MaplePrimes Questions

I have a statement:

print(plots[display](seq(colr2[i],i=1..numplayers, xtickmarks=0,axes=NONE,view=[0..5*numplayers,0..1])):

which outputs a horizontal display of the various "colors" of the players, and it does so.

    However, the height of the output is quite big and makes the output look "chunky".  Inserting the option "scaling=constrained" makes the width of the output much narrower - but there is a bounding box of the output (which is not actually displayed.)  It is displayed when the mouse cursor is clicked on the display.  I was hoping the view= [..] option would provide a solution, but this does not alter the exterior bounding rectangle.

  Is there any means of modifying the dimensions of the rectangular display through Maple code?

Thanks,

   David

Hi,

I use the VectorCalculus package to calcutate derivative formula for geometric functions, and met difficulity simplifying the result expression.

For example, I define some vectors P, S, V like below:

P:=<Px, Py, Pz>, S:=<Sx, Sy, Sz>, V:=<Vx, Vy, Vz>

then define an intermediate variable Q:=P - S,

then define a function d:= sqrt(DotProduct(Q, Q)-(DotProuct(Q,V))^2)

by calculating the function's derivative w.r.t Px I got a very complex result expression:

dpx:=1/2 * (2Px - 2Sx - 2 ( (Px - Sx) Vx + (Py - Sy) Vy + (Pz - Sz)Vz )Vx ) / (sqrt( (Px-Sx)^2 + (Py-Sy)^2 + (Pz-Sz)^2 - .....)

 

Apparently this expression can be simplified by substituting its sub-expression with pre-defined variables like Q and d.

I know I can use subs, eval, and subsalg to do it manually:

subs(1/(sqrt( (Px-Sx)^2 + (Py-Sy)^2 + (Pz-Sz)^2 - .....) = 1/dv, dfdpx)

subs((Px - Sx) Vx + (Py - Sy) Vy + (Pz - Sz)Vz = dotproduct_q_v, dfdpx)

and I can get a simplified expression like this:

(qx-dotproduct_q_v*vx)/d

 

But it's like my brain does the simplification first, and Maple only does the text substitution for me.

Is there any way to do it automatically?

 

Thanks,

-Kai

 

Hi i 2 questions. all pertaining to solving a systems of equations mod 2

First if i have a large set of equations, 11^3 equations in 11 unknowns and i want maple to give me ALL solutions mod 2 how can i do that? Maples msolve is loosing solutions.

Second suppose i want all unique solutions that say 6 of the variables can have but dont care what the solution to the other variables are as long as it is a solution. 

mini example:

say x=1,y=1,z=1 is a solution as well as x=1,y=1,z=0, i just want to know about x=1,y=1.

 

I have produded a 3D plot where I have used the graphic's lighting->user GUI to set the light color, direction, and ambient lighting to my liking.

 

I would like to save the lighting parameters so that I can reproduce the identical lighting in other plots.  I see no way of reading off the lighting parameters from the GUI.  I tried to "lprint(myplot)" to see if it contains that information but apparently it doesn't.

So my question is: Is there a way to retrieve the lighting parameters from a 3D plot?

 

--

Rouben Rostamian

I experienced strange operation of "union" for sets of vectors.

Mt1:=Matrix(2, 4, [[ 0,1, 0, 0], [ 0,  0,  1, 1]]); Ms := Vector[column](4, [8,4,2,1]); St1 := {}:

St1:= `union`(St1, {Mt1 . Ms});

I am surprised, because each execution of union adds new and the same vector <4 | 3> to set St1:

1

But after copying any set in the clipboard and pasting the set St1 has only one instance of vector <4 | 3>:

2

What does it mean?

How to solve the system
{sqrt((x-1)^2+(y-5)^2)+(1/2)*abs(x+y) = 3*sqrt(2), sqrt(abs(x+2)) = 2-y}
over the reals symbolically? Of course, with Maple. Mathematica does the job.

