Maple Questions and Posts

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how to find the contour of time series data? and how to find curvature function of this contour?

In this fuction the maximize is about at t=46 and x=46 but in the plot I look other max at other value, why?because is discontinus fuction?I need not the local max (it is potential energy but I think don't matter)optimization2enerpot.mws

updated:
P := evalm(p2 + c*vector([cos(q1+q2+q3), sin(q1+q2+q3)]));
 
restart:
with(Groebner):
p1 := vector([a*cos(q1), a*sin(q1)]);
p2 := evalm(p1 + b*vector([cos(q1+q2), sin(q1+q2)]));
P := evalm(p2 + c*vector([cos(q1+q2+q3), sin(q1+q2+q3)]));
Pe := map(expand, P);
A := {cos(q1) = c1, sin(q1) =s1, cos(q2)=c2, sin(q2)=s2, cos(q3)=c3, sin(q3)=s3};
P := subs(A, op(Pe));
F1 := [x - P[1], y - P[2], s1^2+c1^2-1, s2^2+c2^2-1, s3^2+c3^2-1 ];
F2 := subs({a=1, b=1, c=1}, F1);
 
g2 := Basis(F2, plex(c3, s3, c2, s2, c1, s1));
LeadingTerm(g2[1], plex(c3, s3, c2, s2, c1, s1));
LeadingTerm(g2[2], plex(c3, s3, c2, s2, c1, s1));
LeadingTerm(g2[3], plex(c3, s3, c2, s2, c1, s1));
LeadingTerm(g2[4], plex(c3, s3, c2, s2, c1, s1));
LeadingTerm(g2[5], plex(c3, s3, c2, s2, c1, s1));
LeadingTerm(g2[6], plex(c3, s3, c2, s2, c1, s1));
LeadingTerm(g2[7], plex(c3, s3, c2, s2, c1, s1));
LeadingTerm(g2[8], plex(c3, s3, c2, s2, c1, s1));
LeadingTerm(g2[9], plex(c3, s3, c2, s2, c1, s1));
 
                                   1, c1
                               2       2    2   2
                           16 y  + 16 x , s1  s2
                                           2
                                 8 x, c1 s2
                                2      2    2  
                             2 y  + 2 x , s1  c2
                                 2 x, c1 c2
                            3            2        
                         2 x  - 2 x + 2 y  x, s2 c2
                                        2
                                   1, c2
                                   2 x, s3
                                    2, c3
originally i think
g2[1], g2[7], g2[9] have single variables c1, c2, c3 respectively
can be used to solve system
 
but without x and y, these equations can not be used
if choose leading term has x and y , but there is no single variable s1 or c1.
 
originally expect solve as follows
g2spec := subs({x=1, y=1/2}, [g2[3],g2[5],g2[6]]);
S1 := [solve([g2spec[1]])];
q1a := evalf(arccos(S1[1]));
q1b := evalf(arccos(S1[2]));
S2 := [solve(subs(s1=S1[1], g2spec[2])), solve(subs(s1=S1[2], g2spec[2])) ];
q2a := evalf(arccos(S2[1]));
q2b := evalf(arccos(S2[2]));
S3 := [solve(subs(s1=S2[1], g2spec[2])), solve(subs(s1=S2[2], g2spec[2])) ];
q2a := evalf(arccos(S3[1]));
q2b := evalf(arccos(S3[2]));
 

 

Books free. Like!!!

Lenin Araujo Castillo

I found from this forum that to plot a 2D array of points use can be made of the Maple procedure surfdata.
 

Does anyone have suggestions on how to plot contours in (preferrably) Maple 16 or Maple 17?

I tried the following

Output := Array(-10 .. 10, -10 .. 10, proc (i, j) options operator, arrow; i^2+j^2 end proc):
F := proc (x, y) -> x^2+y^2 end proc:
surfdata(Output, color = F, dimension = 2);

but "the option dimension = 2" is a Maple 18 addition.

Ideally, I would like also to be able to plot contours with options found in the procedure
contourplot

Please illustrate the answer on the example of a simple wave equation, for instance.

I'm wondering if there is an available command that can evaluate the number of terms required to produce a desired outcome.

Specifically, I am interested in determining the probability of a Poisson distribution, given the parameter (mean) value and the probability outcome. I can obtain the desired result using trial and error / brute force, but I am curious to know if there is a more efficient way. 

Suppose that, lambda = 2.6 and the cumulative sum of the probabilities is 95%. I know that I must add the first 6 terms for P(x) in the series (x=0,1, ..,5) to sum to 0.95. Each term ...  P(x=0)= 0.07, P(x=1)=0.19, and so on.

However, how can we know that desired 95% outcome can be determined from the first 5 terms without trial & error?

will give me

which is indeed a solution of the PDE1

will give me

which is not a solution of the PDE2

However, both differential equations are equal, only the arguments are swapped around. Am I doing something wrong, or is this a bug?

