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[math-fun] Rationalizing the numerator
by James Propp 22 Jan '18

22 Jan '18

Can anyone think of a problem for which the trick is that you should rationalize numerators rather than denominators? Or more generally leave combinations of surds in some nonstandard but tactically helpful form? Jim Propp

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[math-fun] serial development of Engel, pi, G, zeta and representation of functions.
by françois mendzina essomba2 22 Jan '18

22 Jan '18

This morning I was interested in the following formula f(x)=(1+((-1)/3+(1/5+((-1)/7+(1/9+(-1/11+(1/13+((-1)/15+(1/17+((-1)/19+(1/21+((-1)/23+(1/25+(-1/27+(1/29+((-1)/31+(1/33+\cdots )/x^2)/x^2)/x^2)/x^2)/x^2)/x^2)/x^2)/x^2)/x^2)/x^2)/x^2)/x^2)/x^2)/x^2)/x^2)/x^2)/x ; I have deduced the following formulas. Pi/6=(1+((-1)/3+(1/5+((-1)/7+(1/9+(-1/11+(1/13+((-1)/15+(1/17+((-1)/19+(1/21+((-1)/23+(1/25+(-1/27+(1/29+((-1)/31+(1/33+\cdots )/3)/3)/3)/3)/3)/3)/3)/3)/3)/3)/3)/3)/3)/3)/3)/3)/sqrt(3); approximative value Pi/6=(1+((-1)/3+(1/5+((-1)/7+(1/9+(-1/11+(1/13+((-1)/15+(1/17+((-1)/19+(1/21+((-1)/23+(1/25+(-1/27+(1/29+((-1)/31+(1/33 )/3)/3)/3)/3)/3)/3)/3)/3)/3)/3)/3)/3)/3)/3)/3)/3)/sqrt(3); Pi/4=(1+((-3)^(-1)+(5^(-1)+((-7)^(-1)+(9^(-1)+(-(11)^(-1)+((13)^(-1)+((-15)^(-1)+((17)^(-1)+((-19)^(-1)+((21)^(-1)+((-23)^(-1)+((25)^(-1)+(-(27)^(-1)+((29)^(-1)+((-31)^(-1)+((33)^(-1)+\cdots )/1)/1)/1)/1)/1)/1)/1)/1)/1)/1)/1)/1)/1)/1)/1)/1)/1; Pi/6=(1+(1/6+(3/40+(5/112+(35/1152+(63/2816+(231/13312+(143/10240+(6435/557056+(12155/1245184+\cdots )/4)/4)/4)/4)/4)/4)/4)/4)/4)/2; Pi/4=(1+(1/6+(3/40+(5/112+(35/1152+(63/2816+(231/13312+(143/10240+(6435/557056+(12155/1245184+ \cdots )/2)/2)/2)/2)/2)/2)/2)/2)/2)/sqrt(2); Pi/2=(1+(1/6+(3/40+(5/112+(35/1152+(63/2816+(231/13312+(143/10240+(6435/557056+(12155/1245184+\cdots )/1)/1)/1)/1)/1)/1)/1)/1)/1)/1; a_(n):(2n)!/(2^(2n)*(n!)^2*(2n+1)); Catalan number G=(1+(-(3)^(-2)+(5^(-2)+(-(7)^(-2)+(9^(-2)+(-(11)^(-2)+((13)^(-2)+(-(15)^(-2)+((17)^(-2)+(-(19)^(-2)+((21)^(-2)+(-(23)^(-2)+((25)^(-2)+(-(27)^(-2)+((29)^(-2)+(-(31)^(-2)+((33)^(-2)+?)/1)/1)/1)/1)/1)/1)/1)/1)/1)/1)/1)/1)/1)/1)/1)/1)/1; Zeta function zeta(m)=(1+(2^(-m)+(3^(-m)+(4^(-m)+(5^(-m)+(6^(-m)+(7^(-m)+(8^(-m)+(9^(-m)+(10^(-m)+(11^(-m)+(12^(-m)+(13^(-m)+(14^(-m)+(15^(-m)+(16^(-m)+(17^(-m)+\cdots )/1)/1)/1)/1)/1)/1)/1)/1)/1)/1)/1)/1)/1)/1)/1)/1)/1;

