Maple Questions and Posts

These are Posts and Questions associated with the product, Maple

I'm running calculations like this:

    N:=10000;
    f := (i,j)-> (some complicated procedure depending on i and j);
    M:= Matrix([Threads:-Seq([Threads:-Seq( f(i,j), j=1..N)], i=1..N)]);

I have a server with 20 cores, but each core has two threads, so this code should max out all 40 threads. But what I notice is only at most 20 threads being used at a time. 

I checked kernelopts(numcpus) returns 20. 

Does anyone have any advice on how to maximize my resource usage?

Hi All,

 

Where can I download Fourier Series Package? I found other posts related to this but the link is not working.

 

Thanks.

 

I have the following sequence: n2/n2+31n+228.The question asks me to use an appropriate conditional statement to find N such that absolute (xn-x0)<epsilon for every n>=N. I found N but i don't know how to find it using Maple17.Can you please help me?

 

You, I, and others like us, are the beneficiaries of decades of software evolution.

From its genesis as a research project at the University of Waterloo in the early 80s, Maple has continually evolved to meet the challenges of technical computing.

Hi all!

as shown below, how can get a result without 'R':

p_com(z,t):=Re(exp(I*omega*t-I*k*(lambda[r]+I*lambda[i])*z)) assuming omega::real,t::real,k::real,lambda[r]::real,lambda[i]::real,z::real

Thanks very much!

 

I wrote the following procedure that evaluates i-th B-spline basis function of degree n over the knot vector T:

N := proc(i, n, t, T)

if (n = 0) then
    if ((t >= T[i]) and (t < T[i+1])) then
        return 1.0:
    else
        return 0.0:
    fi:
fi:

return (evalf((t-T[i])/(T[i+n]-T[i]))*N(i, n-1, t, T) + evalf((T[i+n+1]-t)/(T[i+n+1]-T[i+1]))*N(i+1, n-1, t, T)):

end proc:

 

Now, I want to compute

int(N(i, n, t, T)*N(j, n, t, T), t=0..1, numeric),

where i, n and T are given. However, Maple evaluates N and, since t is unknown, I get the following error:

"Error, (in N) cannot determine if this expression is true or false: .1 <= t and t < .25"

In http://www.maplesoft.com/support/help/Maple/view.aspx?path=evalf/Int  I read that in order to integrate a procedure, one should write (see the example given there)

evalf(Int(N*N, 0..1));

however, in my case, It will not work because I must pass the known parameters i, n and T. Is there a way to solve my problem?

 

 

 

 

Hi Guys, 

I'm trying to find stationary points of a numeric function. Any help would be appreciated. 

Assume I have a numeric function g(x). I'm attempting the following:

deriv:=(x)->fdiff(g(t),t=x); (I can't use the D operator as it doesn't like g(x))

Now deriv(x) <- Can be computed and computes the value quickly and easily. However plot and other functions require algebraic functions. I can plot this (with difficulty) via the following

plot(deriv,-10..10) <- This avoids converting to algebraic function and runs as a procedure. 

I can't use convert(deriv,algebraic) as it fails. 

I want to use the Student[NumericalAnalysis]Roots command, but it requires an algebraic function and I can't use the procedure trick which I did in the plot command. 

Does anyone know a better way of doing this? or is there a way I can write the numerical differivative as an algebraic function. (I've tried fsolve <- But this guy doesn't give me the correct answers generally). 

Thanks guys. 

 

 

 

Hey,


I want to assign a value to a symbol stored in a vector. I know the position of the symbol in the vector. Is there an easy way to do this?

Here to illsutrate my problem:

> 

restart:

> 

vec:=<a,b,c>:

> 

vec(1):=1;    # expected result: a:=1

vec := Vector(3, {(1) = 1, (2) = b, (3) = c})

(1)
> 

a;            # expected result: 1

a

(2)
> 

 

 

Download point_to_element.mw

 

 


Why does "eval(c, [b=1,l=1])" turn out to be "1" not "1/1+x"??? its driving me cracy. THX

> 

restart

> 

 

> 

a:=(1/(1+x/l))

1/(1+x/l)

(1)
> 

eval(a, [l=1])

1/(1+x)

(2)
> 

eval(a, [x=2,l=1])

1/3

(3)
> 

c:=(1/b(1+x/l))

1/b(1+x/l)

(4)
> 

eval(c, [b=1,l=1])

1

(5)
> 

 



 

Hello,

Still on the thematic on simplification of trigonometric expression.

