(Submitted on 30 Jan 2014 (v1), last revised 10 Mar 2014 (this version, v2))
I'm surprised that there is no other discussion of this in an actual news publication about mathematics (I read those for my occupation, and am a member of several online groups that discuss mathematical research), so I wonder if the recycled press release submitted here is really the only interest that this paper has gathered in the mathematical community.
Ken Ono is relentless in his self-promotion. You will find several similar press releases involving his work, not commensurable with its actual impact.
"Although no other algebraic units are as famous as the golden ratio, they are of central importance to algebra."
Arguably more famous algebraic numbers include:
0, 1, The square root of two, the square root of any integer, i, any integer, the nth root of any integer,...
I realize that you were highlighting some specific, well-known examples, but I'm finding it pretty funny that you could have just said "the nth root of any integer", which encompasses all of the numbers and sets you mentioned before it. (I'm a maths student; it's finals; everything is hilarious now)
I only agree with your first three examples (the third being debatable). Square or higher roots of arbitrary integers are definitely less famous than the golden ratio though...
Yes, I am curious to understand what sort of problems may be solved with this result. I know just enough number theory to be impressed by the finding, but I'm no mathematician and had to stop reading around page 6 of the paper :(
> "In mathematics, an algebraic number is a number that is a root of a non-zero polynomial in one variable with rational coefficients (or equivalently—by clearing denominators—with integer coefficients)."
I almost forgot, the set of algebraic numbers also includes complex numbers.
If you mean that some complex numbers are algebraic, then yes -- the imaginary unit i itself is algebraic, as it satisfies an polynomial equation x^2 + 1 = 0 with rational (in fact integer) coefficients. The set of algebraic numbers does not include all complex numbers, though.
Fun fact: the set of algebraic numbers is countable. If you recall that the set of rational numbers is also countable, then you get that the reals are uncountable only "because" of transcendental numbers (pi, e, Chapernowne's number, etc.).
As Alonzo Church and Alan Turing showed, the computable numbers [0] are countable too. The computable numbers include all the algebraic numbers and some transcendental numbers (including π and e), so the reals are uncountable "because" of those other transcendentals, the uncomputable numbers. Put differently, almost all reals are uncomputable.
This is really interesting. We could take it further and say that, given that some uncomputable reals have a finite definition (e.g. "the probability that a random algorithm halts"), there is a countable number of definable reals (by assigning a Godel number to each definition), so the uncountability of the reals is strictly due to indefinable numbers!
I said uncountably infinite. The parent seemed to be under the impression that a few specific transcendental numbers were the reason that the reals were uncountable. I was pointing out that there are uncountably infinite irrational numbers between any pair of rational numbers. (There being more irrational numbers than real transcendental numbers, as not all irrational numbers are transcendental.)
You are saying rational numbers are countable
but irrational numbers are not countable (uncountably infinite)
This is true.
The parent is saying that algebraic numbers are countable
and transcendental numbers are not countable.
The parent's statement is a level deeper and more remarkable.
Another way to say this, is that "almost all" numbers are transcendental.
The parent was not suggesting that his short list of examples represented the extent of transcendental numbers: despite almost all numbers being transcendental, they are individually hard for us to define, since the set is normally defined by excluding sets of numbers that we can easily define.
To clarify: transcendental numbers are a subset/a type of irrational numbers. https://en.wikipedia.org/wiki/Transcendental_number
If you take out transcendental numbers from the irrationals, those that remain are algebraic and therefore countable. You got it :)
The parent says, and I quote: "...then you get that the reals are uncountable only 'because' of transcendental numbers."
But this is untrue, is it not? Because even if you removed all of the real transcendentals, the real numbers would still be uncountable because of the irrationals.
(Edit: Or are the non-transcendental irrationals countable? I can see how that might be true?)
> Let F be a subfield of a field E. An element \alpha of E is said to be {algebraic} over F if there exist elements a_0, ..., a_n (n >= 1) of F, not all equal to 0, such that a_0 + a_1\alpha^n + ... + a_n\alpha^n = 0.
point being, even a sub-par undergrad knows how to generalize algebraic numbers using some machinery that was just beginning to be built in Ramanujan's time (and of course was completely unavailable to the man himself), and now these guys have plugged fields into something that i'd only heard of in the context of group theory and proved something useful. way to go. but at the same time, if Ramanujan saw this, i have to imagine he'd be thinking something along the lines of, "the game ain't the same." (not that it's a game.)
