November 2018 - mathfun@mailman.xmission.com

[math-fun] Fwd: Rational-coefficient polynomial, applied to integers
by Bill Gosper 17 Nov '18

17 Nov '18

(some trig trivia at end —rwg) ---------- Forwarded message --------- From: Julian Ziegler Hunts <julianj.zh(a)gmail.com> Date: Fri, Nov 16, 2018 at 11:33 PM Subject: Re: [math-fun] Rational-coefficient polynomial, applied to integers To: Bill Gosper <billgosper(a)gmail.com> I'm sure this has already been answered by now, at least the first part, but: (i) Easy proof that if there's an irrational coefficient the polynomial takes irrational values: find a basis over Q, including 1, for the vector space spanned by the coefficients (and 1), then write the polynomial as a linear combination of rational polynomials with coefficients in the basis. An irrational basis element has a nonzero polynomial coefficient, and the value of the polynomial can only be rational when this component is zero, which occurs only finitely many times. (ii) You don't need to test 0 through K-1, just up to the largest prime-power factor of K (minus one), or better yet make it symmetric around zero. (iii) After clearing denominators, you need to check that the polynomial is always zero mod p^m for any prime power p^m dividing K. There are probably general results on how to do this, but there are some obvious tricks to get started: since (x^p-x)^m is always zero, you can mod out by it; the polynomial needs to be a multiple of x(x-1)…(x-p+1) mod p^m, a multiple of x(x-1)…(x-2p+1) mod p^(m-1), …, a multiple of x(x-1)…(x-p^2+1) mod x^(m-p+1), a multiple of x(x-1)…(x-p^2-p+1) mod x^(m-p-1), … . (iv) Further consideration of the above shows that it is necessary and sufficient to test any r consecutive integers, for r large enough that r! is divisible by K. This is, of course, usually much smaller than K. And when it isn't, it's because K has large prime factors with small powers; we can consider those primes separately and use simpler methods which apply to small powers (I think you might be able to make these methods work for larger powers; ask me tomorrow). (v) The above method can give an explicit description of all polynomials of degree at most d which are zero mod p^m; for example, it allows us to count that there are sum(p^(a(j)(d-jp)), j=1,…,l-1)+p^((m-h)(d-lp)), where a(j) is the power of p dividing pj, l=r/p is minimal such that (lp)! is divisible by p^m, and h is the power of p dividing ((l-1)p)!. (I'm not entirely confident about (iv) and (v), and I'd like to double-check them, but I think they should be right.) I think the (somewhat) simpler/faster methods in (iv) apply up to p^p, but fail for p^(p+1), though they may be able to be modified. Julian On Fri, Nov 16, 2018 at 10:06 AM Bill Gosper <billgosper(a)gmail.com> wrote: > On 2018-11-15 15:06, Allan Wechsler wrote: > > For a polynomial P with coefficients in Z, it's trivial that P(n) is > > integer if n is. > > > > The converse is not true. There are lots of polynomials that always give > > integer answers to integer questions, but whose coefficients are not > > integers. For example, n(n+1)/2 = (1/2)n + (1/2)n^2 always takes integer > > values for integer n, even though the coefficients aren't integers. > > > > Is there a way to quickly eyeball a polynomial in general to see if it > is Z > > -> Z? > > > > If the coefficients are rational, one can find K = the LCM of the > > denominators, multiply through by K, and test it for all the integers > from > > 0 to K-1 to see if the result is always divisible by K. But I am hoping > > there is a simpler way. > > > > If any of the coefficients are irrational, my intuition is that the > > polynomial is never Z->Z, but I haven't been able to think of an easy > proof. > > > That's how Macsyma did it, in an internal three-valued Lisp function named > divby1: > (c59) COS(ODD^2*%PI/8); > > 2 > odd > ---- + 7/8 > %pi 8 > (d59) - cos(---) (- 1) > 8 > > (c60) ?LSP("(divby1 '((MEXPT SIMP) $ODD 2)))"); > > (d60) 8 1 true > —rwg > > (c80) SIN(%PI*BERNPOLY(INTEGER,6)/6); 6 5 4 2 2 integer - 6 integer + 5 integer - integer ----------------------------------------------- %pi 4 (d80) sin(---) (- 1) 252 Mathematica partly gets it: In[43]:= FullSimplify[Sin[\[Pi] BernoulliB[6, n]/6], n \[Element] Integers] Out[43]= Sin[1/252 (\[Pi] + 21 (-1 + n)^2 n^2 (-1 + 2 (-1 + n) n) \[Pi])] In[42]:= Table[Sin[\[Pi] BernoulliB[6, n]/6], {n, 9}] Out[42]= {Sin[\[Pi]/252], -Sin[\[Pi]/252], -Sin[\[Pi]/252], Sin[\[Pi]/252], Sin[\[Pi]/252], -Sin[\[Pi]/252], -Sin[\[Pi]/252], Sin[\[Pi]/252], Sin[\[Pi]/252]} Sin[π/7] and Sin[π/9] aren't constructible and thus won't come out in real radicals, but at least you can (c88) HALFANGLES(HALFANGLES(d80)); %pi sqrt(cos(---) + 1) 63 (d88) sqrt(1 - ------------------) sqrt(2) 6 5 4 2 2 integer - 6 integer + 5 integer - integer ----------------------------------------------- 4 (- 1) /sqrt(2) I don't know if Mathematica can partially TrigToRadicals. —rwg (Amazing: OS-X or GMail flags "constructible" and accepted "constructable" for many seconds before flagging that too!)

