MaplePrimes Questions

Dear friends, I have made a doc using embedded components to plot a curve. What commands should I do to animate (and pause and play using a custom play/pause button inserted in the table) the plot. Normally I have to use explore animate command, but it requires the user to use animation tools. There must be a way to do the same using play = true; loop = true commands. I donot know its use. Can any one help. Thanks. 

 

use DocumentTools in

Do(q = %MathContainer0);
Do(p = eval(q));
# Do(%Plot0 = plot(p, x = 0..2*Pi));
Do(%Plot0 = plot(p, x = 0..10));
end use;


 

 

 Write the expression in textarea and click the button to see the plot in plot area and expression in math container

      NULL

 

 

``

NULL


Thanks for helping out. Ramakrishnan V

Download animation_CommandPlease.mw

I am trying to find local extrema for multi variable functions. Using Hessian matrix and eigen values I am able to find the global extrema. Also by using boundary conditions I am able to solve for critical points with in given domain.

But, when there is no maxima or minima inside a local domain, It is believed to be minima/maxima lies on one of the boundaries(that point cannot be a critical point). I want to add this boundary points to the list of critical points

Example:

Term := x^3+x^2*y-2*y^3+6*y;
critical points := [[x = .6928203232, y = -1.039230485], [x = -.6928203232, y = 1.039230485], [x = 0., y = -1.], [x = 0., y = 1.]]
boundaries := [[-1<=x],[ x<=1], [-1<=y], [y<=1]];

now I want to read the boundaries as input and get the output as
boundaries :=[[x = -1,y =0],[x = 1,y =0],[x = 0,y =-1],[x = 0,y =1]];

When I did this manually I observed boundary points are saddle, since eigen values are mixed positive and negitive. But, my interest is to find the function value at boundaries.

Note: I believing value of other variables at perticular boundary is zero. In general I have to deal with multivariable functions with more than 3 variable.


Thank you



 

Hi,

Is it possible to find where a parametric intersects with itself?

For example, lets have the parametric:

[sin(6t),cos(2t)] 

For what value values of t will it yield the same (x,y) cordinate. 

Suppose we have functions dependent on certain variables.  I don't want the dependencies explicitly shown in the output.  How do I hide them?

For example

f(x)+f(x)^2-(diff(y(t), t)) = diff(y(t), t, t)
                         

I would like just f not f(x) and similarily just y instead of y(t) shown.

The third execution of procedure TP returns infinity but it has a real value in the integral's plot. Why is this?

Int_Question.mw


I have an object in 6d I'd like to visualise. The region of 6d space I am interested in is described by these equations:

{f[10] = -(.2000000000*(5.*f[21]*f[20]*f[22]-5.*f[20]*f[22]^2+20.*f[20]*f[21]-20.*f[20]*f[22]+135.*f[20]+46.*f[21]))/(f[21]*(f[21]-1.*f[22])),
f[11] = -1.*f[22]-4.,
f[12] = -(1.*(f[22]^2+4.*f[22]-27.))/f[21],
f[20] = f[20],
f[21] = f[21],
f[22] = f[22]}

clearly the first three variables are dependant, and the latter three are independant. I'd like to graph the first three as the latter three vary between bounds and then colour the points on the output based on where they came from in the input, so i can get some intuition about what these equations mean.
 

Consider f is a polynomial which is constructed from some polynomials. In other words, f=g_1^{k_1}*g_2^{k_2}*...*g_n^{k_n} where  g_1,g_2,..,g_n are some polynomials and also  k_1,k_2,..,k_n are positive integer numbers.

My question: How to define a procedure such that the output of proc(f) is the following list [g_1 , g_2 , ... g_n]. In fact, the procedure separates the factors of the polynomial f into  a list and also removes the powers of the factors. 

For example, suppose that f=x*(x+1)^4*(x^2+x+1)*(x^3+x^2+1)^5. Then, the output of the proc(f) be as follows [x , x+1 , x^2+x+1 , x^3+x^2+1].

Thanks in advance

I wonder if it is possible to automatically obtain the well-known  binomial theorem  for an arbitrary integer and a positive exponent  n  in Maple. The expansion  (1)  below  I want to get in Maple automatically. But all my attempts were unsuccessful:
 

restart;

(a+b)^n=Sum(binomial(n,k)*a^(n-k)*b^k, k=0..n);  # The binomial theorem

(a+b)^n = Sum(binomial(n, k)*a^(n-k)*b^k, k = 0 .. n)

(1)

expand((a+b)^n) assuming n::posint;
convert((a+b)^n, Sum) assuming n::posint;
convert((a+b)^n, polynom) assuming n::posint;
convert((a+b)^n, binomial) assuming n::posint;

 

(a+b)^n

 

(a+b)^n

 

(a+b)^n

 

(a+b)^n

(2)

 


 

Download binomial_theorem.mw

I am trying to price an option with the explicit fnite difference method. The code works up until   but does not run after that. Any help would be greatly appreicated 


I tried to solve this double integral in Maple but didn't get the solution so how can I solve this?

