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[math-fun] Name for function IEEE32(x) LOGAND -2^23 ??
by Henry Baker 04 Apr '17

04 Apr '17

Is there a name for the function that simply kills all of the bits of precision from an IEEE floating point number, leaving it as a simple power of 2 ? For a big-endian machine, pick up the 32 bits of the IEEE32 float and AND it with -2^23. Or pick up 64 bits of IEEE64 float and AND it with -2^52.

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Re: [math-fun] Name for function IEEE32(x) LOGAND -2^23 ??
by Leo Broukhis 03 Apr '17

03 Apr '17

You may benefit from standard functions frexp and ldexp, unless the performance is critical. Leo On Apr 3, 2017 1:52 PM, "Henry Baker" <hbaker1(a)pipeline.com> wrote: Is there a name for the function that simply kills all of the bits of precision from an IEEE floating point number, leaving it as a simple power of 2 ? For a big-endian machine, pick up the 32 bits of the IEEE32 float and AND it with -2^23. Or pick up 64 bits of IEEE64 float and AND it with -2^52. _______________________________________________ math-fun mailing list math-fun(a)mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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Re: [math-fun] Japanese Temple Geometry, and misc comments
by Henry Baker 03 Apr '17

03 Apr '17

At 11:09 PM 4/2/2017, rcs(a)xmission.com wrote: >In a different direction: The IBM 1620 computer (c.1962) used tables >in memory for single-digit addition and multiplication. >This might have been used to do arithmetic in other bases < 10. It's been 58 years; do we have to file a FOIA request to ask NSA if they ever used the 1620 table lookup for bases other than base 10? I hacked the 1620 while still in high school, but didn't know enough math at that time to really utilize the table lookup feature for anything other than base ten. Also, I don't recall whether the machine could even boot if the arithmetic tables got wiped, so trying experiments of this form could be very dangerous. (Core memories retain their data when powered off, so wiping these tables was rare.) BTW, modern GPU's are *really good* at interpolating large tables; Google "texture mapping": https://en.wikipedia.org/wiki/Texture_mapping_unit "In GPGPU, texture maps in 1,2, or 3 dimensions may be used to store arbitrary data. By providing interpolation, the texture mapping unit provides a convenient means of approximating arbitrary functions with data tables."

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[math-fun] Calculating asinh(x)
by Henry Baker 03 Apr '17

03 Apr '17

y = asinh(x) = log(x + sqrt(1+x^2)) the "obvious" formula. We have several choices: * Calculate using the obvious log formula * Calculate using Newton's method on sinh(y) * More complicated methods One more complicated method is to factor the above expression as: y = asinh(x) = f(g(x)), where f(x) = log(x) and g(x) = x+sqrt(1+x^2) Note that if y=g(x), then x=(y-1/y)/2. So we could first invert g(x) using Newton's method, and then apply the log function. The iteration to invert y=g(x) is x -> 2*(y+(x-y)/(1+x^2)) with an initial guess of 2*y. This iteration has quadratic convergence, and 6 iterations provides fpprec=128 precision. It remains to be seen if this is faster and/or more precise than the obvious formula. The other possibility is to get a very good approximation, and finish it off with a very small (1,2,3 ?) iterations of Newton to invert y=sinh(x). The iteration is x -> x-tanh(x)+y/cosh(x), with quadratic convergence: x0+eps -> x0+eps^2*tanh(x0)/2

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[math-fun] Japanese Temple Geometry, and misc comments
by rcs＠xmission.com 03 Apr '17

