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[math-fun] Biologists turn cells into biocomputers
by Henry Baker 29 Mar '17

29 Mar '17

FYI -- The cycle time is a little slow, but what the heck? Perhaps these glow-in-the-dark cells could become a bit-mapped display? Then someone could really say "read my lips"... http://www.sciencemag.org/news/2017/03/scientists-turn-mammalian-cells-comp… Scientists turn mammalian cells into complex biocomputers By Robert F. Service Mar. 27, 2017 , 11:00 AM Computer hardware is getting a softer side. A research team has come up with a way of genetically engineering the DNA of mammalian cells to carry out complex computations, in effect turning the cells into biocomputers. The group hasn't put those modified cells to work in useful ways yet, but down the road researchers hope the new programming techniques will help improve everything from cancer therapy to on-demand tissues that can replace worn-out body parts. Engineering cells to function like minicomputers isn't new. As part of the growing field of synthetic biology, research teams around the globe have been manipulating DNA for years to make cells perform simple actions like lighting up when oxygen levels drop. To date, most such experiments have been done in Escherichia coli and other bacteria, because their genes are relatively easy to manipulate. Researchers have also managed to link multiple genetic circuits together within a single cell to carry out more complex calculations in bacteria. Scientists have tried to extend this to mammalian cells to create genetic circuitry that can help detect and treat human diseases. But efforts to construct large-scale genetic circuits in mammalian cells have largely failed: For complex circuits to work, the individual componentsÂthe turning on and off of different genesÂmust happen consistently. The most common way to turn a gene on or off is by using proteins called transcription factors that bind to and regulate the expression of a specific gene. The problem is these transcription factors "all behave slightly differently," says Wilson Wong, a synthetic biologist at Boston University. To upgrade their DNA "switches," Wong and his colleagues steered clear of transcription factors and instead switched human kidney cell genes on and off using scissorlike enzymes that selectively cut out snippets of DNA. These enzymes, known as DNA recombinases, recognize two target stretches of DNA, each between 30 to 50 or more base pairs long. When a recombinase finds its target DNA stretches, it cuts out any DNA in between, and stitches the severed ends of the double helix back together. To design genetic circuits, Wong and his colleagues use the conventional cellular machinery that reads out a cell' s DNA, transcribes its genes into RNA, and then translates the RNA into proteins. This normal gene-to-protein operation is initiated by another DNA snippet, a promoter, that sits just upstream of a gene. When a promoter is activated, a molecule called RNA polymerase gets to work, marching down the DNA strand and producing an RNA until it reaches another DNA snippetÂa termination sequenceÂthat tells it to stop. To make one of their simplest circuits, Wong's team inserted four extra snippets of DNA after a promoter. The main one produced green fluorescent protein (GFP), which lights up cells when it is produced. But in front of it was a termination sequence, flanked by two snippets that signaled the DNA recombinase. Wong and his team then inserted another gene in the same cell that made a modified recombinase, activated only when bound to a specific drug; without it, the recombinase wouldn't cut the DNA. When the promoter upstream of the GFP gene was activated, the RNA polymerase ran headfirst into the termination sequence, stopped reading the DNA, and didn't produce the fluorescent protein. But when the drug was added, the recombinase switched on and spliced out the termination sequence that was preventing the RNA polymerase from initiating production of GFP. Voila, the cell lit up. As if that Rube Goldbergian feat weren't enough, Wong and his colleagues also showed that by adding additional recombinases together with different target strands, they could build a wide variety of circuits, each designed to carry out a different logical operation. The approach worked so well that the team built 113 different circuits, with a 96.5% success rate, they report today in Nature Biotechnology. As a further demonstration, they engineered human cells to produce a biological version of something called a Boolean logic lookup table. The circuit in this case has six different inputs, which can combine in different ways to execute one of 16 different logical operations. "It's exciting in that it represents another scale at which we can design mammalian genetic circuits," says Timothy Lu, a synthetic biologist at the Massachusetts Institute of Technology in Cambridge. Although the current circuits are a proof of concept, both Lu and Wong say synthetic biologists want to use them to create new medical therapies. For example, scientists could engineer T cells, sentinels of the immune system, with genetic circuits that initiate a response to wipe out tumors when they detect the presence of two or three "biomarkers" produced by cancer cells, Lu says. Another example being explored by Wong and others is to engineer stem cells so they develop into specific cell types when prompted by different signals. This could let synthetic biologists generate tissues on demand, such as insulin-producing cells, or cartilage-producing chondrocytes. DOI: 10.1126/science.aal0963