What's interesting to me is that you can demonstrate this without too much work, and it's true for starting numbers other than 1,1, such as Lucas numbers (1,3,4,7,11,...)
f(n+2)=f(n+1)+f(n)
If you posit that this number has a solution in f(n)=a^x, you see that
a^(n+2)=a^(n+1)+a^n
Dividing by a^n, you get a^2=a+1, a simple quadratic with two roots, (1+sqrt(5))/2 and (1-sqrt(5))/2, or the golden ratio and approximately -0.618.
Because any constant multiple of the function will also solve the equation, a solution to f(n) will be some linear combination of f(n)=c1(1.618..)^n+c2(-0.618)^n, for constants c1 and c2. You calculate c1 and c2 to match your initial conditions.
As n increases, the second term tends to 0, so f(n+1)/f(n) approaches the golden ratio.
> What's interesting to me is that you can demonstrate this without too much work, and it's true for starting numbers other than 1,1, such as Lucas numbers (1,3,4,7,11,...)
Your comment reminded me of chaos theory, small changes in initial conditions can have a radical effect upon the final outcome. In this case it seems to be the opposite.
> Because any constant multiple of the function will also solve the equation, a solution to f(n) will be some linear combination of f(n)=c1(1.618..)^n+c2(-0.618)^n, for constants c1 and c2. You calculate c1 and c2 to match your initial conditions.
Bear with me a second, I have to fish out an old geometry book...
"Geometry", Roger Fenn, Springer-Verlag 2001, page 24:
I've got it, the equation you gave is very similar to Binet's Formula:
Well, that's because it's an awful way to calculate a fibonacci number! At least you could memoize the function, remembering what the (n-1) number was instead of recalculating every time. Or, since you only need the last 2 numbers to keep going, forget everything but those 2:
unsigned long long fib(int a) {
unsigned long long x, y, temp;
while (a > 0) {
temp = x;
x = y;
y = y + temp;
--a;
}
return y;
}
> Well, that's because it's an awful way to calculate a fibonacci number!
To be more precise, it is an inefficient method of calculating a Fibonacci number. Using your function, if you were to measure the execution time of fib(n), then divide it by the execution time of fib(n-1), would you get a more precise value of the Golden ratio?
> This is much faster and won't blow up your stack quite so much.
Or we could use an iterative version ... wait up your code changed to an iterative implementation. Man that's weird.
EDIT:
Your code:
unsigned long long fib(int a) {
unsigned long long x, y, temp;
while (a > 0) {
temp = x;
x = y;
y = y + temp;
--a;
}
return y;
}
I just noticed: x, y and temp have not been initialised.
I have to admit, I did not understand your earlier comment until just now.
Sorry about the code change, the first version was from an old memory and the new version is what happened after I stared at it for a while :) Primitive values are automatically initialized to 0 in C. As for the ratio, it will probably just give a good approximation of 1 as a gets large! Also if you have a good square root function, you can use it to find a Fibonacci number in constant time, so your ratio will always be 1 (until the square root is too big to be computed in a single instruction).
> I have to admit, I did not understand your earlier comment until just now.
From the number of downvotes, I think it was mis-parsed by several HN readers. This happens to me quite often, on many a website, but I have no inkling why.
> Sorry about the code change, the first version was from an old memory and the new version is what happened after I stared at it for a while
No worries, I thought the original slice of code that you posted was more interesting and novel, it wouldn't have occurred to me off the top of my head.
> Primitive values are automatically initialized to 0 in C.
I had to look that up, good old Stack Overflow has again furnished us with an explanation.
Where are you getting at with the the t(n)?
As you said, the golden ratio is defined as the limit of f(n+1)/f(n), and yes, with this implementation (because f ~ t) t(n+1)/t(n) also converges to the golden ratio, but... so what, why not just use f ?
It's been done before. Calculating the golden ratio using your method provides insights into the world of mathematics. My method provides insights into the world of programming.
http://arxiv.org/abs/1401.7718
(Submitted on 30 Jan 2014 (v1), last revised 10 Mar 2014 (this version, v2))
I'm surprised that there is no other discussion of this in an actual news publication about mathematics (I read those for my occupation, and am a member of several online groups that discuss mathematical research), so I wonder if the recycled press release submitted here is really the only interest that this paper has gathered in the mathematical community.