• To: mathgroup at smc.vnet.net
• From: =?tis-620?b?zdTKw+zD0aqq7CCiqMPD2OinyA==?= at smc.vnet.net
• Date: Sun, 5 Dec 2010 21:48:59 -0500 (EST)

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.

Bob Hanlon <hanlonr at cox.net> wrote:

> 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 extracts
> 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
>
> ---- <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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