January 2018 - mathfun@mailman.xmission.com

[math-fun] explicite...
by françois mendzina essomba2 27 Jan '18

27 Jan '18

Limit( (1/2^4)*product((sinh(2*n*k/(2*n+1))/(sinh(k))^(2*n/(2*n+1)))^(4*(-1)^(n)),n=1..infinity) ,k=infinity) = Limit ( (1/2^4)*product((sinh(2*n*k/(2*n+1))/(sinh(k))^(2*n/(2*n+1)))^(4*(-1)^(n)),n=1..infinity) , k=infinity) = 1/2^(Pi) ;

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[math-fun] two small limits
by françois mendzina essomba2 26 Jan '18

26 Jan '18

Hello... p(x)=(1/2^(2/3))*product((cosh(x)^(2/(16*n^2+16*n+3))*cosh((4*n*x)/(4*n+1)))/cosh(((4*n+2)*x)/(4*n+3)),n,1,inf); we have: lim(p(x),x=infinity)=2^(-Pi/4) ; where still under maple: Limit((1/2^(2/3))*product((cosh(x)^(2/(16*n^2+16*n+3))*cosh((4*n*x)/(4*n+1)))/cosh(((4*n+2)*x)/(4*n+3)),n=1..infinity),x=infinity)=1/2^(Pi/4); and Limit((1/2^(2/3))*product((sinh(x)^(2/(16*n^2+16*n+3))*sinh((4*n*x)/(4*n+1)))/sinh(((4*n+2)*x)/(4*n+3)),n=1..infinity),x=infinity)=1/2^(Pi/4); Best regards...

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[math-fun] Muffin problem update
by James Propp 26 Jan '18

26 Jan '18

I just learned that BillGasarch and his students have done a lot of work on the problem: https://arxiv.org/abs/1709.02452 Incidentally, Gasarch says the problem pre-dates Alan Frank’s work and provides citations, but I haven’t checked his sources to see if the pre-Frank work treats the exact same problem. Jim Propp

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Re: [math-fun] Touch Tone frequencies -- SPOILER
by Keith F. Lynch 26 Jan '18

26 Jan '18

Andy Latto <andy.latto(a)pobox.com> wrote: >> Eventually I found an infinite set with this property. Can anyone >> else find one? I'll post my solution in a week. > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > The odd primes work. I missed that solution, but I came close. First I found a solution for just the sums and differences, but not the multiples: Numbers congruent to 1 mod 3: {1,4,7,10,13,16,19,...} Their sums and differences are all congruent to 2 mod 3, so their sums and differences aren't in the set. Then I realized if I restricted it to just the primes in that set, {7,13,19,31,...}, I'd also get the multiples, since no prime is the multiple of another prime. I could prepend 2 and 3 to this set, since the differences between elements (other than the prepended 2 and 3) are always multiples of 6. So {2,3,7,13,19,31,...}. By a theorem by the brother in law of the composer of A Midsummer Night's Dream and Fingal's Cave, that set is infinite. Then I realized that the primes congruent to 2 mod 3, {2,5,11,17,23,...}, would also work. But I completely missed that the set of all odd primes would work. Yours is the best solution because it has the most elements less than N for all large N. Can you find a solution for a dense set of positive real numbers? Odd primes divided by odd primes works for the sums and differences, since the sum (and difference) of two odd-over-odd fractions is always an even-over-odd fraction, and no odd-over-odd fraction can equal any even-over-odd fraction. But what about multiples? An even-over-odd fraction can be an even multiple of an odd-over-odd fraction. For instance 5/3 + 17/3 = 2 * (11/3), so that doesn't work. For extra credit, find a solution for a dense set of positive and negative real numbers. Note that if X is in the set -X can't be in the set, since their sum would be zero, which is of course an integer multiple of every element.

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Re: [math-fun] Touch Tone frequencies
by Keith F. Lynch 26 Jan '18

26 Jan '18

Allan Wechsler <acwacw(a)gmail.com> wrote: > I think the generalization to unrestricted arithmetic is an > overreach. From any integer A, you can generate any other with > an expression of the form (A+A+...+A)/A. Oops. Good catch. What if I restrict it to arithmetic that mentions each element at most once? Another generalization of my original problem (no element is the sum or difference of two elements, and no element is an integer multiple of another element) is to apply it, not to positive integers, but to a finite or better yet infinite dense set of reals.

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Re: [math-fun] Touch Tone frequencies -- SPOILER
by Andy Latto 25 Jan '18

25 Jan '18

On Thu, Jan 25, 2018 at 1:08 AM, Keith F. Lynch <kfl(a)keithlynch.net> wrote: > "Keith F. Lynch" <kfl(a)KeithLynch.net> wrote: > Inspired by that, I tried searching for sets of positive integers such > that the sums and differences of pairs of elements of the set weren't > equal to elements of the set, and such that integer multiples of > elements of the set weren't equal to elements of the set. > > Eventually I found an infinite set with this property. Can anyone > else find one? I'll post my solution in a week. . . . . . . . . . . . . . . . . . . . . . . The odd primes work. Andy > > Next I'll try searching for an infinite set of integers such that no > element can be generated by any arithmetic whatsoever on any of the > other elements. Not even by something like (A-B)*(A-C)/(C+C+C+B^D). > Is this a solved problem? > > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun -- Andy.Latto(a)pobox.com

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Re: [math-fun] Touch Tone frequencies
by Keith F. Lynch 25 Jan '18

25 Jan '18

"Keith F. Lynch" <kfl(a)KeithLynch.net> wrote: > The frequencies were chosen such that the sums or differences of > any two of the eight frequencies are as far as possible from any > of the eight frequencies. Similarly with whole-number multiples > ("harmonics") of the eight frequencies or of their sums or > differences. This is to prevent the phone company equipment from > getting confused, since non-linearities in the circuits could result > in such sum and difference frequencies appearing. Inspired by that, I tried searching for sets of positive integers such that the sums and differences of pairs of elements of the set weren't equal to elements of the set, and such that integer multiples of elements of the set weren't equal to elements of the set. First I tried a greedy algorithm. That gave me various finite sets, such as {2,3,7,11}, {3,4,5,11,13,19}, {4,5,6,7,17,19,27,29}, {5,6,7,8,9,19,22,23,33,34}, and {6,7,8,9,10,11,23,25,26,38,39,41}. Eventually I found an infinite set with this property. Can anyone else find one? I'll post my solution in a week. Next I'll try searching for an infinite set of integers such that no element can be generated by any arithmetic whatsoever on any of the other elements. Not even by something like (A-B)*(A-C)/(C+C+C+B^D). Is this a solved problem?

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Re: [math-fun] Rationalizing the numerator
by Henry Baker 23 Jan '18

23 Jan '18

Don't some integrals work better when you do partial fraction decomposition with rationalized numerators? Unfortunately, I've been relying on symbolic algebra systems for so long, that I don't have an integral table handy to find an example for you! At 06:11 PM 1/22/2018, James Propp wrote: >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] Rationalizing the numerator
by Bill Gosper 23 Jan '18

23 Jan '18

On 2018-01-22 18:11, James Propp wrote: > 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 > 😈 yes, the quadratic formula! In[58]:= Solve[a x x + b x + c == 0, x] Out[58]= {x -> (-b - √(b^2 - 4 a c))/(2 a), x -> (-b + √(b^2 - 4 a c))/(2 a)} Cancellation is often nasty when you need the 2nd form, so just a↔︎c in the reciprocal of the first form. --rwg

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[math-fun] Rationalizing the numerator
by James Propp 23 Jan '18

23 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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