Hi everyone,

 

I have a question regarding the use of the applyrule function. I have an expression that contains a polynomial. The expression looks something like:

 

Y := (a0 + a_1*x + a_2*x^2 + ... a_n*x^n)*f(y) + b_0 + b_1*x + b_2*x^2 + ... b_n*x^n)*g(y):

 

I would like to express this as y(x) = P_1*f(y) + P_2*g(y).

 

So far I have tried applyrule([a0 + a_1*x + a_2*x^2 + ... a_n*x^n = P_1, b_0 + b_1*x + b_2*x^2 + ... b_n*x^n) = P_2],Y):

 

This doesn't seem to work. Any suggestions?

 

 

 

 

I want to solve maximize of equation,but the maximize failed to solve it,who can help me.thanks.

c[1] := (1/8)*w*{(1/((x+y+z)^2+1))^(3/2)+(1/((x+y)^2+1))^(3/2)+(1/((x+z)^2+1))^(3/2)+(1/((y+z)^2+1))^(3/2)+(1/(x^2+1))^(3/2)+(1/(y^2+1))^(3/2)+(1/(z^2+1))^(3/2)+1}+(1/8)*{x/((x+y+z)^2+1)+x/((x+y)^2+1)+x/((x+z)^2+1)+x/(x^2+1)}:

c[2] := (1/8)*w*{(1/((x+y+z)^2+1))^(3/2)+(1/((x+y)^2+1))^(3/2)+(1/((x+z)^2+1))^(3/2)+(1/((y+z)^2+1))^(3/2)+(1/(x^2+1))^(3/2)+(1/(y^2+1))^(3/2)+[1/(z^2+1)]^(3/2)+1}+(1/8)*{y/((x+y+z)^2+1)+y/((x+y)^2+1)+y/((y+z)^2+1)+y/(y^2+1)}:

t[1] := diff(c[1], x);

(1/8)*w*{-(3/2)*(1/((x+y+z)^2+1))^(1/2)*(2*x+2*y+2*z)/((x+y+z)^2+1)^2-(3/2)*(1/((x+y)^2+1))^(1/2)*(2*x+2*y)/((x+y)^2+1)^2-(3/2)*(1/((x+z)^2+1))^(1/2)*(2*x+2*z)/((x+z)^2+1)^2-3*(1/(x^2+1))^(1/2)*x/(x^2+1)^2}+(1/8)*{1/((x+y+z)^2+1)-x*(2*x+2*y+2*z)/((x+y+z)^2+1)^2+1/((x+y)^2+1)-x*(2*x+2*y)/((x+y)^2+1)^2+1/((x+z)^2+1)-x*(2*x+2*z)/((x+z)^2+1)^2+1/(x^2+1)-2*x^2/(x^2+1)^2}

(1)

t[2] := diff(c[2], y);

(1/8)*w*{-(3/2)*(1/((x+y+z)^2+1))^(1/2)*(2*x+2*y+2*z)/((x+y+z)^2+1)^2-(3/2)*(1/((x+y)^2+1))^(1/2)*(2*x+2*y)/((x+y)^2+1)^2-(3/2)*(1/((y+z)^2+1))^(1/2)*(2*y+2*z)/((y+z)^2+1)^2-3*(1/(y^2+1))^(1/2)*y/(y^2+1)^2}+(1/8)*{1/((x+y+z)^2+1)-y*(2*x+2*y+2*z)/((x+y+z)^2+1)^2+1/((x+y)^2+1)-y*(2*x+2*y)/((x+y)^2+1)^2+1/((y+z)^2+1)-y*(2*y+2*z)/((y+z)^2+1)^2+1/(y^2+1)-2*y^2/(y^2+1)^2}

(2)

eliminate({t[1], t[2]}, w);

