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Doug Meade

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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 replies submitted by Doug Meade

printf and nprintf...

Doug Meade 3088
September 22 2008
0 0

In your printf you could use:

printf( "X[%a] = %a\n", i, X[i] );

If you want the output centered, you can replace printf with nprintf. This displays the output as a Maple name, which could be sufficient for your needs.

Before you do too much of this in document mode with 2D Maple input, you might want to consider changing to worksheet mode and using Maple notation. Just a suggestion to keep in mind as you move forward.

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

printf and nprintf...

Doug Meade 3088
September 22 2008

In your printf you could use:

printf( "X[%a] = %a\n", i, X[i] );

If you want the output centered, you can replace printf with nprintf. This displays the output as a Maple name, which could be sufficient for your needs.

Before you do too much of this in document mode with 2D Maple input, you might want to consider changing to worksheet mode and using Maple notation. Just a suggestion to keep in mind as you move forward.

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

Good point, and more...

Doug Meade 3088
September 22 2008
0 0

My previous post was pre-maturely positive when I got Maple to produce the same result for both expressions. I did not stop to notice that the expression was not the correct one.

Let's consider the following. I start by giving each expression a name. And replace q0 and k0 with their negatives (q=-q0 and k=-k0).

a := (-q0)^(-1/2)*(-k0)^(1/2)*q0*k0^2*(q0*k0)^(-1/2):
b := (-q0)^(-1/2)*(-k0)^(-3/2)*q0*k0^4*(q0*k0)^(-1/2):
c := exp(-1/2*ln(-q0)+1/2*ln(-k0)):
d := exp(-1/2*ln(-q0))*exp(1/2*ln(-k0)):
A := q0*k0^2/((-q0)*(-k0))^(1/2):
L := [a,b,c,d,c*A,d*A];
       [                                                             
       [        (1/2)      2                        4                
       [   (-k0)      q0 k0                    q0 k0                 
       [-----------------------, ----------------------------------, 
       [     (1/2)        (1/2)       (1/2)      (3/2)        (1/2)  
       [(-q0)      (q0 k0)       (-q0)      (-k0)      (q0 k0)       

                                            (1/2)  
            /  1           1        \  (-k0)       
         exp|- - ln(-q0) + - ln(-k0)|, ----------, 
            \  2           2        /       (1/2)  
                                       (-q0)       

            /  1           1        \      2                         ]
         exp|- - ln(-q0) + - ln(-k0)| q0 k0           (1/2)      2   ]
            \  2           2        /            (-k0)      q0 k0    ]
         -----------------------------------, -----------------------]
                           (1/2)                   (1/2)        (1/2)]
                    (q0 k0)                   (-q0)      (q0 k0)     ]
L2 := eval( L, [q0=-q,k0=-k] );
    [                                                                    
    [   (1/2)  (5/2)     (1/2)  (5/2)                             (1/2)  
    [  q      k         q      k          /  1         1      \  k       
    [- -------------, - -------------, exp|- - ln(q) + - ln(k)|, ------, 
    [        (1/2)            (1/2)       \  2         2      /   (1/2)  
    [   (q k)            (q k)                                   q       

           /  1         1      \    2                 ]
        exp|- - ln(q) + - ln(k)| q k      (1/2)  (5/2)]
           \  2         2      /         q      k     ]
      - -----------------------------, - -------------]
                      (1/2)                    (1/2)  ]
                 (q k)                    (q k)       ]

Now, when we hit these with various forms of simplify:

map( simplify, L2 );
    [   (1/2)  (5/2)     (1/2)  (5/2)   (1/2)   (1/2)     (1/2)  (5/2)  
    [  q      k         q      k       k       k         q      k       
    [- -------------, - -------------, ------, ------, - -------------, 
    [        (1/2)            (1/2)     (1/2)   (1/2)          (1/2)    
    [   (q k)            (q k)         q       q          (q k)         

