Doug Meade

 

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

MaplePrimes Activity


These are answers submitted by Doug Meade

Is there a question in your posts? These look like statements of exercises from a text, or from another assignment for your course. This is not an appropriate use of MaplePrimes.

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

Except for the fact that you should use MultiInt instead of the now outdated Doubleint, your code looks OK. The problems are in your mathematics.

First, your f represents the plane x+y-z=1, not x+y+z=1.

Next, you do not have the right set for the region of integration. Your region of integration is the unit circle in the x-y plane. Why did you choose this set?

Look at your plot. do the curves ever touch? The points where the plane and paraboloid intersect should be in the solid.

The set where the two curves is not the easiest to find. It's not terrible, but the expressions are not as simple as they could have been.

If you find the intersection of the two surfaces as the curve given by  y0 <= y <- y1  for x0 <= x <= x1 where y0 and y1 dependn on x and x0 and x1 are (somewhat messy) constants, then your code could look something like:

with( Student[MultivariateCalculus] ):
f := ;
g := 4-x^2-y^2;
y0 := ;
y1 := ;
x0 := ;
x1 := ;
plot3d( [f,g], y=y0..y1, x=x0..x1, axes=boxed, color=[red,blue] );
V := MultiInt( g-f, y=y0..y1, x=x0..x1 );

Hoping you find these suggestions helpful,

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

It will be easier to diagnose a cure for your problem if you will share your current code with the group.

But, in advance of that, I can offer a couple suggestions.

As you indicate, there is no "value" option for a Label.  You need to be working with the "caption".

Are you sure "algebraic" is the correct type you expect to have. Is this the full IC,  u(x,0)=1, or just the rhs, 1? If it's the former, then you need to be looking for "equation". If it's the latter, this should pass your test, but maybe you'd do better looking for numeric.

Also, if you are including type-checking on a user-entered field, I recommend also using "corrections=true" as in (assuming Maplets:-Tools is loaded).

Set( 'function_display' = Get( 'h'::numeric, corrections=true );

Here's a simple example showing how you can update a Label's caption:

restart;
with( Maplets[Elements] ): with( Maplets[Tools] ):
UpdateL1 := proc()
  uses Maplets[Tools];
  Set( 'L1'(caption) = Get('TB1'::algebraic, corrections=true) );
end proc:
maplet := Maplet(
[
 [Label['L1']("Enter new label: "),
  TextBox['TB1']()],
 [Button("Change",'onclick'=Evaluate(function=UpdateL1)),
  Button("Close",Shutdown())]
]
):
Maplets[Display](maplet);

Is it really necessary to have a separate proc for the Get? YES! Because Get cannot be used within a Maplet definition and can appear only in a procedure. (It's documented, but somewhat buried within the help page.)

Notice that I included corrections=true in the Get command. This makes it quite easy to fix a syntax problem in your input field.

Hoping this is helpful, and that an easier way can be found to do 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

 

I'll second Alec's question, and add a few of my own (along with some suggestions).

Why do you stop after 10 iterations? why not more? or fewer?

What do you hope to be able to do with the expression that you obtain from this iteration?

Are you interested in the behavior in terms of the parameter A, or in the coefficients of powers of x (and sin(x) and cos(x))?

You can examine some of this structure by using the collect command:

collect( u, A ); will show the expressions in terms of powers of A

collect( u, [x,sin(x),cos(x)] ) will show the coefficients of powers of these 3 factors in terms of A

Here's what I get for the first iteration:

collect( u, [x,cos(x),sin(x)] );
 /35     1\  2   4         3   15         2   /     68  \              697  
 |-- A - -| x  + - A cos(x)  - -- A cos(x)  + |-1 + -- A| cos(x) + 1 - --- A
 \4      2/      9             4              \     3   /              36   
collect( u, A );
 /  697   35  2   68          15       2   4       3\                  1  2
 |- --- + -- x  + -- cos(x) - -- cos(x)  + - cos(x) | A + 1 - cos(x) - - x 
 \  36    4       3           4            9        /                  2   

