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The golden ratio. It has the property that
[1] A/B = (A+B)/A
Opening the brackets, we have
A/B = A/A + B/A
Since A/A=1, we have
A/B = B/A + 1
If we assume B=1, then
A/1 = 1/A + 1
or, more simply,
[2] A = 1/A + 1
Multiplying both sides by A,
A^2 = A/A + A
Replace A/A with 1:
A^2 = A + 1
Subtract one from the other:
A^2 - A - 1 = 0
Applying the usual quadratic solution formula, we obtain two solutions:
A = (1/2) (1 + Sqrt 5) ≈ 1.618 033 988 749 895...
A = (1/2) (1 - Sqrt 5) ≈ -0.618 033 988 749 895...
The first solution is "the golden ratio", commonly denoted by a
lowercase phi (φ).
As required by [2], 1/φ is 0.618... If you add 1 to that, you get φ
again - exactly as [2] demands. (Apparently this latter quantity, 1/φ =
φ - 1, is sometimes denoted by an uppercase phi, Φ. Notice that -Φ is
the other solution to our equation.)
Many people have noticed that the ratios of successive Fibonacci numbers
approach φ. For those that don't know, the Fibonacci numbers are defined as:
Fib[0] = 1
Fib[1] = 1
Fib[n] = Fib[n-1] + Fib[n-2]
1 + 1 = 2
1 + 2 = 3
2 + 3 = 5
3 + 5 = 8
5 + 8 = 13
8 + 13 = 21
13 + 21 = 34
21 + 34 = 55
34 + 55 = 89
55 + 89 = 144
1 / 1 = 1
2 / 1 = 2
3 / 2 = 1.5
5 / 3 = 1.666 666 666 666...
8 / 5 = 1.6
13 / 8 = 1.625
21 / 13 = 1.615 384 615 384...
34 / 21 = 1.619 047 619 047...
55 / 34 = 1.617 647 058 823...
89 / 55 = 1.618 181 818 181...
144 / 89 = 1.617 977 528 089...
φ ≈ 1.618 033 988 749...
Note that φ itself is irrational, so no fraction can ever exactly equal
it. And yet, this sequence approximates it with ever increasing precision.
To many people, this is a seemingly magical property. BUT WAIT! It turns
out that if you apply the same rules but starting with ANY NUMBERS YOU
WANT, the same thing happens!
Pic[0] = 31
Pic[1] = 42
Pic[n] = Pic[n-1] + Pic[n-2]
31 + 42 = 73
42 + 73 = 115
73 + 115 = 188
115 + 188 = 303
188 + 303 = 491
303 + 491 = 794
42 / 31 = 1.354 838 709 677...
73 / 42 = 1.738 095 238 095...
115 / 73 = 1.575 342 465 753...
188 / 115 = 1.634 782 608 695...
303 / 188 = 1.611 702 127 659...
303 / 491 = 1.620 462 046 204...
φ ≈ 1.618 033 988 749...
The initial two terms are randomly chosen. Their ratio is nowhere near φ
(other than that they're roughly the same size, so their ratio is
roughly 1). But a few terms later, and they already agree with φ to
three significant figures.
Even if we start with widely different terms, this happens:
1 + 10 = 11
10 + 11 = 21
11 + 21 = 32
21 + 32 = 53
32 + 53 = 85
53 + 85 = 138
85 + 138 = 223
10 / 1 = 10
11 / 10 = 1.1
21 / 11 = 1.909 090 909 090...
32 / 21 = 1.523 809 523 809...
53 / 32 = 1.656 25
85 / 53 = 1.603 773 584 905...
138 / 85 = 1.623 529 411 764...
223 / 138 = 1.615 942 028 985...
φ ≈ 1.618 033 988 749...
Inexorably, the ratio approaches φ. But why?
Thinking about this intuitively, it becomes (slightly) less surprising.
Instead of looking at the symbols, think about what [1] actually
/means/. We have three quantities, {B, A, A+B}, and the ratio between
successive terms should be the same. We have the Fibonacci numbers,
which build each new term by adding together the previous two. Notice a
similarity there?
