Re: [math-fun] Euler's crazy pi product
interesting how if you start with 1 you get a line of slope 1. you can also include all integers. π/4 = 1/2 × 2/2 × 3/2 × 4/4 × 5/6 × 6/6 × 7/6 ... if I've got that right. For the n-1/2 cases, round up and round down produce divergent products. Round to even gives 4*(1/4)!^2/sqrt(2*pi) . At last, a mathematical justification for round to even! --rwg On Sun, Feb 13, 2011 at 9:36 PM, Bill Gosper <billgosper@gmail.com <http://gosper.org/webmail/src/compose.php?send_to=billgosper%40gmail.com>> wrote:
http://math.ucr.edu/home/baez/week127.html finds in Lennart Berggren, Jonathan Borwein and Peter Borwein, π: A Source Book, Springer-Verlag, New York, 1997, "the following weirdly beautiful formula due to Euler, which unfortunately is not explained:" (Then how can they call it a sourcebook?--rwg)
3 5 7 11 13 17 19 π/2 = - × - × - × -- × -- × -- × -- × ... 2 6 6 10 14 18 18
"Here the numerators are the odd primes, and the denominators are the closest numbers of the form 4n+2." E.g., for 10, 100, ..., 10^6 terms, In[209]:=
Table[Block[{p=2},Nest[N[#*(p=NextPrime[p])/(2+4*Round[(p-2)/4]),9]&,2,10^n]],{n,6}]
Out[209]= {3.10152045, 3.13398462, 3.13772561, 3.14073685, 3.14143290, 3.14157195}
Why the heck?? Was Euler a preincarnation of Ramanujan? --rwg
In www.mathpages.com/home/kmath477.htm, explanation of: 3 5 7 11 13 17 19 π/4 = - × - × - × -- × -- × -- × -- × ... 4 4 8 12 12 16 20 Here the numerators are the odd primes, and the denominators are the closest numbers of the form 4n. Christian. -----Message d'origine----- De : math-fun-bounces@mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com] De la part de Bill Gosper Envoyé : lundi 14 février 2011 08:24 À : math-fun@mailman.xmission.com Objet : Re: [math-fun] Euler's crazy pi product interesting how if you start with 1 you get a line of slope 1. you can also include all integers. π/4 = 1/2 × 2/2 × 3/2 × 4/4 × 5/6 × 6/6 × 7/6 ... if I've got that right. For the n-1/2 cases, round up and round down produce divergent products. Round to even gives 4*(1/4)!^2/sqrt(2*pi) . At last, a mathematical justification for round to even! --rwg On Sun, Feb 13, 2011 at 9:36 PM, Bill Gosper <billgosper@gmail.com <http://gosper.org/webmail/src/compose.php?send_to=billgosper%40gmail.com>> wrote:
http://math.ucr.edu/home/baez/week127.html finds in Lennart Berggren, Jonathan Borwein and Peter Borwein, π: A Source Book, Springer-Verlag, New York, 1997, "the following weirdly beautiful formula due to Euler, which unfortunately is not explained:" (Then how can they call it a sourcebook?--rwg)
3 5 7 11 13 17 19 π/2 = - × - × - × -- × -- × -- × -- × ... 2 6 6 10 14 18 18
"Here the numerators are the odd primes, and the denominators are the closest numbers of the form 4n+2." E.g., for 10, 100, ..., 10^6 terms, In[209]:=
Table[Block[{p=2},Nest[N[#*(p=NextPrime[p])/(2+4*Round[(p-2)/4]),9]&,2 ,10^n]],{n,6}]
Out[209]= {3.10152045, 3.13398462, 3.13772561, 3.14073685, 3.14143290, 3.14157195}
Why the heck?? Was Euler a preincarnation of Ramanujan? --rwg
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Euler knew the factorization of zeta into the infinite product of terms p^s / (p^s -1), and the kmath477 page sketches how he could have reached the formula that Berggren quotes. For the product to converge, we need that the number of 4K+1 primes is close to the number of 4K-1 primes. I don't think this was proved before Dirichlet, and maybe not till the proof of the Prime Number Theorem. Rich -----Original Message----- From: math-fun-bounces@mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Christian Boyer Sent: Monday, February 14, 2011 2:09 AM To: 'math-fun' Subject: Re: [math-fun] Euler's crazy pi product In www.mathpages.com/home/kmath477.htm, explanation of: 3 5 7 11 13 17 19 π/4 = - × - × - × -- × -- × -- × -- × ... 4 4 8 12 12 16 20 Here the numerators are the odd primes, and the denominators are the closest numbers of the form 4n. Christian. -----Message d'origine----- De : math-fun-bounces@mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com] De la part de Bill Gosper Envoyé : lundi 14 février 2011 08:24 À : math-fun@mailman.xmission.com Objet : Re: [math-fun] Euler's crazy pi product interesting how if you start with 1 you get a line of slope 1. you can also include all integers. π/4 = 1/2 × 2/2 × 3/2 × 4/4 × 5/6 × 6/6 × 7/6 ... if I've got that right. For the n-1/2 cases, round up and round down produce divergent products. Round to even gives 4*(1/4)!^2/sqrt(2*pi) . At last, a mathematical justification for round to even! --rwg On Sun, Feb 13, 2011 at 9:36 PM, Bill Gosper <billgosper@gmail.com <http://gosper.org/webmail/src/compose.php?send_to=billgosper%40gmail.com>> wrote:
