sbh

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16 years, 65 days

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These are replies submitted by sbh

@vv Hi again. Well it's not so much what I want as what will work ;)  I want to find these u(z) and v(z), and upon substituting them back into the curve attain zero. That is the only check I can think of...

Anyway, I don't think I explained myself too well. I actually need expansions of the form;

  • u \mapsto u0 + z^2
  • v \mapsto v0 + sum_{i=1}^n a_i * z^i

from which the "u" and "v" come from;

  • alpha = exp(u)
  • beta = exp(v)

So unfortunately I could not use your script exactly. But I did adapt it to the following. (Of course I was unsure of using RootOfs, especially when it was quadratic going the alpha route).

restart;
precurve3 := 1-Q*beta+alpha*beta^3*(Q^2*beta^4-Q*beta^3-2*Q*beta^2+2*beta^2-beta+1)-alpha^2*beta^9*(1-beta);

sol := solve(eval([precurve3 = 0, diff(precurve3, beta) = 0], Q = 2.0), [alpha, beta]);

curve := subs([alpha = exp(u), beta = exp(v)], eval(precurve3, Q = 2));

sol3 := [seq([u = ln(rhs(sol[i][1])), v = ln(rhs(sol[i][2]))], i = 1 .. 8)];

s1 := solve(curve, v);

s11 := allvalues(s1);

vv(i):=sum(a[i,j]*z^(j),j=1..i)+v0;

u0, v0 := (eval([u, v], sol3[1]))[];

vv(3);

vnow := eval(evalf(s11[1]), u = z^2+u0);

series(vnow, z, 8);

Try another "v" branch.

u02, v02 := (eval([u, v], sol3[2]))[];

v2now := eval(evalf(s11[2]), u = z^2+u02);

series(v2now, z, 8);

 

The only problem is that now I have zeros for the odd powers of "z" as coefficients.

 

So, unless I completely missed what you were saying (possible), do you see a method or have any advice? I have reasonable confidence that the first four coefficients (at least) have non-zero powers as that is what Mathematica told me...though I know Mathematica not very well at all...

Thank you much for your time and patience!

 

@vv Hi again. So I did what you suggested, but with unknowns in the power series for v, and yet I still only attain zero coefficients, which hence makes v(z)=0. Maybe I am doing something wrong.... again?

restart;
precurve3 := 1-Q*beta+alpha*beta^3*(Q^2*beta^4-Q*beta^3-2*Q*beta^2+2*beta^2-beta+1)-alpha^2*beta^9*(1-beta);
sol := solve(eval([precurve3 = 0, diff(precurve3, beta) = 0], Q = 2.0), [alpha, beta]);
curve := subs([alpha = exp(u), beta = exp(v)], eval(precurve3, Q = 2));
sol3 := [seq([u = ln(rhs(sol[i][1])), v = ln(rhs(sol[i][2]))], i = 1 .. 8)];
s1 := solve(curve, u);
vv(i):=sum(a[i,j]*z^(j),j=1..9)+rhs(sol3[i][2]);
vv(4);
e1 := eval(s1[2], v = vv(4));
uu := convert(series(e1, z = 0, 6), polynom);
eq0 := coeff(uu, z, 0);
eq1 := coeff(uu, z, 1);
eq2 := coeff(uu, z, 2);
eq3 := coeff(uu, z, 3);
eq4 := coeff(uu, z, 4);
s2 := fsolve(eval([eq1 = 0., eq2 = 0.], Q = 2.0), complex);

Thanks again!

 

@vv Hi, well yes, I do... except that I would like u = u0 + a2*z^2, but if this is too rigid I can hopefully change it easily to your more general suggestion.

Thank you!

PS> I already tried solve(<eqn>,{u(s),v(s)}); but it only returns an empty. If I try solve(curve3(1),v), I get a bunch of RootOf's for which allvalues() can't help.

@acer Hi, well mine is Maple 15 (not 16), but when I copied and pasted yours in it worked! So maybe I had something in effect... Thanks much!

Thanks! Well I don't understand all you wrote but when I implement it I still get multiple terms of the same power e.g. a_1*V^P*V^2 + a_2*V^P*V^2 + \cdots .... I don't need an upper limit as the inversion is up to an order... Maybe there is a better way to solve for V in equation "main"?

Thanks. I was hoping so.... It is just that now the reason for my other post is still unclear. :( Thanks again!
Thanks. I was hoping so.... It is just that now the reason for my other post is still unclear. :( Thanks again!
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