03 Apr '17

Rich Howard passed along a 1998 SciAm article about Japanese Temple Geometry, an interesting kind of art. Google turned up a non-paywalled link at www.math.rutgers.edu/~sussmann/papers/res-japanese-temple-geometry.ps.gz see also https://en.wikipedia.org/wiki/Sangaku Wrt Henry's thread about math function lookup tables, two quick notes: The article includes a list of tables available on a chip, beginning with squares. There's no mention of two ancient methods of multiplication, using quarter-squares, or triangular numbers: QS(N) = floor[N^2 / 4] xy = QS(x+y) - QS(x-y) T(N) = N(N+1)/2 xy = T(x+y) - T(x) - T(y) I've seen references to tables of quarter-squares. As noted previously, the current ratio of compute:memory timing makes this (only) an interesting historical method. But recall that the famous Intel Floating Divide Bug depended on data missing from an internal chip table of reciprocals. In a different direction: The IBM 1620 computer (c.1962) used tables in memory for single-digit addition and multiplication. This might have been used to do arithmetic in other bases < 10. In other news, I didn't see any mention on the list of Raymond Smullyan's passing. Maybe I missed it? Finally, a quick note about a new attack on ECDLP, the Elliptic Curve Discrete Log Problem. The paper is at https://arxiv.org/pdf/1703.07544.pdf. The claim is a new, more efficient algorithm, but the supporting evidence is thin. They mention solving problems in an elliptic curve group of order 129159847, which is too small to be definitive. There's no specific statement of the form "the ECDLP for a curve of order X can be solved in work f(X)", but the ingredients for an analysis are present. They do seem to have a new idea. I haven't worked out a work estimate. Rich

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Re: [math-fun] Lookup-table math
by Henry Baker 02 Apr '17

02 Apr '17

Latency seems to go up as x, while storage size goes up as x^2 (or even x^3), where x is 1/featuresize. Now the constant factors aren't so great, but perhaps caching helps for a lot of this type of arithmetic (e.g., Benford's Law). Also, you can spend some of these spare cycles *interpolating*, since you will likely get back a large number of nearby points after the long latency wait. Unfortunately, I've forgotten how to interpolate certain types of tables -- e.g., modular & EC calculations. At 09:56 AM 4/2/2017, Tomas Rokicki wrote: >Probably latency is much more important than bandwidth for this application. > >The latency of these cards is typically 100 microseconds for a random read (this is for a fast M2 card; secure digital cards and SATA are much slower). > >At 3GHz that's 300,000 cycles. > >Intel's XPoint (now branded Optane) promise much better latency, as low as 9 microseconds, but even this is 27,000 cycles. > >With a GPU and 9 microseconds you can do an awful lot of computation. > >On Sun, Apr 2, 2017 at 9:46 AM, Simon Plouffe <simon.plouffe(a)gmail.com> wrote: >> Hello, >> >> a math table lookup on a SSD micro card ?: why not ... >> >> ok but the reading speed of those are something like >> 90 megs per second on a 'class 10' card, >> >> on the other hand there are those new ssd that can be plugged into >> a PCI express 4x slot (if you have one on your pc), >> >> http://www.materiel.net/ssd/samsung-serie-960-pro-m-2-pcie-vnme-2-to-136308… >> >> the reading speed is 3500 megabytes per second. >> That's another animal, the drawback is of course the price >> and your electricity bill. ;-). and you have 1 TB of space >> to put tables. the price is about 650 us dollars. >> >> best regards, >> >> Simon Plouffe

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[math-fun] Lookup-table math
by Henry Baker 02 Apr '17

02 Apr '17

FYI -- In an age where you can purchase 128GB uSD cards for $50 (or less), who needs to do any calculations anymore? https://www.amazon.com/gp/product/B01M4MW74V Silicon Power 128GB Elite microSDXC UHS-1 Memory Card - with Adapter (SP128GBSTXBU1V20BS) Price: $49.99 & FREE Shipping --- I assume that we'll be back to purchasing (digital) math tables again. That's certainly the case today for modular and EC arithmetic. :-) As you can guess from the discussions, the article below was written quite a while ago, but his argument has gotten much better during the interim. http://wilsonminesco.com/16bitMathTables/ Large Look-up Tables for Hyperfast, Accurate, 16-Bit Scaled-Integer Math (Describes how to interface large -- greater than address space size -- ROMs to 8-bit microprocessors for ADC computations.)

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Re: [math-fun] More power series Q's
by Henry Baker 30 Mar '17

30 Mar '17

At 12:51 PM 3/29/2017, Schroeppel, Richard wrote: >Easy transform for your 3x equations: >Set g(x) = K f(x), with K tbd. >Then >G(3x) = K f(3x) = 3K f(x) + Kb f(x)^3 > = 3 G(x) + (Kb/27) G(x)^3. > >Adjust Kb/27 to be whatever you like; I suggest trying +-4. Indeed! Thanks! (%i1) bar; 3 (%o1) f(3 x) - b f (x) - 3 f(x) (%i2) f(x):=K*sinh(x); (%o2) f(x) := K sinh(x) (%i3) %o1,eval; 3 3 (%o3) - b sinh (x) K + sinh(3 x) K - 3 sinh(x) K (%i4) %/K,expand; 3 2 (%o4) - b sinh (x) K + sinh(3 x) - 3 sinh(x) (%i5) %,K^2=4/b,expand; 3 (%o5) sinh(3 x) - 4 sinh (x) - 3 sinh(x) (%i6) %,trigreduce,expand; (%o6) 0 So our b is simply a weird way to scale sinh().

