From jlwether@alum.mit.edu  Mon May 18 19:13:43 1998
William,
I worked out the X_0(11) information for you.  The short answer is that I
can parameterize X_0(11) in terms of j and a variable I call x by:
  x^12 - 60*x^11 + 1386*x^10 - 15092*x^9 + 72303*x^8 - 59928*x^7
  + (-22*j - 431860)*x^6 + (297*j - 123816)*x^5 + (1309*j + 2299407)*x^4
  + (-6105*j + 5182100)*x^3 + (-12529*j + 5665770)*x^2
  + (-j^2 + 5732*j + 3274332)*x + (16*j^2 + 6969*j + 912673)   = 0
In this paramerization, j' is given by
  j' = (-754*x^15 - 23752655*x^14 - 29613745966*x^13 - 6244036882207*x^12 -
376104349304354*x^11 + (-j - 7114347826251259)*x^10 + (-196707*j -
24131393248600286)*x^9 + (-873240811*j + 264386461447029845)*x^8 +
(-267091946884*j + 1012641259503839482)*x^7 + (-17218222665169*j -
2587403021099187533)*x^6 + (-283597926046359*j - 10441866585623711194)*x^5 +
(-579172383588254*j + 2074752923950835587)*x^4 + (5999011646737623*j +
35578660053077953178)*x^3 + (10293966016026733*j + 20680444732221011671)*x^2
+ (-16583607774597939*j - 13671051080851018906)*x + (-26395208710078913*j -
6903889144381374585))/(-x^15 + 137*x^14 - 7143*x^13 + 114009*x^12 +
5734630*x^11 - 406417441*x^10 + 12628650656*x^9 - 246124449024*x^8 +
3246030815232*x^7 - 29273927843840*x^6 + 175076351672320*x^5 -
634871630266368*x^4 + 1066824779169792*x^3 + 289390601437184*x^2 -
2143360479395840*x - 388127604604928)
The rest of this email is a log of my work.  It contains a lot more
information, including a different parameterization for X_0(11) in terms of
usual functions x,y on the elliptic curve.  I leave it to you to work out
the parameterization in terms of j,j' (if you want it).
Have fun,
Joe

------------------------  LOG OF PARI SESSION --------------------
\\ WARNING: I am using pari 1.39.
\\ From Cremona's tables we find that X_0(11) is the elliptic curve
? e11 = smallinitell([0,-1,1,-10,-20])
%1 = [0, -1, 1, -10, -20, -4, -20, -79, -21, 496, 20008, -161051,
-122023936/161051]
\\ let's get the equation for that for later reference:
? \r reduce

? e11eq = elleq(e11,x,y)
%2 = x^3 - x^2 - 10*x + (-y^2 - y - 20)
\\ Now, I don't know how to directly get the j and j' functions on
\\ X_0(11), so we start by expressing everything in terms of q-expansions.
\\ The easiest thing is to find the q-expansion for the canonical differential
\\ on X_0(11), since this is just the corresponding modular form.
? v11 = anell(e11,200)
%3 = [1, -2, -1, 2, 1, 2, -2, 0, -2, -2, 1, -2, 4, 4, -1, -4, -2, 4, 0, 2,
2, -2, -1, 0, -4, -8, 5, -4, 0, 2, 7, 8, -1, 4, -2, -4, 3, 0, -4, 0, -8,
-4, -6, 2, -2, 2, 8, 4, -3, 8, 2, 8, -6, -10, 1, 0, 0, 0, 5, -2, 12, -14,
4, -8, 4, 2, -7, -4, 1, 4, -3, 0, 4, -6, 4, 0, -2, 8, -10, -4, 1, 16, -6,
4, -2, 12, 0, 0, 15, 4, -8, -2, -7, -16, 0, -8, -7, 6, -2, -8, 2, -4, -16,
0, 2, 12, 18, 10, 10, -2, -3, 8, 9, 0, -1, 0, -8, -10, 4, 0, 1, -24, 8, 14,
-9, -8, 8, 0, 6, -8, -18, -2, 0, 14, 5, 0, -7, -2, 10, -4, -8, 6, 4, 8, 0,
-8, 3, 6, -10, -8, 2, 0, 4, 4, 7, -8, -7, 20, 6, 8, 2, -2, 4, -16, -1, 12,
-12, 0, 3, 4, 0, -12, -6, 0, 8, -4, -5, -30, -15, -4, 7, 16, -12, 0, 3, 14,
-2, 16, -10, 0, 17, 8, 4, 14, -4, -6, -2, 4, 0, 0]