http://math.ucr.edu/home/baez/week127.html finds in Lennart Berggren, Jonathan Borwein and Peter Borwein, π: A Source Book, Springer-Verlag, New York, 1997, "the following weirdly beautiful formula due to Euler, which unfortunately is not explained:" (Then how can they call it a sourcebook?--rwg)
3 5 7 11 13 17 19 π/2 = - × - × - × -- × -- × -- × -- × ... 2 6 6 10 14 18 18
"Here the numerators are the odd primes, and the denominators are the closest numbers of the form 4n+2." E.g., for 10, 100, ..., 10^6 terms, In[209]:=
Table[Block[{p=2},Nest[N[#*(p=NextPrime[p])/(2+4*Round[(p-2)/4]),9]&,2 ,10^n]],{n,6}]
Out[209]= {3.10152045, 3.13398462, 3.13772561, 3.14073685, 3.14143290, 3.14157195}
Why the heck?? Was Euler a preincarnation of Ramanujan? --rwg
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Changing the denominators to the closest multiple of 6, and we have pi*sqrt(3)/6 = (5/6) (7/6) (11/12) (13/12) (17/18) (19/18) (23/24) ... Warut On Mon, Feb 14, 2011 at 4:09 PM, Christian Boyer <cboyer@club-internet.fr> wrote:
In www.mathpages.com/home/kmath477.htm, explanation of:
3 5 7 11 13 17 19 π/4 = - × - × - × -- × -- × -- × -- × ... 4 4 8 12 12 16 20
Here the numerators are the odd primes, and the denominators are the closest numbers of the form 4n.
Christian.
-----Message d'origine----- De : math-fun-bounces@mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com] De la part de Bill Gosper Envoyé : lundi 14 février 2011 08:24 À : math-fun@mailman.xmission.com Objet : Re: [math-fun] Euler's crazy pi product
interesting how if you start with 1 you get a line of slope 1. you can also include all integers.
π/4 = 1/2 × 2/2 × 3/2 × 4/4 × 5/6 × 6/6 × 7/6 ...
if I've got that right.
For the n-1/2 cases, round up and round down produce divergent products. Round to even
gives 4*(1/4)!^2/sqrt(2*pi) . At last, a mathematical justification for round to even! --rwg
On Sun, Feb 13, 2011 at 9:36 PM, Bill Gosper <billgosper@gmail.com <http://gosper.org/webmail/src/compose.php?send_to=billgosper%40gmail.com>> wrote:
http://math.ucr.edu/home/baez/week127.html finds in Lennart Berggren, Jonathan Borwein and Peter Borwein, π: A Source Book, Springer-Verlag, New York, 1997, "the following weirdly beautiful formula due to Euler, which unfortunately is not explained:" (Then how can they call it a sourcebook?--rwg)
3 5 7 11 13 17 19 π/2 = - × - × - × -- × -- × -- × -- × ... 2 6 6 10 14 18 18
"Here the numerators are the odd primes, and the denominators are the closest numbers of the form 4n+2." E.g., for 10, 100, ..., 10^6 terms, In[209]:=
Table[Block[{p=2},Nest[N[#*(p=NextPrime[p])/(2+4*Round[(p-2)/4]),9]&,2 ,10^n]],{n,6}]
Out[209]= {3.10152045, 3.13398462, 3.13772561, 3.14073685, 3.14143290, 3.14157195}
Why the heck?? Was Euler a preincarnation of Ramanujan? --rwg
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I did a numerical experiment in which I rounded the denominators to the nearest 4n+1, and got pi times something close to the continued fraction 0/1A12222... (The A is a 10), which simplifies to (219+V2)/241. (I started with 3, not 5, figuring that the extra factor of 3/1 would just come out in the wash.) When I went to denominators of the form 4n+3, however, things looked a bit murkier. I went up to 10^6 and then 10^7, and the partial product was pi times 0/11152... That's as far as the continued fractions agreed. I'm peering at denominators of the form 5n, and they don't look very promising either. Now I'm starting to think that run of 2's in the cf for the 4n+1 case was a coincidence. On Fri, Feb 18, 2011 at 11:53 AM, Warut Roonguthai <warut822@gmail.com>wrote:
Changing the denominators to the closest multiple of 6, and we have
pi*sqrt(3)/6 = (5/6) (7/6) (11/12) (13/12) (17/18) (19/18) (23/24) ...