         (1/2)  (5/2)]
        q      k     ]
      - -------------]
              (1/2)  ]
         (q k)       ]
map( simplify, L2, symbolic );
                    [           (1/2)   (1/2)          ]
                    [  2    2  k       k         2    2]
                    [-k , -k , ------, ------, -k , -k ]
                    [           (1/2)   (1/2)          ]
                    [          q       q               ]
map( simplify, L2 ) assuming q>0;
                    [           (1/2)   (1/2)          ]
                    [  2    2  k       k         2    2]
                    [-k , -k , ------, ------, -k , -k ]
                    [           (1/2)   (1/2)          ]
                    [          q       q               ]
map( simplify, L2 ) assuming k>0;
                    [           (1/2)   (1/2)          ]
                    [  2    2  k       k         2    2]
                    [-k , -k , ------, ------, -k , -k ]
                    [           (1/2)   (1/2)          ]
                    [          q       q               ]
eval( %, [q=-q0,k=-k0] );
              [                 (1/2)       (1/2)            ]
              [   2     2  (-k0)       (-k0)          2     2]
              [-k0 , -k0 , ----------, ----------, -k0 , -k0 ]
              [                 (1/2)       (1/2)            ]
              [            (-q0)       (-q0)                 ]

Interesting.

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

Good point, and more...

Doug Meade 3088
September 22 2008

My previous post was pre-maturely positive when I got Maple to produce the same result for both expressions. I did not stop to notice that the expression was not the correct one.

Let's consider the following. I start by giving each expression a name. And replace q0 and k0 with their negatives (q=-q0 and k=-k0).

a := (-q0)^(-1/2)*(-k0)^(1/2)*q0*k0^2*(q0*k0)^(-1/2):
b := (-q0)^(-1/2)*(-k0)^(-3/2)*q0*k0^4*(q0*k0)^(-1/2):
c := exp(-1/2*ln(-q0)+1/2*ln(-k0)):
d := exp(-1/2*ln(-q0))*exp(1/2*ln(-k0)):
A := q0*k0^2/((-q0)*(-k0))^(1/2):
L := [a,b,c,d,c*A,d*A];
       [                                                             
       [        (1/2)      2                        4                
       [   (-k0)      q0 k0                    q0 k0                 
       [-----------------------, ----------------------------------, 
       [     (1/2)        (1/2)       (1/2)      (3/2)        (1/2)  
       [(-q0)      (q0 k0)       (-q0)      (-k0)      (q0 k0)       

                                            (1/2)  
            /  1           1        \  (-k0)       
         exp|- - ln(-q0) + - ln(-k0)|, ----------, 
            \  2           2        /       (1/2)  
                                       (-q0)       

            /  1           1        \      2                         ]
         exp|- - ln(-q0) + - ln(-k0)| q0 k0           (1/2)      2   ]
            \  2           2        /            (-k0)      q0 k0    ]
         -----------------------------------, -----------------------]
                           (1/2)                   (1/2)        (1/2)]
                    (q0 k0)                   (-q0)      (q0 k0)     ]
L2 := eval( L, [q0=-q,k0=-k] );
    [                                                                    
    [   (1/2)  (5/2)     (1/2)  (5/2)                             (1/2)  
    [  q      k         q      k          /  1         1      \  k       
    [- -------------, - -------------, exp|- - ln(q) + - ln(k)|, ------, 
    [        (1/2)            (1/2)       \  2         2      /   (1/2)  
    [   (q k)            (q k)                                   q       

           /  1         1      \    2                 ]
        exp|- - ln(q) + - ln(k)| q k      (1/2)  (5/2)]
           \  2         2      /         q      k     ]
      - -----------------------------, - -------------]
                      (1/2)                    (1/2)  ]
                 (q k)                    (q k)       ]

