#!/usr/local/bin/maple # -*- maplev -*- # Nathaniel Shar # HW 5 # Experimental Mathematics # It is okay to link to this assignment on the course webpage. # Stuff from class 5: Help := proc(): print(Di(L), GP(L, x), GQP1(L, x, m), GQP(L, x), GFtoSeq(f,q,m), pmn(m,n), F(m, x), T(n)): end: Di := proc(L): [seq(L[i] - L[i-1], i=2..nops(L))]; end: GP := proc(L, x) local i, d, coeffs, diff: d := nops(L)-1: diff := L[1..d+1]: coeffs := []: if convert(diff, set) = {0} then: return 0: fi: for i from 0 to d do: coeffs := [op(coeffs), diff[1]]: diff := Di(diff): if convert(diff, set) = {0} then: return factor(expand(add(coeffs[j]*binomial(x, j-1), j=1..i+1))): fi: od: return FAIL: end: GQP1 := proc(L, x, m) local i, k, seqs: seqs := [seq([seq(L[i+k*m], k=0..floor((nops(L)-i)/m))], i=1..m)]: return [seq(GP(seqs[i], x), i=2..nops(seqs)), GP(seqs[1], x)]: # Our convention for quasipolynomials requires what probably # should be the first term to be at the end instead. Yet another # issue caused by the fact that lists do not start at 0. end: GQP := proc(L, x) local m, qp, rv: for m from 1 to nops(L/2) do qp := GQP1(L, x, m): if not FAIL in qp then return qp: fi: od: return FAIL: end: ############# # Problem 2 # ############# # Remark 1: There is no sensible reason to start at 1. So I started at # 0 instead. This will come in very handy in problems 6 and 7. GFtoSeq := proc(f,q,m): [seq(coeff(taylor(f, q, m+1), q, i), i=0..m)]: end: ############# # Problem 3 # ############# pmn := proc(m,n): GQP(GFtoSeq(product(1/(1-q^i), i=1..m), q, (m+4)*m!), n): end: # We have: # pmn(2, n) = [n+1, n+1] # pmn(3, n) = [(1+n)(3n+1), (1+n)(3n+2), 3(1+n)^2, (3n+4)(1+n), # (3n+5)(1+n), 1+3n+3n^2]. # The next two are too long to include here. ############# # Problem 4 # ############# # Human mathematics is good enough to figure out this problem: F := proc(m,x): [seq(x-i/m, i=1..m-1), x/m]: end: ############# # Problem 5 # ############# T := proc(n): return round(n^2/12) - floor(n/4)*floor((n+2)/4): end: # Now we just run GQP([seq(T(i), i=0..48)], x) and get # the quasipolynomial # [x(2+3x) , x(1+3x) , 1+3x+3x^2, # x(2+3x) , (1+x)(1+3x), 1+3x+3x^2, # (1+x)(2+3x), (1+x)(1+3x), 3(1+x)^2 , # (1+x)(2+3x), (3x+4)(1+x), 3x^2 ]. ############# # Problem 6 # ############# # Remark: The statement of the problem seems to be wrong. It should # probably say a >= b >= c > 0, and b + c > a (and similarly in the # next two statements of similar problems). Otherwise # degenerate triangles/quadrilaterals are included, which is probably # not what we want. # In any event, suppose a >= b >= c >= d > 0, and b + c + d > a. # We can write the generating function for these partitions as # follows: # \sum_{d > 0} \sum_{c >= d} \sum_{b >= c} \sum_{b+c+d >= a >=b} q^{a+b+c+d}. # Maple can simplify this if we run # simplify(sum(sum(sum(sum(q^(a+b+c+d), a=b..b+c+d-1), b=c..infinity), # c=d..infinity), d=1..infinity)); # We end up with the generating function # (q^3 - q^2 + 1)q^4 / ((q^4-1)(q^2-1)(q-1)(q^6-1)) # The form of the denominator indicates that the coefficients can be # given by a quasipolynomial of order lcm(1,2,4,6) = 12. The degree of the # quasipolynomial should be at most 4-1 = 3 (since there are 4 factors # in the denominator). Hence we only need to check the first 49 # terms. Taking 200 or so just to be safe, we get the following # quasipolynomial # (by running GQP(GFtoSeq((q^3-q^2+1)q^4/((q^4-1)(q^2-1)(q-1)(q^6-1))))): # AnswerToProblem6 := [1/2*x*(1+9*x+12*x^2), 3/2*x*(1+3*x+4*x^2), # 1/2*x*(5+15*x+12*x^2), 1+7/2*x+15/2*x^2+6*x^3, # 1/2*(1+x)*(12*x^2+9*x+2), 1+13/2*x+21/2*x^2+6*x^3, # 1/2*(1+x)*(12*x^2+15*x+4), 3/2*(1+x)*(4*x^2+5*x+2), # 1/2*(1+x)*(12*x^2+21*x+8), 1/2*(1+x)*(12*x^2+21*x+10), # 1/2*(1+x)*(12*x^2+27*x+14), 1/2*x*(1+3*x+12*x^2)]: # That is a mess. # If we really do want to allow degenerate quadrilaterals, then (by a # very similar process) we instead get the quasipolynomial # AnswerToProblem6D := [1/2*x*(5+15*x+12*x^2), 1+13/2*x+21/2*x^2+6*x^3, # 1/2*(1+x)*(12*x^2+9*x+2), 3/2*(1+x)*(4*x^2+5*x+2), # 1/2*(1+x)*(12*x^2+15*x+4), 1/2*(1+x)*(12*x^2+21*x+10), # 1/2*(1+x)*(12*x^2+21*x+8), 1/2*(1+x)*(12*x^2+27*x+16), # 1/2*(1+x)*(12*x^2+27*x+14), 3/2*(1+x)*(4*x^2+11*x+8), # 1/2*(1+x)*(12*x^2+33*x+22), 1+7/2*x+15/2*x^2+6*x^3]: ############# # Problem 7 # ############# # Now we would like to do the same thing we did in problem 6, but with # one additional term. This is going to take more computer time but # should otherwise be the same process. # The generating function is produced by: # simplify(sum(sum(sum(sum(sum(q^(a+b+c+d), a=b..b+c+d+e-1), b=c..infinity), # c=d..infinity), d=e..infinity), e=1..infinity)); # which yields # -(q^10+q^9+q^8+q^7+q^6+q^5+q^4+q^3+q^2+q+1)*q^5/(q^6-1)/(q^8-1)/(q-1)^3/(q+1)/(q^3+q^2+q+1)/(q^4+q^3+q^2+q+1) # Massaging the denominator a little, we can put it in the form # (q^6-1)(q^8-1)(q^2-1)(q^4-1)(q^5-1) # so we are looking for a quasipolynomial of order lcm(6,8,2,4,5) = 120 # and degree at most 4, and that will require at least 601 terms. # I don't have the patience to nicely format an order-120 # quasipolynomial so I have put it in a separate file, hw5p7.txt. # Note also that if we allow degenerate pentagons, then the whole # process has to be done again. This time the generating function will # be # simplify(sum(sum(sum(sum(sum(q^(a+b+c+d), a=b..b+c+d+e), b=c..infinity), # c=d..infinity), d=e..infinity), e=0..infinity)) # The resulting quasipolynomial is, of course, different, but the # method is the same.