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