[{w = -{1/((x+y+z)^2+1)-y*(2*x+2*y+2*z)/((x+y+z)^2+1)^2+1/((x+y)^2+1)-y*(2*x+2*y)/((x+y)^2+1)^2+1/((y+z)^2+1)-y*(2*y+2*z)/((y+z)^2+1)^2+1/(y^2+1)-2*y^2/(y^2+1)^2}/{-(3/2)*(1/(x^2+2*x*y+2*x*z+y^2+2*y*z+z^2+1))^(1/2)*(2*x+2*y+2*z)/((x+y+z)^2+1)^2-(3/2)*(1/(x^2+2*x*y+y^2+1))^(1/2)*(2*x+2*y)/((x+y)^2+1)^2-(3/2)*(1/(y^2+2*y*z+z^2+1))^(1/2)*(2*y+2*z)/((y+z)^2+1)^2-3*(1/(y^2+1))^(1/2)*y/(y^2+1)^2}}, {{1/((x+y+z)^2+1)-x*(2*x+2*y+2*z)/((x+y+z)^2+1)^2+1/((x+y)^2+1)-x*(2*x+2*y)/((x+y)^2+1)^2+1/((x+z)^2+1)-x*(2*x+2*z)/((x+z)^2+1)^2+1/(x^2+1)-2*x^2/(x^2+1)^2}*{-(3/2)*(1/(x^2+2*x*y+2*x*z+y^2+2*y*z+z^2+1))^(1/2)*(2*x+2*y+2*z)/((x+y+z)^2+1)^2-(3/2)*(1/(x^2+2*x*y+y^2+1))^(1/2)*(2*x+2*y)/((x+y)^2+1)^2-(3/2)*(1/(y^2+2*y*z+z^2+1))^(1/2)*(2*y+2*z)/((y+z)^2+1)^2-3*(1/(y^2+1))^(1/2)*y/(y^2+1)^2}-{1/((x+y+z)^2+1)-y*(2*x+2*y+2*z)/((x+y+z)^2+1)^2+1/((x+y)^2+1)-y*(2*x+2*y)/((x+y)^2+1)^2+1/((y+z)^2+1)-y*(2*y+2*z)/((y+z)^2+1)^2+1/(y^2+1)-2*y^2/(y^2+1)^2}*{-(3/2)*(1/(x^2+2*x*y+2*x*z+y^2+2*y*z+z^2+1))^(1/2)*(2*x+2*y+2*z)/((x+y+z)^2+1)^2-(3/2)*(1/(x^2+2*x*y+y^2+1))^(1/2)*(2*x+2*y)/((x+y)^2+1)^2-(3/2)*(1/(x^2+2*x*z+z^2+1))^(1/2)*(2*x+2*z)/((x+z)^2+1)^2-3*(1/(x^2+1))^(1/2)*x/(x^2+1)^2}}]

(3)

w = -(1/((x+y+z)^2+1)-y*(2*x+2*y+2*z)/((x+y+z)^2+1)^2+1/((x+y)^2+1)-y*(2*x+2*y)/((x+y)^2+1)^2+1/((y+z)^2+1)-y*(2*y+2*z)/((y+z)^2+1)^2+1/(y^2+1)-2*y^2/(y^2+1)^2)/(-(3/2)*sqrt(1/(x^2+2*x*y+2*x*z+y^2+2*y*z+z^2+1))*(2*x+2*y+2*z)/((x+y+z)^2+1)^2-(3/2)*sqrt(1/(x^2+2*x*y+y^2+1))*(2*x+2*y)/((x+y)^2+1)^2-(3/2)*sqrt(1/(y^2+2*y*z+z^2+1))*(2*y+2*z)/((y+z)^2+1)^2-3*sqrt(1/(y^2+1))*y/(y^2+1)^2);

w = -(1/((x+y+z)^2+1)-y*(2*x+2*y+2*z)/((x+y+z)^2+1)^2+1/((x+y)^2+1)-y*(2*x+2*y)/((x+y)^2+1)^2+1/((y+z)^2+1)-y*(2*y+2*z)/((y+z)^2+1)^2+1/(y^2+1)-2*y^2/(y^2+1)^2)/(-(3/2)*(1/(x^2+2*x*y+2*x*z+y^2+2*y*z+z^2+1))^(1/2)*(2*x+2*y+2*z)/((x+y+z)^2+1)^2-(3/2)*(1/(x^2+2*x*y+y^2+1))^(1/2)*(2*x+2*y)/((x+y)^2+1)^2-(3/2)*(1/(y^2+2*y*z+z^2+1))^(1/2)*(2*y+2*z)/((y+z)^2+1)^2-3*(1/(y^2+1))^(1/2)*y/(y^2+1)^2)