Thanks

I have a module with quite a few procedures and it is getting too long and complex. Basicially I write each procedure in a seperate document, them copy and paste it into the module. I want to improve matters as save each proc and read it in to the module

e.g.  Qdim:=proc(A,B).........end proc

        save Qdim , "Qdim.?"   have tried .txt ,.mla , .m  They save fine.

in the module have tried

read "Qdim.txt" etc.   I have included Qdim in export but Qdim doesnt work Qdim(A,B) returns Qdim(A,B)

read "C:\Users\Ronan\Documents\MAPLE\Rational Trinonometry\Qdim.m";

which procuces an error

Error, (in unknown) could not open `C:UsersRonanDocumentsMAPLERational TrinonometryQdim.m` for reading

 

Not sure if this is documented explicitly to Maple users, but normally (prior to Maple 2017) interface(typesetting=extended) was all that was required to output display diff(y(x),x) as y'(x)

With Maple 2017 typesetting=extended is default and one must use with(Typesetting)  Settings(typesetprime=true) in order to output the display of diff(y(x),x) as y'(x).

Perhaps this is well documented somewhere, however I was unable to find the change. 

I am new to Maple and I am trying to add units to the "Flow Through an Expansion Valve" Application Demonstration.  I was trying pressure in [PSI], temperature in [degC] and flow rate in [kg/hour] everything else in SI units.  I included with(Units:-Standard) but had no luck with the fsolve function.

Any chance someone could make a version of this demonstration applicaton that includes units?

 

Thanks

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> restart; with(PDETools), with(plots);
> n := .3; Pr := 7; Da := 0.1e-4; Nb := .1; Nt := .1; tau := 5;
> Eq1 := (1-n)*(diff(f(x, y), `$`(y, 3)))+(1+x*cot(x))*f(x, y)*(diff(f(x, y), `$`(y, 2)))-(diff(f(x, y), y))/Da+(diff(f(x, y), y))^2+n*We*(diff(f(x, y), `$`(y, 2)))*(diff(f(x, y), `$`(y, 3)))+sin(x)*(theta(x, y)+phi(x, y))/x = x*((diff(f(x, y), y))*(diff(f(x, y), y, x))+(diff(f(x, y), `$`(y, 2)))*(diff(f(x, y), x)));
> Eq2 := (diff(theta(x, y), `$`(y, 2)))/Pr+Nt*(diff(theta(x, y), y))^2/Pr+Nb*(diff(phi(x, y), y))*(diff(theta(x, y), y))/Pr+(1+x*cot(x))*f(x, y)*(diff(theta(x, y), y)) = x*((diff(f(x, y), y))*(diff(theta(x, y), x))+(diff(theta(x, y), y))*(diff(f(x, y), x)));
> Eq3 := Nb*(diff(phi(x, y), `$`(y, 2)))/(tau*Pr)+Nt*(diff(theta(x, y), `$`(y, 2)))/(tau*Pr)+(1+x*cot(x))*f(x, y)*(diff(phi(x, y), y)) = x*((diff(f(x, y), y))*(diff(phi(x, y), x))+(diff(phi(x, y), y))*(diff(f(x, y), x)));
> ValWe := [0, 5, 10];
> bcs := {Nb*(D[2](phi))(x, 0)+Nt*(D[2](theta))(x, 0) = 0, f(0, y) = ((1/12)*y)^2*(6-8*((1/12)*y)+3*((1/12)*y)^2), f(x, 0) = 0, phi(0, y) = -.5*y, phi(x, 12) = 0, theta(0, y) = (1-(1/12)*y)^2, theta(x, 0) = 1, theta(x, 12) = 0, (D[2](f))(x, 0) = Da^(1/2)*(D[2, 2](f))(x, 0)+Da*(D[2, 2, 2](f))(x, 0), (D[2](f))(x, 12) = 0};
> pdsys := {Eq1, Eq2, Eq3}; for i to 3 do We := ValWe[i]; ans[i] := pdsolve(pdsys, bcs, numeric) end do;
> p1 := ans[1]:-plot(theta(x, y), x = 1, color = blue); p2 := ans[2]:-plot(theta(x, y), x = 1, color = green); p3 := ans[3]:-plot(theta(x, y), x = 1, color = black);
> plots[display]({p1, p2, p3});

I am trying to solve a system of equations (I'm using MapleTA< but I'm pretty sure that this applies to any Maple product).  I have successfully solved the system, and obtain a set of solutions, which has name Soln.  I can access the element Soln[1], which is an expression:

vn2 = 12/7

Now, I just want that 12/7, as a decimal.  I try evalf(Soln[1]), but again I end up with vn2 = 12/7.  How do I get the decimal number out, without it being an expression?

on the basis of Fundamental theorem of calculus, the function " f " has Antiderivative F , whichi F is continuous and differentiabe in the range of integration, i have a question ? if function f has an Antiderivative F, should it be also continuous and differentiabe in that range? in the picture below it says "when f is continuous", can f be discontinuous? how it can be ? any examples?

Using convert(f,elsymfun) when f is a symmetric polynomial will write it in terms of elementary symmetric polynomials. For example, x^2+y^2 would become (x+y)^2-2x*y. For this command not to return an error, f must be symmetric, equivalently of type symmfunc(all indeterminants in expression).

I'd like to use this in a broader sense, when there are extra variables hanging around. For example, suppose f is a polynomial in a,b,c,x,y,z, and symmetric in x,y,z, so that type(f,symmfunc(x,y,z)) returns true. Then it is still possible to write this in terms of the elementary symmetric polynomials on x,y,z (with coefficients taken being rational polynomials in a,b,c).

For example, a+x^2+y^2 can be written as a+(x+y)^2-2x*y.

Is there a command available for this? Or is there a roundabout way to make Maple forget temporarily that a,b,c are indeterminants, before putting them back in after?

Thanks

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