February 2010 - mathfun@mailman.xmission.com

[math-fun] Correction Re: parametrizing the Koch snowflake curve
by Dan Asimov 25 Feb '10

25 Feb '10

Sorry, that should be {4t} instead of {t}, as fixed below: Yes, if you let T_k, k in Z/4Z denote the transformation C -> C of the form z |-> az+b. a,b in C, that takes [0,1] to the kth segment of the initial "witch's hat" stage of the snowflake (4 intervals of length 1/3 each), then a parametrization S: [0,1] -> C of the snowflake curve should satisfy S({4t}) = T_[4t](S(t)) (where {} is fractional part and [] is floor). Does this determine S uniquely? --Dan ________________________________________________________________________________________ "Outside of a dog, a book is man's best friend. Inside of a dog, it's too dark to read." --Groucho Marx

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Re: [math-fun] parametrizing the Koch snowflake curve
by Dan Asimov 25 Feb '10

25 Feb '10

Yes, if you let T_k, k in Z/4Z denote the transformation C -> C of the form z |-> az+b. a,b in C, that takes [0,1] to the kth segment of the initial "witch's hat" stage of the snowflake (4 intervals of length 1/3 each), then a parametrization S: [0,1] -> C of the snowflake curve should satisfy S({t}) = T_[4t](S(t)) (where {} is fractional part and [] is floor). Does this determine S uniquely? --Dan Rich wrote: << If you view the starting triangle as running from 0 to 3, and write the fraction base 4, you can use the fraction digits to select which part of the (sub)edge you are on. The orientation of the edge can be calculated from the 1s and 2s of the high order portion of the fraction, and that should give a somewhat kludgy sum for the (x,y) coordinates of the image point. RWG likely has some nice FFT clock sum expression >> ________________________________________________________________________________________ "Outside of a dog, a book is man's best friend. Inside of a dog, it's too dark to read." --Groucho Marx

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Re: [math-fun] parametrizing the Koch snowflake curve
by N. J. A. Sloane 25 Feb '10

25 Feb '10

Jim Propp asked: > Does anyone (Bill Gosper, maybe?) know an explicit parametrization of the Koch snowflake curve? The Cantor set consists of the points 0 <= x <= 1 which have a base 3 "decimal" expansion that does not contain a 1. Is there a similar characterization of the Koch snowflake? Neil Sloane

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Re: [math-fun] What is the name of [mod 9 digitsum]
by Dan Asimov 25 Feb '10

25 Feb '10

<< Casting out nines ? http://mathforum.org/library/drmath/view/72752.html >> No. Casting out nines is the (oldish) name for the *procedure* used to obtain the digital root, not that digit itself. << "check digit" ? http://en.wikipedia.org/wiki/Check_digit www.math.unl.edu/~mbrittenham2/classwk/203f07/quiz/203words1.pdf "http://math.boisestate.edu/~tconklin/MATH124/Main/Readings/3 Number Theory/PDFs/Number Theory/Modular Arithmetic in the Marketplace.pdf" >> No. The remainder mod 9 is only one of many types of "check digits", so that is not its name. --Dan ________________________________________________________________________________________ "Outside of a dog, a book is man's best friend. Inside of a dog, it's too dark to read." --Groucho Marx

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Re: [math-fun] What is the name of [mod 9 digitsum]
by Henry Baker 25 Feb '10

25 Feb '10

Casting out nines ? http://mathforum.org/library/drmath/view/72752.html "check digit" ? http://en.wikipedia.org/wiki/Check_digit www.math.unl.edu/~mbrittenham2/classwk/203f07/quiz/203words1.pdf "http://math.boisestate.edu/~tconklin/MATH124/Main/Readings/3 Number Theory/PDFs/Number Theory/Modular Arithmetic in the Marketplace.pdf" At 10:44 AM 2/25/2010, Dan Asimov wrote: >Eric wrote: > ><< >Does the operation going from 178 to 7 have >a name (other than "modulus 9 digitsum")? >>> > >Besides "digital root", it's also known as >the remainder mod 9. > >--Dan

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[math-fun] parametrizing the Koch snowflake curve
by James Propp 25 Feb '10

25 Feb '10

Does anyone (Bill Gosper, maybe?) know an explicit parametrization of the Koch snowflake curve? Jim Propp

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Re: [math-fun] What is the name of [mod 9 digitsum]
by Eric Angelini 25 Feb '10

25 Feb '10

Thank you, Dave -- this is it! (rest in peace, MJ) Best, É. -----Message d'origine----- De : math-fun-bounces(a)mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com] De la part de Dave Blackston Envoyé : jeudi 25 février 2010 18:53 À : math-fun Objet : Re: [math-fun] What is the name of [mod 9 digitsum] I've seen it referred to as a "digital root" http://en.wikipedia.org/wiki/Digital_root Dave On Thu, Feb 25, 2010 at 9:29 AM, Eric Angelini <Eric.Angelini(a)kntv.be> wrote: > > Hello Math-Fun, > is there a name for this operation : > > - start with 178 for instance; > - add the digits: 1+7+8 => 16 > - add the digits again: 1+6 => 7 > > Does the operation going from 178 to 7 have > a name (other than "modulus 9 digitsum")? > > Thank you, > Best, > É. > > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > _______________________________________________ math-fun mailing list math-fun(a)mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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[math-fun] What is the name of [mod 9 digitsum]
by Eric Angelini 25 Feb '10

25 Feb '10

Hello Math-Fun, is there a name for this operation : - start with 178 for instance; - add the digits: 1+7+8 => 16 - add the digits again: 1+6 => 7 Does the operation going from 178 to 7 have a name (other than "modulus 9 digitsum")? Thank you, Best, É.

