July 2020 - mathfun@mailman.xmission.com

[math-fun] Can this recurrence be solved in closed form?
by James Propp 12 Jul '20

12 Jul '20

Kelly DeRango writes (on Twitter): <https://twitter.com/Kderango> [image: Baseball] Kelly DeRango @Kderango <https://twitter.com/Kderango> · Jul 11 <https://twitter.com/Kderango/status/1282007791041904640> Math Twitter we need your help. Our high school math club would like to find an explicit formula for the recursive relationship g[n+1] = (g[n])² - g[n] + 5. Can anyone help? The initial term is g[0] = 3. The students in question believe (and may have proved) that the sum of arctan(2/g[n]) as n goes from 0 to infinity is Pi/4, so they'd like a formula for g[n]. Jim Propp

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Re: [math-fun] High schoolers: simplify this expression
by Henry Baker 12 Jul '20

12 Jul '20

True. Now do it like a high schooler... At 09:17 AM 7/10/2020, Simon Plouffe wrote: >Hello, > >the expression is simplified in a fraction of a second (sqrt(108)+10)^(1/3)-(sqrt(108)-10)^(1/3); to 2 by Maple. > >Best regards, > >Simon Plouffe > >Le ven. 10 juil. 2020 Ã 17:47, Henry Baker <hbaker1(a)pipeline.com> a ÃÂ©crit : >> (sqrt(108)+10)^(1/3)-(sqrt(108)-10)^(1/3) >> >> [I thought this was a pretty cool problem that came up on the Maxima email >> list.]

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Re: [math-fun] OEIS A001414 research project for an undergrad
by Steve Witham 12 Jul '20

12 Jul '20

> From: Allan Wechsler <acwacw(a)gmail.com> > Date: 7/4/20, 4:35 PM > > If it is easy for you to do, I would like to see a linear/linear version. Belatedly, I've added the following bits at http://www.mac-guyver.com/switham/2020/07/OEIS_A001414/ o some linear/linear plots, o cut-off histograms of the a() values, o the (short) Python code (but also the necessary factoring code), o well it's a primitive web page now. Stopping n in the histogram is traumatic-- at a given N cutoff, early values of a(n < N) have appeared as many times as they ever will, and that section of the plot is very smooth, but beyond that it's like the trail of a rocket that exploded during ascent... oh yeah use OEIS... The full histogram is http://oeis.org/A000607 That math looks interesting in itself. > My theory has to do with A056239; this is very similar in concept > to A001414, but instead of summing the primes themselves, you sum the > *indices* of the primes, so for 84 = 2 * 2 * 3 * 7, the value is 1 + 1 + > 2 + 4 = 8. It's possible that the "feathers" are formed by classes of > integers which share the same A056239 value, so you might try doing a plot > where points are assigned colors based on A056239. I have not done this. I imagine one can just keep rotating the hue by 137.508 degrees... --Steve

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Re: [math-fun] High schoolers: simplify this expression
by Hilarie Orman 12 Jul '20

12 Jul '20

It's been all downhill since the end of the digital age (that horrid crutch, the abacus). Hilarie Sat, 11 Jul 2020 at 18:25:11 +0100 Fred Lunnon: > Hey --- lighten up, gramps! > >WFL On 7/11/20, Joerg Arndt <arndt(a)jjj.de> wrote: > * Henry Baker <hbaker1(a)pipeline.com> [Jul 11. 2020 16:49]: >> True. >> >> Now do it like a high schooler... > > Search the internet, you mean? > And I wish this was true only for high schoolers. > The concept of "book" (or, god forbid, "journal article") > appears to be disappearing young people. > > Best regards, jj > > >> >> At 09:17 AM 7/10/2020, Simon Plouffe wrote: >> >Hello, >> > >> >the expression is simplified in a fraction of a second >> > (sqrt(108)+10)^(1/3)-(sqrt(108)-10)^(1/3); to 2 by Maple. >> > [...]