Warut
On Mon, Feb 14, 2011 at 4:09 PM, Christian Boyer <cboyer@club-internet.fr> wrote:
In www.mathpages.com/home/kmath477.htm, explanation of:
3 5 7 11 13 17 19 π/4 = - × - × - × -- × -- × -- × -- × ... 4 4 8 12 12 16 20
Here the numerators are the odd primes, and the denominators are the closest numbers of the form 4n.
Christian.
-----Message d'origine----- De : math-fun-bounces@mailman.xmission.com [mailto: math-fun-bounces@mailman.xmission.com] De la part de Bill Gosper Envoyé : lundi 14 février 2011 08:24 À : math-fun@mailman.xmission.com Objet : Re: [math-fun] Euler's crazy pi product
interesting how if you start with 1 you get a line of slope 1. you can also include all integers.
π/4 = 1/2 × 2/2 × 3/2 × 4/4 × 5/6 × 6/6 × 7/6 ...
if I've got that right.
For the n-1/2 cases, round up and round down produce divergent products. Round to even
gives 4*(1/4)!^2/sqrt(2*pi) . At last, a mathematical justification for round to even! --rwg
On Sun, Feb 13, 2011 at 9:36 PM, Bill Gosper <billgosper@gmail.com < http://gosper.org/webmail/src/compose.php?send_to=billgosper%40gmail.com>> wrote:
http://math.ucr.edu/home/baez/week127.html finds in Lennart Berggren, Jonathan Borwein and Peter Borwein, π: A Source Book, Springer-Verlag, New York, 1997, "the following weirdly beautiful formula due to Euler, which unfortunately is not explained:" (Then how can they call it a sourcebook?--rwg)
3 5 7 11 13 17 19 π/2 = - × - × - × -- × -- × -- × -- × ... 2 6 6 10 14 18 18
"Here the numerators are the odd primes, and the denominators are the closest numbers of the form 4n+2." E.g., for 10, 100, ..., 10^6 terms, In[209]:=
Table[Block[{p=2},Nest[N[#*(p=NextPrime[p])/(2+4*Round[(p-2)/4]),9]&,2 ,10^n]],{n,6}]
Out[209]= {3.10152045, 3.13398462, 3.13772561, 3.14073685, 3.14143290, 3.14157195}
Why the heck?? Was Euler a preincarnation of Ramanujan? --rwg
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If the denominators are rounded down to the nearest 4n+1, the product we have is just \prod p/(p-2) = \prod (1-2/p)^(-1), where primes p = 3 (mod 4). This product diverges because \sum 1/p diverges. The same goes for the case 4n+3. Warut On Sat, Feb 19, 2011 at 2:22 AM, Allan Wechsler <acwacw@gmail.com> wrote:
I did a numerical experiment in which I rounded the denominators to the nearest 4n+1, and got pi times something close to the continued fraction 0/1A12222... (The A is a 10), which simplifies to (219+V2)/241. (I started with 3, not 5, figuring that the extra factor of 3/1 would just come out in the wash.)
When I went to denominators of the form 4n+3, however, things looked a bit murkier. I went up to 10^6 and then 10^7, and the partial product was pi times 0/11152... That's as far as the continued fractions agreed.
I'm peering at denominators of the form 5n, and they don't look very promising either. Now I'm starting to think that run of 2's in the cf for the 4n+1 case was a coincidence.
