MaplePrimes - answers and comments on Question, Assign value on array entries

One approach: assign

Doug Meade — Mon, 17 Sep 2007 06:00:04 Z

One way to do make this type of assignment is with the assign command. The argument to assign needs to be an equation, or a list or set of equations. There are many ways to construct the equations. You can use a loop (or seq) but I like to promote the zip command. The first argument of zip is a mapping from elements of two different lists/sets/arrays.

> restart;
> Ecce:=array(1..2,[x,y]);
                              Ecce := [ x y ]
> ZERO:=array(1..2,[0,0]);
                              ZERO := [ 0 0 ]
> eqs := zip( (n,v)->n=v, Ecce, ZERO );
                          eqs := [ x = 0  y = 0 ]
> zip( (n,v)->assign(n=v), Ecce, ZERO ):
> 'x'=x,'y'=y;
                                x = 0, y = 0

Note that the output from the first zip is an array (there is no comma between the elements). If we were working with lists, we could work as follows:

> unassign( 'x', 'y' );
> eqs := zip( (n,v)->n=v, convert(Ecce,list), convert(ZERO,list) );
                               [x = 0, y = 0]
> assign( eqs );
> 'x'=x,'y'=y;
                                x = 0, y = 0

Note the comman between the elements in eqs. Also, it was necessary to unassign the previous assignments before we could repeat this step. An alternate here would have been to use restart to completely erase everything. I hope this is instructive, Doug

---------------------------------------------------------------------
Douglas B. Meade
Math, USC, Columbia, SC 29208  E-mail: mailto:meade@math.sc.edu       
Phone:  (803) 777-6183         URL:    http://www.math.sc.edu/~meade/

Thank you

sasomao — Tue, 18 Sep 2007 01:25:27 Z

Hi, thank you for your rapid and useful help. I wish to ask you about innumerable questions! I've to use maple to study some asymptotic expansion for the Gauss-Codazzi eq. That is R[ijkm] = eta[A,B] (∇[i]∇[k] z^A ∇[j]∇[m] z^B - (A->B)) (I'm not sure that the maple tag work so I write the source: R[ijkm] = eta[A,B] (∇[i]∇[k] z^A ∇[j]∇[m] z^B - (A->B))) in which I consider expansion for the z^A functions, e for the Christoffel symbols (so Nabla also is an expansion) (eta is a high dimensional minkowski diagonal matrix, and is implicit the sum A=1..6, B=1..6) That I would is to isolate the term of the riemann tensor proportional to the several power of the expansion parameter. The problem is that if I let maple to do the explicit sum I've not a symbolic and compact answer because I can't "recognize" the summation terms (this will be excessively time expensive, nearly that do the calculus by hand!!!). On the other hand if I use the add command, maple don't expand the summation, and there is no simplifications!) Thank you for any help!!!! S.V.

I don't see the difficulty, or the problem (more info requested)

Doug Meade — Tue, 18 Sep 2007 16:42:20 Z

S.V., I'm glad my previous response was useful. Let's see what I can do with your newest question. First, there appears to be a mismatched single quote. After adding a single quote before doublederivcov, here is what I get from Maple 11 for add, sum, and Sum:

restart;
addRiemann:=Array(1..4,1..4,1..4,1..4,(i,j,k,m)->add(add(''eta''[A,B]*('doublederivcov'(Z(X[])[A],Chris(X[]),i,k)*'doublederivcov'(Z(X[])[B],Chris(X[]),j,m) -'doublederivcov'(Z(X[])[A],Chris(X[]),j,k)*'doublederivcov'(Z(X[])[B],Chris(X[]),i,m)),A=1..6),B=1..6)):
sumRiemann:=Array(1..4,1..4,1..4,1..4,(i,j,k,m)->sum(sum(''eta''[A,B]*('doublederivcov'(Z(X[])[A],Chris(X[]),i,k)*'doublederivcov'(Z(X[])[B],Chris(X[]),j,m) -'doublederivcov'(Z(X[])[A],Chris(X[]),j,k)*'doublederivcov'(Z(X[])[B],Chris(X[]),i,m)),A=1..6),B=1..6)):
SumRiemann:=Array(1..4,1..4,1..4,1..4,(i,j,k,m)->Sum(Sum(''eta''[A,B]*('doublederivcov'(Z(X[])[A],Chris(X[]),i,k)*'doublederivcov'(Z(X[])[B],Chris(X[]),j,m) -'doublederivcov'(Z(X[])[A],Chris(X[]),j,k)*'doublederivcov'(Z(X[])[B],Chris(X[]),i,m)),A=1..6),B=1..6)):
addRiemann[1,1,1..4,1..4]; convert(%,listlist);
                           Array(%id = 164471836)
          [[0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0]]
sumRiemann[1,1,1..4,1..4]; convert(%,listlist);
                           Array(%id = 164517788)
          [[0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0]]
SumRiemann[1,1,1..4,1..4]; convert(%,listlist);
                           Array(%id = 164580792)
  [[  6   /  6    \    6   /  6    \    6   /  6    \    6   /  6    \]  
  [[----- |-----  |  ----- |-----  |  ----- |-----  |  ----- |-----  |]  
  [[ \    | \     |   \    | \     |   \    | \     |   \    | \     |]  
  [[  )   |  )    |    )   |  )    |    )   |  )    |    )   |  )    |]  
  [[ /    | /    0|,  /    | /    0|,  /    | /    0|,  /    | /    0|], 
  [[----- |-----  |  ----- |-----  |  ----- |-----  |  ----- |-----  |]  
  [[B = 1 \A = 1  /  B = 1 \A = 1  /  B = 1 \A = 1  /  B = 1 \A = 1  /]  