I would like to simplify this equation. Normally, for a mecanical point of view, this equation could be simplified a lot and namely the psi[1](t) and theta[1](t) variables should disappear.

The difference with the former posts is the fact that now each term (for example  2*sin(gamma0(t))*z0(t)*cos(beta0(t))*xb[1]) can regroup 2 terms in factor with the trigonometric part.

eq:=l2[1]^2 = 2*sin(gamma0(t))*z0(t)*cos(beta0(t))*xb[1]-2*sin(gamma0(t))*zp[1](t)*cos(beta0(t))*xb[1]+2*sin(gamma0(t))*y0(t)*sin(alpha0(t))*zb[1]-2*sin(gamma0(t))*yp[1](t)*sin(alpha0(t))*zb[1]+2*sin(gamma0(t))*x0(t)*cos(alpha0(t))*zb[1]-2*sin(gamma0(t))*xp[1](t)*cos(alpha0(t))*zb[1]-2*cos(gamma0(t))*z0(t)*cos(beta0(t))*zb[1]+2*cos(gamma0(t))*zp[1](t)*cos(beta0(t))*zb[1]+2*cos(gamma0(t))*y0(t)*sin(alpha0(t))*xb[1]-2*cos(gamma0(t))*yp[1](t)*sin(alpha0(t))*xb[1]+2*cos(gamma0(t))*x0(t)*cos(alpha0(t))*xb[1]-2*cos(gamma0(t))*xp[1](t)*cos(alpha0(t))*xb[1]+2*y0(t)*cos(alpha0(t))*cos(beta0(t))*yb[1]-2*yp[1](t)*cos(alpha0(t))*cos(beta0(t))*yb[1]-2*x0(t)*sin(alpha0(t))*cos(beta0(t))*yb[1]+2*xp[1](t)*sin(alpha0(t))*cos(beta0(t))*yb[1]-2*sin(psi[1](t))*cos(theta[1](t))*l3[1]*xb[1]+2*sin(psi[1](t))*sin(theta[1](t))*l3[1]*zb[1]-2*cos(theta[1](t))*cos(psi[1](t))*l3[1]*zb[1]-2*cos(psi[1](t))*sin(theta[1](t))*l3[1]*xb[1]-2*sin(gamma0(t))*y0(t)*sin(alpha0(t))*cos(theta[1](t))*cos(psi[1](t))*l3[1]-2*sin(gamma0(t))*yp[1](t)*sin(alpha0(t))*sin(psi[1](t))*sin(theta[1](t))*l3[1]+2*sin(gamma0(t))*yp[1](t)*sin(alpha0(t))*cos(theta[1](t))*cos(psi[1](t))*l3[1]+2*sin(gamma0(t))*x0(t)*cos(alpha0(t))*sin(psi[1](t))*sin(theta[1](t))*l3[1]-2*sin(gamma0(t))*x0(t)*cos(alpha0(t))*cos(theta[1](t))*cos(psi[1](t))*l3[1]-2*sin(gamma0(t))*xp[1](t)*cos(alpha0(t))*sin(psi[1](t))*sin(theta[1](t))*l3[1]+2*sin(gamma0(t))*xp[1](t)*cos(alpha0(t))*cos(theta[1](t))*cos(psi[1](t))*l3[1]-2*cos(gamma0(t))*z0(t)*cos(beta0(t))*sin(psi[1](t))*sin(theta[1](t))*l3[1]+2*cos(gamma0(t))*z0(t)*cos(beta0(t))*cos(theta[1](t))*cos(psi[1](t))*l3[1]+2*cos(gamma0(t))*zp[1](t)*cos(beta0(t))*sin(psi[1](t))*sin(theta[1](t))*l3[1]-2*cos(gamma0(t))*zp[1](t)*cos(beta0(t))*cos(theta[1](t))*cos(psi[1](t))*l3[1]-2*cos(gamma0(t))*y0(t)*sin(alpha0(t))*sin(psi[1](t))*cos(theta[1](t))*l3[1]-2*cos(gamma0(t))*y0(t)*sin(alpha0(t))*cos(psi[1](t))*sin(theta[1](t))*l3[1]+2*cos(gamma0(t))*yp[1](t)*sin(alpha0(t))*sin(psi[1](t))*cos(theta[1](t))*l3[1]+2*cos(gamma0(t))*yp[1](t)*sin(alpha0(t))*cos(psi[1](t))*sin(theta[1](t))*l3[1]-2*cos(gamma0(t))*x0(t)*cos(alpha0(t))*sin(psi[1](t))*cos(theta[1](t))*l3[1]-2*cos(gamma0(t))*x0(t)*cos(alpha0(t))*cos(psi[1](t))*sin(theta[1](t))*l3[1]+2*cos(gamma0(t))*xp[1](t)*cos(alpha0(