March 2013 - mathfun@mailman.xmission.com

Re: [math-fun] this HAS to be known
by Bill Gosper 02 Mar '13

02 Mar '13

Wouter> rwg>And, as Neil empiricizes, Sum[2^k/(1 + z^2^k), {k, 0, Infinity}] == 1/(z-1) 1. Why did Julian change his name? 2. do I understand your "it telescopes" correctly as f[z_,n_]:=Sum[2^k/(1 + z^2^k), {k, 0, n}] implies f[z, n+1] == 1/(1 + z) + 2*f[z^2, n] (1) 3. it is easy to show that 1/(z-1) satisfies (1) for n->inf and thus dropped, but how do you generaly solve a recursion like that? Look at it until the solution occurs to you? Wouter. 1. I have three tutors: Neil Bickford, http://nbickford.wordpress.com/ (who is currently trying to 3Dprint a largish biroller, and should soon announce his sliding block results) Julian Ziegler Hunts, the "evil child" who cooked DanA's extension of Alon Amit's vector puzzle, coauthor of http://www.blurb.com/b/2172660-minskys-trinskys-3rd-edition Corey Ziegler Hunts, author of the Mma version of my path invariant matrix tools and a really handy radical denester, and coauthor of Minskys & Trinskys. 2. It's simpler than that: FullSimplify[ 2^(1 + n)/(1 - x^2^(1 + n)) -(2^n/(1 - x^2^n))]== 2^n/(1 + x^2^n), so f[z,n]==1/(1-z) - 2^(n+1)/(1-z^2^(n+1)), giving both infinite sums. And the above identity follows from (f[z,oo]==1/(z-1)) - (f[z^2^(n+1),oo] == 1/(z^2^(n+1)-1), so it practically proves itself. 3. If we knew how to do this, we would teach our computer algebra systems. WHOA! Mma (9.0.1) can do this! In[80]:= RSolve[f[n + 1] - f[n] == 2^n/(1 + x^2^n), f[n], n] Out[80]= {{f[n] -> 1/(-1 + x) - 2^n/(-1 + x^2^n) + C[1]}} Now all they need to do is call RSolve from Sum. And all we need to do is figure out how they did it. DanA>Both of these are incredibly beautiful identities, especially if true. Which, given their source, seems very likely. If anyone has proved this already, my guess is it would be Euler. Have his complete works been collected yet, and even better, are they freely available in English translation on the Web? --Dan On 2013-02-27, at 3:05 PM, Bill Gosper wrote: Sum[2^k/(1 + z^2^k), {k, -Infinity, Infinity}] == 1/Log[z] Can somebody tell me where? --rwg And, as Neil empiricizes, Sum[2^k/(1 + z^2^k), {k, 0, Infinity}] == 1/(z-1) I clearly remember confronting several shelf-feet of Euler's Opera Omnia in the Stanford Math Library, while working for Knuth. The Latin is only half the problem, since Euler was only part way through inventing all our familiar notation! That guy was inhuman. He even knew about the motion of a freely tumbling brick. So I'd be surprised if he *didn't* know these z^2^n identities. But it would be nice to know for sure their origin. --rwg

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[math-fun] this HAS to be known
by Bill Gosper 01 Mar '13

01 Mar '13

Sum[2^k/(1 + z^2^k), {k, -Infinity, Infinity}] == 1/Log[z] Can somebody tell me where? --rwg And, as Neil empiricizes, Sum[2^k/(1 + z^2^k), {k, 0, Infinity}] == 1/(z-1)

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[math-fun] "printing" 3D fractals
by Warren Smith 01 Mar '13

01 Mar '13

> There are lots of examples of structural applications at the research level, such as this: http://physicsworld.com/cws/article/news/2012/nov/27/ultralight-fractal-str… but I haven't heard of any commercial CAD software yet incorporating the capability to make fractal supporting elements. FYI, I've illustrated a variety of algorithms for 3D printed fractals here, but they are for aesthetic, not structural applications: http://www.georgehart.com/rp/rp.html --1. that was interesting. You will recall I suggested pretty much the same "fractal structures" idea in 11 Nov 2011 math-fun post. --2. I am interested in whether one can fabricate complicated shape but ultra-precise structures made of glass. Is there any hope to do that? I want smooth surfaces and accuracy better than 1 micron. If the shapes are 2D (with some constant thickness) would that help? -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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[math-fun] Casino and unshuffled cards
by Eric Angelini 01 Mar '13

01 Mar '13

Hello Math-Fun, this news is dated August 2012: does someone know the end of the story? Best, É. http://www.dailymail.co.uk/news/article-2191343/Casino-sues-gamblers-won-1-…

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[math-fun] Wolfram Alpha large factorial bug
by Bill Gosper 01 Mar '13

01 Mar '13

MRob> I discovered a calculation error problem in Wolfram Alpha. When given "20000!" (the factorial of 20000) it gives the answer in two forms: * a "Decimal approximation", 1.8192063202303451348276... x 10^77337 * a "Power of 10 representation", 10^(10^4.888364627124846). My own calculations (using the Stirling series for the log Gamma function) indicate that the first answer is correct. One can see that the second answer disagrees with the first, by computing log(77337+log(1.81920632023)) using base-10 logarithms. The exponent should be 4.888388... not 4.888364... For "LogGamma[20001]", Wolfram Alpha gives 178075.621737198700312867928177..., which agrees with its first 20000! answer and with my calculations. I also checked this in GP/PARI, which of course gives all the digits of 20000!, but you can multiply by 1.0 to put it in the more familiar scientific notation, and it agrees with Wolfram Alpha's first answer. I'd be curious to find out what Mathematica gives (I suspect it will be the correct first answer). I sent feedback to WoldramAlpha to report the problem. -- Robert Munafo -- mrob.com Follow me at: gplus.to/mrob - fb.com/mrob27 - twitter.com/mrob_27 -mrob27.wordpress.com - youtube.com/user/mrob143 - rilybot.blogspot.com In[487]:= N[20000!, 22] Out[487]= 1.819206320230345134828*10^77337 In[488]:= Log[10, Log[10, %]] Out[488]= 4.888388780625234145206724465 In[489]:= N[Log[10, Log[10, 20000!]], 23] Out[489]= 4.8883887806252341452067 --rwg

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[math-fun] Mathpages
by David Wilson 01 Mar '13

01 Mar '13

I worry that someday www.mathpages.com will disappear from the web. Are the pages archived somewhere?

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[math-fun] "printing" 3D fractals
by Henry Baker 01 Mar '13

01 Mar '13

I've seen some examples of 3D fractals "printed" by these new 3D printers. Is anyone making CAD software for doing this sort of thing -- i.e., designing parts that incorporate 3D fractals as part of their structure?

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