? f11 = series(v11,q);
\\ OK, now we find q-expansions for x and y.
\\ We do this one term at a time.
\\ We know the order of poles of x,y, so now we want to find
\\ the next coefficient (represented by ax, ay):
? xq = q^-2 + ax*q^-1 + o(q^0)
%19 = q^-2 + ax*q^-1 + O(q^0)
? yq = -q^-3 + ay*q^-2 + o(q^1)
%23 = -q^-3 + ay*q^-2 + O(q)
\\ These need to satisfy the equation for X_0(11)
? subst(subst(e11eq,y,yq),x,xq)
%24 = (3*ax + 2*ay)*q^-5 + O(q^-4)
\\ And they need to give the right canonical differential
? deriv(xq,q)/(2*yq+1) - f11
%26 = (1/2*ax + (ay + 2))*q + O(q^2)
\\ And the solution is...
? gauss([3,2; 1/2,1], [0,-2])
%27 = [2, -3]
? xq = subst(xq,ax,2)
%28 = q^-2 + 2*q^-1 + O(q^0)
? yq = subst(yq,ay,-3)

%29 = -q^-3 - 3*q^-2 + O(q)
\\ and so on....
\\ we formalize the above with this function:
? {extend(xvyv,
xv,yv, xq,yq,eq1,eq2,v,m,r)=
  xv = xvyv[1];
  yv = xvyv[2];
  xq = series(concat(xv,[ax]),q)/q^2;
  yq = series(concat(yv,[ay]),q)/q^3;
  eq1 = vec(subst(subst(e11eq,y,yq),x,xq))[1];
  eq2 = vec(deriv(xq,q)/(2*yq+1)-f11)[1];
  v = [ -subst(subst(eq1,ax,0),ay,0),
        -subst(subst(eq2,ax,0),ay,0) ];
  m = [ subst(eq1+v[1],ay,0)/ax, subst(eq1+v[1],ax,0)/ay;
subst(eq2+v[2],ay,0)/ax, subst(eq2+v[2],ax,0)/ay ];
  r = gauss(m,v);
  xv = concat(xv,[r[1]]);
  yv = concat(yv,[r[2]]);
  [xv,yv];
}
\\ OK, first we double-check against the coefficients we did by hand:
? extend([1,-1])
%34 = [[1, 2], [-1, -3]]
\\ Yes, that checked.  Now we do this a few more times
? xvyv = %34;
? for(n=3,10, print(xvyv=extend(xvyv)))
[[1, 2, 4], [-1, -3, -7]]
[[1, 2, 4, 5], [-1, -3, -7, -13]]
[[1, 2, 4, 5, 8], [-1, -3, -7, -13, -17]]
[[1, 2, 4, 5, 8, 1], [-1, -3, -7, -13, -17, -26]]
[[1, 2, 4, 5, 8, 1, 7], [-1, -3, -7, -13, -17, -26, -19]]
[[1, 2, 4, 5, 8, 1, 7, -11], [-1, -3, -7, -13, -17, -26, -19, -37]]
[[1, 2, 4, 5, 8, 1, 7, -11, 10], [-1, -3, -7, -13, -17, -26, -19, -37, 15]]
[[1, 2, 4, 5, 8, 1, 7, -11, 10, -12], [-1, -3, -7, -13, -17, -26, -19, -37,
15, 16]]
\\ OK, let's do it silently for a while now
? for(n=11,190, xvyv=extend(xvyv))
? xq = series(xvyv[1],q)/q^2;
? yq = series(xvyv[2],q)/q^3;
\\ let's double-check:
? subst(subst(e11eq,x,xq),y,yq)
%38 = O(q^184)