t))*sin(psi[1](t))*cos(theta[1](t))*l3[1]+2*cos(gamma0(t))*xp[1](t)*cos(alpha0(t))*cos(psi[1](t))*sin(theta[1](t))*l3[1]+yb[1]^2+xb[1]^2+zb[1]^2+l3[1]^2+z0(t)^2+zp[1](t)^2+y0(t)^2+yp[1](t)^2+x0(t)^2+xp[1](t)^2+2*z0(t)*sin(beta0(t))*yb[1]-2*zp[1](t)*sin(beta0(t))*yb[1]-2*z0(t)*zp[1](t)-2*y0(t)*yp[1](t)-2*x0(t)*xp[1](t)-2*sin(gamma0(t))*y0(t)*cos(alpha0(t))*sin(beta0(t))*xb[1]+2*sin(gamma0(t))*yp[1](t)*cos(alpha0(t))*sin(beta0(t))*xb[1]+2*sin(gamma0(t))*x0(t)*sin(alpha0(t))*sin(beta0(t))*xb[1]-2*sin(gamma0(t))*xp[1](t)*sin(alpha0(t))*sin(beta0(t))*xb[1]+2*cos(gamma0(t))*y0(t)*cos(alpha0(t))*sin(beta0(t))*zb[1]-2*cos(gamma0(t))*yp[1](t)*cos(alpha0(t))*sin(beta0(t))*zb[1]-2*cos(gamma0(t))*x0(t)*sin(alpha0(t))*sin(beta0(t))*zb[1]+2*cos(gamma0(t))*xp[1](t)*sin(alpha0(t))*sin(beta0(t))*zb[1]-2*sin(gamma0(t))*z0(t)*cos(beta0(t))*sin(psi[1](t))*cos(theta[1](t))*l3[1]-2*sin(gamma0(t))*z0(t)*cos(beta0(t))*cos(psi[1](t))*sin(theta[1](t))*l3[1]+2*sin(gamma0(t))*zp[1](t)*cos(beta0(t))*sin(psi[1](t))*cos(theta[1](t))*l3[1]+2*sin(gamma0(t))*zp[1](t)*cos(beta0(t))*cos(psi[1](t))*sin(theta[1](t))*l3[1]+2*sin(gamma0(t))*y0(t)*sin(alpha0(t))*sin(psi[1](t))*sin(theta[1](t))*l3[1]+2*sin(gamma0(t))*y0(t)*cos(alpha0(t))*sin(beta0(t))*sin(psi[1](t))*cos(theta[1](t))*l3[1]+2*sin(gamma0(t))*y0(t)*cos(alpha0(t))*sin(beta0(t))*cos(psi[1](t))*sin(theta[1](t))*l3[1]-2*sin(gamma0(t))*yp[1](t)*cos(alpha0(t))*sin(beta0(t))*sin(psi[1](t))*cos(theta[1](t))*l3[1]-2*sin(gamma0(t))*yp[1](t)*cos(alpha0(t))*sin(beta0(t))*cos(psi[1](t))*sin(theta[1](t))*l3[1]-2*sin(gamma0(t))*x0(t)*sin(alpha0(t))*sin(beta0(t))*sin(psi[1](t))*cos(theta[1](t))*l3[1]-2*sin(gamma0(t))*x0(t)*sin(alpha0(t))*sin(beta0(t))*cos(psi[1](t))*sin(theta[1](t))*l3[1]+2*sin(gamma0(t))*xp[1](t)*sin(alpha0(t))*sin(beta0(t))*sin(psi[1](t))*cos(theta[1](t))*l3[1]+2*sin(gamma0(t))*xp[1](t)*sin(alpha0(t))*sin(beta0(t))*cos(psi[1](t))*sin(theta[1](t))*l3[1]+2*cos(gamma0(t))*y0(t)*cos(alpha0(t))*sin(beta0(t))*sin(psi[1](t))*sin(theta[1](t))*l3[1]-2*cos(gamma0(t))*y0(t)*cos(alpha0(t))*sin(beta0(t))*cos(theta[1](t))*cos(psi[1](t))*l3[1]-2*cos(gamma0(t))*yp[1](t)*cos(alpha0(t))*sin(beta0(t))*sin(psi[1](t))*sin(theta[1](t))*l3[1]+2*cos(gamma0(t))*yp[1](t)*cos(alpha0(t))*sin(beta0(t))*cos(theta[1](t))*cos(psi[1](t))*l3[1]-2*cos(gamma0(t))*x0(t)*sin(alpha0(t))*sin(beta0(t))*sin(psi[1](t))*sin(theta[1](t))*l3[1]+2*cos(gamma0(t))*x0(t)*sin(alpha0(t))*sin(beta0(t))*cos(theta[1](t))*cos(psi[1](t))*l3[1]+2*cos(gamma0(t))*xp[1](t)*sin(alpha0(t))*sin(beta0(t))*sin(psi[1](t))*sin(theta[1](t))*l3[1]-2*cos(gamma0(t))*xp[1](t)*sin(alpha0(t))*sin(beta0(t))*cos(theta[1](t))*cos(psi[1](t))*l3[1]

