May 2010 - mathfun@mailman.xmission.com

Re: [math-fun] Terminology for exponentiation
by Dan Asimov 06 May '10

06 May '10

I wrote (mildly edited): << When we write a base A to a power c to get the number A^c we say we "exponentiate A to the power c". I seek an appropriate verb so the the power is the object, reversing the emphasis: I.e., we use A *to do what* to c ??? Any help will be appreciated. >> A friend has pointed out that the word "exponentiate" already has this meaning, and that the default base is e. The online OED confirms this: << To raise e or some other base (BASE n.1 18) to the power of; to subject to exponentiation It's not entirely clear to me what's the best preposition to use when the base is not e and so must be mentioned. If it were a logarithm, to get log_b(x) from x is standardly "take the logarithm to the base b of x", or just "take the base-b logarithm of x". In the present case, maybe "from" is appropriate -- as in "exponentiate c from the base A" ? --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] Terminology for exponentiation
by Dan Asimov 05 May '10

05 May '10

Nice one, Joshua! Gosh, I haven't come across that word for years! But it's not really a verb. To go from c to x^c I'd have to say, "Find the base-x antilog of c." Or more formally, "Find the base-x antilogarithm of c." 12 syllables, as compared with only 7 for "Raise x to the power c". Maybe that's the best possible, though. --Dan << antilog? I wrote: << When we write a base x to a power c to get the number x^c we say we "exponentiate x to the power c". I seek an appropriate verb so the the power is the object, reversing the emphasis: I.e., we use x *to do what* to c ??? Any help will be appreciated. >> >> ________________________________________________________________________________________ "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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[math-fun] Terminology for exponentiation
by Dan Asimov 05 May '10

05 May '10

When we write a base x to a power c to get the number x^c we say we "exponentiate x to the power c". I seek an appropriate verb so the the power is the object, reversing the emphasis: I.e., we use x *to do what* to c ??? Any help will be appreciated. --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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[math-fun] hyperbolic paraboloid xy=z
by Bill Gosper 03 May '10

03 May '10

ParametricPlot3D[{x, y, x*y}, {x, -1, 1}, {y, -1, 1}] was a popular shape for "modernistic" public attractions in former decades--a glass walled, saddle shaped shell standing on two corners. ("Times table architecture".) What's its area? (Answer at end.) Now ParametricPlot3D[{{x, y, x*y}, {x, Sqrt[2]*Cos[Pi/4*x]*y, Sqrt[2]*Sin[Pi/4*x]*y}}, {x, -1, 1}, {y, -1, 1}] . Do these surfaces really coincide? What should the Cos and Sin really be? Now try ParametricPlot3D[{{x, y, x*y},r*x*{Cos[Pi*y/2], Sin[Pi*y/2], r*x*Sin[Pi*y]/2}}, {x, -1, 1}, {y, -1, 1}] for various 1<r<2. What are the areas of these round ones? --rwg Ans 1: A surprisingly complicated 4/Sqrt[3] - 2*Pi/9 + 6*ArcCsch[Sqrt[2]] + 2/3*Log[(Sqrt[3]-1)/Sqrt[2]] ~ 5.12316 . Ans 2: 1/Sqrt[1+x^2], x/Sqrt[1+x^2], Ans 3: 2 2 3/2 - Pi ((1 + r ) - 1) . Scales as r^3? 3 Also check out ParametricPlot3D[{{x, Sqrt[2]/Sqrt[1 + x^2]*y,Sqrt[2]*x/Sqrt[1 + x^2]*y}, {-Sqrt[2]/Sqrt[1 + x^2]*y, x, -Sqrt[2]*x/Sqrt[1 + x^2]*y}}, {x, -3, 3}, {y, -3, 3}] Voices tell me that the Shanghai-based Clank foundation is offering big bucks for the capture and conviction of the fabled parabolic hyperboloid--hyperbolas in sections parallel to two coordinate planes and parabolic intersections with the 3rd set of planes. Omigod, I've been using Vaseline five years past expiration!

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Re: [math-fun] Slanted coverage
by Bill Gosper 02 May '10

02 May '10

Further playing with the big hammer dA = dt du |V_t x V_u| gives the area of a 1x1xc spheroid as 2 pi (1 + c^2 asec(c)/sqrt(c^2-1)), with the axaxb general case following easily. This agrees with Mathworld's prolate (c>1) formula, but for oblate, Weisstein gives the logarithmic equivalent that avoids imaginary/imaginary. Something must be wrong with our notation if we need to switch at c = 1. The function there is smooth as a marble!-) A = 2 pi (1 + c 2F1[1,1,3/2,1/2-1/2c]) . (Rotsa ruck coaxing this out of Mma. Or even checking it.) A simpler example might be Plot[ArcCos[x]^2,{x,-1,5}], where the plotter needed to call a complex-valued function to get real results. Should we put acos^2 on pocket calculators? --rwg Joerg: The images are now (again, for me) all visible. Thanks for fixing this. The thanks go to Neil for rescuing the .nb and creating the working .html with his Windows 7 Machine.

