October 2006 - mathfun@mailman.xmission.com

Re: [math-fun] Coriolism
by Daniel Asimov 14 Oct '06

14 Oct '06

Mike Stay: << So you're saying a 350 lb person standing on an awl; that kind of pressure would be useful in a machine shop. >> No, an awl would have "area" of almost 0, perhaps 10^(-4)" at the utmost. Lean on that with 100 lbs and you have one million psi easy. Rather, imagine a 250 lb person (I can beat that with both hands tied behind my back and blindfolded), standing on one stiletto heel that's (1/15)" x (1/15)". That would exceed 55,000 psi. (The whips and chains would add even a little more weight.) Sure, I exaggerated, but not enormously. --Dan

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Re: [math-fun] Coriolism
by Daniel Asimov 14 Oct '06

14 Oct '06

Mike Stay wrote: << On 10/14/06, Daniel Asimov <dasimov(a)earthlink.net> wrote: > Don't know about the physical effects, but 55,000 psi sounds close to the > pressure exerted by a stiletto heel (depending on who's wearing it), so not > that hard to imagine. Perhaps an elephant in stilettos would exert that much pressure. I think you're off by at least an order of magnitude. >> Well, I figured that if a 100-lb person stood on one stiletto heel that measures 1/10" x 1/10", that would give 10,000 psi, right? For someone a little heavier and a heel a little smaller, that could easily come close to 55,000 psi. --Dan

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Re: [math-fun] Coriolism
by Daniel Asimov 14 Oct '06

14 Oct '06

<< My laser shop claims to have a 55,000 psi waterjet. That's 24 miles deep in a hypothetical ocean. Few materials could contain such pressure, let alone create it. Wouldn't rocks 24 mi underwater turn to mush from effects tending to change their surface area? Or would the water turn solid? >> Don't know about the physical effects, but 55,000 psi sounds close to the pressure exerted by a stiletto heel (depending on who's wearing it), so not that hard to imagine. --Dan

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[math-fun] Coriolism
by R. William Gosper 14 Oct '06

14 Oct '06

Traveling equator to pole on a rotating sphere, I expected maximal Coriolis force (and more tornadoes) at 45 degrees latitude. But with d latitude/dt = c, I got maximal |acceleration| at the endpoints, except constant |acceleration| if c = w/sqrt(2), where w = Earth's angular velocity (d longitude/dt). Subtracting out the normal (local vertical) component introduces the expected extremum, but at latitude 2 2 c 1 asin sqrt(---- + -) 2 2 w instead of asin sqrt 1/2. Does this sound right? --rwg (DISTINGUISHEDLY THINLY-DISGUISED) Minor paradox: It gets too hot to mine about two miles deep. Even the Antarctic icecap melts underneath. But it's near freezing six miles deep in the ocean. My laser shop claims to have a 55,000 psi waterjet. That's 24 miles deep in a hypothetical ocean. Few materials could contain such pressure, let alone create it. Wouldn't rocks 24 mi underwater turn to mush from effects tending to change their surface area? Or would the water turn solid?

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Re: [math-fun] "Prime zeta function"
by Daniel Asimov 13 Oct '06

13 Oct '06

Franklin wrote: << Well, to start with, it's called the "prime zeta function" (what a coincidence!). See MathWorld: http://mathworld.wolfram.com/PrimeZetaFunction.html for a start. I don't know anything about your alternating prime zeta function, but the specific sum you mention is A078437; also mentioned in http://mathworld.wolfram.com/PrimeSums.html, equation (7). >> Thanks, Franklin. (Sometimes it gets very tiresome to learn of all the mathematicians who have pre-plagiarized me!) --Dan

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[math-fun] "Prime zeta function"
by Daniel Asimov 13 Oct '06

13 Oct '06

Is anything known about (let's dub it) "the prime zeta function" defined for Re(s) > 1 via Pzeta(s) := Sum{n=1..oo} of 1 / (p_n)^s, and elsewhere by analytic continuation, where as usual (p_n)^s := exp(s*ln(p_n)) ??? There's likewise the "alternating prime zeta function" defined for Re(s) > 1 via aPzeta(s) := Sum{n=1..oo} of (-1)^(n+1) / (p_n)^s I wonder what kind of a number aPzeta(1) = 1/2 - 1/3 + 1/5 - 1/7 + 1/11 - 1/13 + . . . = 0.26960+ is. --Dan

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[math-fun] First consecutive run of composites of a given length
by Daniel Asimov 13 Oct '06

13 Oct '06

A well-known guaranteed way to get a run of N consecutive composite integers is to multiply together all primes <= N+1, and then add 2, 3, 4, . . ., N+1, respectively, to the product. E.g., to get 9 consecutive composites, we could use [210+2 . . 210+10] = [212 . . 220] (since 210 = 2*3*5*7 with 7 <= 9+1 < 11). Call this sequence S(N). Which suggests a few questions: 1. How often, does S(N) give the *first* run of N composites? ([114 . . 122] is the first run of 9, rather than S(9) = [212 . . 220].) 2. How much farther can S(N) be extended to include more consecutive composites? (S(9) extends 2 farther to [212 . . 222].) And especially 3. {4 (1), 8 (3), 24 (5), 30 (7), 114 (13), . . . } shows the first number of each new record-breaking (in length) run of composites followed by the (length) of this run. The latter sequence 1, 3, 5, 7, 13,... is A066963 in OEIS, but 4, 8, 24, 30, 114, . . . -- the first numbers of each such run -- does not seem to be there. So I wonder if there is an asymptotic expression for A066963, or the companion sequence 4, 8, 24, 30, 114, . . . (or more likely their logs). (Or maybe a "fairer" companion sequence should be the *centers* of each record run of composites: 4, 9, 26, 33, 120, . . . ?) --Dan

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Re: [math-fun] First consecutive run of composites of a given length
by Daniel Asimov 13 Oct '06

13 Oct '06

Thanks to Jon Ziegler's sharp eyes re << And especially 3. {4 (1), 8 (3), 24 (5), 90 (7), 114 (13), . . . } shows the first number of each new record-breaking (in length) run of composites followed by the (length) of this run. The latter sequence 1, 3, 5, 7, 13,... is A066963 in OEIS, . . . >> it's clear that << . . . but 4, 8, 24, 90, 114, . . . -- the first numbers of each such run -- does not seem to be there. is wrong: 4,8,24,90,114,... is indeed in OEIS as A008950 (with the same definition). (The midpoints of each record run: 4,9,26,93,120,... don't appear, though that's not too important.) --Dan

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[math-fun] order of 3 in Mersenne-prime group
by Don Reble 11 Oct '06

11 Oct '06

Math-funners: I observe that, for many Mersenne primes Mp > 3, the co-order* of 3 (in Mp's multiplicative group) is a power of 3. Indeed, it's true for any such Mp-group, whose order (Mp - 1) has been fully factored. Does anyone know a related theorem? *co-order: group order divided by element order -- Don Reble djr(a)nk.ca -- This message has been scanned for viruses and dangerous content by MailScanner, and is believed to be clean.

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Re: [math-fun] To Rich: Finding primes between multiples of a fixednumber
by Daniel Asimov 10 Oct '06

10 Oct '06

Thanks for the link, Josh. (I was unable to find it myself.) I doubt any more terms are needed! Dan

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