6 ms·

Mathematicians trace source of Rogers-Ramanujan identities, find algebraic gold

136 points by zoowar 12y ago

carstimon 12y ago
"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,...
- tromp 12y ago
  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...
  - NAFV_P 12y ago
    > Square or higher roots of arbitrary integers are definitely less famous than the golden ratio though... Out of the list of algebraic numbers in question, zero is the most infamous.
- jdpage 12y ago
  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 12y ago
    Yeah, I guess those are the only ones I can think of that are more famous than the golden ratio :)
- NAFV_P 12y ago
  An obscure algebraic number: http://en.wikipedia.org/wiki/Plastic_number http://en.wikipedia.org/wiki/Plastic_number
iandanforth 12y ago
Can someone enlighten me as to how, as the article states, the Rogers-Ramanujan identities have played a role in Physics?
- anigbrowl 12y ago
  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 :(
- IvyMike 12y ago
  Considering statistical mechanics as a branch of physics: https://en.wikipedia.org/wiki/Hard_hexagon_model https://en.wikipedia.org/wiki/Hard_hexagon_model
NAFV_P 12y ago
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.
- xyzzyz 12y ago
  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.
  - deleted 12y ago
    [deleted]
- auvrw 12y ago
  more generally, from Lang: > 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.)
- surement 12y ago
  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.).
  - 12y ago
NAFV_P 12y ago
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.
- ipsin 12y ago
  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 12y ago
    > 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.html http://mathworld.wolfram.com/BinetsFibonacciNumberFormula.ht...
tokenadult 12y ago
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 12y ago
  Ken Ono is relentless in his self-promotion. You will find several similar press releases involving his work, not commensurable with its actual impact.
  - deleted 12y ago
    [deleted]

136 points by zoowar 12y ago

"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,...

tromp 12y ago
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...
- NAFV_P 12y ago
  > Square or higher roots of arbitrary integers are definitely less famous than the golden ratio though... Out of the list of algebraic numbers in question, zero is the most infamous.
jdpage 12y ago
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 12y ago
  Yeah, I guess those are the only ones I can think of that are more famous than the golden ratio :)
NAFV_P 12y ago
An obscure algebraic number: http://en.wikipedia.org/wiki/Plastic_number http://en.wikipedia.org/wiki/Plastic_number

iandanforth 12y ago

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

anigbrowl 12y ago
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 :(
IvyMike 12y ago
Considering statistical mechanics as a branch of physics: https://en.wikipedia.org/wiki/Hard_hexagon_model https://en.wikipedia.org/wiki/Hard_hexagon_model

NAFV_P 12y ago

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.

xyzzyz 12y ago
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.
- deleted 12y ago
  [deleted]
auvrw 12y ago
more generally, from Lang: > 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.)
surement 12y ago
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.).
- 12y ago

NAFV_P 12y ago

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.

ipsin 12y ago
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 12y ago
  > 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.html http://mathworld.wolfram.com/BinetsFibonacciNumberFormula.ht...

tokenadult 12y ago

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 12y ago
Ken Ono is relentless in his self-promotion. You will find several similar press releases involving his work, not commensurable with its actual impact.
- deleted 12y ago
  [deleted]