Do you need to keep the sin(x) and cos(x), or would you be satisfied with a series expansion in powers of x? If so, then you might be able to use

restart;
u := convert(series((1-cos(x)),x,10),polynom);
for i to 10 do
 u := convert(series(u+int(subs(x = t, diff(u, [`$`(x, 2)])+u-A*(1+2*u+3*u^2+4*u^3))*(t-x), t = 0 .. x),x,max(10,2*i)),polynom)
end do:

This runs pretty fast, and the collect options are useful as well, The last iteration I get from the above, when collected in terms of powers of A is

collect( u, A );
  3371356577    18  9   /  2860547    16     153919429    18\  8
-------------- x   A  + |----------- x   - ------------- x  | A 
50018544576000          \14859936000       3572753184000    /   

     /    191431    16    3067303    14     85329911     18\  7
   + |- ---------- x   + ---------- x   + ------------- x  | A 
     \  1667952000       5448643200       7145506368000    /   

     /   240923    14    45533    12     635483     16      2142109     18\  6   
   + |- --------- x   + -------- x   + ----------- x   - ------------- x  | A  + 
     \  908107200       29937600       27243216000       1563079518000    /      

  /    9763    12    1973   10       9898139      18      3141647     16
  |- -------- x   + ------ x   + --------------- x   - ------------- x  
  \  14968800       453600       114328101888000       1307674368000    

       44323    14\  5   /  2203    12    73   8      27887      16
   + --------- x  | A  + |-------- x   + ---- x  + ------------ x  
     908107200    /      \14968800       5040      104613949440    

         441751      18    461    10      9853     14\  4   /11   6
   - -------------- x   - ------ x   - ---------- x  | A  + |--- x 
     57164050944000       226800       1362160800    /      \360   

         1367      16     257     12    1   8     17    10
   - ------------ x   - -------- x   - --- x  + ------ x  
     871782912000       59875200       336      120960    

          65567        18      1369      14\  3   /      1       14    1   6
   + ---------------- x   + ----------- x  | A  + |- ---------- x   - --- x 
     3201186852864000       14529715200    /      \  7264857600       180   

           1        16      1      12    1    8          1         18
   + ------------- x   + -------- x   + ---- x  - --------------- x  
     1494484992000       47900160       6720      400148356608000    

       1     10   1   4\  2   /1  2   1   4       1      12    1   6
   - ------ x   + -- x | A  + |- x  - -- x  - --------- x   + --- x 
     453600       12   /      \2      24      479001600       720   

          1       14     1    8         1         16      1     10
   + ----------- x   - ----- x  - -------------- x   + ------- x  
     87178291200       40320      20922789888000       3628800    

            1          18\  
   + ---------------- x  | A
     6402373705728000    /  

I hope some of this is of use to you. Please let us know if you have more questions, or more information about your needs that will help us to find a solution to your problem.

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

I don't see that this is an easy problem to solve, with Maple.

The problem is a little simpler to understand when the specific bases are replaced by parameters a and b, as follows:

eq := (a,b) -> log[a](y) = log[b](y-1);
(a, b) -> log[a](y) = log[b](y - 1)

Then, your equation is:

eq(2*sqrt(2+sqrt(3)),2+sqrt(3));
                           ln(y)            ln(y - 1)   
                    ------------------- = --------------
                      / (1/2)    (1/2)\     /     (1/2)\
                    ln\6      + 2     /   ln\2 + 3     /

I agree that Maple's solution to this is not very nice:

solve( %, y );
                  /      /                            /      _Z      \    \\
                  |      |                            |--------------|    ||
                  |      |                            |  /     (1/2)\|    ||
                  |      |                            \ln\2 + 3     //    ||
                  |      |           / (1/2)    (1/2)\                    ||
                  |RootOf\-exp(_Z) + \6      + 2     /                 - 1/|
                  |--------------------------------------------------------|
                  |                       /     (1/2)\                     |
                  \                     ln\2 + 3     /                     /
 / (1/2)    (1/2)\                                                          
 \6      + 2     /                                                          

