November 2007 - mathfun@mailman.xmission.com

Correction: Re: [math-fun] The Axiom of Choice for roots of z^2 + 1
by Dan Asimov 23 Nov '07

23 Nov '07

I wrote: << P.S. In a faintly related vein, if we define i^z := exp(pi*z*i/2), then iterating this function on the starting value of z = i approaches a limit L of approx. [[[erroneous value was here]]]. Hmmm, Mathematica seems to give a rather different answer. . . . . . . >> I just had a stupid bug in the program. Duh. Now it's fixed and the program finds that the limit is approx. L = 0.4382829363310479 + i 0.3605924719745690, and this is confirmed by other calculation systems. (It took 183 iterations of i^z on the starting value z = i before successive values agreed within taxicab distance |x-xlast| + |y-ylast| < 10^(-9)). --Dan

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[math-fun] near identity for Catalan
by wouter meeussen 23 Nov '07

23 Nov '07

or possibly a true identity (?) negaodd posieven Catalan == {0.9992596423851822204, 0.9999976893644401925, 0.9999999241377495063, 0.9999999985217359313, 0.9999999999896304167} with "negaodd" defined as sum(p prime, p mod 4=1, 1-1/p^2) and "posieven" defined as sum(p prime, p mod 4=3, 1+1/p^2) negaodd= Table[N[ Product[p=Prime[q];w=If[Mod[p+1,4]===2,p^2, Infinity];1-1/w,{q,1,10^w}] ,20], {w,5}] posieven= Table[N[ Product[p=Prime[q];w=If[Mod[p+1,4]===0,p^2, Infinity];1+1/w,{q,1,10^w}] ,20], {w,5}] nothing under http://mathworld.wolfram.com/CatalansConstant.html Wouter.

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[math-fun] a(n) is the n-th integer with digit sum = n
by Eric Angelini 23 Nov '07

23 Nov '07

Hello Math-Fun, Could someone help me compute S with this rule: a(n) is the n-th integer of N with digit sum = n S starts like this, I think: S = 1,11,21,31,41,51,61,71,81,109,... 1 is the 1st integer with digit sum = 1 [1,10,100,1000,10000,...] 11 is the 2nd integer with digit sum = 2 [2,11,20,101,200,1001,2000,...] 21 is the 3rd integer with digit sum = 3 [3,12,21,30,102,120,201,210,300,...] 31 is the 4th integer with digit sum = 4 [4,13,22,31,40,103,112,121,130,...] ... 109 is the 10th integer with digit sum = 10 [19,28,37,46,55,64,73,82,91,109,127,...] Best, É.

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Re: [math-fun] Doubly true equalities
by N. J. A. Sloane 23 Nov '07

23 Nov '07

Eric said: >>>those equalities are doubly true : >>>2+3=5 >>>5+1!=6 >>>because they have respectively 5 and 6 symbols, altogether. How about: Twenty three has 11 letters (which has 23 symbols (not counting spaces)). I'm not proud of this and would sign it with an alias if sendmail would allow it!

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[math-fun] Doubly true equalities
by Eric Angelini 22 Nov '07

22 Nov '07

Hello Math-fun, those equalities are doubly true : 2+3=5 5+1!=6 because they have respectively 5 and 6 symbols, altogether. I'm looking for a general formula, always true for any integer "a". Here is one for any odd "a" (with a > 1): a + 0 + 0 + ... [k times the string "+0"]... + 0 = a with k=(a-1)/2 - L(a) and L(a) being the "length of a" (quantity of digits of integer a). Examples: 97 + 0 + 0 + 0 ... [46 times "+0"] ... + 0 = 97 99 + 0 + 0 + 0 ... [47 times "+0"] ... + 0 = 99 101 + 0 + 0 + 0 ... [47 times "+0"] ... + 0 = 101 103 + 0 + 0 + 0 ... [48 times "+0"] ... + 0 = 103 Here is one for any even "a" (with a > 6): a-1+1! + 0 + 0 ... [k times "+0"] ... + 0 = a with k=(a-6)/2 - L(a) Examples: 98-1+1!+0+0 ... [44 times "+0"] ... +0=98 100-1+1!+0+0 ... [44 times "+0"] ... +0=100 102-1+1!+0+0 ... [45 times "+0"] ... +0=102 The general formula could be of any shape, of course -- no need to play with "+0" strings or factorials. Best, É.

