't rational, and it is just as far from rational numbers as all other irrational numbers (distance = 0).

I have no doubt that you know better than this statement. So I am curious what you really meant by it.)

Agnishom · 2014-03-29 13:58:30

It seems to work pretty much the same no matter how many more digits you add.

That is the limitation of the applet. Within such a small place, there is not much hope that you can distinguish between such insignificant differences.

Last edited by Agnishom (2014-03-29 13:59:25)

eigenguy · 2014-03-29 14:04:08

I think to understand it, I would need to know a bit more about how the animation works. How fast do the seeds grow as you move out? how far do you move out with each new seed. I suspect that with these parameters, you can put together a polynomial equation whose roots include the golden ratio, but also something close to this number.

Agnishom · 2014-03-29 14:05:44

Only MIF can answer those questions

bobbym · 2014-03-29 14:11:51

Hi eigenguy;

But the Golden Ratio (its symbol is the Greek letter Phi, shown at left) is an expert at not being any fraction. It is an Irrational Number (meaning you cannot write it as a simple fraction), but more than that ... it is as far as you can get from being near any fraction.

Who said that? Link please.

To PSLQ it I would need more digits. eigenguy's answer is possible as well as a lot more.

eigenguy · 2014-03-29 14:15:10

It is straight off that page, just a little below the animation (next to the big phi).

bobbym · 2014-03-29 15:04:49

Yes, it is irrational and irrational numbers can be approximated to any degree of accuracy we desire by a rational number so the limit of the distance is 0 because we can make epsilon as small as we like.

I think what he meant was in character it is very different from a rational number.

A point the page makes is that the number they found like any other we can enter into the input box, will eventually veer off from the spiral. Each one will be a truncated decimal and thus a fraction.

eigenguy · 2014-03-29 15:19:12

Any number you enter actually will be a fraction. The real question is why some fractions work better than others, and this relates to which irrational numbers they approximate. It appears that algebraic numbers in general do better than others, but the reason isn't clear to me.

Another number that works well (though not quite as well) is sqrt(2) - 1 = 0.414.
Sqrt(3), sqrt(5), sqrt(7) are not as good, but still tighter than most numbers.

eigenguy · 2014-03-29 15:45:06

As an interesting tidbit, trying to explore what MIF intended by his furthest from fractions, I looked at | nx - round(nx) | for various inputs x to the animation and integers n. What I found was that the lowest value of n for which that distance drops below approximately 0.03 seems to be the number of "seeds" on the outside when animation stops +/- 1 when run on input x.

( It is also interesting to note that when x = golden ratio, the values of n for which the distance hits a new record low are 1, 2, 3, 5, 8, 13, ... . Not really surprising, but interesting just the same.)

bobbym · 2014-03-29 16:36:49

Looked at some literature and It is very complicated.

Bob · 2014-03-29 19:41:11

Try Fibonacci but with new seed numbers (no pun intended )

Bob

eigenguy · 2014-03-30 05:55:17

Fibonacci with new seed numbers just gives you the golden ratio again. It isn't the initial values that makes the Fibonacci ratios converge to phi, but rather the particular recursion rule F[sub]n[/sub] = F[sub]n-1[/sub] + F[sub]n-2[/sub]. To get a different ratio for the limit, you need a different recursion. For example:

A[sub]n[/sub] = 2A[sub]n-1[/sub] + A[sub]n-2[/sub] implies A[sub]n+1[/sub]/A[sub]n[/sub] ---> 1 + √2
B[sub]n[/sub] = 6B[sub]n-1[/sub] - 2B[sub]n-2[/sub] implies B[sub]n+1[/sub]/B[sub]n[/sub] ---> 3 + √7

MathsIsFun · 2014-03-30 09:04:06

So perhaps it is just a byproduct of my app's algorithm.

ShivamS · 2014-03-30 09:07:28

Works very well!

eigenguy · 2014-03-31 10:36:10

MathsIsFun wrote:

So perhaps it is just a byproduct of my app's algorithm.

Well, of course it's a product of the algorithm. That doesn't mean it isn't interesting. It appears to be a fundamental property of the algorithm itself, not just an artifact of computing limitations (though I could be wrong about that).

I would still like to know what exactly you meant when you wrote that phi was farthest from fractions. And an outline of the algorithm.

Bob · 2014-03-31 19:50:41

If the angular rotation required to go from one 'seed' to the next is a simple fraction of 360, then it creates lines of weakness and maybe space complications in fitting the next seed in.

As I understand it atan(phi) is significant in this respect because it is the irrational that 'avoids' being simplified to such a fraction. There's a proof somewhere although I haven't managed to track it down yet. The presumption is that evolution has 'found' this rotation and that is why sunflower seed heads (and many other similar plant properties) display the Fibonacci sequence as they grow.

But some plants have similar, but not standard, Fibonacci properties. After all, these characteristics have arisen as a result of natural selection and, in nature, other solutions also work.

I calculated 180 - atan(0.15656) and started to make the diagram below by rotating points around the circle. I can do many steps quickly, but not letter them at the same time, so what you see here is just the first few steps. Essentially, it's the same as MIF's program, but not so slick. I needed to try it for myself to see the result slowly emerge. If you choose an 'easy fraction' of 360, like 90 degrees, you get seeds landing along a straight line. On a seed head this would create a line of weakness and there would be a high risk of the head falling apart.