(4)

sub(w = -(1/((x+y+z)^2+1)-y*(2*x+2*y+2*z)/((x+y+z)^2+1)^2+1/((x+y)^2+1)-y*(2*x+2*y)/((x+y)^2+1)^2+1/((y+z)^2+1)-y*(2*y+2*z)/((y+z)^2+1)^2+1/(y^2+1)-2*y^2/(y^2+1)^2)/(-(3/2)*(1/(x^2+2*x*y+2*x*z+y^2+2*y*z+z^2+1))^(1/2)*(2*x+2*y+2*z)/((x+y+z)^2+1)^2-(3/2)*(1/(x^2+2*x*y+y^2+1))^(1/2)*(2*x+2*y)/((x+y)^2+1)^2-(3/2)*(1/(y^2+2*y*z+z^2+1))^(1/2)*(2*y+2*z)/((y+z)^2+1)^2-3*(1/(y^2+1))^(1/2)*y/(y^2+1)^2), c[1]);

sub(w = -(1/((x+y+z)^2+1)-y*(2*x+2*y+2*z)/((x+y+z)^2+1)^2+1/((x+y)^2+1)-y*(2*x+2*y)/((x+y)^2+1)^2+1/((y+z)^2+1)-y*(2*y+2*z)/((y+z)^2+1)^2+1/(y^2+1)-2*y^2/(y^2+1)^2)/(-(3/2)*(1/(x^2+2*x*y+2*x*z+y^2+2*y*z+z^2+1))^(1/2)*(2*x+2*y+2*z)/((x+y+z)^2+1)^2-(3/2)*(1/(x^2+2*x*y+y^2+1))^(1/2)*(2*x+2*y)/((x+y)^2+1)^2-(3/2)*(1/(y^2+2*y*z+z^2+1))^(1/2)*(2*y+2*z)/((y+z)^2+1)^2-3*(1/(y^2+1))^(1/2)*y/(y^2+1)^2), (1/8)*w*{(1/((x+y+z)^2+1))^(3/2)+(1/((x+y)^2+1))^(3/2)+(1/((x+z)^2+1))^(3/2)+(1/((y+z)^2+1))^(3/2)+(1/(x^2+1))^(3/2)+(1/(y^2+1))^(3/2)+(1/(z^2+1))^(3/2)+1}+(1/8)*{x/((x+y+z)^2+1)+x/((x+y)^2+1)+x/((x+z)^2+1)+x/(x^2+1)})

(5)

subs(w = -(1/((x+y+z)^2+1)-y*(2*x+2*y+2*z)/((x+y+z)^2+1)^2+1/((x+y)^2+1)-y*(2*x+2*y)/((x+y)^2+1)^2+1/((y+z)^2+1)-y*(2*y+2*z)/((y+z)^2+1)^2+1/(y^2+1)-2*y^2/(y^2+1)^2)/(-(3/2)*(1/(x^2+2*x*y+2*x*z+y^2+2*y*z+z^2+1))^(1/2)*(2*x+2*y+2*z)/((x+y+z)^2+1)^2-(3/2)*(1/(x^2+2*x*y+y^2+1))^(1/2)*(2*x+2*y)/((x+y)^2+1)^2-(3/2)*(1/(y^2+2*y*z+z^2+1))^(1/2)*(2*y+2*z)/((y+z)^2+1)^2-3*(1/(y^2+1))^(1/2)*y/(y^2+1)^2), c[2]);

