Re: [Please Help] How to get coefficient list from a

• To: mathgroup at smc.vnet.net
• Subject: [mg114487] Re: [Please Help] How to get coefficient list from a
• From: =?tis-620?b?zdTKw+zD0aqq7CCiqMPD2OinyA==?= at smc.vnet.net
• Date: Sun, 5 Dec 2010 21:54:51 -0500 (EST)

```Thank you very much Bob. I am very appreciate for all of your answer.

Bob Hanlon <hanlonr at cox.net> wrote:

>
> n == 4;
>
> u1 == Normal[
>    Series[u[Ca[t], T[t]] - u[3.2066, 317.5529], {Ca[t], 3.2066,
>      n}, {T[t], 317.5529, n}]] // Simplify;
>
> d == {};
>
> q[0] == {};
>
> For[k == 1, k <== n - 1, k++,
>  For[i == 0, i <== k, i++,
>  p[i, k - i] == (i!*(k - i)!)*
>    SeriesCoefficient[u1, {Ca[t], 3.2066, i}, {T[t], 317.5529, k - i}];
>  q[k] == Append[q[k - 1], p[i, k - i]];
>  d == Union[Join[d, q[k]]];]]
>
> d == d /. 1. -> 1
>
> {Derivative[0, 1][u][3.2066, 317.5529],
>  Derivative[0, 2][u][3.2066, 317.5529],
>   Derivative[0, 3][u][3.2066, 317.5529],
>  Derivative[1, 0][u][3.2066, 317.5529],
>   Derivative[1, 1][u][3.2066, 317.5529],
>  Derivative[1, 2][u][3.2066, 317.5529],
>   Derivative[2, 0][u][3.2066, 317.5529],
>  Derivative[2, 1][u][3.2066, 317.5529],
>   Derivative[3, 0][u][3.2066, 317.5529]}
>
>
> d ==== Flatten[
>   Table[D[u[x, y], {x, j}, {y, k}], {j, 0, n - 1}, {k, 1 - Sign[j],
>     n - j - 1}], 1] /. {x -> 3.2066, y -> 317.5529}
>
> True
>
>
> Bob Hanlon
>
>" <kajornrungsilp.i at gmail.com> wrote:
>
> ==========================
> From previous equation, I want to know  if i want to change the value in
> p[i,k-i] into sets of coefficient. How to do that?
> Any hint is greatly appreciated.
>
> kajornrungsilp.i at gmail.com> wrote:
>
> > Thank you for all of your answer. I very appreciate you. but i have some
> > question that i confuse in my code.
> > My code is
> > n == 4;
> > u1 == Normal[
> >   Series[u[Ca[t], T[t]] - u[3.2066, 317.5529], {Ca[t], 3.2066,
> >     n}, {T[t], 317.5529, n}]]
> > d == {};
> > q[0] == {};
> > For[k == 1 , k <== n - 1, k++,
> >  For[i == 0 , i <== k , i++,
> >   p[i, k - i] == (i!*(k - i)!)*
> >     SeriesCoefficient[u1, {Ca[t], 3.2066, i}, {T[t], 317.5529, k - i}];
> >   q[k] == Union[q[k - 1], p[i, k - i]];
> >   d==Union [d,q[k]];
> >   ]]d
> > I want to know,why it doesn't collect the coefficient of my equation?
> what
> > is something wrong in my code?
> >
> > Best Regard.
> >
> > =E0=C1=D7=E8=CD 4 =B8=D1=B9=C7=D2=A4=C1 2553, 7:10, Bob Hanlon <hanlonr=
@cox.net> =E0=A2=D5=C2=B9=C7=E8=D2:
> >
> > A simpler form:
> >>
> >> expr1 == a1 + a2*x + a3*y + a4*x^2 + a5*x*y + a6*y^2;
> >>
> >> myCoef[expr_, var_List] :==
> >>  SortBy[List @@ expr1, Total[Exponent[#, var]] &] /. Thread[var -> 1]