To understand what this is telling us, let's look at the more general problem:

eq(a,b);
                              ln(y)   ln(y - 1)
                              ----- = ---------
                              ln(a)     ln(b)  
solve( eq(a,b), y );
                    /      /             /_Z ln(a)\    \\
                    |RootOf|exp(_Z) - exp|--------| + 1||
                    |      \             \ ln(b)  /    /|
                    |-----------------------------------|
                    \               ln(b)               /
                   a                                     

To make sense of this, notice the exp(_Z) in the RootOf. Call this W, then Maple is telling us that the solution to this equation is a^(W/ln(b)) where W is the solution to W-W^(ln(a)/ln(b)+1=0. Unless a and b have some special relationship, this is going to be almost impossible to solve explicitly.

In your case, a=2*sqrt(b) and b=2+sqrt(3). This suggests another special form to consider, with b as the parameter:

eq(2*sqrt(b),b);
                             ln(y)       ln(y - 1)
                          ------------ = ---------
                            /   (1/2)\     ln(b)  
                          ln\2 b     /            
solve( %, y );
                    /      /             /2 _Z ln(b)\    \\
                 exp|RootOf|exp(_Z) - exp|----------| - 1||
                    \      \             \ ln(4 b)  /    //

This means we need to be looking at the solution to W - W^(2*ln(b)/ln(4*b))-1=0.

I'm not going any further down this path. I'll close with an examination of the exact solution. First, it's easy to verify that y=8+4*sqrt(3) does satisfy the original equation:

eval( eq(2*sqrt(2+sqrt(3)),2+sqrt(3)), y=8+4*sqrt(3) );
                      /       (1/2)\       /       (1/2)\
                    ln\8 + 4 3     /     ln\7 + 4 3     /
                   ------------------- = ----------------
                     / (1/2)    (1/2)\      /     (1/2)\ 
                   ln\6      + 2     /    ln\2 + 3     / 
simplify(%);
                                    2 = 2

If we look at the last special case considered above, we might guess that the solution has the form y=4*b. Let's check that:

eval( eq(2*sqrt(b),b), y=4*b );
                           ln(4 b)      ln(4 b - 1)
                         ------------ = -----------
                           /   (1/2)\      ln(b)   
                         ln\2 b     /              
simplify(%);
                                   ln(4 b - 1)
                               2 = -----------
                                      ln(b)   

This does not look look like it's going to be true in general. In fact, this condition is satisfied when

solve( %, b );
                                (1/2)       (1/2)
                           2 + 3     , 2 - 3     

So, the case with b=2+sqrt(3) is one of only 2 times when this problem works out so nicely.

This has been fun, and interesting, but I'm not sure it really gets you any closer to the solution you are seeking.

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

 

I assume your function is f(x)=sin(x)

Your definition of Dn does not take into account the fact that n is an integer (and positive). I would modify your approach by defining Dn to be a procedure with arguments n and L, as follows:

Dn := (n,L) -> (2/L)* int(f*sin(n*Pi*x/L), x=0..L);

Then, for example:

Dn(n,Pi);
                                 2 sin(n Pi) 
                               - ------------
                                    /      2\
                                 Pi \-1 + n /
Dn(1,Pi);
                                      1
Dn(2,Pi);
                                      0
Dn(n,Pi) assuming n::posint;
                                      0

Look closely at the last of these. The assumption that n is a positive integer should NOT give 0 as the result. As shown above, if n=1, the result should be 1. As you have already indicated, you can get this by taking the limit as n->1, but Maple won't do this when you tell it that n is a positive integer. In this case I believe it is seeing that sin(n*Pi)=0 for all positive integers and not taking into consideration the fact that the denominator is also 0 when n=1. I'd call this a bug.