Now, when we hit these with various forms of simplify:

map( simplify, L2 );
    [   (1/2)  (5/2)     (1/2)  (5/2)   (1/2)   (1/2)     (1/2)  (5/2)  
    [  q      k         q      k       k       k         q      k       
    [- -------------, - -------------, ------, ------, - -------------, 
    [        (1/2)            (1/2)     (1/2)   (1/2)          (1/2)    
    [   (q k)            (q k)         q       q          (q k)         

         (1/2)  (5/2)]
        q      k     ]
      - -------------]
              (1/2)  ]
         (q k)       ]
map( simplify, L2, symbolic );
                    [           (1/2)   (1/2)          ]
                    [  2    2  k       k         2    2]
                    [-k , -k , ------, ------, -k , -k ]
                    [           (1/2)   (1/2)          ]
                    [          q       q               ]
map( simplify, L2 ) assuming q>0;
                    [           (1/2)   (1/2)          ]
                    [  2    2  k       k         2    2]
                    [-k , -k , ------, ------, -k , -k ]
                    [           (1/2)   (1/2)          ]
                    [          q       q               ]
map( simplify, L2 ) assuming k>0;
                    [           (1/2)   (1/2)          ]
                    [  2    2  k       k         2    2]
                    [-k , -k , ------, ------, -k , -k ]
                    [           (1/2)   (1/2)          ]
                    [          q       q               ]
eval( %, [q=-q0,k=-k0] );
              [                 (1/2)       (1/2)            ]
              [   2     2  (-k0)       (-k0)          2     2]
              [-k0 , -k0 , ----------, ----------, -k0 , -k0 ]
              [                 (1/2)       (1/2)            ]
              [            (-q0)       (-q0)                 ]

Interesting.

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

Strange, but more than likely true...

Doug Meade 3088
September 22 2008
0 0

I don't see a problem with the first example:

evalb( A^(1/2) *A^2 = A^(-3/2)*A^4 );
                                    true

The second one is not a problem if you tell Maple that q0 and k0 are negative (as you did in your third problem):

simplify((-q0)^(-1/2)*(-k0)^(1/2)*q0*k0^2*(q0*k0)^(-1/2)) assuming q0<0,k0<0;
                                       2
                                    -k0 
simplify((-q0)^(-1/2)*(-k0)^(-3/2)*q0*k0^4*(q0*k0)^(-1/2)) assuming q0<0,k0<0;
                                       2
                                    -k0

The important thing here is that you have to understand the rules of powers, and worry about branch cuts, when working with complex-valued variables (as Maple uses by default). The assumptions help here.

The third problem is the most interesting. I note that there is a common expression in each:

A := q0*k0^2/(q0*k0)^(1/2);
                                        2   
                                   q0 k0    
                                ------------
                                       (1/2)
                                (q0 k0)    

Then, extracting this common expression from both terms we are left with:

simplify(exp(-1/2*ln(-q0)+1/2*ln(-k0))) assuming q0<0,k0<0;
                                      (1/2)
                                 (-k0)     
                                 ----------
                                      (1/2)
                                 (-q0)     
simplify(exp(-1/2*ln(-q0))*exp(1/2*ln(-k0))) assuming q0<0,k0<0;
                                      (1/2)
                                 (-k0)     
                                 ----------
                                      (1/2)
                                 (-q0)    

This is as you would expect. Note that the assumptions do nothing in this case:

simplify(exp(-1/2*ln(-q0)+1/2*ln(-k0)));
                                      (1/2)
                                 (-k0)     
                                 ----------
                                      (1/2)
                                 (-q0)     
simplify(exp(-1/2*ln(-q0))*exp(1/2*ln(-k0)));
                                      (1/2)
                                 (-k0)     
                                 ----------
                                      (1/2)
                                 (-q0)    

But, when you include the common expression (in exactly the same literal form) the two expressions are now different:

simplify(exp(-1/2*ln(-q0)+1/2*ln(-k0))*A) assuming q0<0,k0<0;
                                       2
                                     k0 
simplify(exp(-1/2*ln(-q0))*exp(1/2*ln(-k0))*A) assuming q0<0,k0<0;
                                       2
                                    -k0