\\ Finally, we need j and j'.
\\ But we already know the q-expansions for j and j':
? \serieslength = 200
   series precision = 200 significant terms
? j = jell(q);
? j + o(q^20)
%40 = q^-1 + 744 + 196884*q + 21493760*q^2 + 864299970*q^3 +
20245856256*q^4 + 333202640600*q^5 + 4252023300096*q^6 + 44656994071935*q^7
+ 401490886656000*q^8 + 3176440229784420*q^9 + 22567393309593600*q^10 +
146211911499519294*q^11 + 874313719685775360*q^12 +
4872010111798142520*q^13 + 25497827389410525184*q^14 +
126142916465781843075*q^15 + 593121772421445058560*q^16 +
2662842413150775245160*q^17 + 11459912788444786513920*q^18 +
47438786801234168813250*q^19 + O(q^20)
? jj = subst(j,q,q^11+o(q^211));
? jj + o(q^20)
%42 = q^-11 + 744 + 196884*q^11 + O(q^20)
? length(j)
%43 = 199
? length(jj)
%44 = 200

\\ OK, we are now at the point where we can compute j and j'
\\ in terms of x and y.  Let's do j' first.  Note that j' is a degree 12
\\ function on X_0(11).  So we can write j' = (a(x)+y*b(x))/c(x) where
\\ deg(a)<=12, deg(b)<=9, deg(c)<=12.  (Note deg(y)=3.)
\\ But from the q-expansion of j' we know it has a pole of order 11 at
\\ \infty, so actually deg(c)=1.
\\ (This why we do j' first.  After we do j' we will know the other cusp
\\ of X_0(11), which will help for computing j.
\\ number of coefficients:
? 12+1 + 9+1 + 1+1
%45 = 25
\\ let's make a place to store the coefficients for coefficient matching.
? m = matrix(150,25,r,c,0);
\\ coefficients for a:
? for(n=0,12, xnq=xq^n; for(k=1,150, m[k,n+1]=coeff(xnq,k-13)))
\\ coefficients for b:
? for(n=0,9,  xnq=yq*xq^n; for(k=1,150, m[k,n+1+12+1]=coeff(xnq,k-13)))

\\ coefficients for c:
? for(n=0,1,  xnq= -1*jj*xq^n; for(k=1,150,
m[k,n+1+12+1+9+1]=coeff(xnq,k-13)))
? bigprime=nextprime(10^20)
%47 = 100000000000000000039
? length(m)
%48 = 25
? rank(m*mod(1,bigprime))
%49 = 24
? centerlift(ker(m*mod(1,bigprime)))~
%51 =
[-3308 -1781 6780 2704 -643 -148 11 0 0 0 0 0 0 353 2170 1031 -697 23 1 0 0
0 0 16 -1]
? (m*%~)~
%52 =
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
\\ so, we have found the coefficients of the expression for j' in terms
\\ of x and y!

? ajj = polyrev(vector(13,n,%51[1,n]),x)
%59 = 11*x^6 - 148*x^5 - 643*x^4 + 2704*x^3 + 6780*x^2 - 1781*x - 3308
? bjj = polyrev(vector(10,n,%51[1,n+13]),x)
%60 = x^5 + 23*x^4 - 697*x^3 + 1031*x^2 + 2170*x + 353
? cjj = polyrev(vector(2,n,%51[1,n+13+10]),x)
%61 = -x + 16
\\ In other words:
\\
\\ j' = (11*x^6 - 148*x^5 - 643*x^4 + 2704*x^3 + 6780*x^2 - 1781*x - 3308
\\        + y*(x^5 + 23*x^4 - 697*x^3 + 1031*x^2 + 2170*x + 353)) / (-x + 16)
\\
\\ By the way, we really should have been able to predict that ajj and bjj
\\ would have degree 6 and 5, in the following way: jj has a pole of order 11
\\ at \infty, while x and y have poles of order 2 and 3.  Since we know the
\\ denominator is linear (ie: pole of order 2) the numerator has to have a
pole
\\ of order 13.  So 6 and 5 are the maximum possible degrees.
\\ On the other hand, it didn't hurt to be a bit conservative.

