MaplePrimes Questions

I solve 100 differential equations to get the numerical solutions for specific independent variable x=1, how to plot those 100 points?

Hello,

I am wondering if anyone knows why Maple gives different results calling this directly and applying the definition.  For example:

 

int(1/(4*sin(x)-3*cos(x)), x = -(1/2)*Pi .. (1/2)*Pi, 'CauchyPrincipalValue')

-(2/5)*ln(2)

 

vs.

 

 

limit(Int(1/(4*sin(x)-3*cos(x)), x = -(1/2)*Pi .. arctan(3/4)-e)+Int(1/(4*sin(x)-3*cos(x)), x = arctan(3/4)+e .. (1/2)*Pi), e = 0, right)

-(2/5)*ln(2)-(1/5*I)*Pi

 

hey, does anyone know if I am using the solver and would like to make sure that all my outcomes are positive real number, is there a way to put in a constraint? thanks

I'm a new user, and here is what puzzled me.

The system of two linear equations are solved below.  Great!

Hi, Could anybody help me with the command that can check group multiplication of permutations in Maple? For instance, alpha=(1234)(56) and beta=(125)(36) So, alpha*beta=(135264) or beta*alpha=(153426) Not sure how to check it in Maple. thanx a lot. antonio

I've made this proc to solve the absolute orientation problem, but I'm not sure if its correct...

I've got two sets of points, one called 'Actual' and one called 'Ideal'. Each set has n points.
I want to find rotation matrix R and translation matrix t which I can apply to the points in Actual so that the mean squared error, e, between the points in Actual and Ideal would be minimum.
e = 1/n (Summation from i = 1 to n) ||Ideal[i] - R.Actual[i]||^2


centroid := proc(l::list)
> local i,Centroid;

For which value of c is the area between f(x) and g(x) minimal?  f(x) and g(x) are both continuous   ( in the interval [0,1] )        

> g := x-> piecewise(0<=x and x<=c, m1*x+b1, c<x and x<=1, m2*x+b2);

f := x-> x^3;

 

 

Good Morning everyone and happy easter

I have a short question on Maples Eigenvectors procedure. I want to diagonalize a Matrix A, which works works perfectly using the following code:

A:= some quadratic (complex) float matrix

(l,L):=Eigenvectors(A):

B:=L . DiagonalMatrix(l) . MatrixInverse(L):

 

Excecuting this code tells me, that A=B, so it works.

 

I recently installed Maple 11 and have just been looking at the Maplet Builder.  I'm having trouble with MathMLViewer.  In the simplest case, I created a maplet with 2 elements, a MathMLViewer and a button.  I set clickButton1 to evaluate the expression MathML[Export](1) with MathMLViewer1 as the target.  Then I ran the maplet and clicked the button, expecting to see 1 in the viewer.  Instead an error box titled Dispatch popped up saying, invalid input: XMLTools:-NSXML:-Parser:-processOutputString uses a 1st argument, str, which is missing. 

dsolve(diff(r(t),t$1)=-r(t)*(-1+r(t)^2));

                       1                                  1           
     r(t) = ------------------------, r(t) = - ------------------------

Here's an interesting problem from a colleague. Consider the following two improper integrals; the only difference is the absolute value in the integrand. I think this is another instance in which Maple mis-applies the FTOC.

restart;
with( IntegrationTools ):
#infolevel[all]:=3:
q := x*sin(1/x^2)-cos(1/x^2)/x;
                                            /1 \
                                         cos|--|
                                            | 2|
                                  /1 \      \x /
                             x sin|--| - -------
                                  | 2|      x   
                                  \x /          
I0 := Int( q, x=0..1 ):
I1 := Int( abs(q), x=0..1 ):

Maple gives values for both integrals.

value( I0 );
                                  1       
                                  - sin(1)
                                  2       
value( I1 );
                                  1       
                                  - sin(1)
                                  2       

It's a little odd that these two answers came out the same. It's easy to see that q does change signs infinitely often in [0,1]. A nice visualization of this can be obtained with:

plot([1/x,abs(q)],x=0..1,y=0..50);

We can dig a little deeper by looking at the values of the definite integrals on [A,1] and taking a limit as A->0.