In this case, the assumptions seem to make matters worse. Without the assumptions, Maple sees these two expressions as being identical:

simplify(exp(-1/2*ln(-q0)+1/2*ln(-k0))*A);
                                 (1/2)      (5/2)
                            (-q0)      (-k0)     
                            ---------------------
                                       (1/2)     
                                (q0 k0)          
simplify(exp(-1/2*ln(-q0))*exp(1/2*ln(-k0))*A);
                                 (1/2)      (5/2)
                            (-q0)      (-k0)     
                            ---------------------
                                       (1/2)     
                                (q0 k0)         

Strange, very strange. I agree that there are some real inconsistencies here. But, I still believe all of this makes sense when viewed from the point of view of powers of complex quantities. (All of my results are with Maple 12.)

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

Strange, but more than likely true...

Doug Meade 3088
September 22 2008

I don't see a problem with the first example:

evalb( A^(1/2) *A^2 = A^(-3/2)*A^4 );
                                    true

The second one is not a problem if you tell Maple that q0 and k0 are negative (as you did in your third problem):

simplify((-q0)^(-1/2)*(-k0)^(1/2)*q0*k0^2*(q0*k0)^(-1/2)) assuming q0<0,k0<0;
                                       2
                                    -k0 
simplify((-q0)^(-1/2)*(-k0)^(-3/2)*q0*k0^4*(q0*k0)^(-1/2)) assuming q0<0,k0<0;
                                       2
                                    -k0

The important thing here is that you have to understand the rules of powers, and worry about branch cuts, when working with complex-valued variables (as Maple uses by default). The assumptions help here.

The third problem is the most interesting. I note that there is a common expression in each:

A := q0*k0^2/(q0*k0)^(1/2);
                                        2   
                                   q0 k0    
                                ------------
                                       (1/2)
                                (q0 k0)    

Then, extracting this common expression from both terms we are left with:

simplify(exp(-1/2*ln(-q0)+1/2*ln(-k0))) assuming q0<0,k0<0;
                                      (1/2)
                                 (-k0)     
                                 ----------
                                      (1/2)
                                 (-q0)     
simplify(exp(-1/2*ln(-q0))*exp(1/2*ln(-k0))) assuming q0<0,k0<0;
                                      (1/2)
                                 (-k0)     
                                 ----------
                                      (1/2)
                                 (-q0)    

This is as you would expect. Note that the assumptions do nothing in this case:

simplify(exp(-1/2*ln(-q0)+1/2*ln(-k0)));
                                      (1/2)
                                 (-k0)     
                                 ----------
                                      (1/2)
                                 (-q0)     
simplify(exp(-1/2*ln(-q0))*exp(1/2*ln(-k0)));
                                      (1/2)
                                 (-k0)     
                                 ----------
                                      (1/2)
                                 (-q0)    

But, when you include the common expression (in exactly the same literal form) the two expressions are now different:

simplify(exp(-1/2*ln(-q0)+1/2*ln(-k0))*A) assuming q0<0,k0<0;
                                       2
                                     k0 
simplify(exp(-1/2*ln(-q0))*exp(1/2*ln(-k0))*A) assuming q0<0,k0<0;
                                       2
                                    -k0

In this case, the assumptions seem to make matters worse. Without the assumptions, Maple sees these two expressions as being identical:

simplify(exp(-1/2*ln(-q0)+1/2*ln(-k0))*A);
                                 (1/2)      (5/2)
                            (-q0)      (-k0)     
                            ---------------------
                                       (1/2)     
                                (q0 k0)          
simplify(exp(-1/2*ln(-q0))*exp(1/2*ln(-k0))*A);
                                 (1/2)      (5/2)
                            (-q0)      (-k0)     
                            ---------------------
                                       (1/2)     
                                (q0 k0)         

Strange, very strange. I agree that there are some real inconsistencies here. But, I still believe all of this makes sense when viewed from the point of view of powers of complex quantities. (All of my results are with Maple 12.)