That's as far as I understand it. I'm eager to learn more about this, especially that proof.

Bob

eigenguy · 2014-04-01 09:06:51

The continued fraction for phi is [1; 1, 1, 1, ...]
The continued fraction for 0.15656 approximates [0; 6, 2, 1, 1, 2, 1, 1, ...] (I didn't expand it all the way out, so I don't know exactly where it breaks off the pattern.)
If we assume that this is the "real" value, it comes out to be (16 - √10) / 82 = 0.156558...
√2 has continued fraction [1; 2, 2, 2, ...]

I haven't tried some of the others, but it looks like maybe low values in the continued fraction expression is best, or may near constant, or maybe both.

Bob · 2014-04-01 09:15:28

hi eigenguy,

I've only got a vague idea what continued fractions are, so I'll have to do some studying.

Bob

ShivamS · 2014-04-01 09:16:24

Don't say that word! Continued fractions are the most horrid thing I have ever encountered!

Bob · 2014-04-01 09:25:38

Oh! Thanks for the encouragement.

Bob

eigenguy · 2014-04-01 14:37:47

They are horrible for computing, but they have some interesting properties. A continued fraction is an expression of the form:

This is denoted by [a[sub]0[/sub]; a[sub]1[/sub], a[sub]2[/sub], a[sub]3[/sub], ..., a[sub]n[/sub]] (note the leading term is separated by a ";:, the remainder by "," - a favored convention these days, but one I find foolish). The extension to infinite continued fractions is obvious. Canonically, a[sub]0[/sub] is an integer, and all the other a[sub]i[/sub] are positive integers. Under that condition, every rational number has exactly two continued fraction expansions, both of them finite (one of which ends in 1, while the other has a[sub]n[/sub] > 1). Every irrational number has exactly 1 canonical continued fraction, which is infinite. Quadratic numbers (solutions of quadratic equations with rational coefficients) have, eventually, a repeating pattern in their expansions. All other irrational numbers do not repeat. But e = [2; 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8, ...], continuing in the same pattern.

You can allow the a[sub]i[/sub] to be other than integer, in which case you don't get unique expansions anymore, but which also allows you to express some useful relationships that are undefined if you demand positive integers only. For example, [a[sub]0[/sub]; a[sub]1[/sub], a[sub]2[/sub], a[sub]3[/sub], ...] = [a[sub]0[/sub]; a[sub]1[/sub], a[sub]2[/sub], a[sub]3[/sub], ..., a[sub]n-1[/sub], [a[sub]n[/sub]; a[sub]n+1[/sub], ...]]

To find the continued fraction expansion of a number x: let x[sub]0[/sub] = x, a[sub]0[/sub] = floor(x), the greatest integer <= x. Then for all n > 0, x[sub]n[/sub] = 1/(x[sub]n-1[/sub] - a[sub]n-1[/sub]) and a[sub]n[/sub] = floor(x[sub]n[/sub]). Then x = [a[sub]0[/sub]; a[sub]1[/sub], a[sub]2[/sub], ...].

To show the usefulness of continued fractions, consider our mystery number x = 0.15656. Following the procedure gives x = [0; 6, 2, 1, 1, 2, 1, 1, ...] (as a rational number, its continued fraction will terminate, but the spreadsheet I used to calculate this went screwy at this point, as the error grew out of control). Looking at this, I decided to look at the quadratic number whose continued fraction follows that pattern of repeating 2, 1, 1 forever. Let y = [2; 1, 1, 2, 1, 1, ...], then x = [0; 6, y], and y = [2; 1, 1, y]. That latter equation is:

which simplifies to (noting that y > 0)

Now, x = [0; 6, y], so

Which turns out to be 0.156558..., very close to our original number.

Math Is Fun Forum

#1 2014-03-29 10:50:13

Why does this number work?

#2 2014-03-29 13:44:22

Re: Why does this number work?

#3 2014-03-29 13:49:08

Re: Why does this number work?

#4 2014-03-29 13:52:17

Re: Why does this number work?

#5 2014-03-29 13:57:20

Re: Why does this number work?

#6 2014-03-29 13:58:30

Re: Why does this number work?

#7 2014-03-29 14:04:08

Re: Why does this number work?

#8 2014-03-29 14:05:44

Re: Why does this number work?

#9 2014-03-29 14:11:51

Re: Why does this number work?

#10 2014-03-29 14:15:10

Re: Why does this number work?

#11 2014-03-29 15:04:49

Re: Why does this number work?

#12 2014-03-29 15:19:12

Re: Why does this number work?

#13 2014-03-29 15:45:06

Re: Why does this number work?

#14 2014-03-29 16:36:49

Re: Why does this number work?

#15 2014-03-29 19:41:11

Re: Why does this number work?

#16 2014-03-30 05:55:17

Re: Why does this number work?

#17 2014-03-30 09:04:06

Re: Why does this number work?

#18 2014-03-30 09:07:28

Re: Why does this number work?

#19 2014-03-31 10:36:10

Re: Why does this number work?

#20 2014-03-31 19:50:41

Re: Why does this number work?

#21 2014-04-01 09:06:51

Re: Why does this number work?

#22 2014-04-01 09:15:28

Re: Why does this number work?

#23 2014-04-01 09:16:24

Re: Why does this number work?

#24 2014-04-01 09:25:38

Re: Why does this number work?

#25 2014-04-01 14:37:47

Re: Why does this number work?

Board footer