? jjxynum(xx,yy)=(subst(ajj,x,xx)+yy*subst(bjj,x,xx))
? jjxy(xx,yy)=jjxynum(xx,yy)/(-xx+16)
? ordell(e11,16)
%62 = [60, -61]
? jjxynum(16,60)
%63 = 0
? jjxynum(16,-61)
%64 = -121
\\ So we see that jjxy(16,-61) is infinity, so this is the other cusp.
\\ by the way, the cusp has order 5 in the Jacobian
? cusp=[16,-61];
? torsell(e11)
%66 = [5, [5], [[5, 5]]]
? powell(e11,cusp,5)
%67 = [0]

\\ by the way, we should check this actually works:
? jjxy(xq,yq)
%72 = q^-11 + 744 + 196884*q^11 + 21493760*q^22 + 864299970*q^33 +
20245856256*q^44 + 333202640600*q^55 + 4252023300096*q^66 +
44656994071935*q^77 + 401490886656000*q^88 + 3176440229784420*q^99 +
22567393309593600*q^110 + 146211911499519294*q^121 +
874313719685775360*q^132 + 4872010111798142520*q^143 +
25497827389410525184*q^154 + 126142916465781843075*q^165 +
593121772421445058560*q^176 + O(q^179)
? %-jj
%73 = O(q^179)
\\ OK, I'm convinced.
\\ We can do j in the same way:  j = (aj(x) + y*bj(x))/(x-16)^11
\\ Since j has a pole of order 1 at \infty and the denominator has a pole
\\ of order 22, the numerator must have a pole of order 23, so deg(aj)<=11,
\\ deg(bj)<=10.

? 11+1+10+1+0+1
%74 = 24
? m = matrix(150,24,r,c,0);
? for(n=0,11, xnq=xq^n; for(k=1,150, m[k,n+1]=coeff(xnq,k-23)))
? for(n=0,10, xnq=yq*xq^n; for(k=1,150, m[k,n+1+11+1]=coeff(xnq,k-23)))
? xnq=-j*(xq-16)^11; for(k=1,150, m[k,24]=coeff(xnq,k-23))
? ker1 = centerlift(ker(m*mod(1,bigprime)))[,1]
%92 = [-65571886394223652, 91576505827906298, 84887294145843033,
6419026712245052, -15124441025165095, -2179206296697095, 558818644494837,
81898270819143, 2535876517772, 19161808636, 21559873, 743,

-26395208710078913, -16583607774597939, 10293966016026733,
5999011646737623, -579172383588254, -283597926046359, -17218222665169,
-267091946884, -873240811, -196707, -1, 1]~
? biggerprime3 = nextprime(10^30)
%93 = 1000000000000000000000000000057
? ker1 = centerlift(ker(m*mod(1,biggerprime)))[,1]
%94 = [-65571886394223652, 91576505827906298, 84887294145843033,
6419026712245052, -15124441025165095, -2179206296697095, 558818644494837,
81898270819143, 2535876517772, 19161808636, 21559873, 743,
-26395208710078913, -16583607774597939, 10293966016026733,
5999011646737623, -579172383588254, -283597926046359, -17218222665169,
-267091946884, -873240811, -196707, -1, 1]~
? m*%
%95 = [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0]~
\\ YUCK!  what completely ugly coefficients!
\\ OK, we have to deal with it
? aj = polyrev(vector(12,n,ker1[n]),x)
%96 = 743*x^11 + 21559873*x^10 + 19161808636*x^9 + 2535876517772*x^8 +
81898270819143*x^7 + 558818644494837*x^6 - 2179206296697095*x^5 -
15124441025165095*x^4 + 6419026712245052*x^3 + 84887294145843033*x^2 +
91576505827906298*x - 65571886394223652
? bj = polyrev(vector(11,n,ker1[n+12]),x)
%97 = -x^10 - 196707*x^9 - 873240811*x^8 - 267091946884*x^7 -
17218222665169*x^6 - 283597926046359*x^5 - 579172383588254*x^4 +
5999011646737623*x^3 + 10293966016026733*x^2 - 16583607774597939*x -
26395208710078913
? jxy(xx,yy)=(subst(aj,x,xx)+yy*subst(bj,x,xx))/(xx-16)^11
\\ cross-check
? jxy(xq,yq)-j
%98 = O(q^189)

\\ OK, how do you use this to find isogenous j-invariants?
\\ suppose you want to find the j-invariants 11-isogenous to jj=12 mod 1117.
? jjxy(x,y)-12
%100 = (11*x^6 + (y - 148)*x^5 + (23*y - 643)*x^4 + (-697*y + 2704)*x^3 +
(1031*y + 6780)*x^2 + (2170*y - 1769)*x + (353*y - 3500))/(-x + 16)