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Re: [math-fun] about a certain François Mendzina Essomba and surprising pi formulas
by Simon Plouffe 29 Mar '17

29 Mar '17

Hello, I programmed the first one, is is not like the AGM, the number of valid digits is proportional to n, something like approx. (5/2)^n, here is the maple routine : # F.M. Essomba formula es:=proc(n) option remember; if n = 0 then 1/2 elif n = 1 then sqrt(3.0)/2 else sqrt(2.0*es(n-1) + 2)/2.0 end if: end; a:=proc(n) 3*2.0^n/315*sqrt(1.0-es(n)^2)*(256 + 597*es(n) - 1024*es(n)^2 + 542*es(n)^3-56*es(n)^5);end; a(10000) differs from Pi at the 3979'th position. a (4000) differs from Pi at the 1591'th position, confirming that it converges rather slowly. It is not a record calculation, but certainly original (IMHO). I have not tried the other ones yet. Have a nice day and bonne journée.

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Re: [math-fun] posits, IEEE floats, asinh numbers
by Henry Baker 29 Mar '17

29 Mar '17

At 07:16 PM 3/27/2017, W. Edwin Clark wrote: >On Mon, Mar 27, 2017 at 8:41 PM, Gareth McCaughan <gareth.mccaughan(a)pobox.com> wrote: >The book is apparently a "recent" ("revolutionary" ><https://www.amazon.com/gp/product/1482239868?keywords=the%20end%20of%20erro…>) > book by the author of the slides. http://johngustafson.net/ Has anyone been able to find a valid email for John Gustafson? The only one I found (john.gustafson(a)nus.sg.edu) doesn't work.

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[math-fun] More power series Q's
by Henry Baker 28 Mar '17

28 Mar '17

We have sin(3*x)=3*sin(x)-4*sin(x)^3 and sinh(3*x)=3*sinh(x)+4*sinh(x)^3 These give us recursive approximations to sin(x), sinh(x), thanks to RWG/Hakmem. So we have f(3*x)=3*f(x)+-4*f(x)^3 What are the functions f(3*x)=3*f(x)+b*f(x)^3, for some real b ? They seem to live in the land between sin(x) and sinh(x). Actually, I'm more interested in f(x), s.t. f(2*x) = 2*f(x)+b*f(x)^3, for some real (probably integer?) constant b, because I want a representation in which binary shifting is easy(er). If f(x) = x + ..., then we have f(x) = 8 17 7 15 6 13 5 11 4 9 3 7 3035153449 b x 32516717 b x 151 b x 163 b x 37 b x b x --------------------- + ----------------- + ---------- + ---------- + -------- + ----- 137019759699497875200 29401699409082000 3204726525 97389600 771120 945 2 5 3 b x b x + ----- + ---- + x 60 6 How to easily compute these coefficients? If b=4, the coefficient of x^3 becomes 2/3, which matches that in sinh(2x)/2, although f(x) doesn't converge quite as fast. If b=1, f(x) converges somewhat faster, but how fast? Does it converge everywhere? What is its radius of convergence? If f(2*x)=2*x+b*x^3, what is the/a differential equation for f(x)?