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Re: [math-fun] Tutorial embarrassment
by Bill Gosper 24 Feb '10

24 Feb '10

wfl>You're solving y^2 = x(x+1)/2, and the solutions y comprise a >second-order linear recurrence (and y^2 third-order), as any >fule no --- but the x evidently do not (and of course, there's >actually no reason to suppose that they might). But notice I imposed recurrencehood on x(x+1)/2 ! >I'm not sure I'd have recognised the distinction in good time, either ... Nice elephant trap! WFL Seriously! --rwg On 2/23/10, Bill Gosper <billgosper(a)gmail.com <http://gosper.org/webmail/src/compose.php?send_to=billgosper%40gmail.com>> wrote: > A few years ago I mentioned an effective way to list the square triangular > numbers by brute force finding the first few, and then computing each new > one > from the previous six so as to annihilate the determinant of a 4x4 Toeplitz > matrix. > > How can this fail: > > (c256) subset(-9..9,lambda([x],integerp(sqrt(binomial(x+1,2))))) > > (d256) [- 9, - 2, - 1, 0, 1, 8] > > (c257) genmatrix(lambda([i,j],(''(endcons(x,%*(%+1)/2)))[i+j-1]),4) > > [ 36 1 0 0 ] > [ ] > [ 1 0 0 1 ] > (d257) [ ] > [ 0 0 1 36 ] > [ ] > [ 0 1 36 x ] > > (c258) sqrt(solve(det(%))) > > (d258) [sqrt(x) = 6 sqrt(35)] > > while this succeeds? > > (c261) delete(-1,subset(-50..9,lambda([x],integerp(sqrt(binomial(x+1,2)))))) > > (d261) [- 50, - 9, - 2, 0, 1, 8] > > (c262) genmatrix(lambda([i,j],(''(endcons(x,%*(%+1)/2)))[i+j-1]),4) > > [ 1225 36 1 0 ] > [ ] > [ 36 1 0 1 ] > (d262) [ ] > [ 1 0 1 36 ] > [ ] > [ 0 1 36 x ] > > (c263) sqrt(solve(det(%))) > > (d263) [sqrt(x) = 35] > > (c280) > genmatrix(lambda([i,j],(''(append(d261*(d261+1)/2,[35^2,x])))[i+j]),4) > > [ 36 1 0 1 ] > [ ] > [ 1 0 1 36 ] > (d280) [ ] > [ 0 1 36 1225 ] > [ ] > [ 1 36 1225 x ] > > (c281) sqrt(solve(det(%))) > > (d281) [sqrt(x) = 204] > etc. > --rwg

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[math-fun] Tutorial embarrassment
by Bill Gosper 24 Feb '10

24 Feb '10

A few years ago I mentioned an effective way to list the square triangular numbers by brute force finding the first few, and then computing each new one from the previous six so as to annihilate the determinant of a 4x4 Toeplitz matrix. How can this fail: (c256) subset(-9..9,lambda([x],integerp(sqrt(binomial(x+1,2))))) (d256) [- 9, - 2, - 1, 0, 1, 8] (c257) genmatrix(lambda([i,j],(''(endcons(x,%*(%+1)/2)))[i+j-1]),4) [ 36 1 0 0 ] [ ] [ 1 0 0 1 ] (d257) [ ] [ 0 0 1 36 ] [ ] [ 0 1 36 x ] (c258) sqrt(solve(det(%))) (d258) [sqrt(x) = 6 sqrt(35)] while this succeeds? (c261) delete(-1,subset(-50..9,lambda([x],integerp(sqrt(binomial(x+1,2)))))) (d261) [- 50, - 9, - 2, 0, 1, 8] (c262) genmatrix(lambda([i,j],(''(endcons(x,%*(%+1)/2)))[i+j-1]),4) [ 1225 36 1 0 ] [ ] [ 36 1 0 1 ] (d262) [ ] [ 1 0 1 36 ] [ ] [ 0 1 36 x ] (c263) sqrt(solve(det(%))) (d263) [sqrt(x) = 35] (c280) genmatrix(lambda([i,j],(''(append(d261*(d261+1)/2,[35^2,x])))[i+j]),4) [ 36 1 0 1 ] [ ] [ 1 0 1 36 ] (d280) [ ] [ 0 1 36 1225 ] [ ] [ 1 36 1225 x ] (c281) sqrt(solve(det(%))) (d281) [sqrt(x) = 204] etc. --rwg

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