Re: Question about subscripts and polynomial

• To: mathgroup at smc.vnet.net
• Subject: [mg107369] Re: [mg107355] Question about subscripts and polynomial
• From: DrMajorBob <btreat1 at austin.rr.com>
• Date: Thu, 11 Feb 2010 05:16:49 -0500 (EST)
• References: <201002100836.DAA21317@smc.vnet.net>

```I'm not entirely sure what ordering you want, but here's something to get
you on the way.

You can tweak the "order" function if needed.

poly = d^2*v;
Subscript[q, 1] =
poly /. {d -> Sum[Subscript[d, i], {i, 1, 3}]} /. {v ->
Sum[Subscript[v, i], {i, 1, 3}]};
basicSum = Expand[Subscript[q, 1]]

(ugly output omitted; it looks fine in Mathematica)

Here's the same thing transformed to a List and sorted by the total order
of each product:

order[term_] :=
term /. Times -> Plus /. Power -> Times /. Subscript[_, k_] :> k
listSum = SortBy[basicSum /. Plus -> List, order]

(ugly output omitted; it looks fine in Mathematica)

That's a List, not a sum; the following is also NOT a sum, but it looks
like one, sorted the same as listSum:

displaySum = Infix[listSum, "+"]

(ugly output omitted; it looks fine in Mathematica)

To get back the original:

backToBasics = Plus @@ displaySum[[1]];
basicSum === backToBasics

True

Bobby

On Wed, 10 Feb 2010 02:36:40 -0600, Luca Zanotti Fragonara
<Luca.Zanottifragonara at polito.it> wrote:

> Hello everybody,
>
> I would like to write a Polynomial, in this way:
>
> Poly = d^2*v
> Subscript[q, 1] =
>  Poly /. {d -> Sum[Subscript[d, i], {i, 1, 3}]} /. {v ->Sum[Subscript[v,
> i], {i, 1, 3}]}
> Expand[Subscript[q, 1]]
>
> In this way I will obtain a polynomial in this form:
>
> d_1^2 v_1+2 d_1 d_2 v_1+d_2^2 v_1+2 d_1 d_3 v_1+2 d_2 d_3 v_1+d_3^2
> v_1+d_1^2 v_2+2 d_1 d_2 v_2+d_2^2 v_2+2 d_1 d_3 v_2+2 d_2 d_3 v_2+d_3^2
> v_2+d_1^2 v_3+2 d_1 d_2 v_3+d_2^2 v_3+2 d_1 d_3 v_3+2 d_2 d_3 v_3+d_3^2
> v_3
>
> I would like to reorder the expanded polynomial in a way such that the
> terms with lower subcripts indexes will be at the beginning of the
> polynomial, and the terms with higher order of subscripts will be at the
> end (a sort of order due to the subscript instead of the power terms).
> So the order should be something:
>
> Order 3: d_1^2 v_1
> Order 4: 2 d_1 d_2 v_1+d_1^2 v_2+...
> Order 5: d_2^2 v_1+2 d_1 d_3 v_1+...
> Order 6: 2 d_2 d_3 v_1+...
>
> I've tried to figure it out but I don't know which way to turn!!!
>
>
> Luca
>
>
>

--
DrMajorBob at yahoo.com

```

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