> >>
> >> myCoef[expr1, {x, y}]
> >>
> >> {a1, a2, a3, a4, a5, a6}
> >>
> >>
> >> Bob Hanlon
> >>
> >> ---- Bob Hanlon <hanlonr at cox.net> wrote:
> >>
> >> ==========================
> >>
> >> expr1 == a1 + a2*x + a3*y + a4*x^2 + a5*x*y + a6*y^2;
> >>
> >> Since you apparently want to ignore zero coefficients then this extrac=
ts
> >> the non-zero coefficients
> >>
> >> (List @@ expr1) /. {x -> 1, y -> 1}
> >>
> >> {a1, a2, a4, a3, a5, a6}
> >>
> >> However, this standard ordering is different from yours. Presumably,
> your
> >> ordering is
> >>
> >> myCoef[expr_, var_List] :== Module[
> >>   {coef == (List @@ expr) /. Thread[var -> 1]},
> >>   Last /@ Sort[
> >>     Cases[List @@ expr,
> >>      a_?(MemberQ[coef, #] &)*z_. ->
> >>       {Total[Exponent[z, var]], a}]]];
> >>
> >> myCoef[expr1, {x, y}]
> >>
> >> {a1, a2, a3, a4, a5, a6}
> >>
> >>
> >> Bob Hanlon
> >>
> >> ---- "=CD=D4=CA=C3=EC=C3=D1=AA=AA=EC =A2=A8=C3=C3=D8=E8=A7=C8=D4=C5=BB=
=EC" <kajornrungsilp.i at gmail.com> wrote:
> >>
> >> ==========================
> >> Thank you for your answer. But I would like only the coefficient of
> series
> >> equation in the form {a1, a2 , a3 ,a4,a5,a6} not to be in the form of
> >> {{a1,
> >> a3, a6}, {a2, a5, 0}, {a4, 0, 0}}. How to get there?
> >> Best regard
> >>
> >>
> >> 2010/12/3 Bob Hanlon <hanlonr at cox.net>
> >>
> >> >
> >> > expr1 == a1 + a2*x + a3*y + a4*x^2 + a5*x*y + a6*y^2;
> >> >
> >> > coef == CoefficientList[expr, {x, y}]
> >> >
> >> > {{a1, a3, a6}, {a2, a5, 0}, {a4, 0, 0}}
> >> >
> >> > From example in documentation on CoefficientList
> >> >
> >> > expr2 == Fold[FromDigits[Reverse[#1], #2] &, coef, {x, y}]
> >> >
> >> > a1 + a2 x + a4 x^2 + (a3 + a5 x) y + a6 y^2
> >> >
> >> > expr1 ==== expr2 // Simplify
> >> >
> >> > True
> >> >
> >> >
> >> > Bob Hanlon
> >> >
> >> > ---- Autt <kajornrungsilp.i at gmail.com> wrote:
> >> >
> >> > ==========================
> >> > Greeting,
> >> > I've a list issued from  multivariate series like : K=={a1+a2*x +a3*=
y
> >> > +a4*x^2*+a5*x*y +a6*y^2 }
> >> > I need to have {a1, a2 , a3 ,a4,a5,a6}
> >> >
> >> >  How to do that please?
> >> > e.g.
> >> > n == 3
> >> > u == Normal[Series[x^4 + y^4 + x^2, {x, 5, n}, {y, 5, n}]]
> >> > c[0] == {};
> >> > d == {};
> >> > q[0] == {};
> >> > For[k == 1 , k <== n - 1, k++,
> >> >  For[i == 0 , i <== k , i++,
> >> >  p[i, k - i] ==
> >> >   SeriesCoefficient[u, {x, 5, i}, {y, 5, k - i}];
> >> >  q[k] == Union[{q[k - 1], {p[i, k - i]}}];
> >> >  d == Union[d, q[k]];
> >> >
> >> >  Print[d];
> >> >
> >> > Best regard,
>
>

```

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