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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 $$H = \sum_i e^{a_iv_i + b_iz_i^2}=0,$$ where $a_i$ and $b_i$ are complex, and replace the $v_i$ with $\sum_{i=low}^{i=high} h_i z^i$. 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!

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=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 $$H = \sum_i e^{a_iv_i v + b_iz_i^2}=0,$$ z^2}=0,$$ where $a_i$ and $b_i$ are complex, and replace the $v_i$ $v$ with $\sum_{i=low}^{i=high} h_i z^i$. 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!

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 $\sum_{i=low}^{i=high} h_i z^i$. 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!

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 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.<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 t = M.gens()
f = sum(t[n]*z^n for n in range(8))
eq=e^(4*v+3*z)+...; eq = e^(4*v+3*z)+...; 
eq.substitute(v=f)
eqZ1=eq.solve(z)
eqZ2=eq.solve
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!