    [  6   /  6    \    6   /  6    \    6   /  6    \    6   /  6    \]  
    [----- |-----  |  ----- |-----  |  ----- |-----  |  ----- |-----  |]  
    [ \    | \     |   \    | \     |   \    | \     |   \    | \     |]  
    [  )   |  )    |    )   |  )    |    )   |  )    |    )   |  )    |]  
    [ /    | /    0|,  /    | /    0|,  /    | /    0|,  /    | /    0|], 
    [----- |-----  |  ----- |-----  |  ----- |-----  |  ----- |-----  |]  
    [B = 1 \A = 1  /  B = 1 \A = 1  /  B = 1 \A = 1  /  B = 1 \A = 1  /]  

    [  6   /  6    \    6   /  6    \    6   /  6    \    6   /  6    \]  
    [----- |-----  |  ----- |-----  |  ----- |-----  |  ----- |-----  |]  
    [ \    | \     |   \    | \     |   \    | \     |   \    | \     |]  
    [  )   |  )    |    )   |  )    |    )   |  )    |    )   |  )    |]  
    [ /    | /    0|,  /    | /    0|,  /    | /    0|,  /    | /    0|], 
    [----- |-----  |  ----- |-----  |  ----- |-----  |  ----- |-----  |]  
    [B = 1 \A = 1  /  B = 1 \A = 1  /  B = 1 \A = 1  /  B = 1 \A = 1  /]  

    [  6   /  6    \    6   /  6    \    6   /  6    \    6   /  6    \]]
    [----- |-----  |  ----- |-----  |  ----- |-----  |  ----- |-----  |]]
    [ \    | \     |   \    | \     |   \    | \     |   \    | \     |]]
    [  )   |  )    |    )   |  )    |    )   |  )    |    )   |  )    |]]
    [ /    | /    0|,  /    | /    0|,  /    | /    0|,  /    | /    0|]]
    [----- |-----  |  ----- |-----  |  ----- |-----  |  ----- |-----  |]]
    [B = 1 \A = 1  /  B = 1 \A = 1  /  B = 1 \A = 1  /  B = 1 \A = 1  /]]
value( SumRiemann[1,1,1..4,1..4] ); convert(%,listlist);
                           Array(%id = 149420696)
          [[0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0]]

Because it is not possible to copy the "Array" contents from the worksheet to another application without losing the contents (WYSInotWYG), I have uploaded the worksheet to MaplePrimes. You can access it from the following links:
View 178_RiemannTensor.mw on MapleNet or Download 178_RiemannTensor.mw
View file details So, I do not see any problem with any of the three forms for constructing this tensor. Of course, I do not have your definitions for doublederivcov, Chris, X, Z, ... so my results are not very interesting. In general, because you have a fixed finite sum, I would think that you would want to use add. You should be using sum only if you think the individual terms involved will include expressions that could benefit from a summation formula. Only after the specifics of this problem are known can I make a final suggestion about this. Doug

---------------------------------------------------------------------
Douglas B. Meade
Math, USC, Columbia, SC 29208  E-mail: mailto:meade@math.sc.edu       
Phone:  (803) 777-6183         URL:    http://www.math.sc.edu/~meade/

My worksheet

sasomao — Tue, 18 Sep 2007 18:04:11 Z

Hi another thank you for the reply. I think that this worked for you because you don't have the definitions for the functions and operator. So I've charged the ws. It's here View 4492_secondo ordine2.mw on MapleNet or Download 4492_secondo ordine2.mw
View file details I've asked some questions on the worksheet Tk Salvatore

In all cases, what I want is

sasomao — Tue, 18 Sep 2007 18:35:16 Z

In all cases, what I want is a formal result for the riemann tensor. That is, I would know the ijkm component without fix the actual value for i j k m (after all the possible simplification). I don't understand if Maple can do it. Can you make clearer this point? Tk Salvatore

About sum/add

sasomao — Tue, 18 Sep 2007 13:36:49 Z

The problem is that if I let maple to do the explicit sum I've not a symbolic and compact answer because I can't "recognize" the summation terms (this will be excessively time expensive, nearly that do the calculus by hand!!!). On the other hand if I use the add command, maple don't expand the summation, and there is no simplifications!)

Obviously I've reversed the role of "add" and "sum" About the use of add ad sum: if I create this array: Riemann:=Array(1..4,1..4,1..4,1..4,(i,j,k,m)->add(add(''eta''[A,B]*('doublederivcov'(Z(X[])[A],Chris(X[]),i,k)*'doublederivcov'(Z(X[])[B],Chris(X[]),j,m) -doublederivcov'(Z(X[])[A],Chris(X[]),j,k)*'doublederivcov'(Z(X[])[B],Chris(X[]),i,m)),A=1..6),B=1..6)) it work. But if I do the same declaration with sum (or Sum) at the place of the two "add", maple print Error, bad index into Array So, how can I do create an array which entries are unevalued sum, without bad index error? Thank you S.V.