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[math-fun] ries strikes again. Mathematica answers!
by Bill Gosper 28 Mar '17

28 Mar '17

Nice! Tnx! --rwg Date: 2017-03-28 10:15 From: Dan Asimov <asimov(a)msri.org> To: math-fun <math-fun(a)mailman.xmission.com> Reply-To: math-fun <math-fun(a)mailman.xmission.com> I took this as a personal challenge. First note that 1/2 ((-I)^(1 + √2) + I^(1 + √2)) = cos(pi(1+sqrt(2))/2) . Let T = pi(1+sqrt(2))/2. Also note that 1/4 ((-I)^(1 + √2) + I^(1 + √2))^2 = (cos T)^2 so that sqrt(-1 + 1/4 ((-I)^(1 + √2) + I^(1 + √2))^2) = i sin T . Then the whole expression is (cos T - i sin T)^(1 + sqrt(2)) + (cos T + i sin T)^(1 + sqrt(2)) or exp( -i T (1 + sqrt(2)) + exp( i T (1 + sqrt(2)) = 2 cos(T(1 + sqrt(2))) = 2 cos(pi(1 + sqrt(2))^2 / 2) = 2 cos((pi/2) (3 + 2 sqrt(2)) = 2 cos(3 pi/2 + pi sqrt(2)) = 2 [cos(3 pi/2) cos(pi sqrt(2)) - sin(3 pi/2) sin(pi sqrt(2))] = 2 sin(pi sqrt(2)) Q.E.D. —Dan Bill Gosper wrote (March 27, 2017 at 5:25 pm): ----- 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??) -----

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[math-fun] ries strikes again. Mathematica answers!
by Bill Gosper 28 Mar '17

28 Mar '17

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

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Re: [math-fun] posits, IEEE floats, asinh numbers
by W. Edwin Clark 27 Mar '17

27 Mar '17

> > Is there a write-up of posits that actually specifies how the regime bits > work--I couldn't really tell from the Gustafson slides. Is there > aparameter besides the total number of bits that needs to be specified? It should be in this book: The End of Error: Unum Computing (Chapman & Hall/CRC Computational Science) – February 5, 2015 by John L. Gustafson (Author) <https://www.amazon.com/End-Error-Computing-Chapman-Computational/dp/1482239…>

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Re: [math-fun] posits, IEEE floats, asinh numbers
by Leo Broukhis 27 Mar '17

27 Mar '17

I've explained how the regime bits work in one of my recent messages. There are two parameters: total size S and the size of the exponent tail E. The most significant bits of the exponent are expressed in unary ((N+1) 1s, then 0 for non-negative, -N 0s then 1 for negative). The value of the exponent is N*2^E + tail (if abs(N) is too large for the whole tail to fit in the representation, the trailing bits of the tail are assumed to be 0). The mantissa is 1.{remaining bits to the right of the exponent tail} Leo On Mar 27, 2017 12:17, "Mike Speciner" <ms(a)alum.mit.edu> wrote: You're correct if the integer addition produces "mathematically correct" results and we're talking about 32-bit integers and 64-bit IEEE floats. And the point about javascript is right on. I agree that asinh numbers (particularly when scaled so that they handle denormalized floats) would be problematic for integers. I do like the idea of using asinh in place of log for mapping axes on some graphs/plots. Is there a write-up of posits that actually specifies how the regime bits work--I couldn't really tell from the Gustafson slides. Is there a parameter besides the total number of bits that needs to be specified? --ms On 27-Mar-17 15:06, Dave Dyer wrote: > At 11:50 AM 3/27/2017, Mike Speciner wrote: > >> No, e.g., int32 x = -1<<31, y = -1 >> In that case x+y == -1 >> is definitely not equal to (double)x+(double)y >> > That's not well behaved in the integer domain either - you > ought to get a range exception, but of course you never do. > > In cases where the pure integer arithmetic is well behaved, > is there any case where the same calculation in IEEE doubles > gives a different result? > > > > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > > _______________________________________________ math-fun mailing list math-fun(a)mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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[math-fun] Pafnuti piles on.
by Bill Gosper 27 Mar '17