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

Human vs. Maple...

Doug Meade 3088
September 16 2008

Acer,

As a human, RootOf( cos(_Z)-_Z*sin(_Z)+1 ) is optimal.

But, Maple is not human and can't tell if this equation might have a simpler solution. In the process of putting the problem into standard form, extra solutions are allowed. This needs to be eliminated.

Robert Israel's post shows a quick way to bring g into the problem. My step-by-step derivation provides a clearer picture of the precise nature of the problems. When the equation is squared, an additional condition that f and g have the same sign at the solution. And, when the equation is divided by sin(x), care has to be taken to ensure that sin(x)<>0. So, maybe I'd expect Maple to return a result that says something like

RootOf( { sin(_Z)*_Z^2+sin(_Z)-2*_Z)
                  and signum(cos(_Z))=signum(_Z*sin(_Z)-1)
                  and sin(_Z)<>0 } )

I know this is not valid Maple syntax, but I think the idea is clear. There's probably a concise way to represent this, but as I stated at the outset this is not an area where I have much expertise.

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

I know I tried to be careful...

Doug Meade 3088
September 16 2008

I know I tried to be careful about what I wrote. Didn't I say that the roots of f that are not integer multiples of Pi are also roots of g? This does not say that all roots of g are also roots of f. In fact, my later analysis showing how g is obtained from f shows the origins of these extra roots. Look at the step where the equation is squared.

Subtle, I admit, but mathematically essential.

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

Relationship between f and g...

Doug Meade 3088
September 16 2008

I hadn't really given this any thought before now.

My first idea was to see what Maple could tell me about what it's doing when it's looking for roots of f.

f := cos(x)-x*sin(x)+1:
infolevel[all]:=10:
solve( f=0, x );
radnormal: entering radnormal at time 1.982
solve/rec/trig: trigonometric substitutions _S01 = sin(x) _S02 = cos(x)
Apply: solving for linear equation in _S02
factor/polynom: polynomial factorization: number of terms 3
factor/polynom: polynomial factorization: number of terms 3
PolynomialSystemMain: Polynomial system split into parts  1
PolynomialSystemMain: Skipping Groebner and trying resultant methods.
dosubs: normalize equations, length= 42
AdHocSolver: 52085 [1 3999999986 _S01 1] 2 1 42 0 3 0
[ + several more screens full of similar output ]

I'm not going to comment on the details in here. But, I do see that Maple appears to look at trigonometric substitutions. This got me to thinking, what if we:

isolate( f=0, cos(x) );
                            cos(x) = x sin(x) - 1           (1)

map( a->a^2, (1) );               # square both sides of equation
                                2                 2
                          cos(x)  = (x sin(x) - 1)          (2)

eval( (2), cos(x)^2=1-sin(x)^2 ); # eliminate all cos terms
                                  2                 2
                        1 - sin(x)  = (x sin(x) - 1)        (3)
expand( (3) );
                            2        2       2             
                  1 - sin(x)  = 1 + x  sin(x)  - 2 x sin(x) (4)
(4) - lhs((4));                   # move everything to one side
                         2       2                      2
                    0 = x  sin(x)  - 2 x sin(x) + sin(x)    (5)
factor( (5) ) / sin(x);           # simplify
                             2                      
                        0 = x  sin(x) + sin(x) - 2 x        (6)
And, voila, there is g.

g;
                           2                      
                          x  sin(x) + sin(x) - 2 x

The problems with this derivation are quite obvious. Squaring and simplifying can introduce spurious solutions. Here, it appears that the only real problem is avoiding the zeroes where sin(x)=0, i.e. all multiples of Pi. (The odd multiples are, in fact, solutions as is easily seen by inspection of the original problem: f(x)=0.)

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

 

graphical evidence...