Do you have some ideas so as to simplify this equation ?

N.B : Former posts on the topic of trigonometric simplification

http://www.mapleprimes.com/questions/209884-Simplification-Of-Trigonometric-Expression-II

http://www.mapleprimes.com/questions/209721-Simplification-Of-Trigonometric-Expressions

I put a worksheet attached in order to facilitate the troubleshooting.

Thanks a lot for your help

trigonometric_simplification.mw

 

Solution.mw 
 
 

Download Solution.mw

 

Hi all, if there anyone eho couyld help me with this difficult problem. I couldn't solve the attached nonhomogeneous equation...

But I found one series solution which doesn't satisfy the command: odetest (Solution, ode)

Hi. I'm hacing trouble writing a maple procedure for the question below, can anyone help?

 

Write a maple procedure which takes as its input the vectoeat u1 and u2 and the eigenvectors lambda1 and lambda2 where u1,u2 are element of R^2 and the lambdas are real numbers.

If u1,U2 is linearly independent then the output is the matrix A an element of R^2x2 with the property that Au1= lambda1u1 and AU2=lambda2u2;

if u1,u2 is linearly dependent then the output is the statement "not an eigenbasis".

 

I I then have two inputs which I have to do but I'm not sure on how to write the procedure. Any help will be much appreciated.  

 

Thanks :)

 

 

Hi dear Maple masters:

Excuse-me. I have a stupid question to ask: how to extract symbolic coefficient in maple? For example, I would like to get the coefficient before sin(Ωt) and cos(Ωt) in the following equation:

eq := (-Omega^2*a*A[2]-Omega^2*m*B[1]+Omega*A[1]*c[1]+B[1]*k[1])*cos(Omega*t)+(Omega^2*a*B[2]-Omega^2*m*A[1]-Omega*B[1]*c[1]+A[1]*k[1])*sin(Omega*t) = 0;

Thank you in adavance for taking a look, wish you a nice day!

Best regards,

Zihan

Hi dear friends. Recently, I am troubled by a question of how to draw an exictation-response diagram in Maplesoft.

I would like to draw an excitation-response diagram with following equation, Ω is greatrer or equal to 0:

(0.1299e-4*(17.1740*Omega^2-1.570200000*10^6))*Omega^2/(-196.1270800*Omega^4+3.954121290*10^7*Omega^2-1.877174100*10^12);

 

But, when I would like to use Bode diagram in Maple to draw it, it looks very differently.

Maybe I use a wrong function to draw it.


In the book, the diagram takes the form:

 

Thanks a lot for taking a lookDiagram in book.

Looking everywhere for this.  Hard to find.  Trying to shade under the curve of $2=x^{.5}+y^{.5}$ from x=0 to 1.  Here is what I found and tried:

 

A:=implicitplot(x^.5+y^.5=2,x=-1..2,y=-1..2,thickness=3,color=blue):
B:=implicitplot(x^.5+y^.5=2,x=0..1,filled=true,color=yellow,view[0..1,-1..1]):
plots[display](A,B, scaling=constrained);

 

 

no idea what the view does.  Any help would be much appreciated. 

 

Nick

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