On Fri, Feb 18, 2011 at 11:53 AM, Warut Roonguthai <warut822@gmail.com>wrote:
Changing the denominators to the closest multiple of 6, and we have
pi*sqrt(3)/6 = (5/6) (7/6) (11/12) (13/12) (17/18) (19/18) (23/24) ...
Warut
On Mon, Feb 14, 2011 at 4:09 PM, Christian Boyer <cboyer@club-internet.fr> wrote:
In www.mathpages.com/home/kmath477.htm, explanation of:
3 5 7 11 13 17 19 π/4 = - × - × - × -- × -- × -- × -- × ... 4 4 8 12 12 16 20
Here the numerators are the odd primes, and the denominators are the closest numbers of the form 4n.
Christian.
-----Message d'origine----- De : math-fun-bounces@mailman.xmission.com [mailto: math-fun-bounces@mailman.xmission.com] De la part de Bill Gosper Envoyé : lundi 14 février 2011 08:24 À : math-fun@mailman.xmission.com Objet : Re: [math-fun] Euler's crazy pi product
interesting how if you start with 1 you get a line of slope 1. you can also include all integers.
π/4 = 1/2 × 2/2 × 3/2 × 4/4 × 5/6 × 6/6 × 7/6 ...
if I've got that right.
For the n-1/2 cases, round up and round down produce divergent products. Round to even
gives 4*(1/4)!^2/sqrt(2*pi) . At last, a mathematical justification for round to even! --rwg
On Sun, Feb 13, 2011 at 9:36 PM, Bill Gosper <billgosper@gmail.com < http://gosper.org/webmail/src/compose.php?send_to=billgosper%40gmail.com>> wrote:
http://math.ucr.edu/home/baez/week127.html finds in Lennart Berggren, Jonathan Borwein and Peter Borwein, π: A Source Book, Springer-Verlag, New York, 1997, "the following weirdly beautiful formula due to Euler, which unfortunately is not explained:" (Then how can they call it a sourcebook?--rwg)
3 5 7 11 13 17 19 π/2 = - × - × - × -- × -- × -- × -- × ... 2 6 6 10 14 18 18
"Here the numerators are the odd primes, and the denominators are the closest numbers of the form 4n+2." E.g., for 10, 100, ..., 10^6 terms, In[209]:=
Table[Block[{p=2},Nest[N[#*(p=NextPrime[p])/(2+4*Round[(p-2)/4]),9]&,2 ,10^n]],{n,6}]
Out[209]= {3.10152045, 3.13398462, 3.13772561, 3.14073685, 3.14143290, 3.14157195}
Why the heck?? Was Euler a preincarnation of Ramanujan? --rwg
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Not rounded down. Rounded to nearest. Which I only this moment realized was not well-defined; I was measuring whatever Common Lisp considered "nearest". Sigh. On Sat, Feb 19, 2011 at 12:11 PM, Warut Roonguthai <warut822@gmail.com>wrote:
If the denominators are rounded down to the nearest 4n+1, the product we have is just
\prod p/(p-2) = \prod (1-2/p)^(-1),
where primes p = 3 (mod 4). This product diverges because \sum 1/p diverges.
The same goes for the case 4n+3.
Warut
On Sat, Feb 19, 2011 at 2:22 AM, Allan Wechsler <acwacw@gmail.com> wrote:
I did a numerical experiment in which I rounded the denominators to the nearest 4n+1, and got pi times something close to the continued fraction 0/1A12222... (The A is a 10), which simplifies to (219+V2)/241. (I started with 3, not 5, figuring that the extra factor of 3/1 would just come out in the wash.)
When I went to denominators of the form 4n+3, however, things looked a bit murkier. I went up to 10^6 and then 10^7, and the partial product was pi times 0/11152... That's as far as the continued fractions agreed.
I'm peering at denominators of the form 5n, and they don't look very promising either. Now I'm starting to think that run of 2's in the cf for the 4n+1 case was a coincidence.
On Fri, Feb 18, 2011 at 11:53 AM, Warut Roonguthai <warut822@gmail.com wrote:
Changing the denominators to the closest multiple of 6, and we have
pi*sqrt(3)/6 = (5/6) (7/6) (11/12) (13/12) (17/18) (19/18) (23/24) ...
Warut
On Mon, Feb 14, 2011 at 4:09 PM, Christian Boyer <cboyer@club-internet.fr> wrote:
In www.mathpages.com/home/kmath477.htm, explanation of:
3 5 7 11 13 17 19 π/4 = - × - × - × -- × -- × -- × -- × ... 4 4 8 12 12 16 20
Here the numerators are the odd primes, and the denominators are the closest numbers of the form 4n.