With the golden mean, we ask for a ratio such that successive steps
produced by adding always have the same ratio. With the Fibonacci
numbers, we /force/ each successive step to be produced by adding, and
the ratio then approaches the golden mean - seemingly regardless of the
starting point!
It's not dissimilar to the iterative blind deconvolution algorithm; you
start with random data, transform it, apply known constraints to the
result, inverse transform it, apply known constraints to that result,
transform, constrain, inverse transform, constrain... Eventually, the
system converges to a solution that satisfies all the constraints -
hopefully your original, unblurred image. ;-)
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Relevant: http://vihart.com/
But I like your analysis better.
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On 16/01/2012 06:05 PM, Darren New wrote:
> Relevant: http://vihart.com/
>
> But I like your analysis better.
Why thank you honey. :-)
Also, I gather Vi has recently joined Khan Academy. Having never heard
of it, I of course decided to see what it's about. The introductory
video is very amusing, and Mr Khan seems like a nice guy. A lot of the
things he says make a lot of sense.
And then I picked a random YouTube video from Khan Academy. As it
happens, "Introduction to Matricies". The results were... unimpressive.
The video is 11 minutes long, and in that time I learned the following
facts:
1. A matrix is a grid of numbers. (I'd feel happier if he said
*rectangular* grid of numbers.)
2. To add two matrices, they must be the same size, and you add them
element by element to produce a result of the same size.
OK, so there were some examples in there. But 11 minutes to say that? I
could read out the above two sentences in about 11 *seconds*. Sure, if
you're trying to explain something for the first time, taking it slowly
and having a few examples is all good. But 11 minutes??
The reason seems to be that rather than having a tightly written script,
the guy seems to be just ad libbing the whole thing. He dithers, he
interrupts himself, he's indecisive, he rambles a lot. That's why it
takes him so long to explain so little.
Unlike Vi, who draws stuff on paper (and talks like a machine gun), this
guy laboriously doodles stuff with some kind of electronic whiteboard
software. The results are utterly illegible. So many of his videos have
little comments on them such as "this is a 3, not a 5". Quite apart from
being hard to read, it looks *ugly*.
Now, to be clear, I only looked at, like, 3 videos. Maybe that's not a
representative sample. And, if the blurb is to be believed, there are a
vast number of such videos available, covering a wide range of topics,
/and/ nicely categorised and arranged. I gather there's some sort of
interface where you can track your progress, find related topics, and so
forth. All of which /is/ very cool... but I can't help feeling that the
actual videos could be done better.
It'll be interesting to see what happens with Vi there. The video with
him and her seemed... awkward. But we'll see. It's a sensible alliance;
nobody does slight video quite like Vi.
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http://www.youtube.com/watch?v=wRnSnfiUI54
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On 20/01/2012 12:18 PM, nemesis wrote:
> http://www.youtube.com/watch?v=wRnSnfiUI54
GAH! >_<
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On 1/16/2012 11:05 AM, Invisible wrote:
> To many people, this is a seemingly magical property. BUT WAIT! It turns
> out that if you apply the same rules but starting with ANY NUMBERS YOU
> WANT, the same thing happens!
>
> Pic[0] = 31
> Pic[1] = 42
> Pic[n] = Pic[n-1] + Pic[n-2]
>
> 31 + 42 = 73
> 42 + 73 = 115
> 73 + 115 = 188
> 115 + 188 = 303
> 188 + 303 = 491
> 303 + 491 = 794
>
> 42 / 31 = 1.354 838 709 677...
> 73 / 42 = 1.738 095 238 095...
> 115 / 73 = 1.575 342 465 753...
> 188 / 115 = 1.634 782 608 695...
> 303 / 188 = 1.611 702 127 659...
> 303 / 491 = 1.620 462 046 204...
> φ ≈ 1.618 033 988 749...
If you start with any two integers, at least one of which is not zero,
and proceed, the series that follows can be constructed from some
combination of the original Fibonacci series, with some of the input
series being shifted, and some being multiplied by some factor.
For instance, if you start with 1 and 3, the series you get is:
1 3 4 7 11 18 29 47 76 123
Now subtract from that the same series beginning with 0 and 1, and you get:
1 3 4 7 11 18 29 47 76 123
0 1 1 2 3 5 8 13 21 34
---------------------------
1 2 3 5 8 13 21 34 55 89
Which is the (0,1) series, shifted to the left by two members.