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[math-fun] triadic dragon reminiscence
by Bill Gosper 11 Jul '20

11 Jul '20

Half a triadic dragon <http://gosper.org/semitriadic.png> self-similarly trisects into knobby canes (blue, orange+green, red+purple). But the whole triadic dragon self-similarly trisects, so two thirds of this half-dragon (blue+orange+green) is itself a triadic dragon. Deleting the blue cane leaves a (not self similar) congruently bisected backwards Z, orange+green, red+purple. But it is also in fact self-similarly trisected! (orange, green+red, purple). The Triadic Dragon joining 0 to 1 has bounding box West: -1/16, East: 17/16 (with index 239(!)/240), and N|S bounds ±9*i*/(16√3) (less certain here). Is there a plane figure that can be dissected into two, as well as three, but not six congruent pieces? —rwg

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[math-fun] Counting vs. cross-products vs. tree traversal
by Steve Witham 11 Jul '20

11 Jul '20

Counting in decimal is analogous to walking a tree.Â A tree of unbounded depth for that matter.Â "Carrying" is like popping the stack and going to another [[great...] grand] parent node.Â The higher digits are further *up* the tree and it just keeps going up... I'm getting hypoxia just thinking about it. Grade-school kids (me included) eventually learn to count without a lot of thinking. In contrast I find tree-traversals (usually implicit) one of the most awkward tasks in programming.Â I mean depth-first.Â Even when the number of levels is fixed, even when it's shallow enough to use recursion in a language with a limited stack.Â The bookkeeping, the program structure are just awkward. Even in Prolog (where depth-first tree search is everything) it might make a program a bit scattered-looking.Â Well let's try. (I'm making up a dialect of Prolog here. "(car : cdr)" shall mean cons.) Â Â Â (Decimal (d))Â ifÂ (Digit d). Â Â Â (Digit d)Â ifÂ (Element d (0 1 2 3 4 5 6 7 8 9)). Â Â Â (Decimal (d : n)) if (NZDigit d) and (ZDecimal n). Â Â Â (NZDigit d)Â ifÂ (Element d (1 2 3 4 5 6 7 8 9)). Â Â Â (ZDecimal (d))Â ifÂ (Digit d). Â Â Â (ZDecimal (d : n))Â ifÂ (Digit d) and (ZDecimal n). But all that aside, the digit-counting analogy sounds like a nice way to organize a tree search or a very-nested loop.Â Have an array of digit positions, and at each position a sub-array of possibilities.Â Then have a parallel array of indexes into those arrays.Â So, each time through the loop you increment the bottom digit with carry.Â Easy peazy, heh. (I'm currently writing a test for a module with a big cross-product of test cases.) Â --Steve

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[math-fun] Draft of July 17 essay
by James Propp 10 Jul '20

10 Jul '20

The current draft of my July 17 Mathematical Enchantments essay is online at http://mathenchant.org/063-draft1.pdf . Comments are welcome! Jim Propp

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[math-fun] High schoolers: simplify this expression
by Henry Baker 10 Jul '20

10 Jul '20

(sqrt(108)+10)^(1/3)-(sqrt(108)-10)^(1/3) [I thought this was a pretty cool problem that came up on the Maxima email list.]

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[math-fun] Amusing math problem
by Veit Elser 10 Jul '20

10 Jul '20

For some reason I only got Gareth’s cool solution. Below is the obvious one I sent just to Dan. -Veit On 10/07/2020 12:06, J.P. Grossman wrote: Another short solution: [...] Oh, that's _very_ nice. -- g Begin forwarded message: From: Veit Elser <ve10(a)cornell.edu<mailto:ve10@cornell.edu>> Subject: Re: [math-fun] Amusing math problem Date: July 8, 2020 at 4:10:07 PM EDT To: Dan Asimov <dasimov(a)earthlink.net<mailto:dasimov@earthlink.net>> g(x)=log f(x) * = int_0^1 exp(g(x+c) - g(x)) dx >= int_0^1 (1 + g(x+c) - g(x)) dx = 1 -Veit

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[math-fun] Amusing math problem
by Dan Asimov 10 Jul '20

10 Jul '20

I recently found this problem online somewhere and with a little thought and a little luck managed to solve it. I think. ----- Let f : R —> R be a continuous function that is positive and periodic of period 1. That is, f(x+1 ) = f(x) for all x. Prove that for any real number c Integral_{0 ≤ x ≤ 1} f(x+c)/f(x) dx ≥ 1 ----- —Dan

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