int( q, x=A..1 );
                            1  2    /1 \   1       
                          - - A  sin|--| + - sin(1)
                            2       | 2|   2       
                                    \A /           
int(abs(q),x=A..1);
                        / 2    /1 \      /1 \\                   
                        |A  sin|--| - cos|--||                   
                        |      | 2|      | 2||                   
             1  2       |      \A /      \A /|    /1 \   1       
           - - A  signum|--------------------| sin|--| + - sin(1)
             2          \         A          /    | 2|   2       
                                                  \A /           

This starts to show where Maple is missing some important information. More is seen by looking at the antiderivatives of q and abs(q):

Q := int(q,x);
                                1  2    /1 \
                                - x  sin|--|
                                2       | 2|
                                        \x /
Q2 := int( abs(q), x );
                             / 2    /1 \      /1 \\        
                             |x  sin|--| - cos|--||        
                             |      | 2|      | 2||        
                  1  2       |      \x /      \x /|    /1 \
                  - x  signum|--------------------| sin|--|
                  2          \         x          /    | 2|
                                                       \x /

plot( [q, Q], x=0..1, view=[DEFAULT,-5..5] ):

plot( [abs(q), Q2], x=0..1, view=[DEFAULT,-1..5] );


plot( [Q,Q2], x=0..0.6, thickness=[3,1], discont=true );

Clearly, the two improper integrals at the start of this post should not have the same values.

Let's convert the problem to an improper integral on [1,infinity) by making a substitution x=1/u. When Maple makes this change of variable, it finds:

I2 := Change( I1, x=1/u );
                      /infinity       / 2\      / 2\  2   
                     |            -sin\u / + cos\u / u    
                   - |          - --------------------- du
                     |                      3             
                    /1                     u              

It's a little interesting that the absolute values have disappeared. (A plot quickly shows that this integrand is not always an increasing function.

q2 := GetIntegrand( Combine(I2) );
                                / 2\      / 2\  2
                            -sin\u / + cos\u / u 
                            ---------------------
                                      3          
                                     u           
plot( q2, u=1..50 );

What gets really interesting is the value of this integral:

 

value( I2 );
                                   1       
                                 - - sin(1)
                                   2       

Maple returns the same value for the definite integral with the correct (absolute value) in the integrand. Once again it is surprising that Maple reports the same values for these two integrals. But, the fact that these values are negative - even when the integrand is non-negative - is alarming.

It does appear that in some cases Maple is able to detect and correctly handle an infinite number of discontinuities in the antiderivative:

int(abs(cos(v)/v),v=1..infinity);
                /1   \
     -Ci(1) + Ci|- Pi|
                \2   /

          /infinity                                                   \
          | -----                                                     |
          |  \                                                        |
          |   )    /  /1           \     _k     /3           \     _k\|
        + |  /     |Ci|- Pi + Pi _k| (-1)   - Ci|- Pi + Pi _k| (-1)  ||
          | -----  \  \2           /            \2           /       /|
          \ _k = 0                                                    /
evalf( % );
                               Float(infinity)

 

 

Okay, I have a simple 5 X 5 matrix.  How do I view it as an image?

 

With a very large image 1000 X 1000, I can left click on it and view the matrix in a spreadsheet and then click on a tab that says image and view that matrix as an image.  How do I do that with small maticies?  I can't figure it out.

I have an  ode system, and use  2 D Math  (I don't know how to copy here, so I type by hand)

x'(t) = -y(t) + x(t) (1 - x^2(t)  -  y^2(t)),

y'(t) = x(t) + y(t) (1 - x^2(t) - y^2(t)).

I would like to introduce  x(t) = r(t) cos(Theta(t)), y(t) = r(t) sin(Theta(t)) and manipulate the equations further.

The problem is the left hand side. In one case I obtain Dx(t), in the other case (r(t) cos(Theta(t))' without calculation on it.

I ask some help. Thanks.

Hi guys

 

How can i construct the standard vector e[i]. Where e[i] is a 4 dimensional vector with 0 in all entries apart from the ith which has ?

Hello there, Would appreciate if someone could illustrate the procedure to obtain, using the plot3d command, simply the solid bound by the given set of graphs. In other words, all superfluous areas are substracted. For example, for the cone z=sqrt(x^2 + y^2) & planes z=2 & z=4, I'd like to see just the solid bound by the planes and not what appears when the following commands are executed: a:=implicitplot3d(z = sqrt(x^2+y^2), x = -5 .. 5, y = -5 .. 5, z = -5 .. 5); b:=implicitplot3d(z = 2, x = -5 .. 5, y = -5 .. 5, z = -5 .. 5);
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