Hi
I wrote a loop which is too long but I need its continuity so I wrote all of them in one loop.
The problem is too much time is needed to run it,even maple cannot run it.
I do not know whether my software has a problem or the way I wrote the program is wrong.
I attached the file.
thanks
result2_with_op90_to.mw

Hi! I have been searching the forum, but with no luck.  How do I get maple to handle graphs with multiple edges between vertices?  I can't seem to find a function for this.  Does one exist?  I tried to do this by coloring the edges, but they just get re-colored.

EE1:=[1, 2], [3, 1], [2, 3], [4, 5], [6, 4], [5, 6], [7, 8], [9, 7], [8, 9];
EE2:=[2, 3], [4, 2], [3, 4], [5, 6], [7, 5], [6, 7], [1, 8], [9, 1], [8, 9];
EE3:=[3, 4], [5, 3], [4, 5], [6, 7...

Not sure how to do this.  What I want to do is use maple to grab an RSS feed from the internet and use maple to put it into a readable format

In particular from this website http://tomorrowsgaspricetoday.com/city/Hamilton/ and getting the rss feed xml style webpage from here http://tomorrowsgaspricetoday.com/rssfeed.php?city=16

phi:=array(1..3):

for q to 3 do

     phi[q] := x->p q x;

end do

Output

       phi[1]:= 

          phi[2]:=

How to solve the following system of differentials equations using Maple:

f'' - g' + g =sin x

g' - g + f = cos x

subject to the initial conditions:

f(0)= 2, g(0)=1 , f'(0)=-1

Earlier this week a colleagues approached me with a problem he was trying to solve with Maple.

The basic problem is to compute the asymptotic expansion of an expression. It's not difficult to see the leading term of this expansion, but Maple's result does not match this expectation. Let me demonstrate:

B := (n,h) -> (n+h)!/r^(n+h) * (1+(P0+h*P1)*r/n);
                           /    (P0 + h P1) r\
          factorial(n + h) |1 + -------------|
                           \          n      /
(n, h) -> ------------------------------------
                         (n + h)              
                        r                     

simplify( asympt( B(n,1)/B(n,0)-1, n, 3 ) );
                            n + O(1)

asympt( simplify( B(n,1)/B(n,0)-1 ), n, 3 );
                                   2                          
 n   1 + r P1 - r   P0 r + r P1 - r  P0 + (-1 - r P1 + r) P0 r
 - + ------------ + ------------------------------------------
 r        r                            r n                    

      /            2   \           
      \-r P1 + P0 r  P1/ P0    /1 \
    + --------------------- + O|--|
                2              | 3|
               n               \n /

The real issue is the drastically different results from the previous two commands.  The only difference in the commands is the explicit simplify before or after the call to asympt.

It would be best if this last expression could be seen in the form:

(n+1)/r + (P1-1) + (1-r*P0)*P1/n - r*(1-r*P0)*P0*P1/n^2 + O(1/n^3);
  n + 1            (1 - P0 r) P1   r (1 - P0 r) P0 P1    /1 \
  ----- + P1 - 1 + ------------- - ------------------ + O|--|
    r                    n                  2            | 3|
                                           n             \n /

What's the difference in the two uses of asympt? This simpler example appears to show the essence of what is going on. On one hand,

B := (n,h) -> 1/r^(n+h);
             1    
(n, h) -> --------
           (n + h)
          r       

simplify( asympt( B(n,1)/B(n,0)-1, n, 3 ) );
Error, (in asympt) unable to compute series

while on the other hand

asympt( simplify( B(n,1)/B(n,0)-1 ), n, 3 );
                              r - 1
                            - -----
                                r  

Hey guys, I'm going to go ahead and post the whole problem along with some of the work that I've done so that it won't look like I'm letting somebody else complete the entire problem. Any help would be appreciated. Here goes:

Tom and Jerry are two snail brothers that need to travel five miles to a lake. Tom travels 1 mile on the first day, 1/2 on the second day, 1/3 on the third day and progresses 1/n miles each day. Jerry travels 2 miles on the first...

Hi, everyboyd I wan to draw a vertical line at theta:=Pi  for any function.

How can i do it.

Amr

hi.my program is this=

a:=[[2.5,0.05],[-2.5,1],[0,1],[-1.5,1],[-1,2],[1,2],[1.5,1],[0,1],[2.5,1],[2.5,0.05],[-2.5,0.05]]

plot(a)

r:=plot(a)

b:=[[0,2],[0,1]]

plot(b)

v:=plot(b)

c:=plot([-1+1/2*sin(t),1/2*cos(t),t=0..2Pi],scaling=constrained])

d:=  plot([1+1/2*sin(t),1/2*cos(t),t=0..2Pi],scaling=constrained]) ...

Hello,

I have this problem. I would like to construct clothoid as a transition curve between two arcs. The length of this curve has to be 30 meters. The start azimuth is 0.3773 (rad) and the start radius is -355.835 m.

How could I do that simply in Maple?

Thank you

zakyn

Hi everyone.. Now, i learn Maplesim.. I want measures force between 2 frame(a and b). which are connected by prismatic joint.

But a force and moment sensor acts as a weld joint.. so althought i measured force but frame b can't move.

who can solve my problem??

thank very much!!!

PS: i want measure force 2 frame to calculate the friction between them.

Hi,

I would like to apply watson lemma on f(t) = 1/((t+1)*((t/a)-1)), with a a constant, and x tending to infinity.

The problem is that my integral I have to compute is not from 0, but from a....

My integral is : I = int (exp(-x*t)*(t^(λ))*f(t), t=a..∞), with λ>-1, x->∞.

I know that if a=0, and (taylor expansion) f(t) = ∑ b_n*(t^n), n=0..∞, then

I = ∑ b_n*Γ(λ+n+1)/(x^(λ+n+1)), n=0..∞.

Hi,

I'm trying to do get the Inverse of Matrix A, I tried this:

> restart;
> with(LinearAlgebra);
> A := Matrix(4, 4, [[1, 0, 0, 0], [0, 1, 0, 0], [1, l, l^2, l^3], [0, 1, l, l^2]]);
                     Matrix(%id = 83774380)
> B := MatrixInverse(A);
Error, (in LinearAlgebra:-MatrixInverse) singular matrix

 And got an error, could someone help me.

Thanks

I have a matrix m with 9 rows and 4 columns and when I enter the command below I get an error message:

num := product((z-(m[s,1]+I*m[s,2])),s = 1..9);
Error, Matrix index out of range

However, there is no problem if I use brute force:

If I extract the denominator of the following expression

> s:=(1/2/Pi+Q*(1-u^2)/(1+u^2))^2/((1+beta^2*(1/2/Pi+Q*(1-u^2)/(1+u^2))^2)*(1+u^2));

maple expands it automatically

s2:=denom(s);

but the result of solving the expanded denominator

solve(s2=0,u);

is different to the result of solving the non-expanded expression