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[math-fun] article about kaplansky
by Michael Reid 21 Nov '07

21 Nov '07

there is a nice article about kaplansky (by hyman bass and t.y. lam) in the december issue of the ams notices: http://www.ams.org/notices/200711/tx071101477p.pdf (it is possible that the url requires a subscription or proxy access; i'm not sure about that.) mike

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[math-fun] Re: n-D median?
by Steve Witham 20 Nov '07

20 Nov '07

>From: "Fred lunnon" <fred.lunnon(a)gmail.com> > >Using the term "median" suggests to me that you intend the result to be a >member of the original point set. Is that what you meant? No more than that's true of the 1-D median, which sometimes averages two points. With the sum-of-directions-approaches-zero definition, original points would be more likely candidates at least, because the direction function has a step as you pass through a point, making it a zero-crossing magnet. I guess if there are zero-crossing curves in the areas between original points, they have to lead to original points. --Steve

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[math-fun] Four consecutive abundant numbers
by Richard Schroeppel 19 Nov '07

19 Nov '07

[From the NMBRTHRY list, slightly edited. I reformatted the factorizations. No arithmetic actually checked. Odd abundant numbers are sparse, and finding 4 consecutive abundants means two of them must be odd. -- rcs] Date: Thu, 1 Nov 2007 15:38:16 -0400 From: Bruno Mishutka <bruno.mishutka(a)googlemail.com> Subject: consecutive abundant numbers To: NMBRTHRY(a)LISTSERV.NODAK.EDU Snyam Sunder Gupta (http://www.shyamsundergupta.com/canyoufind.htm) has recently asked for the smallest set of four consecutive abundant numbers. I suspect that this is not the smallest m1 such that (m1, m1+1, m1+2, m1+3) are all abundant, but it's the smallest set that I could find. m1 := 141363708067871564084949719820472453374; It is approximately 1.41*10^38. igcd(m1,6);ifactor(m1+1);ifactor(m1+2);ifactor(m1+3); 6 5^3 7^2 11^2 17 23^2 37 43 53 59 73 79 83 281 31696364109343 2^8 337 1638581556795618092601883807266233 3^4 13^2 19 29 31 41 47 61 67 71 151 7160300747074201457 sigma(m1+1)/1.0/(m1+1);sigma(m1+2)/1.0/(m1+2);sigma(m1+3)/1.0/(m1+3); 2.005172014 2.002016877 2.003079741 Note that Paul Erdos ("Note on consecutive abundant numbers", J. London Math. Soc. 10, 128-131 (1935)) has shown that there are two absolute constants c1, c2 such that for all large n there are at least c1 log log log n but not more than c2 log log log n consecutive abundant numbers less than n. Also, the smallest set of two consecutive abundant numbers is (5775, 5776). And the smallest set of three consecutive abundant numbers is (171078830, 171078832, 171078832), due apparently to Laurent Hodges and Michael Reid in 1975 (does anyone have a reference to this?). Yours Sincerely, Bruno

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Re: [math-fun] The shortest successful Maths Ph.D. theses
by Richard Schroeppel 19 Nov '07

19 Nov '07

[This note is from Victor Miller, from about two weeks ago. --rcs] Date: Sun, 4 Nov 2007 09:40:54 -0400 From: "victor miller" <victorsmiller(a)gmail.com> Barry Mazur's was only 5 pages long. In it he proved the "Generalized Schoenflies Conjecture". The proof took essentially 1/2 page, but there was another 5 pages or so, in giving the technical ideas in the proof. Victor On 11/4/07, Dan Asimov <dasimov(a)earthlink.net> wrote: > > Guy asks: > > << > Apologies if this has been asked before. I'm wondering what have been > the shortest successfully submitted Ph.D. theses in mathematics. > >> > > Urk, mine was pretty short -- just 28 double-spaced typewritten pages. > Google claims there's something about this in the Nov. 2003 > College Mathematics Journal, which I don't have access to at the moment. > > (Can someone please get ahold of that and report what it says on the > subject?) > > On the other hand, someone claimed on some discussion board that > (presumably William Rowan) Hamilton's Ph.D. thesis was only one page long. > > --Dan > > _______________________________________________ > 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] Maths, Security and Poisoned Chips
by Guy Haworth 19 Nov '07

19 Nov '07

A similar issue came up when the (DES) Data Encryption Standard became a (FIPS) Federal Information Processing Standard, q.v. http://en.wikipedia.org/wiki/Data_Encryption_Standard There were suspicions that NSA had back-door access to the keys, possibly via the chips with which IBM were involved. I do not know how this was resolved: presumably nothing was proved, or I would have heard about it. A DES 56-bit key was cracked, and the aware world has I think moved on to Triple-DES. http://en.wikipedia.org/wiki/Backdoor_%28computing%29 discusses backdoors but not the DES/NSA scenario. Ken Thompson's 'Reflections on Trusting Trust', his Turing Award address (CACM 27.8 1984), is an 'ACM Classic' and worth a read. Guy

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