Now, back to your problem. If I try to do the general sum as you have it, I get zero.

Sum( Dn(n,Pi)*sin(n*x)*cos(n*t), n=1..infinity );
                  infinity                                 
                   -----                                   
                    \                                      
                     )    /  2 sin(n Pi) sin(n x) cos(n t)\
                    /     |- -----------------------------|
                   -----  |             /      2\         |
                   n = 1  \          Pi \-1 + n /         /
value( % );
                                      0

To get the result you want, I have to separate all of the special terms:

Dn(1,Pi)*sin(x)*cos(t) + Sum( Dn(n,Pi)*sin(n*x)*cos(n*t), n=2..infinity );
                         /infinity                                 \
                         | -----                                   |
                         |  \                                      |
                         |   )    /  2 sin(n Pi) sin(n x) cos(n t)\|
         sin(x) cos(t) + |  /     |- -----------------------------||
                         | -----  |             /      2\         ||
                         \ n = 2  \          Pi \-1 + n /         //
value( % );
                                sin(x) cos(t)

If you want to automate the identification of the singular terms, you can probably make use of the singular command:

singular( Dn(n,Pi), n );
                              {n = -1}, {n = 1}

I hope this is helpful.

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

Are you sure your system of 12 equations in 9 unknowns is consistent?

The fact that Maple returns no solution to the full system of 12 equations and one solution when the 12th equation is omitted suggests (to me) that the 12th equation is not consistent with the others.

Here are some quick illustrations to support these statements:

sys := [
exp(-z[1][1]/t[1])*(sum(p[2][j]*exp(-z[2][j]/t[j]), j = 1 .. 3))
*(sum(p[3][j]*exp(-z[3][j]/t[j]), j = 1 .. 3)) = t[1](1-w)/(p[1][1]*w),
exp(-z[1][2]/t[2])*(sum(p[2][j]*exp(-z[2][j]/t[j]), j = 1 .. 3))
*(sum(p[3][j]*exp(-z[3][j]/t[j]), j = 1 .. 3)) = t[2](1-w)/(p[1][2]*w),
exp(-z[1][3]/t[3])*(sum(p[2][j]*exp(-z[2][j]/t[j]), j = 1 .. 3))
*(sum(p[3][j]*exp(-z[3][j]/t[j]), j = 1 .. 3)) = t[3](1-w)/(p[1][3]*w),
exp(-z[2][1]/t[1])*(sum(p[1][j]*exp(-z[1][j]/t[j]), j = 1 .. 3))
*(sum(p[3][j]*exp(-z[3][j]/t[j]), j = 1 .. 3)) = t[1](1-w)/(p[2][1]*w),
exp(-z[2][2]/t[2])*(sum(p[1][j]*exp(-z[1][j]/t[j]), j = 1 .. 3))
*(sum(p[3][j]*exp(-z[3][j]/t[j]), j = 1 .. 3)) = t[2](1-w)/(p[2][2]*w),
exp(-z[2][3]/t[3])*(sum(p[1][j]*exp(-z[1][j]/t[j]), j = 1 .. 3))
*(sum(p[3][j]*exp(-z[3][j]/t[j]), j = 1 .. 3)) = t[3](1-w)/(p[2][3]*w),
exp(-z[3][1]/t[1])*(sum(p[1][j]*exp(-z[1][j]/t[j]), j = 1 .. 3))
*(sum(p[2][j]*exp(-z[2][j]/t[j]), j = 1 .. 3)) = t[1](1-w)/(p[3][1]*w),
exp(-z[3][2]/t[2])*(sum(p[1][j]*exp(-z[1][j]/t[j]), j = 1 .. 3))
*(sum(p[2][j]*exp(-z[2][j]/t[j]), j = 1 .. 3)) = t[2](1-w)/(p[3][2]*w),
exp(-z[3][3]/t[3])*(sum(p[1][j]*exp(-z[1][j]/t[j]), j = 1 .. 3))
*(sum(p[2][j]*exp(-z[2][j]/t[j]), j = 1 .. 3)) = t[3](1-w)/(p[3][3]*w),
z[1][1]+z[1][2]+z[1][3] = l[1],
z[2][1]+z[2][2]+z[2][3] = l[2],
z[3][1]+z[3][2]+z[3][3] = l[3]
];
vars := [
  z[1][1],  z[1][2],  z[1][3],
  z[2][1],  z[2][2],  z[2][3],
  z[3][1],  z[3][2],  z[3][3]
];