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Re: [math-fun] Bell 202 1200 baud modem emulation
by Henry Baker 22 Jan '18

22 Jan '18

The approach outlined below basically works, but you do need to *normalize* the data first. When correlating a real signal with a complex signal, the real signal has to be multiplied by sqrt(2) to get its power up to the same level. In an actual radio, one would require some sort of real-time *automatic gain control* to keep the input signal at the appropriate level. With symbolic algebra systems like Maxima, we can calculate the solutions for the continuous (sine wave forms) case exactly. Correlating a 1200Hz real signal with a 2200Hz complex signal isn't identically zero over only one bit time, but is 2% +- 1% in the continuous case; the average here is over all possible phases. If one transmits an alternating sequence of "0" and "1" bits, both the "0" filter and the "1" filter will each put out a sawtooth wave, but perfectly out of phase; thus we could subtract one from the other to improve robustness of detection. This output needs to be sampled at the *end* of each bit time, at which point the correlator reaches its maximum distinction between a "0" and a "1". There still needs to be some sort of *phase locked loop* to synchronize the receiver clock with the transmitter clock; this requires that the digital data have enough transitions to allow for such synchronization. At 06:04 AM 1/20/2018, Henry Baker wrote: >Ok, now we have to *receive* these 2-tone signals. > >Below is my high school "crystal radio" sophistication >level version of a software Bell 202 1200 baud receiver. > >As in our transmitter, we have digital samples >s(t) of (real) numbers coming in at a rate of >1200*6=13.2kHz. > >We'd like to decide if the last 6 samples denote >a 1200Hz frequency or a 2200Hz frequency. We >can't guarantee strict clock synchronization, >so we're going to have to deal with that, too. > >Recall that w is a primitive 66'th root of 1: >w^66=1. So (w^11)^6=1 and (w^6)^11=1. > >We'd like to take the input stream s(t) and >*convolve* it with 6 samples of the sinusoidal >stream (w^11)^i. I.e., we'd like to keep a >running (complex) sum of the last 6 samples: > >S(i) = s(i)+s(i-1)*w^11+s(i-2)*(w^11)^2 > +s(i-3)*(w^11)^3+s(i-4)*(w^11)^4 > +s(i-5)*(w^11)^5 > >So when we get s(i+1), we need to subtract >s(i-5)*(w^11)^5 from S(i) and then multiply it >by w^11: > >S(i+1) = s(i+1) + [S(i)-s(i-5)*(w^11)^5]*(w^11) > >We also compute |S(i+1)|^2 = S(i+1)*S(i+1)^* > >We do a similar calculation for w^6, which will >help us detect 1's: > >T(i+1) = s(i) + [T(i)-s(i-5)*(w^6)^5]*(w^6) > >and compute |T(i+1)|^2 = T(i+1)*T(i+1)^* > >The reason for performing *convolutions* is that >a suitably long sample of a 1200Hz sinusoid are >*orthogonal* to a similarly long sample of a 2200Hz >sinusoid *no matter what the phase* of each. And by >doing the convolution using *complex numbers* instead >of just *real numbers*, we don't have to know the >phase in advance, but can perform the convolution >and then are free to later *throw away* that phase >information by computing the absolute value. > >Hopefully, with only 6 iterations of each calculation >on a new sample, we should know whether the state S >has a stronger claim on 0 or T has a stronger claim >on 1. > >Note that if we utilize a floating point representation >for the (real,imaginary) components of w, S and T, >we will likely get *drift*, since the cancellation of >the obsolete samples will not be exact. One obvious >way to deal with this is to simply *reset* S=T=0 at >the end of each bit time, which allows the receiver >to *forget* which bit preceded the current bit being >received. > >When floating point representation isn't convenient >or precise enough, we can dispense with almost all >error by keeping S,T as high-precision numbers in >*Chinese Remainder Theorem* (CRT) representation. >With modern cheap memory, we can