Doug Meade 3088
September 16 2008

My response last night was rushed. If I had been thinking clearly I would have included a plot and not used so many words. Without further ado, please consider:

f := cos(x)-x*sin(x)+1:
g := x^2*sin(x)+sin(x)-2*x:
S := [[(1+2*k)*Pi,0] $ k=-4..4]:
P := plot( [f,g,S], x=-30..30, style=[line,line,point],
           color=[blue,green,red], symbol=circle,
           symbolsize=24, numpoints=200 ):
P;


plots[display]( P, view=[9..16,-5..5] );

The "zoom tool" is rather nice to use on the first plot. It certainly appears as though the solutions to f(x)=0 that are not integer multiples of Pi are also solutions to g(x)=0.

The error appears to be that g(x)=0 has other solutions, such as x=0, that are not solutions to f(x)=0.

This appears to be a very significant error. I hope it gets some attention at Maplesoft. But, now I'm starting to use to many words again.

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

More details for solve bug...

Doug Meade 3088
September 16 2008

Sure looks strange to me. Here's what I believe you did, without the troublesome images. First, without _EnvAllSolutions:

restart;
f := x->cos(x)-x*sin(x)+1;
x -> cos(x) - x sin(x) + 1
solve( f(x)=0, x );
                            /          2                 \
                  Pi, RootOf\sin(_Z) _Z  + sin(_Z) - 2 _Z/
evalf( %[2] );
                                     0.

and now with:

restart;
f := x->cos(x)-x*sin(x)+1;
x -> cos(x) - x sin(x) + 1
_EnvAllSolutions := true;
                                    true
solve( f(x)=0, x );
                                  /          2                 \
             Pi + 2 Pi _Z1, RootOf\sin(_Z) _Z  + sin(_Z) - 2 _Z/
evalf( %[2] );
                                     0.

The odd multiples of pi are correct. The other roots of f do occur at roots of x^2*sin(x)+sin(x)-2*x, but there are some roots of this equation that are not roots of f. So, yes, this does appear to be a bug.

Now, fsolve does do a decent job with this:

fsolve( f(x)=0, x );
                                -1.306542374
fsolve( f(x)=0, x, avoid={x=%} );
                                 1.306542374

I hope this helps others who might be better able to explain what is happening in this example.

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

you have what you need...

Doug Meade 3088
September 13 2008
0 0

Thank you for the additional information about your problem. My original response contains all of the information that you need. Since you are interested in finite sums, you might want to change sum to add (see the online help for the difference between the two commands). Also, since your function has two arguments, its defintion will start: f := (N,x) -> add( ..., n=1..N );

To create the plot of the function with N=1 and N=3 use: plot( [ f(1,x), f(3,x) ], x = lo .. hi ); where lo and hi are the endpoints of the domain on which you want to plot the functions.

If you need more help, show us what you have done.

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

you have what you need...

Doug Meade 3088
September 13 2008

Thank you for the additional information about your problem. My original response contains all of the information that you need. Since you are interested in finite sums, you might want to change sum to add (see the online help for the difference between the two commands). Also, since your function has two arguments, its defintion will start: f := (N,x) -> add( ..., n=1..N );

To create the plot of the function with N=1 and N=3 use: plot( [ f(1,x), f(3,x) ], x = lo .. hi ); where lo and hi are the endpoints of the domain on which you want to plot the functions.

If you need more help, show us what you have done.

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

overkill (infolevel[int]:=1000)...

Doug Meade 3088
September 12 2008
0 0

The maximum infolevel is 5, at least for library routines (including int).

If you are like me, you know that setting printlevel to large number is often productive, but infolevel is supposed to max out at 5 (see ?infolevel).

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

overkill (infolevel[int]:=1000)...

Doug Meade 3088
September 12 2008

The maximum infolevel is 5, at least for library routines (including int).

If you are like me, you know that setting printlevel to large number is often productive, but infolevel is supposed to max out at 5 (see ?infolevel).

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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