? numer(%)
%101 = 11*x^6 + (y - 148)*x^5 + (23*y - 643)*x^4 + (-697*y + 2704)*x^3 +
(1031*y + 6780)*x^2 + (2170*y - 1769)*x + (353*y - 3500)
\\ Note that this is linear in y.  We can make this explicit:
? subst(%101,x,xx)
%104 = (xx^5 + 23*xx^4 - 697*xx^3 + 1031*xx^2 + 2170*xx + 353)*y + (11*xx^6
- 148*xx^5 - 643*xx^4 + 2704*xx^3 + 6780*xx^2 - 1769*xx - 3500)
\\ Anyway, we combine this with equation for X_0(11)
? result(%101,e11eq,y)
%149 = -x^13 + 76*x^12 - 2346*x^11 + 37268*x^10 - 313775*x^9 + 1216776*x^8
- 526724*x^7 - 6793732*x^6 - 4239147*x^5 + 31933000*x^4 + 76226018*x^3 +
84903780*x^2 + 52488947*x + 15977680
\\ This has an extra factor of x=16 in it...
? %149/(-x+16)
%150 = x^12 - 60*x^11 + 1386*x^10 - 15092*x^9 + 72303*x^8 - 59928*x^7 -
432124*x^6 - 120252*x^5 + 2315115*x^4 + 5108840*x^3 + 5515422*x^2 +
3342972*x + 998605
\\ now we factor this.
? factmod(%150,1117)
%157 =
[mod(1, 1117)*x^10 + mod(129, 1117)*x^9 + mod(789, 1117)*x^8 + mod(372,
1117)*x^7 + mod(341, 1117)*x^6 + mod(560, 1117)*x^5 + mod(624, 1117)*x^4 +
mod(429, 1117)*x^3 + mod(577, 1117)*x^2 + mod(205, 1117)*x + mod(788, 1117) 1]
[mod(1, 1117)*x + mod(249, 1117) 1]
[mod(1, 1117)*x + mod(679, 1117) 1]
\\ let's take -249 mod 1117:
? subst(%101,x,mod(-249,1117))
%158 = mod(1040, 1117)*y + mod(158, 1117)
? jxy(mod(-249,1117),mod(-158/1040, 1117))
%159 = mod(883, 1117)
\\ so that's it, one isogenous j-invariant is 883.

\\ Let's do that in a bit more generality.
\\ Suppose the j-invariant we want to start with is jval
? jjxy(x,y)-jval
%160 = (11*x^6 + (y - 148)*x^5 + (23*y - 643)*x^4 + (-697*y + 2704)*x^3 +
(1031*y + 6780)*x^2 + (2170*y + (jval - 1781))*x + (353*y + (-16*jval -