use table lookup >to interconvert between CRT numbers and other >representations. > >At 10:38 AM 1/16/2018, Henry Baker wrote: >>Here's an even better way to think about this >>using complex numbers and roots of unit. >> >>Let w be a primitive 66'th root of unity: w^66=1, >>e.g., w=exp(2*pi*i/66). >> >>Then w^11 is a 6'th root of unity: (w^11)^6=1. >> >>And w^6 is an 11'th root of unity: (w^6)^11=1. >> >>Now consider a clock frequency of 1200*N Hz. >> >>At any point in time, the modem is transmitting >>nothing, it is transmitting a "1" bit, or it is >>transmitting a "0" bit. >> >>At any point in time, the current state of the >>transmitter is a complex number S, s.t. |S|=1. >>If the transmitter is transmitting either a "0" >>bit or a "1" bit, the state S of the transmitter >>is updated after every clock by multiplying S by >>another complex number. >> >>During the transmission of a full "1" bit, the >>state S is updated like so: S<-S*(w^6)^N=S*w^(6*N). >> >>During the transmission of a full "0" bit, the >>state S is updated like so: S<-S*(w^11)^N=S*w^(11*N). >> >>We can see that N=11 is a convenient choice, so >>that the clock frequency is 1200*11=13.2kHz. (Of >>course, we might also have N be some multiple of >>11 and thereby gain some improvement in the >>quality of the generated sinusoids; details are >>left to the reader.) >> >>A "1" transmission will last 11 clock cycles, >>during which time we will output a complete >>sinusoid. In order to achieve this, after each >>clock during a "1" transmission, we update S: >>S<-S*(w^6)=S*exp(12*pi*i/66)=S*exp(2*pi*i/11). >> >>Similarly, a "0" transmission will also last >>11 clock cycles, during which time we will >>output 1 5/6'ths of a sinusoid. In order to >>achieve this, after each clock during a "0" >>transmission, we update S: >>S<-S*(w^11)=S*exp(22*pi*i/66)=S*exp(2*pi*i/6). >> >>We can verify that after 11 clock cycles of >>a "1" transmission, S<-S*(w^6)^11=S*(w^66)=S, >>which is correct, since a single "1" bit >>always transmits one complete sinusoid cycle. >> >>But after 11 clock cycles of a "0" transmission, >>S<-S*(w^11)^11=S*w^121=S*w^55, which is correct, >>since a "0" transmission consists of 1+5/6's of >>a complete sinusoid cycle. >> >>We note that even if we initialize S=1, before >>starting to transmit a packet of bits, an >>unfortunate number of "0" bits will *not* >>return S to 1, so when we stop transmitting, >>we get a small glitch in the output. >> >>With floating point being almost ubiquitous in >>small CPU chips, these calculations may even be >>faster even than using a RAM table lookup. One >>thing to keep in mind: S should be *normalized* >>from time to time -- S<-S/|S| -- in order to >>preserve a constant waveform envelope. >> >>At 05:35 PM 1/15/2018, Henry Baker wrote: >>>The Bell 202 modem (1976) was the first digital modem in >>>widespread use. >>> >>>https://en.wikipedia.org/wiki/Bell_202_modem >>> >>>It uses "audio frequency shift keying" with the two >>>frequencies being 1200Hz ("1") and 2200Hz ("0"). It is >>>still used to transmit Caller ID on wireline phones and >>>by ham radio operators in the ham radio bands. >>> >>>The choice of the frequencies 1200&2200 was a tiny >>>bit unfortunate, because although a "1" consists of a >>>single complete sinusoidal cycle, a "0" consists of *not >>>quite* 2 complete cycles. >>> >>>Here's the rub: when you transition from a 0->1 or from >>>a 1->0, you want the waveform to be *continuous*. So >>>this means that you're going to have to adjust the >>>*phase* of the sinusoid that you generate when you make >>>the transition; i.e., you can't simply start at zero for >>>every bit, else these discontinuities will generate high >>>frequency noise which will widen the overall bandwidth >>>used by the modem. >>> >>>Assuming that you're going to implement the "0" and the >>>"1" by means of *tables* of *digital samples* of the >>>appropriate sinusoids, how many tables do you need?