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[math-fun] Candela
by Adam P. Goucher 17 Nov '18

17 Nov '18

There seems to be a lot of recent excitement about the redefinition of the kilogram, but I'm more worried about the candela. On the face of it, it seems to have a relatively simple definition: "The candela is the luminous intensity, in a given direction, of a source that emits monochromatic radiation of frequency 540*10^12 Hz and that has a radiant intensity in that direction of 1/683 watt per steradian." where luminous intensity is defined as: I_v = 683 V(gamma) I_e where I_e is the radiant intensity (in watts per steradian) and V(gamma) is the 'standard luminosity function' (which must be equal to exactly 1 when gamma = 540*10^12 Hz, for these definitions to be consistent). But this raises the question: what is V(gamma) in general? I've found a tabulation of values for integer * 10^-9 metre wavelengths in the visible interval: http://donklipstein.com/photopic.html but no indication as to how to compute V(gamma) for any of the 2^(aleph_null) other wavelengths beside the 401 provided. Any ideas? Best wishes, Adam P. Goucher

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[math-fun] something for cos(x) and cosh(x)
by françois mendzina essomba2 17 Nov '18

17 Nov '18

the ratio (x/sinh (x)) : (cos(x))^2=(sum(((-1)^n/(2*n)!)*(x/sinh(x))^(2*n)*cosh(2*n*x),n=0..infinity))^2+(sum(((-1)^n/(2*n+1)!)*(x/sinh(x))^(2*n+1)*cosh((2*n+1)*x),n=0..infinity))^2; and with x=i*x (cosh(x))^2=(sum(((-1)^n/(2*n)!)*(x/sin(x))^(2*n)*cos(2*n*x),n=0..infinity))^2+(sum(((-1)^n/(2*n+1)!)*(x/sin(x))^(2*n+1)*cos((2*n+1)*x),n=0..infinity))^2; FME...

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Re: [math-fun] Rational-coefficient polynomial, applied to integers
by Dan Asimov 17 Nov '18

17 Nov '18

All this talk of polynomials suggested something I hadn't noticed before: Let P(X) = Sum_{1 <= k <= n} a_n X^n for any (say) reals a_k. Then think of P(X) as the point (a_0, a_1, ..., a_n) in R^(n+1). Polynomials add just like vectors in R^(n+1). Using this we can also multiply vectors v in R^(n+1) and w in R(m+1) to get one in R^(n+m+1). So if we take the union of all R^n, 1 <= n < oo (call this R^oo), we get a (commutative, associative) ring with unit. In fact it's an integral domain (the product of nonzero elements is nonzero). Hence R^oo has a field of fractions, which I suppose is equivalent to the field of rational functions over R. It's even a topological field, since addition and multiplication are continuous functions of the coefficients. Is this not a real division algebra? I've read that Kaplansky and others proved that the only real division algebras are in 1, 2, 4, and 8 dimensions. In particular, there are *none* in infinitely many dimensions (excepting the case of vector spaces over purely inseparable fields, which I don't think applies to the reals). Whatever it is, it's a pretty cool object. —Dan

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[math-fun] Rational-coefficient polynomial, applied to integers
by Allan Wechsler 16 Nov '18

16 Nov '18

For a polynomial P with coefficients in Z, it's trivial that P(n) is integer if n is. The converse is not true. There are lots of polynomials that always give integer answers to integer questions, but whose coefficients are not integers. For example, n(n+1)/2 = (1/2)n + (1/2)n^2 always takes integer values for integer n, even though the coefficients aren't integers. Is there a way to quickly eyeball a polynomial in general to see if it is Z -> Z? If the coefficients are rational, one can find K = the LCM of the denominators, multiply through by K, and test it for all the integers from 0 to K-1 to see if the result is always divisible by K. But I am hoping there is a simpler way. If any of the coefficients are irrational, my intuition is that the polynomial is never Z->Z, but I haven't been able to think of an easy proof.