When we solve the full set of 12 equations, we get no solution:

solve( sys, vars );

But, if we omit the last equation, Maple does return a solution:

sol2 := solve( sys[1..11], vars );

To check that this is, in fact, a solution, it's easier to compare the difference of the left- and right-hand sides to zero, so we massage the original equations (all 12 of them):

sys2 := map( e->lhs(e)-rhs(e)=0, sys );

Next, put the solution to the first 11 equations back into the full set of 12 equations, and simplify:

eval( sys2, sol2[] ):
simplify( %, symbolic );
[0 = 0, 0 = 0, 0 = 0, 0 = 0, 0 = 0, 0 = 0, 0 = 0, 0 = 0, 0 = 0, 0 = 0, 0 = 0, 

  t[1] ln(p[3][1]) + t[1] ln(p[1][1])

   + 2 t[1] ln(t[3](1 - w) + t[2](1 - w) + t[1](1 - w)) + t[1] ln(w)

   + t[1] ln(p[2][1]) - 3 t[1] ln(t[1](1 - w)) - l[1] - l[2] - l[3]

   + t[3] ln(w) + 2 t[2] ln(t[3](1 - w) + t[2](1 - w) + t[1](1 - w))

   + t[2] ln(w) + t[2] ln(p[3][2]) + t[2] ln(p[1][2]) - 3 t[3] ln(t[3](1 - w))

   + t[3] ln(p[1][3]) + 2 t[3] ln(t[3](1 - w) + t[2](1 - w) + t[1](1 - w))

   + t[3] ln(p[2][3]) + t[3] ln(p[3][3]) + t[2] ln(p[2][2])

   - 3 t[2] ln(t[2](1 - w)) = 0]
map( evalb, % );
  [true, true, true, true, true, true, true, true, true, true, true, false]

So, this solution does not satisfy the last equation.

When I try to solve the last 11 equations (omitting the first equation), Maple again is unable to find a solution. I really looks to me as though there is something about the 12th equation that makes this system inconsistent.

I have not checked all of your inputs to see if there might be more variables or even a mistake in one of the original equations. Unless there are more unknowns floating around, it looks to me as though the full system is not consistent.

Should t[1], t[2], and t[3] be included in the list of variables? When I add these to make a system of 12 equations in 12 unknowns, Maple reports that "solutions may have been lost".

I hope this is helpful to you,

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

 

Here is how I would start your problem, including some alternatives for plotting the rows of a matrix.

restart;
with( LinearAlgebra ):
A := <<1|1>,<4|1>>;
B := MatrixExponential(A,t);
C := Vector(2,'c');
B.C;
plot( B[1, .. ], t=-1..1 );
plot( [B[2,1],B[2,2], t=-1..1] );

Notice some of the different ways to refer to a row of a matrix. I'm not exactly sure what type of plot you are looking to obtain, so I included the commands to create a couple different views.

I hope this is useful to you,

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

 

This sounds as though it might succomb to the Shooting Method. This has been discussed in this forum, about 2 years ago, see http://www.mapleprimes.com/blog/glenn-carlson/dsolve-using-set-output-system-of-odes .

In addition, I would refer you to Lesson 17 in the collection of worksheets for a complete course in Differential Equations that is on Maple's Application Center. The URL is http://www.maplesoft.com/applications/app_center_view.aspx?AID=1523 .