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[math-fun] 5-fold symmetry with just one tile
by Joerg Arndt 01 May '10

01 May '10

I always thought that 5-fold symmetry can only be obtained by sophisticated constructions (as with Penrose tiles, see the image on the top right of http://en.wikipedia.org/wiki/Penrose_tiling ) Now a childishly simple contruction gives 5-fold symmetry with just one tile: Our tile is a triangle: /\ / \ /====\ The two angles at the base shall equal 2*Pi/k for k>=5. (Drawing obviously not to scale). Assemble tiles as follows: /\====/\====/\====/\== / \ / \ / \ / \ etc. ad inf. /====\/====\/====\/====\/ This is "plank A" Now arrange planks into an infinite wedge: / /\====/\ etc ad inf. to right and top / \ / \ / \/ /\====/\====/\====/\== / \ / \ / \ / \ / \/ \/ \/ /\====/\====/\====/\== / \ / \ / \ / \ / \/ \/ \/ /\====/\====/\====/\=== / \ / \ / \ / \ /====\/====\/====\/====\/== This is "wedge A" Arrange k such wedges around the origin to fill the plane. This arrangement has k-fold rotational symmetry but (except for k=6) no translational symmetry. Rotational symmetry can be avoided by taking for one wegde the "wedge B" obtained by reflecting wedge A over its horizontal base (we still use one tile only, reflection of tiles is obtained by rotation them by 180 degrees). Wedge B cannot be transformed to wedge A by a rotation. This is surely known since about Silurian years, but why doesn't it appear anywhere, e.g. in http://tilings.math.uni-bielefeld.de/ ?

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[math-fun] 10 millions joules ? how to avoid it.
by Simon Plouffe 01 May '10

01 May '10

Hello, There is this news about a man found in India that claims he has not eaten or drink for the past 70 years. He is now under strict medical attention. http://www.tdg.ch/loisirs/buzz/70-ans-boire-mager-2010-04-30 (the video is in english), But ? as far as I know, we need 10 million joules to move around, so how can this be possible. Ok, the guy does not move much since he does a lot of yoga and has a very strange look. I am skeptic. Strictly speaking how can he has any energy at all even if he does not do extreme sport ? In terms of physics, this is plain impossible, isn't ? 10^6 joules per day for 70 years = 2.5 x 10^10 joules. ...!!! Does anybody has any idea now the body can be alive for so long without any source of food or water ? Best regards, Simon Plouffe

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[math-fun] Iterated Modulus
by Meeussen Wouter (bkarnd) 01 May '10

01 May '10

for any number n and m<n, define the list m, Mod[n,m], Mod[n,Mod[n,m]], Mod[n,Mod[n,Mod[n,m]]], Mod[n,Mod[n,Mod[n,Mod[n,m]]]], .... stopping on Mod[n, 1 ] as the iterated modulus list on n started from m. It has the property that it is a strictly decreasing list since Mod[n,m] < m. This descent can only proceed until Mod[n, 1 ] is reached. Call the length of the resulting list modlen[n,m] example: the iterated modulus list of n=1439 started from m=20 is with length 12. Now, look at the modlen(n,k) with k running from 1 up to n. Call it the "fullmodlist". This list (length n) is not symmetric, but it has : n - 2 * S = 3 for n odd and n - 2 * S = 2 for n even where S equals the sum over last half of fullmodlist(n) - first half of fullmodlist(n) and that, I think, is curious. Wouter. -------------Mathematica---------------------------- mo[n_,m_]:= NestWhileList[Mod[n,#]&,m,#>1&] modlen[n_,m_]:=Length[NestWhileList[Mod[n,#]&,m,#>1&]] Table[p=2n+1;it=modlen[p,#]&/@Range[p];p-2 Plus@@Take[Reverse[it]-it,Floor[Length[it]/2]],{n,99}] Table[p=2n ;it=modlen[p,#]&/@Range[p];p-2 Plus@@Take[Reverse[it]-it,Floor[Length[it]/2]],{n,99}] =============================== This email is confidential and intended solely for the use of the individual to whom it is addressed. If you are not the intended recipient, be advised that you have received this email in error and that any use, dissemination, forwarding, printing, or copying of this email is strictly prohibited. You are explicitly requested to notify the sender of this email that the intended recipient was not reached.

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