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Re: [math-fun] Rational-coefficient polynomial, applied to integers
by Bill Gosper 16 Nov '18

16 Nov '18

On 2018-11-15 15:06, Allan Wechsler wrote: > For a polynomial P with coefficients in Z, it's trivial that P(n) is > integer if n is. > > The converse is not true. There are lots of polynomials that always give > integer answers to integer questions, but whose coefficients are not > integers. For example, n(n+1)/2 = (1/2)n + (1/2)n^2 always takes integer > values for integer n, even though the coefficients aren't integers. > > Is there a way to quickly eyeball a polynomial in general to see if it is Z > -> Z? > > If the coefficients are rational, one can find K = the LCM of the > denominators, multiply through by K, and test it for all the integers from > 0 to K-1 to see if the result is always divisible by K. But I am hoping > there is a simpler way. > > If any of the coefficients are irrational, my intuition is that the > polynomial is never Z->Z, but I haven't been able to think of an easy proof. > That's how Macsyma did it, in an internal three-valued Lisp function named divby1: (c59) COS(ODD^2*%PI/8); 2 odd ---- + 7/8 %pi 8 (d59) - cos(---) (- 1) 8 (c60) ?LSP("(divby1 '((MEXPT SIMP) $ODD 2)))"); (d60) 8 1 true —rwg

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[math-fun] Drake formula's
by françois mendzina essomba2 16 Nov '18

16 Nov '18

Drake's formula for the number of planets in the universe that can support life is reasonable ... What formula can be found to evaluate the number of planet that can shelter an intelligent life? The logarithm of Drake's formula???

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[math-fun] Identity for Pi
by françois mendzina essomba2 16 Nov '18

16 Nov '18

Hello, dear all... 1/2*Pi = t+sum(1/2*(cos(t)/sinh(t))^(2*n)/n*sin(2*n*t)*cosh(2*n*t),n = 1 .. infinity)+sum((cos(t)/sinh(t))^(2*n-1)/(2*n-1)*sin((2*n-1)*t)*sinh((2*n-1)*t),n = 1 .. infinity) ; t in [1,2] FME

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Re: [math-fun] Rational-coefficient polynomial, applied to integers
by Dan Asimov 16 Nov '18

16 Nov '18

Which immediately answers in the affirmative another question I had: (If P(Z) = Z, does its derivative P'(X) also take Z —> Z ?) —Dan Victor Miller wrote: ----- ... The theorem is as follows: for a positive integer n define f_n(x) = x(x-1)...(x-n+1)/n!, and f_0(x) = 1. These are clearly is integer valued at all integers. Every polynomial with rational coefficients which is integer valued at all integers is an linear combination of the f_n(x) with integer coefficients. ... -----

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Re: [math-fun] Rational-coefficient polynomial, applied to integers
by Dan Asimov 16 Nov '18

16 Nov '18

I this is a terrific question. Very interesting! I'll have to delay thinking about this till later, but meanwhile it suggests a few other related questions: Suppose the nth degree polynomial P(X) = Sum_{1 <= j <= n} a_j X^j has coefficients in some nice ring S containing Z. ((( Like S = Gaussian integers Z[i] (a_j = K + L*i), or S = Eisenstein integers Z[exp(2πi/3)] (a_j = K + L*exp(2πi/3), or S = Hurwitz quaternions Z[1, i, j, k, ±1/2 ± i/2 ± j/2 ± k/2] (a_j = integer linear combinations of those 24 quaternions), or even S = integral octonions (see http://math.ucr.edu/home/baez/octonions/integers/integers_1.html) The important thing is, they're power associative, so polynomials P(X) are well-defined when evaluated on octonions. (Question to algebraists: Are these "maximal orders" ?) ))) Suppose for any one of these S we have P(Z) = Z (what Allan calls Z —> Z). When are the a_j actually in Z ? When don't they need to be? Each of these rings S can be tensored with the rationals (S tensor Q just allows a rational linear combinations where you see an integer in those linear combinations above). Then: Same question for these rings S(x)Q (S tensor Q) —Dan ----- For a polynomial P with coefficients in Z, it's trivial that P(n) is integer if n is. The converse is not true. There are lots of polynomials that always give integer answers to integer questions, but whose coefficients are not integers. For example, n(n+1)/2 = (1/2)n + (1/2)n^2 always takes integer values for integer n, even though the coefficients aren't integers. Is there a way to quickly eyeball a polynomial in general to see if it is Z -> Z? If the coefficients are rational, one can find K = the LCM of the denominators, multiply through by K, and test it for all the integers from 0 to K-1 to see if the result is always divisible by K. But I am hoping there is a simpler way. If any of the coefficients are irrational, my intuition is that the polynomial is never Z->Z, but I haven't been able to think of an easy proof. -----

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