Christian.
-----Message d'origine----- De : math-fun-bounces@mailman.xmission.com [mailto: math-fun-bounces@mailman.xmission.com] De la part de Bill Gosper Envoyé : lundi 14 février 2011 08:24 À : math-fun@mailman.xmission.com Objet : Re: [math-fun] Euler's crazy pi product
interesting how if you start with 1 you get a line of slope 1. you can also include all integers.
π/4 = 1/2 × 2/2 × 3/2 × 4/4 × 5/6 × 6/6 × 7/6 ...
if I've got that right.
For the n-1/2 cases, round up and round down produce divergent products. Round to even
gives 4*(1/4)!^2/sqrt(2*pi) . At last, a mathematical justification for round to even! --rwg
On Sun, Feb 13, 2011 at 9:36 PM, Bill Gosper <billgosper@gmail.com <
http://gosper.org/webmail/src/compose.php?send_to=billgosper%40gmail.com>>
wrote:
http://math.ucr.edu/home/baez/week127.html finds in Lennart Berggren, Jonathan Borwein and Peter Borwein, π: A Source Book, Springer-Verlag, New York, 1997, "the following weirdly beautiful formula due to Euler, which unfortunately is not explained:" (Then how can they call it a sourcebook?--rwg)
3 5 7 11 13 17 19 π/2 = - × - × - × -- × -- × -- × -- × ... 2 6 6 10 14 18 18
"Here the numerators are the odd primes, and the denominators are the closest numbers of the form 4n+2." E.g., for 10, 100, ..., 10^6 terms, In[209]:=
Table[Block[{p=2},Nest[N[#*(p=NextPrime[p])/(2+4*Round[(p-2)/4]),9]&,2
,10^n]],{n,6}]
Out[209]= {3.10152045, 3.13398462, 3.13772561, 3.14073685, 3.14143290, 3.14157195}
Why the heck?? Was Euler a preincarnation of Ramanujan? --rwg
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Hi all, Just applied to join and Rich mentioned that he invites people to post a bio, which I had effectively created already in the mail to validate my address, so here it is for all: My name is David Makin and a fellow fractal artist and mathematician (Kerry Mitchell) suggested I join this list - I am a Fractal Artist and programmer and a three times winner of the International Benoit Mandelbrot Fractal Art Contest: http://www.fractalartcontests.com/2006/ http://www.fractalartcontests.com/2007/ http://www.fractalartcontests.com/2009/ You can see my work here: http://www.fractalgallery.co.uk/ http://makinmagic.deviantart.com/ http://www.youtube.com/user/MakinMagicFractals As for the maths, my primary interest is in anything fractal related but IMHO that actually includes just about any math I can think of ;) I am only formally educated to UK "A" level in maths (back in 1980) and by trade am a programmer - I was a late starter to programming as I only came into contact with it at age 20 when I read a BASIC manual and was immediately hooked (my first thought was "I should be doing this, it's so easy") I'm essentially self-taught with respect to programming, my most prominent commercial software was "Crystal Dragon" on the Amiga (a Dungeon Master style RPG) in the late 90's (unfortunately released just as Commodore disappeared due to their stupid decision to go to a console instead of improving the Amiga). The next life-changing event was getting hooked on fractals courtesy of Fractint in April 1999, boy do I wish I'd known more about fractals when I first started programming - I don't think the render times involved or low resolution would have put me off even on my old Dragon 32 and 64 (6809cpu). The main (formal) gaps in my maths relate to more recent stuff like group theory, not something even included at degree level in the early 80's I think (or so my cousin told me - he did maths at Oxford). I am particularly interested in anything related to the mathematics of 3+ dimensions - especially in finding a true "field" of 3 or more dimensions (I've been told all possible 3D fields have been shown to be merely holomorphic? to complex numbers but I don't know the details). For anything public with respect to my maths and programming there are my posts at http://www.fractalforums.com/ where I am a moderator and my formulas for Ultra Fractal in the UF formula database (all "mmf") here: http://formulas.ultrafractal.com/ Oh - also before moving to using Ultra Fractal I wrote my own 32-bit DOS fractal generator "MMFrac", a rather clunky interface and only single layer rendering (except for bump-map mode) but pretty advanced for 2000-2002 in that it eventually included 3D fractals (parallel and perspective projection to 2D) :) Still available on the web but only of interest for those who can run full-screen dos without using dosbox. OK enough about me, does *anyone* know of a 3D or higher mathematical "field" satisfying all the axions on Wolfram here: http://mathworld.wolfram.com/FieldAxioms.html Or does anyone know that there's a proof that no such field exists beyond 2D (i.e. complex numbers). Thanks for reading. Dave