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Re: [math-fun] Trig identity
by Henry Baker 22 Jan '18

22 Jan '18

Interesting! Thanks! I don't see any "trigcontract" in Maxima (as opposed to Macsyma). I guess this feature never made it into Maxima. At 06:42 PM 1/21/2018, Bill Gosper wrote: >Macsyma: >(c1) TRIGCONTRACT(SIN(X+2*THETA)-SIN(X)); > >(d1) 2 sin(theta) cos(x + theta) > >(c2) BLOCK([FANCY_DISPLAY : FALSE],PLAYBACK([1,2])); > >Mathematica: >In[1]:= TrigFactor[Sin[X + 2*THETA] - Sin[X]] > >Out[1]= 2 Cos[THETA + X] Sin[THETA] > >On 2018-01-21 15:18, Henry Baker wrote: >> I can use Maxima to show that the rhs simplifies to the lhs, >> but I can't go the other way (except by hand). >> >> Any ideas how to force Maxima to derive the rhs from the lhs? >> >> Can Mathematica do this? >> >> (This is a generalization of sin(2*theta)=2*sin(theta)*cos(theta).) >> >> (%i1) sin(x+2*theta)-sin(x) = 2*sin(theta)*cos(x+theta); >> (%o1) sin(x + 2 theta) - sin(x) = 2 sin(theta) cos(x + theta) >> (%i2) %,trigexpand,expand; >> 2 2 >> (%o2) - sin (theta) sin(x) + cos (theta) sin(x) - sin(x) >> + 2 cos(theta) sin(theta) cos(x) = 2 cos(theta) sin(theta) cos(x) >> 2 >> - 2 sin (theta) sin(x) >> (%i3) %,trigreduce,expand; >> (%o3) sin(x + 2 theta) - sin(x) = sin(x + 2 theta) - sin(x) > >--rwg

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Re: [math-fun] Trig identity
by Bill Gosper 21 Jan '18

21 Jan '18

Macsyma: (c1) TRIGCONTRACT(SIN(X+2*THETA)-SIN(X)); (d1) 2 sin(theta) cos(x + theta) (c2) BLOCK([FANCY_DISPLAY : FALSE],PLAYBACK([1,2])); Mathematica: In[1]:= TrigFactor[Sin[X + 2*THETA] - Sin[X]] Out[1]= 2 Cos[THETA + X] Sin[THETA] On 2018-01-21 15:18, Henry Baker wrote: > I can use Maxima to show that the rhs simplifies to the lhs, > but I can't go the other way (except by hand). > > Any ideas how to force Maxima to derive the rhs from the lhs? > > Can Mathematica do this? > > (This is a generalization of sin(2*theta)=2*sin(theta)*cos(theta).) > > (%i1) sin(x+2*theta)-sin(x) = 2*sin(theta)*cos(x+theta); > (%o1) sin(x + 2 theta) - sin(x) = 2 sin(theta) cos(x + theta) > (%i2) %,trigexpand,expand; > 2 2 > (%o2) - sin (theta) sin(x) + cos (theta) sin(x) - sin(x) > + 2 cos(theta) sin(theta) cos(x) = 2 cos(theta) sin(theta) cos(x) > 2 > - 2 sin (theta) sin(x) > (%i3) %,trigreduce,expand; > (%o3) sin(x + 2 theta) - sin(x) = sin(x + 2 theta) - sin(x) > --rwg

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[math-fun] Trig identity
by Henry Baker 21 Jan '18

21 Jan '18

I can use Maxima to show that the rhs simplifies to the lhs, but I can't go the other way (except by hand). Any ideas how to force Maxima to derive the rhs from the lhs? Can Mathematica do this? (This is a generalization of sin(2*theta)=2*sin(theta)*cos(theta).) (%i1) sin(x+2*theta)-sin(x) = 2*sin(theta)*cos(x+theta); (%o1) sin(x + 2 theta) - sin(x) = 2 sin(theta) cos(x + theta) (%i2) %,trigexpand,expand; 2 2 (%o2) - sin (theta) sin(x) + cos (theta) sin(x) - sin(x) + 2 cos(theta) sin(theta) cos(x) = 2 cos(theta) sin(theta) cos(x) 2 - 2 sin (theta) sin(x) (%i3) %,trigreduce,expand; (%o3) sin(x + 2 theta) - sin(x) = sin(x + 2 theta) - sin(x)

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[math-fun] Bell 202 1200 baud modem emulation
by Henry Baker 20 Jan '18