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[math-fun] Newman = -1/Oldman
by Bill Gosper 30 Mar '17

30 Mar '17

This must have been found by Newman himself but I only just noticed. Enumerate the nonnegative rationals by iterating "Newman's function": In[96]:= NestList[1/(2*Floor[#] + 1 - #) &, 0, 9] Out[96]= {0, 1, 1/2, 2, 1/3, 3/2, 2/3, 3, 1/4, 4/3} "Oldman's function" runs the sequence back to 0, counting the rationals: In[97]:= NestList[2*Ceiling[1/#] - 1 - 1/# &, 4/3, 9] Out[97]= {4/3, 1/4, 3, 2/3, 3/2, 1/3, 2, 1/2, 1, 0} E.g., 0 is the zeroth rational, 4/3 is the ninth. Oldman is completely unnecessary! Just negate and reciprocate the last value, and Newman runs backwards: In[100]:= NestList[1/(2*Floor[#] + 1 - #) &, -3/4, 9] Out[100]= {-3/4, -4, -1/3, -3/2, -2/3, -3, -1/2, -2, -1, ComplexInfinity} (with a much louder completion announcement). Or if you prefer, Oldman runs the negative rationals forwards: In[99]:= NestList[2*Ceiling[1/#] - 1 - 1/# &, -1, 9] Out[99]= {-1, -2, -1/2, -3, -2/3, -3/2, -1/3, -4, -3/4, -5/3} obviating Newman. --rwg

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[math-fun] ries strikes again. Mathematica answers!
by Bill Gosper 29 Mar '17

29 Mar '17

Date: 2017-03-27 18:21 From: "David Wilson" <davidwwilson(a)comcast.net> rwg, apologies for filling your head with all these √2's. > -----Original Message----- > From: math-fun [mailto:math-fun-bounces@mailman.xmission.com] On > Behalf Of Bill Gosper > Sent: Monday, March 27, 2017 8:25 PM > To: math-fun(a)mailman.xmission.com > Subject: [math-fun] ries strikes again. Mathematica answers! > > In[69]:= ((1/2 ((-I)^(1 + √2) + I^(1 + √2)) - > Sqrt[-1 + 1/4 ((-I)^(1 + √2) + I^(1 + √2))^2])^(-1 + √2) + > (1/2 ((-I)^(1 + √2) + I^(1 + √2)) + > Sqrt[-1 + 1/4 ((-I)^(1 + √2) + I^(1 + √2))^2])^(-1 + √2)) // > FullSimplify > > Out[69]= 2 Sin[π 2√2] > > (How??) --rwg CNS Crash Report: Brain capacity exceeded. OS version: √2 Reboot. OS version √3 In[518]:= %367 /. m -> 2 - Sqrt[3] /. n -> Sqrt[3] + 2 // Simplify // tim During evaluation of In[518]:= 20.750373,2 Out[518]= 1/2 ((1/2 ((x - Sqrt[-1 + x^2])^(2 - Sqrt[3]) + (x + Sqrt[-1 + x^2])^( 2 - Sqrt[3])) - Sqrt[-1 + 1/4 ((x - Sqrt[-1 + x^2])^( 2 - Sqrt[3]) + (x + Sqrt[-1 + x^2])^(2 - Sqrt[3]))^2])^( 2 + Sqrt[3]) + (1/ 2 ((x - Sqrt[-1 + x^2])^(2 - Sqrt[3]) + (x + Sqrt[-1 + x^2])^( 2 - Sqrt[3])) + Sqrt[-1 + 1/4 ((x - Sqrt[-1 + x^2])^( 2 - Sqrt[3]) + (x + Sqrt[-1 + x^2])^(2 - Sqrt[3]))^2])^( 2 + Sqrt[3])) == x ∀ complex x ! (according to Plot3D, which can miss isolated point discontinuities, which I think can only occur along ridges of infinite curvature. --rwg)

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