
Mathematicians trace source of Rogers-Ramanujan identities, find algebraic gold - zoowar
http://phys.org/news/2014-04-mathematicians-source-rogers-ramanujan-identities-algebraic.html
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tokenadult
This appears to be the arXiv.org link to the paper in question, "A framework
of Rogers-Ramanujan identities and their arithmetic properties."

[http://arxiv.org/abs/1401.7718](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.

~~~
ninguem2
Ken Ono is relentless in his self-promotion. You will find several similar
press releases involving his work, not commensurable with its actual impact.

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carstimon
"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,...

~~~
jdpage
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)

~~~
carstimon
Yeah, I guess those are the only ones I can think of that are more famous than
the golden ratio :)

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iandanforth
Can someone enlighten me as to how, as the article states, the Rogers-
Ramanujan identities have played a role in Physics?

~~~
IvyMike
Considering statistical mechanics as a branch of physics:
[https://en.wikipedia.org/wiki/Hard_hexagon_model](https://en.wikipedia.org/wiki/Hard_hexagon_model)

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NAFV_P
From wikipedia on algebraic numbers:

> "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.

~~~
surement
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.).

~~~
pash
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.

0\.
[https://en.wikipedia.org/wiki/Computable_number](https://en.wikipedia.org/wiki/Computable_number)

~~~
NAFV_P
Like Omega:

[http://en.wikipedia.org/wiki/Chaitin's_constant](http://en.wikipedia.org/wiki/Chaitin's_constant)

~~~
GregBuchholz
Can't mention Omega without linking to:

Meta Math!

[http://arxiv.org/abs/math/0404335](http://arxiv.org/abs/math/0404335)

~~~
Ind007
Such a good read.Thanks for sharing this.

~~~
JonnieCache
Check him out on youtube too, he's a fun speaker.

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NAFV_P
Regarding the golden ratio ....

The nth number in Fibonacci sequence is denoted by f(n)

The golden ratio is equal to the limit of f(n)/f(n-1) as n increases without
limit.

A recursive implementation of the nth number in the Fibonacci sequence in C:

    
    
      unsigned long long fib(int a) {
        return a>1 ? fib(a-1)+fib(a-2) : 1;
      }
    

The above function can take a while to execute if a is sufficiently large, on
my machine fib(40) takes about a second to return a value.

Time taken to execute the function fib(a) is denoted by t(a).

As n increases without limit t(n)/t(n-1) approaches the golden ratio.

In practise this should give you a rough value of the golden ratio.

The reason this happens is because the function can only increase the return
value by unity, one call at a time.

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ipsin
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.

~~~
NAFV_P
> _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:

[http://mathworld.wolfram.com/BinetsFibonacciNumberFormula.ht...](http://mathworld.wolfram.com/BinetsFibonacciNumberFormula.html)

