       • To: mathgroup at smc.vnet.net
• From: Bob Hanlon <hanlonr at cox.net>
• Date: Sat, 4 Dec 2010 06:16:45 -0500 (EST)

```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}

e.g.
n = 3
u = Normal[Series[x^4 + y^4 + x^2, {x, 5, n}, {y, 5, n}]]
c = {};
d = {};
q = {};
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,

--

Bob Hanlon

```

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