20 Jan '18

The Bell 202 modem (1976) was the first digital modem in widespread use. https://en.wikipedia.org/wiki/Bell_202_modem It uses "audio frequency shift keying" with the two frequencies being 1200Hz ("1") and 2200Hz ("0"). It is still used to transmit Caller ID on wireline phones and by ham radio operators in the ham radio bands. The choice of the frequencies 1200&2200 was a tiny bit unfortunate, because although a "1" consists of a single complete sinusoidal cycle, a "0" consists of *not quite* 2 complete cycles. Here's the rub: when you transition from a 0->1 or from a 1->0, you want the waveform to be *continuous*. So this means that you're going to have to adjust the *phase* of the sinusoid that you generate when you make the transition; i.e., you can't simply start at zero for every bit, else these discontinuities will generate high frequency noise which will widen the overall bandwidth used by the modem. Assuming that you're going to implement the "0" and the "1" by means of *tables* of *digital samples* of the appropriate sinusoids, how many tables do you need? * * * * * * * * * * * * * * * * ** * ------ I believe that 6 tables will be needed, because 11 full cycles of 2200Hz fit exactly 6 full cycles of 1200Hz. We need to keep a modulo 6 counter ("mod6") which keeps track of the number of "0" bits encountered. With each "0" bit, the mod6 counter is incremented, and we use this counter to index which sinusoid table. How come we don't have to keep track of the number of "1" bits? Extra credit: If the overall sampling rate is 6*1200=7200, then we can overlap all of the "0" tables together and index the samples using table0[mod(mod6+i,6)]; ditto for table1. Of course, 6 samples/sinusoid is pretty crude, but we can choose a sampling rate N*6*1200 for integer N>1 and then index the samples accordingly.

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[math-fun] pdf files have been disappearing from my computer in PAPERS
by Neil Sloane 19 Jan '18

19 Jan '18

Dear Funsters, For decades I have been using the program PAPERS to save and organize my pdf files. I have thousands of them, and PAPERS has always done a good job. But lately I have tried to look at articles that I know I had saved there, only to be told that the pdf file has disappeared. Does anyone know what is going on? It has happened twice in the last couple of months. I guess I have to give up on PAPERS. It is like an Itunes program, but for pdf files not music. Does anyone have any recommendations for a replacement?

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Re: [math-fun] Theta polynomials
by Dan Asimov 19 Jan '18

19 Jan '18

The 240 "integral octonions" of unit length can be arranged to coincide with the first shell of the E_8 lattice. These form an alternative monoid with inverses, in fact a "Moufang loop (https://en.wikipedia.org/wiki/Moufang_loop) Thorold Gosset discovered ca. 1900 what's now called the Gosset polytope, which Coxeter calls the 4_21 polytope, basically the convex hull of these 240 points in 8-space (more precisely: its boundary). —Dan ----- It has occurred to me that (vertex-transitive) spherical codes have `theta polynomials' in the same way that lattices have `theta series'. Specifically, the coefficient of q^n gives the number of points at a Euclidean distance of sqrt(n). I was somewhat surprised to see that the coefficients of the theta polynomial of the first shell of the E8 lattice -- namely 1,56,126,56,1 -- do not feature in the OEIS. When you only have three non-zero coefficients, you can define a regular graph on the vertices of the spherical code. If the spherical code has sufficiently many symmetries -- specifically, that the point stabiliser of one vertex is transitive on each of the 'shells' whose cardinalities are specified in the coefficients of the polynomial -- then this graph is strongly regular. The standard construction of the Higman-Sims graph comes from such a spherical code [living in the Leech lattice]. -----

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Re: [math-fun] Touch Tone frequencies
by Dan Asimov 19 Jan '18

19 Jan '18

Thanks! —Dan P.S. I've also heard that Indian Carnatic music divides the octave into 40+ parts, but am not familiar with it. >From: Cris Moore <moore(a)santafe.edu> >You should check out the electronic etudes of Easley Blackwood, who divided the octave into N tones for 12 < n <= 24. > >Cris > >> On Jan 18, 2018, at 1:55 PM, Dan Asimov <dasimov(a)earthlink.net> wrote: >> >> I'm curious to hear low-integer ratios that don't commonly occur >> in Western music. >> >> (Anyone for 7/K, 2 <= K <= 5 ???) >> >> —Dan >> >> _______________________________________________ >> 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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