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Re: Question about evaluation with Sum

• To: mathgroup at smc.vnet.net
• Subject: [mg19613] Re: Question about evaluation with Sum
• From: "Allan Hayes" <hay at haystack.demon.co.uk>
• Date: Sat, 4 Sep 1999 21:09:17 -0400
• References: <7qqbom$4ge@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

CORNIL Jack Michel <jmcornil at club-internet.fr> wrote in message
news:7qqbom$4ge at smc.vnet.net...
> Hello,
>
>
> I have a big problem of evaluation with the function Sum.
>
> With the very simple following example,
>
>    p=k
>
>    Sum[p,{k, 1, n}]
>
>      p k
>
>    Sum[Evaluate[p], {k, 1, n}]
>
>      1
>      - n (1+n)
>      2
>
>    Sum[p,{k, 1, 5}]
>
>      15
>
> how can I explain my students that
>
>    in the first case p is left unevaluate and it needs Evaluate,
>
>    in the last case p is evaluate correctly  ?
>
> I do not understand how the "Attributes" of Sum can explain this.
>
>
> Thank You in advance
>
>
> Jack Michel CORNIL
>
> Versailles FRANCE
>

Jack,
Interesting examples.
Here is my account of what is happening
I come back to your examples in part [C]

[A] EVALUATION OF
                   Sum[expr,{j, a,b}]
(see notes  below)
Because Sum has the attribute Holdall, the order of evaluation need not
follow the standard head then elements pattern, and instead it preceeds as
follows:

(notice how the kind of iterator affects the evaluation and the special
treatment of the summation variable, j)

Evaluate a,b to get
                   Sum[expr,{j, a*,b*}]
---  If the iterator {j, a*,b*} does not give an explicit finite sequence of
values for j,  then replace j and any occurences of j in expr with a local
variable K$r where r is a unique  integer index
                   Sum[expr',{K$n, a*,b*}]
Evaluate expr' to get
                   Sum[expr'*,{K$n, a*,b*}]
If this can be summed then the sum is returned; if not thenthe  original
input is returned.

--- If the iterator {j, a*,b*} gives an explicit sequence of values,
j1,j2,... for j, then evaluate expr to expr* and sum this over these values
of j.

NOTES:
(1) The evaluation can be altered using Evaluate, Hold etc.
For example, with
                     Sum[Evaluate[expr],{j, a,b}]
the first change is to
                     Sum[expr*,{j, a,b}]
(where expr* is the value of  expr) - then we procede as above (to evaluate
a and b etc).

(2) More generally, the iterator can be of the form {i, a, b, d}, in which
case we start with evluating a, b and d.

[B] GENERAL EXAMPLES

1)

ClearAll["`*"]

p = j; a = A; b = B;

Sum[f[p, j], {j, a, b}]

                Sum[f[p, j], {j, a, b}]

Evaluation steps

Evaluate the limits a, b,  in the iterator {j, a, b}

                 Sum[f[p, j], {j, A, B}]

{j, A,B}, does not give an explicit finite list   j1,j2, .... so introduce a
local variable K$n in place of the summation variable, j.

                 Sum[f[p,K$n],{K$r,a,b}]

Evaluate f[p,K$r]

                 Sum[f[i,K$r],{K$r,A,B}]

This cannot be summed so return the original  input

2 )

Add another assignment

f[j, r_] := j r

Sum[f[p, j], {j, a, b}]

                -(1/2)*(-1 + A - B)*(A + B)*j

Evaluation steps

Evaluate the limits a, b in the iterator {j, a, b}

                 Sum[f[p, j], {j, A, B}]

{j, A,B} does not give an explicit finite list   j1,j2, .... so introduce a
local variable K$r in place of the summation variable, j.

                 Sum[f[p,K$r],{K$r,a,b}]

Evaluate f[p,K$r]

                 Sum[j K$r,{K$r,A,B}] (*note the new j introduced*)

Sum this assuming that B-A is non negative integer

                -(1/2)(-1 + A - B)(A + B) j

3)

Clear["`*"]

a = A; b = A + 2; p = j;

Sum[f[p, j], {j, a, b}]

                f[A, A] + f[1 + A, 1 + A] + f[2 + A, 2 + A]

Evaluation steps

Evaluate the limits a, b,  in the iterator {j, a, b}

                 Sum[f[p, j], {j, A, B}]

{j, A,B} gives an explicit sequance of values for j,  A,A+1,A+2;  so do not
introduce a local variable K$r in place of the summation symbol, j.

Evaluate f[p,j] to get f[j,j]
Sum this over the sequence of values for j

                 f[A,A]+f[1+A,1+A]+f[2+A,2+A]

[C] YOUR EXAMPLES

Clear["`*"]

p = k;
Sum[p, {k, 1, n}]

                k n

( not p k)

Evaluation steps

                 Sum[p,{K$r,1,n}]
                 Sum[k,{K$r,1,n}]
                 k n

2)

Sum[Evaluate[p], {k, 1, n}]

                   1/2*n*(1 + n)

Evaluation steps

Because of Evaluate the first change is to

                    Sum[k, {k, 1, n}]

Then we follow the general pattern

                    Sum[K$r, {K$r, 1, n}]

Which can be summed.

3)

Sum[p, {k, 1, 5}]

                        15

Evaluation steps

Because the iterator gives the explicit list 1,2,3,4,5, no local variable is
introduced for k and we simply sum k over the value 1,2,3,4,5

Allan
---------------------
Allan Hayes
Mathematica Training and Consulting
Leicester UK
www.haystack.demon.co.uk
hay at haystack.demon.co.uk
Voice: +44 (0)116 271 4198
Fax: +44 (0)870 164 0565