3308)))/(-x + 16)
? numer(%)
%161 = 11*x^6 + (y - 148)*x^5 + (23*y - 643)*x^4 + (-697*y + 2704)*x^3 +
(1031*y + 6780)*x^2 + (2170*y + (jval - 1781))*x + (353*y + (-16*jval - 3308))
? result(%,e11eq,y)
%162 = -x^13 + 76*x^12 - 2346*x^11 + 37268*x^10 - 313775*x^9 + 1216776*x^8
+ (22*jval - 526988)*x^7 + (-649*jval - 6785944)*x^6 + (3443*jval -
4280463)*x^5 + (27049*jval + 31608412)*x^4 + (-85151*jval + 77247830)*x^3 +
(jval^2 - 206196*jval + 87377988)*x^2 + (-32*jval^2 + 84743*jval +
51476639)*x + (256*jval^2 + 111504*jval + 14602768)
? %/(-x+16)
%163 = x^12 - 60*x^11 + 1386*x^10 - 15092*x^9 + 72303*x^8 - 59928*x^7 +
(-22*jval - 431860)*x^6 + (297*jval - 123816)*x^5 + (1309*jval +
2299407)*x^4 + (-6105*jval + 5182100)*x^3 + (-12529*jval + 5665770)*x^2 +
(-jval^2 + 5732*jval + 3274332)*x + (16*jval^2 + 6969*jval + 912673)
\\ So far this has worked very well.  In fact this is an alternate equation
\\ for X_0(11) in terms of x and j'.  From here we can recover y:
? subst(subst(%161,x,xx),y,x)
%164 = (xx^5 + 23*xx^4 - 697*xx^3 + 1031*xx^2 + 2170*xx + 353)*x + (11*xx^6
- 148*xx^5 - 643*xx^4 + 2704*xx^3 + 6780*xx^2 + (jval - 1781)*xx +
(-16*jval - 3308))
? subst(-coeff(%,0)/coeff(%,1),xx,x)
%165 = (-11*x^6 + 148*x^5 + 643*x^4 - 2704*x^3 - 6780*x^2 + (-jval +
1781)*x + (16*jval + 3308))/(x^5 + 23*x^4 - 697*x^3 + 1031*x^2 + 2170*x + 353)
\\ here's the function for y:
? yxjj(xval,yval)=(-11*xval^6 + 148*xval^5 + 643*xval^4 - 2704*xval^3 -
6780*xval^2 + (-jval + 1781)*xval + (16*jval + 3308))/(xval^5 + 23*xval^4 -
697*xval^3 + 1031*xval^2 + 2170*xval + 353)
\\ and since we have x and y we have j:
? jxy(x,yxjj(x,jval))
%166 = (-754*x^15 - 23752655*x^14 - 29613745966*x^13 - 6244036882207*x^12 -
376104349304354*x^11 + (-jval - 7114347826251259)*x^10 + (-196707*jval -
24131393248600286)*x^9 + (-873240811*jval + 264386461447029845)*x^8 +
(-267091946884*jval + 1012641259503839482)*x^7 + (-17218222665169*jval -
2587403021099187533)*x^6 + (-283597926046359*jval -
10441866585623711194)*x^5 + (-579172383588254*jval +
2074752923950835587)*x^4 + (5999011646737623*jval +
35578660053077953178)*x^3 + (10293966016026733*jval +
20680444732221011671)*x^2 + (-16583607774597939*jval -
13671051080851018906)*x + (-26395208710078913*jval -
6903889144381374585))/(-x^15 + 137*x^14 - 7143*x^13 + 114009*x^12 +
5734630*x^11 - 406417441*x^10 + 12628650656*x^9 - 246124449024*x^8 +
3246030815232*x^7 - 29273927843840*x^6 + 175076351672320*x^5 -

634871630266368*x^4 + 1066824779169792*x^3 + 289390601437184*x^2 -
2143360479395840*x - 388127604604928)
\\ So, there you go.  The short answer is to plug the initial value for j
\\ into
\\        x^12 - 60*x^11 + 1386*x^10 - 15092*x^9 + 72303*x^8 - 59928*x^7 +
(-22*j - 431860)*x^6 + (297*j - 123816)*x^5 + (1309*j + 2299407)*x^4 +
(-6105*j + 5182100)*x^3 + (-12529*j + 5665770)*x^2 + (-j^2 + 5732*j +
3274332)*x + (16*j^2 + 6969*j + 912673)
\\
\\ to get values of (x,j) and then evaluate
\\
\\  j' = (-754*x^15 - 23752655*x^14 - 29613745966*x^13 - 6244036882207*x^12
- 376104349304354*x^11 + (-j - 7114347826251259)*x^10 + (-196707*j -
24131393248600286)*x^9 + (-873240811*j + 264386461447029845)*x^8 +
(-267091946884*j + 1012641259503839482)*x^7 + (-17218222665169*j -
2587403021099187533)*x^6 + (-283597926046359*j - 10441866585623711194)*x^5
+ (-579172383588254*j + 2074752923950835587)*x^4 + (5999011646737623*j +
35578660053077953178)*x^3 + (10293966016026733*j +
20680444732221011671)*x^2 + (-16583607774597939*j - 13671051080851018906)*x
+ (-26395208710078913*j - 6903889144381374585))/(-x^15 + 137*x^14 -
7143*x^13 + 114009*x^12 + 5734630*x^11 - 406417441*x^10 + 12628650656*x^9 -
246124449024*x^8 + 3246030815232*x^7 - 29273927843840*x^6 +
175076351672320*x^5 - 634871630266368*x^4 + 1066824779169792*x^3 +
289390601437184*x^2 - 2143360479395840*x - 388127604604928)
___________________________________________________________________________
  Joseph Loebach Wetherell                    jlwether@alum.mit.edu
                                              wetherel@math.usc.edu
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