On Sun, Feb 20, 2011 at 7:30 PM, David Makin <makinmagic@tiscali.co.uk> wrote:
OK enough about me, does *anyone* know of a 3D or higher mathematical "field" satisfying all the axions on Wolfram here:
Well, there's a long history of quaternionic fractals. http://www.google.com/images?q=quaternion+fractals There are also finite fields of arbitrary degree, though I don't know how you'd talk about convergence in a finite field. You could color the points by their period under the update function, I suppose. -- Mike Stay - metaweta@gmail.com http://www.cs.auckland.ac.nz/~mike http://reperiendi.wordpress.com
It's not entirely clear what you have in mind by the "dimension" of a field in this context. If you're looking for an algebraic framework in which to do Euclidean and related geometry in higher space, the closest it's possible to get to a field is a generalisation of complex numbers and quaternions called "Clifford algebra" --- see http://en.wikipedia.org/wiki/Clifford_algebra http://en.wikipedia.org/wiki/Geometric_algebra A Clifford algebra is a ring, but not in general a field: multiplication is non- commutative, and nonzero elements may be isotropic (singular; lacking reciprocal). But it is still possible to do some analysis (calculus) in this environment. The situation is analogous to matrices, which provide a more traditional (though much clunkier) approach to geometric computation. The geometric algebra specific to Euclidean 3-space is Cl(3,0,1) or DCQ's. Curiously little information is available about these --- a brief intro and toy application is posted at https://docs.google.com/leaf?id=0B6QR93hqu1AhNDdmOGRhM2QtZThjNy00ZTc4LWI4MzY... Most practitioners prefer to use the larger algebra Cl(4,1,0) or CGA, partly because it is slightly more mathematically respectable, but mostly on the grounds of incorporating the Poincare group for Minkowski spacetime (aka conformal or Moebius 3-space). There seems no obvious reason why the quaternionic version of fractals should not generalise further --- given sufficient algorithmic dedication and computational muscle! Fred Lunnon On 2/21/11, Mike Stay <metaweta@gmail.com> wrote:
On Sun, Feb 20, 2011 at 7:30 PM, David Makin <makinmagic@tiscali.co.uk> wrote:
OK enough about me, does *anyone* know of a 3D or higher mathematical "field" satisfying all the axions on Wolfram here:
Well, there's a long history of quaternionic fractals. http://www.google.com/images?q=quaternion+fractals
There are also finite fields of arbitrary degree, though I don't know how you'd talk about convergence in a finite field. You could color the points by their period under the update function, I suppose. -- Mike Stay - metaweta@gmail.com http://www.cs.auckland.ac.nz/~mike http://reperiendi.wordpress.com
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On 2/21/11, Mike Stay <metaweta@gmail.com> wrote:
... There are also finite fields of arbitrary degree, though I don't know how you'd talk about convergence in a finite field. You could color the points by their period under the update function, I suppose. --
p-adic numbers & analysis --- (very) brief intros at http://en.wikipedia.org/wiki/P-adic_analysis http://www.maths.gla.ac.uk/~ajb/dvi-ps/padicnotes.pdf WFL
From: Fred lunnon <fred.lunnon@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Sent: Sun, February 27, 2011 12:51:53 PM Subject: Re: [math-fun] Newbie intro/bio On 2/21/11, Mike Stay <metaweta@gmail.com> wrote:
... There are also finite fields of arbitrary degree, though I don't know how you'd talk about convergence in a finite field. You could color the points by their period under the update function, I suppose. --
p-adic numbers & analysis --- (very) brief intros at http://en.wikipedia.org/wiki/P-adic_analysis http://www.maths.gla.ac.uk/~ajb/dvi-ps/padicnotes.pdf WFL _______________________________________________ Watch out here -- p-adic fields are characteristic zero, not characteristic p, and thus are necessarily infinite. -- Gene
participants (9)
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Allan Wechsler -
Bill Gosper -
Christian Boyer -
David Makin -
Eugene Salamin -
Fred lunnon -
Mike Stay -
Schroeppel, Richard -
Warut Roonguthai