Since the two component series have member-to-member ratios that
converge to phi, intuition suggests that the sums of individual series
members will also have a ratio converging to this same value.
The rigorous proof that all such series will have a member-to-member
ratio approaching phi will entail showing that for any two non-negative
integers, A and B, with B>A, that the value of B/(A+B) is closer to phi
(1.618~) than the value A/B.
|A/B - phi| > |B/(A+B) - phi| for all A and B, 0 < A < B
The full proof is left as an exercise for the interested student.
Regards,
John
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On 23/01/2012 10:02 AM, John VanSickle wrote:
> The rigorous proof that all such series will have a member-to-member
> ratio approaching phi will entail showing that for any two non-negative
> integers, A and B, with B>A, that the value of B/(A+B) is closer to phi
> (1.618~) than the value A/B.
>
> |A/B - phi| > |B/(A+B) - phi| for all A and B, 0 < A < B
>
> The full proof is left as an exercise for the interested student.
I've always found an intuitive understanding to be more interesting than
a rigorous proof.
For example, the square root of 2 is irrational. The standard way to
prove this is to demonstrate that root 2 = A/B implies that A/B can be
cancelled down an infinite number of times, which is clearly impossible.
This shows that the number *is* irrational. But it gives no indication
as to *why* it is impossible.
By contrast, root 2 = A/B implies that A^2/B^2 = 2. The only way this
can be true is if A^2 is twice as large as B^2. And /that/ would mean
that if you look at the prime factors of each number, A^2 would have one
more 2 factor than B^2 does. However, the prime factors of X^2 are
exactly the prime factors of X, but repeated twice. This entails that
EVERY SQUARE NUMBER HAS AN _EVEN_ NUMBER OF PRIME FACTORS. But for A^2
to have *one* extra factor of 2, either A^2 or B^2 would have to have an
ODD number of 2 factors - which is impossible.
Here we clearly see /why/ you cannot make this equation work. We also
see exactly which numbers can and cannot work in the equation. Namely,
A^2/B^2 can be equal to any number having EVEN numbers of each prime
factor. In other words, SQUARE NUMBERS. So the square root of any
integer is either an integer or an irrational number. It can never be
some non-integer rational.
Of course, from a rigorous point of view, proving that a fraction
cancels down forever merely involves some trivial algebra, whereas
proving the properties of unique factorisation and so forth is a major
undertaking. I still find it much more intuitively illuminating.
Likewise, the Fibonacci numbers demonstrate how repeatedly applying a
constraint to something causes it to approximate the unique solution to
that constraint. That's a much more intuitive idea than any amount of
fidgeting with equations.
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On 1/23/2012 2:17, Invisible wrote:
> 2, either A^2 or B^2 would have to have an ODD number of 2 factors - which
> is impossible.
I've never seen that proof before. Very cool.
--
Darren New, San Diego CA, USA (PST)
People tell me I am the counter-example.
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On 27/01/2012 04:08 AM, Darren New wrote:
> On 1/23/2012 2:17, Invisible wrote:
>> 2, either A^2 or B^2 would have to have an ODD number of 2 factors -
>> which
>> is impossible.
>
> I've never seen that proof before. Very cool.
Meh. Just /think/ about it for a moment or two... Nothing I've said is
non-obvious if you play with the ideas for a moment.
Indeed, the non-obvious thing is that by performing algebraic
manipulations on A^2/B^2=2, you can prove that A and B are both even...
Would you like an encore of my derivation of the quadratic solution
formula? ;-)
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On 1/23/2012 5:02 AM, John VanSickle wrote:
> The rigorous proof that all such series will have a member-to-member
> ratio approaching phi will entail showing that for any two non-negative
> integers, A and B, with B>A, that the value of B/(A+B) is closer to phi
> (1.618~) than the value A/B.
>
> |A/B - phi| > |B/(A+B) - phi| for all A and B, 0 < A < B
>
> The full proof is left as an exercise for the interested student.
Although on second look it appears that you will have better results if
you try proving this instead:
|B/A - phi| > |(A+B)/B - phi| for all A and B, 0 < A < B
Regards,
John
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