# Want to create a generic series with indexed coeffs, e.g. a[1], a[2],...

Hello, I'm very new to Sage and would like to take a horrible equation, $H(e^v,z)=0$, and substitute into $v$ a power series, like $v(z)=\sum_{i=l}^{u} a[i] z^i,$ where maybe $l<0$. I'd then like to solve for the coefficients, at each power of $z$, or all at once like Mathematica does (or so I'm told). The Mathematica version is here in case it helps clear up my question.

ord = 10;
eq2 = (-(1 - Exp[v])*Exp[2*(z^2 + Log[-Sqrt[3]] + 2*Log[3]) + 9*v]) -
3^(5/2)*(9*Exp[4*v] - 3*Exp[3*v] - 4*Exp[2*v] - Exp[v] + 1)*
Exp[z^2 + 3*v] - 3*Exp[v] + 1
VV = Series[Sum[a[i] z^i, {i, 0, $ord}], {z, 0,$ord}]
a[0] = (-1/2)*Log[3]
ef2 = Block[{$MaxExtraPrecision=1000},Solve[Series[(eq2 /. v -> VV), {z, 0,$ord}] == 0,
Table[a[i], {i, 1, $ord}]]//FullSimplify] VV/.ef2  Everything is complex and I was thinking of something like this pseudo-script: M.<v,z> = AsymptoticRing(growth_group='exp(v^CC)*exp(z^CC)*CC',coefficient_ring=CC); M t = M.gens() f = sum(t[n]*z^n for n in range(8)) eq = e^(4*v+3*z)+...; eq.substitute(v=f) eqZ1 = eq.solve(z) eqZ2 = eq.solve  But this is very scattered, rough, and not even meaningful. I'm not sure if using this 'experimental' AsymptoticRing is a good idea, nor if my growth group is correct. I also see things like .substitute, etc. from the /ref/Calculus. All I really want to do is take a generic expression $$H = \sum_i e^{a_iv + b_iz^2}=0,$$ where$a_i$and$b_i$are complex, and replace the$v$with$\sum_{i=low}^{i=high} h_i z^i$-- actually I am working on 'eq2', of several, in the first Mathematica block. Then take Taylor series and solve for the coefficients$a_i$in ascending powers of$z$. This is much like Hinch's book, Perturbation Methods, in the very beginning, for "expansion methods". Thank you for your time and help! edit retag close merge delete ## Comments I do not quite understand your subtitution$v_i$by$\sum_{i=low}^{i=high} h_i z^i$. The left term has$i$has an index and you replaced it by something which uses$i\$ as an index of sumation.

( 2016-01-15 05:37:07 -0600 )edit

Ahh you'r right, I was conflating two things. I'll fix it, Thanks

( 2016-01-15 12:30:30 -0600 )edit