Please help

restart

a := .5

.5

(1)

b := .6

.6

(2)

R := 0.1e-1

0.1e-1

(3)

E := 210000000000/(1-.3^2)

0.2307692308e12

(4)

h := a/(2*sqrt(-2*y^2+1))

.2500000000/(-2*y^2+1)^(1/2)

(5)

F1 := sqrt(tan((1/2)*Pi*h)/((1/2)*Pi*h))*[.752+2.02*h+.37*(1-sin((1/2)*Pi*h))^3]/cos((1/2)*Pi*h)

2.828427125*(tan(.1250000000*Pi/(-2*y^2+1)^(1/2))*(-2*y^2+1)^(1/2)/Pi)^(1/2)*[.752+.5050000000/(-2*y^2+1)^(1/2)+.37*(1-sin(.1250000000*Pi/(-2*y^2+1)^(1/2)))^3]/cos(.1250000000*Pi/(-2*y^2+1)^(1/2))

(6)

F4 := sqrt(tan((1/2)*Pi*h)/((1/2)*Pi*h))

2.828427125*(tan(.1250000000*Pi/(-2*y^2+1)^(1/2))*(-2*y^2+1)^(1/2)/Pi)^(1/2)

(7)

C1 := (1/2)*u*(F1^2+1.3*F4^2)/(Pi*E*R)

0.2166666666e-9*u*(8.000000001*tan(.1250000000*Pi/(-2*y^2+1)^(1/2))*(-2*y^2+1)^(1/2)*[.752+.5050000000/(-2*y^2+1)^(1/2)+.37*(1-sin(.1250000000*Pi/(-2*y^2+1)^(1/2)))^3]^2/(Pi*cos(.1250000000*Pi/(-2*y^2+1)^(1/2))^2)+10.40000000*tan(.1250000000*Pi/(-2*y^2+1)^(1/2))*(-2*y^2+1)^(1/2)/Pi)/Pi

(8)

B := int(int(C1, u = 0 .. a), y = -b .. b)

int(0.2195292312e-10*tan(.3926990817/(-2.*y^2+1.)^(1/2))*(-2.*y^2+1.)^(1/2)*[.7520000000+.5050000000/(-2.*y^2+1.)^(1/2)+.3700000000*(1.-1.*sin(.3926990817/(-2.*y^2+1.)^(1/2)))^3]^2/cos(.3926990817/(-2.*y^2+1.)^(1/2))^2+0.2853880005e-10*tan(.3926990817/(-2.*y^2+1.)^(1/2))*(-2.*y^2+1.)^(1/2), y = -.6 .. .6)

(9)

evalf(B)

int(0.2195292312e-10*tan(.3926990817/(-2.*y^2+1.)^(1/2))*(-2.*y^2+1.)^(1/2)*[.7520000000+.5050000000/(-2.*y^2+1.)^(1/2)+.3700000000*(1.-1.*sin(.3926990817/(-2.*y^2+1.)^(1/2)))^3]^2/cos(.3926990817/(-2.*y^2+1.)^(1/2))^2+0.2853880005e-10*tan(.3926990817/(-2.*y^2+1.)^(1/2))*(-2.*y^2+1.)^(1/2), y = -.6 .. .6)

(10)

 

(11)

``


 

Download mapleprime.mw

Can Maple perform huge calculations via parallel-pooling (i.e working on multiple or all CPUs) similar to Matlab?

verify(2*cos(phi)^2-1, cos(2*phi), equal); returns FAIL even though is a trig identify. Any ideas why?
 

Maple currently can't integrate things like sin(x)^n or cos(x)^n. These have antiderivative in terms of  hypergeometric functions.

Is there a technical reason why Maple at version 18 still can't integrate these? Will it be able to in next version?

int(cos(x)^k, x) 
int(sin(x)^k, x) 

fyi, These are the antiderivatives

Same problem for tan(x)^n....  I am asking because Maple is able to solve many ODE's but some results contains unevaluated integrals such as the above. 

 

When making this call to dsolve

 

restart;
ode:=diff(y(x),x)=1:
stopat(dsolve);
dsolve(ode,y(x));

Then in the debugger, I see that dsolve signature is

dsolve := proc(ODEs::{anything} := NULL, 
                {atomizenames::truefalse := true, build::truefalse := false, type::name := 'none'})

When typing print(ODEs) inside the debugger, it gives "diff(y(x),x)=1" which is correct.

But where is the second argument I passed to dsolve in the above call, which is "y(x)" gone?  I also do print(_nparams) inside the debugger and Maple returns 1 and not 2.

I expected to see 2 since I passed in two arguments to dsolve.

dsolve actually works without passing y(x), as follows

dsolve(ode);

And it seems Maple figure inside what is the dependent variable.

But my question is, is the second argument being thrown away before calling dsolve? If not, why it does not show in the debugger?   Is there some other pre-processing being done between the time the user calls dsolve() and the time the debugger is called?

 

 

 

So i have beeing working with finite elements and i have the stifness matrix that i came with Maple as you can see here:

Ke := Matrix(4, 4, {(1, 1) = (12*I)*E/l^3, (1, 2) = (6*I)*E/l^2, (1, 3) = -(12*I)*E/l^3, (1, 4) = (6*I)*E/l^2, (2, 1) = (6*I)*E/l^2, (2, 2) = (4*I)*E/l, (2, 3) = -(6*I)*E/l^2, (2, 4) = (2*I)*E/l, (3, 1) = -(12*I)*E/l^3, (3, 2) = -(6*I)*E/l^2, (3, 3) = (12*I)*E/l^3, (3, 4) = -(6*I)*E/l^2, (4, 1) = (6*I)*E/l^2, (4, 2) = (2*I)*E/l, (4, 3) = -(6*I)*E/l^2, (4, 4) = (4*I)*E/l})

How can i re-write this matrix to put in this form:


Thanks

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