#hw3.txt, January 30, 2014, Frank Wagner


#1. 
#Here are some examples from pages 60-90 of Garvan's Maple booklet

#with(plots):
#implicitplot3d(y^2-x^2=z, x=-2..2, y=-2..2, z=-4..4);
#with(plots):
#animate3d([r*cos(t+a),r*sin(t+a),r^2*cos(2*t)],r=0..1,t=0..2*Pi,a=0..3,
#frames=10,style=patch,title=`The Rotating Saddle`);
#animate3d([r*cos(t+a),r*sin(t+a),r^2*cos(2*t)+sin(a)],r=0..1,t=0..2*Pi,a=0..2*Pi,
#frames=10,style=patch,title=`The Flying Pizza`);
#with(linalg):
#B:=matrix(2,2);
#entermatrix(B);
#enter element 1,1 > 12;
#enter element 1,2 > 13;
#enter element 2,1 > 14;
#enter element 2,2 > 15;
#matadd(B,B);
#f := (i,j) -> x^(i*j);
#A := matrix(4,4,f);
#factor(det(A));
#with(linalg):
#A:=matrix(2,2,[1,2,3,4]):
#B:=matrix(2,2,[-2,3,-5,1]):
#A+B;
#evalm(%);


#2. 

#For Hil(50), the lowest number of Digits that produces a good
#agreement (<1% difference) is 75
#Considering Hil(n):
#For n=10, this number is 14
#For n=20, this number is 29
#For n=30, this number is 44
#For n=40, this number is 59
#For n=60, this number is 90
#For n=70, this number is 105
#For n=80, this number is 120
#For n=90, this number is 135
#For n=100, this number is 151

#Clearly, the relationship between n and the lowest number of
#Digits that produces a good agreement seems to be
#Digits=1.5*n



#3.

with(combinat):

inv1:=proc(p) local i,j,n,s:
s:=0:
n:=nops(p):

for i from 1 to n do
 for j from i+1 to n do

     if p[i] > p[j] then
       s:=s+1:
     fi:

 od:
od:

s:

end:


gf:=proc(n,q) local P:
P:=permute(n):
add ( q^inv1(p) , p in P):

end:


#For n=1, gf(n,q)=1
#For n=2, gf(n,q)=1+q
#For n=3, gf(n,q)=(1+q)*(q^2+q+1)
#For n=4, gf(n,q)=(1+q)^2*(q^2+q+1)*(q^2+1)
#For n=5, gf(n,q)=(1+q)^2*(q^2+q+1)*(q^2+1)*(q^4+q^3+q^2+q+1)
#For n=6, gf(n,q)=(1+q)^3*(q^2+q+1)^2*(q^2+1)*(q^4+q^3+q^2+q+1)*(q^2-q+1)
#For n=7, gf(n,q)=(1+q)^3*(q^2+q+1)^2*(q^2+1)*(q^4+q^3+q^2+q+1)*(q^2-q+1)*(q^6+q^5+q^4+q^3+q^2+q+1)
#For n=8, gf(n,q)=(1+q)^4*(q^2+q+1)^2*(q^2+1)^2*(q^4+q^3+q^2+1)*(q^2-q+1)*(q^6+q^5+q^4+q^3+q^2+q+1)*(q^4+1)

#A conjecture of a "nice" formula for this factored form could be:
gfc:=proc(n,q):

mul(((1+q^(2^i))^(floor(n/(2^(i+1))))),i=0..floor(log[2](n/2)))*mul((add(q^i,i=0..2*j))^floor(n/(2*j+1)),j=0..floor((n-1)/2))*mul(add(((-1)^i)*q^i,i=0..2*j),j=0..floor(n/6)):

end:


#Upon experimentation, this works for n=1,..,8, but fails for n=9.