-(1/8)*(1/((x+y+z)^2+1)-y*(2*x+2*y+2*z)/((x+y+z)^2+1)^2+1/((x+y)^2+1)-y*(2*x+2*y)/((x+y)^2+1)^2+1/((y+z)^2+1)-y*(2*y+2*z)/((y+z)^2+1)^2+1/(y^2+1)-2*y^2/(y^2+1)^2)*{(1/((x+y+z)^2+1))^(3/2)+(1/((x+y)^2+1))^(3/2)+(1/((x+z)^2+1))^(3/2)+(1/((y+z)^2+1))^(3/2)+(1/(x^2+1))^(3/2)+(1/(y^2+1))^(3/2)+[1/(z^2+1)]^(3/2)+1}/(-(3/2)*(1/(x^2+2*x*y+2*x*z+y^2+2*y*z+z^2+1))^(1/2)*(2*x+2*y+2*z)/((x+y+z)^2+1)^2-(3/2)*(1/(x^2+2*x*y+y^2+1))^(1/2)*(2*x+2*y)/((x+y)^2+1)^2-(3/2)*(1/(y^2+2*y*z+z^2+1))^(1/2)*(2*y+2*z)/((y+z)^2+1)^2-3*(1/(y^2+1))^(1/2)*y/(y^2+1)^2)+(1/8)*{y/((x+y+z)^2+1)+y/((x+y)^2+1)+y/((y+z)^2+1)+y/(y^2+1)}

(6)

"#"Iwant to maximize the equation (5)and (6),under the conditon of x,y,z are negative or positive at the same time.

 

NULL

 

Download maximize.mw

 

I recently downloaded a Maple reader software program from

http://www.crystaloffice.com/

I'mpuzzled as I'm not sure if this has anything to do with the Maple mathematics software,  On cursory observation it just looks like a text editor.

  I was curious to know if there was any software available which would allow people to read & execute a Maple program on the Internet. 

Cheers

   David

 

I have some triangles ABC with vertices

1) A(-13,-5,5), B(-5,11,-11), C(-3,-9,15) has centre of out circle is (3, 3, 3), orthocentre (-27, -9, 3) and centroid (-7, -1, -3). 

2) A(-6,6,-1), B(-5,-1,-3), C(2,10,7) has centre of out circle is (1, 2, 3), orthocentre (-11, 11, -3) and centroid (-3, 5, 1). 

How can I write a program to find a triangle with integer coordinates of vertices, centroid, orthocenter and center of the triangle in geometry 3D? 

Does anyone know how to incorporate the tetrad with the directional derivative? I tried using the SumOverIndices, but get crazy results. I know Maple can find the answer easily because I have done the same thing by hand. What am I missing?

The directional derivative should take the form f,1 = eaμ df/dxμ . The answer is Y,1 = dY/dζ – Ybar dY/du.  I obviously do not get this result.

 


restart; with(Physics); with(Tetrads)

0, "%1 is not a command in the %2 package", Tetrads, Physics

(1)

`#msup(mi("ds",mathcolor = "#af00af"),mn("2",mathcolor = "#af00af"))` := Physics:-`*`(Physics:-`*`(2, dzeta), dzetabar)+Physics:-`*`(Physics:-`*`(2, du), dv)+Physics:-`*`(Physics:-`*`(2, H(zetabar, zeta, v, u)), (du+Physics:-`*`(Ybar(zetabar, zeta, v, u), dzeta)+Physics:-`*`(Y(zetabar, zeta, v, u), dzetabar)-Physics:-`*`(Physics:-`*`(Y(zetabar, zeta, v, u), Ybar(zetabar, zeta, v, u)), dv))^2)

2*dzeta*dzetabar+2*du*dv+2*H(zetabar, zeta, v, u)*(du+Ybar(zetabar, zeta, v, u)*dzeta+Y(zetabar, zeta, v, u)*dzetabar-Y(zetabar, zeta, v, u)*Ybar(zetabar, zeta, v, u)*dv)^2

(2)

X = [zetabar, zeta, v, u]

X = [zetabar, zeta, v, u]

(3)

PDEtools:-declare(`#msup(mi("ds",mathcolor = "#af00af"),mn("2",mathcolor = "#af00af"))`)

Ybar(zetabar, zeta, v, u)*`will now be displayed as`*Ybar

(4)