27 Mar '17

We can play D Wilson's game with Chebyshevs because T_m ○T_n = T_mn . Are these just conjugates of x² - 2? Then why can't I guess the formula for the hyperbolish boundary gosper.org/chebcheb.png ? A possible approach to simplification: Even when it fails at a specific point, e.g., In[432]:= %370 /. x -> -1/Sqrt[5] // FullSimplify Out[432]= 1/10 (6 + (1/ 2 ((-1 - 2 I)^Sqrt[2] + (-1 + 2 I)^Sqrt[2] - . . . (where %370 is T_√2 ○T_√2(x) - 2 x² +1, as in the png), Series can sometimes cough up any number of 0 terms: In[438]:= Series[%370,{x,-1/Sqrt[5],4}]//tim During evaluation of In[438]:= Series::ztest: Unable to decide whether numeric quantities {3/5+1/2 (1/2 5^ . . . (-1+2 I)^Sqrt[2]]} are equal to zero. Assuming they are. During evaluation of In[438]:= 12.774193,6 Out[438]= O[x+1/Sqrt[5]]^5 But if you accept that the expression vanishes at specific points, you might believe that it does for a region peppered with such points. David's remark that there should be branch choices extending the validity of these identities must be true, because of analytic continuation, But it's not clear to me where to switch branches. --rwg gosper.org/cheb1.png shows T_2 ○T_½(x) - x with a point discontinuity at the origin, and T_½ ○T_2(x) - x with an isolated ray discontinuity along the canonical cut. Bruiser: Using the radicals for In[451]:= ChebyshevT[m, x] Out[452]= 1/2 ((x - Sqrt[-1 + x^2])^m + (x + Sqrt[-1 + x^2])^m) show T_mn = T_nm.

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[math-fun] about Viète formula and patterns in the binary expansion of the series
by Simon Plouffe 27 Mar '17

27 Mar '17

Hello everybody, here is something peculiar about Viète formula for Pi. Vieta or Viète formula here : https://proofwiki.org/wiki/Vieta%27s_Formula_for_Pi Viète was the most famous person from Vendée (France). as you may know the formula converges to Pi (or Pi/2 or 2/Pi depending on which side you put the square roots in the expression), Here is the peculiar thing : there is a definite pattern in the successive terms of the binary expansion of those numbers when they converge to Pi. Here is a sample code for Pari-gp for doing this at high precision : \p 4000 p0 = 2000 p1 = sqrt(2^p0) p2 = 2.0 v = p1/p2 for (n=1, 8000 , p1=sqrt(2.0+p1); v=v*p1/2) print(binary(v)) By doing so, it produces a stream of binary digits at each step of the formula. from these lines of bits, I can do 1 image , colourized for a better view of the contrast. And here is the result : http://plouffe.fr/dessins/vi%c3%a8te.png that image is 20000 x 5000 , which is 20000 binary digits of the first 5000 terms of Viète formula. we see the convergence of the formula, if we look from a distance : we see nothing at all BUT if we take a closer look at the patterns in the binary expansion we see this : http://plouffe.fr/dessins/formule%20de%20vi%c3%a8te.PNG Many things can be said about this : - There is a definite pattern in the binary expansion of those numbers that we do not see individually but when they are grouped we see something. - This pattern appears also with the obvious generalization of Viète formula replacing the '2' by 1, 2, 3, 4, 5, ... the pattern is the same. so, numbers like 4*Pi*sqrt(3)/2, Pi/2, 4/5*log(phi)*sqrt(5) , where phi is the golden ratio and sqrt(3)/2*log(2+sqrt(3)). etc. shows similar patterns. - If we do the same method for series like the Newton-Leibniz for Pi/4 : nothing at all appears, the same with sum(1/(n*2^n), n=1..infinity) = log(2), nothing to see either. - Viète process appears so far as being the only one able to produce a pattern, I have tried many , this is the only example so far. - What is surprising is the way the pattern persist, in diagonal as well as vertical. NOTE : to produce your own image : 1- run the small program with Pari-GP. 2- Clean the output to have only ones and zeros for each line 3- Find a way to transform this text file into a pure Unix file with LF at the end of each line OR a pure binary file of let 's say 10000000 million bits. 4- rename the file by adding .raw at the end. 5- Open Photoshop and open the file, the program will detect that it is something in binary and propose to open it in raw format :, just specify the width and depth of the file properly, 6- convert the file to 'indexed colors', 7- change it to a black body color scheme, this will color the ones and the zeros in blue and black or light brown and white depending of your choise, and voilà the image. Any comment on this ? best regards, Simon Plouffe

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