# here is what I saw that confirmed the fraction
# for e^x

get_fraction:= proc(ff)
   local gg,nn,hh;

   # get the coefficient of x^0
   gg:=coeff(ff,x,0):

   # subtract the x^0 and do the division
   hh:=taylor(x/(ff-gg),x=0,30):

   # pull out the coeff of x^0
   gg:=simplify(coeff(hh,x,0)):

   # return the coeff of x^0 and the current fraction
   return (gg,hh);
end proc;

iterate_sequence := proc(aa)
   local ii, bb,cc,dd;
   bb:=aa:
   
   # iterate through 15 divisions
   for ii from 1 by 1 to 15 do 
      (dd,cc) := get_fraction(bb):
      bb := cc:
      print(ii,"===",dd);
   end do;
end proc;

w:=taylor( exp(x), x=0, 30 );
iterate_sequence (w);


#########################################################
# this is a general case.  
# you can see the determinants in the first
# few terms.


restart;

y:=sum(a[i]*x^i,i=0..10);


trya := proc(ff)
   local gg,nn,hh;
   gg:=coeff(ff,x,0):
   hh:=taylor(x/(ff-gg),x=0,10):
   gg:=simplify(coeff(hh,x,0)):
   return (gg,hh);
end proc;

tryb := proc(aa)
   local ii, bb,cc,dd;
   bb:=aa:
   for ii from 1 by 1 to 12 do 
      (dd,cc) := trya(bb):
      bb := cc:
      print(ii,"===",dd);
   end do;
end proc;

tryb(y);



################################################
# This is a test of the general case where the
# actual determinants are computed.
# This is an implementation of my hypothesis.
#

restart;
with(LinearAlgebra);

# Generate the functions which are the determinants
fun := proc(i,j)
   local A;
   if i=0 and j=0 then
     return 1;
   end if;
   if i=1 and j=0 then
     return a[0];
   end if;
   if i=1 and j=1 then
     return a[1];
   end if;
   if i=2 and j=1 then
     return a[2];
   end if;

   A:=(x,y) -> a[x+y-2+i];
   return Determinant(Matrix(j,A));
   # return (Matrix(j,A));
end proc;

coef := proc(n)
  local C,size;
 
  if n=0 then
    return a[0];
  end if;

  if n=1 then
    return 1/a[1];
  end if;

  if type(n,even) then
    size:=n/2;
    return -(fun(1,size))^2/(fun(2,size-1)*fun(2,size));
  else
    size:=(n-1)/2;
    return (fun(2,size))^2/(fun(1,size)*fun(1,size+1));
  end if;
end proc;


# print the 
for i from 0 by 1 to 10 do
  v[i]:=coef(i);
end do;

# test with e^x
# set the values for a[i] to be the values
# for the taylor series for e^x 

for i from 0 by 1 to 30 do
  a[i] := 1/i!;
end do;


# now, let's see what the values are for the 
# continued fraction
for i from 0 by 1 to 10 do
  v[i];
end do;