In the examples you have posted, you can make some simplifications by noting that G'/G = (log(G))' and making a change of variable. For the BC you might also benefit from writing G'(alpha)/G(alpha)=A as G'(alpha)-A*G(alpha)=0. The latter form is likely to be more numerically stable, particularly if G(alpha) is expected to be close to zero.

I hope something in here is of use to you,

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.ed

Did you try to simplify eqn?


simplify( eqn );
   1    /               2                    
------- \D(Ubt[0])(1) h1  &eta;bt[0](x, t, T)
h1 + h2                                      

                                    2
   - D[2](&phi;bt[0])(x, 1, t, T) h1 

   + 2 D(Ubt[0])(1) h1 &eta;bt[0](x, t, T) h2

   + 4 h1 h2 D(Ubc[0])(1) &eta;bc[0](x, t, T)

                                                                           2
   - 2 D[2](&phi;bt[0])(x, 1, t, T) h1 h2 - D[2](&phi;bt[0])(x, 1, t, T) h2 

                    2                    \
   + D(Ubt[0])(1) h2  &eta;bt[0](x, t, T)/


If you prefer to do this simplification without typing commands, position the cursor over the expression to be simplified, press the right mouse button, select simplify, then simplify from the sublist that this shows. When you release you the mouse, the simplified expression is displayed.

I hope this helps,

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.ed

To the best of my knowledge, there is no difference in the software, or with the worksheets.

The differences are in the license agreements - length of use, and who is allowed to use the software.

Many years ago there was a Maple Student Edition that had restrictions on the size of objects that could be used. There was a maximum size for matrices. The number of terms in an expression was limited, .... But, this has not existed for many years.

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.ed

While Joe is right that there is no way to change the background color of a worksheet, this does not mean you can't achieve the same affect with a little ingenuity.

You can change the color of a cell in a "table". So, insert a table with one row and column. In the "table" menu (either contxt menu (right click) or from the menu bar) select "cell color" and change the background color of the cell.  Put everything in the worksheet, including section markers, etc., and you will have a worksheet whose background is the color you have selected.

This approach allows you to be even fancier. You can have a worksheet with different background colors in different sections, etc.

I hope this is helpful,

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.ed

Depending on the size and/or structure of your matrix, you might find that the Map command is more efficient than map.

The difference between map and Map (and map2 and Map2) is that Map (and Map2) work "in-place". For more details, see the online hellp for Map (?Map)

Note that Map and Map2 are part of the LinearAlgebra package. Here's one way to temporarily load the package without having to explicitly refer to LinearAlgebra:-Map and LinearAlgebra:-MatrixFunction. The output is the same as shown above.

use LinearAlgebra in
   Map( limit, MatrixFunction( W, x^t, x ), t=infinity );
end use;

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.ed

In recent releases Maple has made this about as natural as you could ask it to be.

A := Array(1..10,1..10,1..10, (i,j,k)->i+j/k ):
A[1,1,1..10]

And, if you want to get all of these columns at once, I'd suggest using:

for i from 1 to 10 do
  for j from 1 to 10 do
    a||i||j := A[i,j,1..10];
  end do;
end do;

(You could also do this with seq and assign, but this is not as transparent.)

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.ed

 

By default the BodePlot command returns an array of 2 plots. (Where do you get 3 plots?) The help for BodePlot tells that there is an output parameter that can control the output. Here's what I tried:

> with(DynamicSystems);
> sys := TransferFunction(1/(s-10)):
> bp := BodePlot(sys, output = [magnitudeplot, phaseplot]);
> plots:-display( bp );

Now, you might want to override the default labels with something like:

> plots:-display( bp, labels=["Freq [rad/s]", "Mag & Phase [dB]"] );

It might take a little more effort to get Maple to plot the curves in different colors and to include a legend. (This can be done by modifying the Plot structure, but this is not for the faint of heart. Suggestions for the next release?)

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.ed
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