Setup(automaticsimplification = true, coordinatesystems = (X = [zetabar, zeta, v, u]), metric = 2*dzeta*dzetabar+2*du*dv+2*H(zetabar, zeta, v, u)*(du+Ybar(zetabar, zeta, v, u)*dzeta+Y(zetabar, zeta, v, u)*dzetabar-Y(zetabar, zeta, v, u)*Ybar(zetabar, zeta, v, u)*dv)^2)

[automaticsimplification = true, coordinatesystems = {X}, metric = {(1, 1) = 2*H(X)*Y(X)^2, (1, 2) = 1+2*H(X)*Y(X)*Ybar(X), (1, 3) = -2*H(X)*Y(X)^2*Ybar(X), (1, 4) = 2*H(X)*Y(X), (2, 2) = 2*H(X)*Ybar(X)^2, (2, 3) = -2*H(X)*Ybar(X)^2*Y(X), (2, 4) = 2*H(X)*Ybar(X), (3, 3) = 2*H(X)*Y(X)^2*Ybar(X)^2, (3, 4) = 1-2*H(X)*Y(X)*Ybar(X), (4, 4) = 2*H(X)}]

(5)

g_[]

g_[mu, nu] = (Matrix(4, 4, {(1, 1) = 2*H(X)*Y(X)^2, (1, 2) = 1+2*H(X)*Y(X)*Ybar(X), (1, 3) = -2*H(X)*Y(X)^2*Ybar(X), (1, 4) = 2*H(X)*Y(X), (2, 1) = 1+2*H(X)*Y(X)*Ybar(X), (2, 2) = 2*H(X)*Ybar(X)^2, (2, 3) = -2*H(X)*Ybar(X)^2*Y(X), (2, 4) = 2*H(X)*Ybar(X), (3, 1) = -2*H(X)*Y(X)^2*Ybar(X), (3, 2) = -2*H(X)*Ybar(X)^2*Y(X), (3, 3) = 2*H(X)*Y(X)^2*Ybar(X)^2, (3, 4) = 1-2*H(X)*Y(X)*Ybar(X), (4, 1) = 2*H(X)*Y(X), (4, 2) = 2*H(X)*Ybar(X), (4, 3) = 1-2*H(X)*Y(X)*Ybar(X), (4, 4) = 2*H(X)}))

(6)

``

NULL

NULL

eqn3 := SumOverRepeatedIndices(Physics:-`*`(d_[mu](Y(X)), e_[1, `~mu`]))

((Y(X)*Ybar(X)-1)*(diff(Y(X), zetabar))+(Y(X)*Ybar(X)-1)*(diff(Y(X), zeta))+(diff(Y(X), u)+diff(Y(X), v))*(Y(X)+Ybar(X)))*2^(1/2)/((-(Ybar(X)^2+1)*(Y(X)^2+1)/(Y(X)+Ybar(X))^2)^(1/2)*(2*Y(X)+2*Ybar(X)))

(7)

NULL

``

NULL


Download Directional_Derivative.mw

Dear All,

I'm trying to solve the following in Maple.

minimize(int(0.1e-3+.5*t+0.2e-2*t^2-b*t-a, t = 0 .. 300), location = true)

But Maple told me that the answer is

Float(-infinity), {[{a = Float(infinity), b = Float(infinity)}, Float(-infinity)]}.

I really need to get a kind of numerical answer. Would it be possible? Please Help me!!

https://social.msdn.microsoft.com/Forums/vstudio/en-US/cc2a85ad-30ec-44ed-8c75-636ff71eade2/how-to-convert-integer-or-decimal-number-into-any-base-number?forum=csharpgeneral

1. for example how to convert decimal or integer number into base 3 number, base 5 number etc.

2.how to do logical operation with custom logic table for example,

 

120 special operator 235 

01111000

11101011

 

special operator according to logical table is

1st op 2nd op output
0 0 1
0 1 0
1 0 1
1 1 0

 

  01111000

  11101011

=00010100 = 20

Hi everybody,

is it possible to define an homokinetic joint in MapleSim for multibody modeling? How can I do that?

 

thanks.

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