MaplePrimes Questions


Consider the worksheet below.

restart

with(Physics[Vectors])

`dτ_` := -I*z*dz*B__0*x^2*_j/d^2-I*y*dz*B__0*x^2*_k/d^2

-I*z*dz*B__0*x^2*_j/d^2-I*y*dz*B__0*x^2*_k/d^2

(1)

I want to integrate the above expression. The differential is already present in the expression. Currently, I have to manually remove the differential to use the int procedure as below.

`τ_` := int(-I*z*B__0*x^2*`#mover(mi("j"),mo("∧"))`/d^2-I*y*B__0*x^2*`#mover(mi("k"),mo("∧"))`/d^2, z = l .. -l)

(2*I)*y*B__0*x^2*_k*l/d^2

(2)

Since I want to put this operation inside a procedure, this removal of dz needs to happen programatically.


For even more context, dtau is a differential torque. 

I am integrating these infinitesimal torques.

dtau is calculated based on a magnetic field B.

I would like to have a procedure that accepts a field B and computes the torque integral.

Download RemoveDifferential.mw

Here is a "toy" instance. 
As the title suggests, I would like to simplify the following boolean combinations: (Note that the  should be taken into account.) 

formula__0 := 
 convert(And(
   Non(`xor`(Or(0 >- 0, y ** 3=x), 
     And(((y*x - 0^0)^2 + (y*y - x*x)^2)*(y^2 + x^2) > 
       0, 0 <- (y**3 - x)*(y - x**3)**3, 
      `implies`(y + x*2 >= y ** 3*2 + x**3, 
       Or(y + x < y*y*y + x*x*x)), Not((y**3 - x)*(y - x**3) = 0), 
      0 >= 0), y <> x*x*x))), 'boolean_function'):

Regretfully, the built-in command simplify is unable to simplify : 

simplify(formula__0, assume = real); # only rudimentary simplifications 
 = 



The library function SMTLIB:-Simplify seems to simplify it, but the result is incorrect

(SMTLIB:-Simplify(formula__0) assuming real); # check {y = 1, x = 0} 
 = 
                    /   /   / 3           3\\\
                 And\Not\And\y  = x, y = x ///

The only procedure I can find that is capable of rewriting it appears to be RealDomain:-solve

RealDomain:-solve(formula__0, 'allsolutions'); # see below 
 = 
Warning, solutions may have been lost
    /            3\    /     3       \    /     3        \   
   { x = x, y = x  }, { x = y , y = y }, { y = x , x < -1 }, 
    \             /    \             /    \              /   

      /     3              \    /     3        \   
     { y = x , 0 < x, x < 1 }, { x = y , y < -1 }, 
      \                    /    \              /   

      /     3              \ 
     { x = y , 0 < y, y < 1 }
      \                    / 


Nevertheless, this is more or less overkill, since a complete solution set is somewhat unnecessary, and in practice, an simplified and compact but presumably unsolved form is more applicable to further manipulations. (For example, the simplest form of And(y*y + x*x > 0^0): (over ℝ²) should at least be x2y2>1, yet  simply returns sort(convert(Or,function,map2(`?()`,And,RealDomain[solve](And(y*y+x*x>0^0),[x,y]))),[x,y]), which is definitely unsuitable here.) (A more real example can be found in the Ex3 of this compressed file.)

I believe that this is a common problem; curiously, I cannot find any related questions in this forum. So, are such simplifications (similar to SLFQ) available in Maple? 


 

I have the following expressions:

m^3*r*(cos(theta)^6*a^6-9*cos(theta)^4*a^4*r^2+11*cos(theta)^2*a^2*r^4-(1/3)*r^6)*(cos(theta)^2*a^2-(1/3)*r^2)/(r^2+cos(theta)^2*a^2)^9

m^3*r*(cos(theta)^6*a^6-9*cos(theta)^4*a^4*r^2+11*cos(theta)^2*a^2*r^4-(1/3)*r^6)*(cos(theta)^2*a^2-(1/3)*r^2)/(r^2+cos(theta)^2*a^2)^9

(1)

(a^8*(r^10-10*m*r^9+(5*(a^2+8*m^2))*r^8-80*m^3*r^7+(10*(a^4+12*a^2*m^2+8*m^4))*r^6+(4*(-15*a^4*m-40*a^2*m^3-8*m^5))*r^5+(10*(a^6+12*a^4*m^2+8*a^2*m^4))*r^4+(40*(-a^6*m-2*a^4*m^3))*r^3+(5*(a^8+8*a^6*m^2))*r^2-10*a^8*m*r+a^10)*cos(theta)^8-(1/3)*(28*(-30*sin(theta)^2*a^4*m*r^5*(1/7)+r^10-10*m*r^9+(5*(a^2+8*m^2))*r^8+(40*(-a^2*m-2*m^3))*r^7+(10*(a^4+12*a^2*m^2+8*m^4))*r^6+(32*(-5*a^2*m^3-m^5))*r^5+(10*(a^6+12*a^4*m^2+8*a^2*m^4))*r^4+(40*(-a^6*m-2*a^4*m^3))*r^3+(5*(a^8+8*a^6*m^2))*r^2-10*a^8*m*r+a^10))*a^6*r^2*cos(theta)^6+(14*(-260*sin(theta)^2*a^6*m*r^3*(1/7)+r^10-10*m*r^9+(5*(a^2+8*m^2))*r^8+(40*(-a^2*m-2*m^3))*r^7+(10*(a^4+12*a^2*m^2+8*m^4))*r^6+(4*(-15*a^4*m-40*a^2*m^3-8*m^5))*r^5+(10*(a^6+12*a^4*m^2+8*a^2*m^4))*r^4-80*a^4*m^3*r^3+(5*(a^8+8*a^6*m^2))*r^2-10*a^8*m*r+a^10))*a^4*r^4*cos(theta)^4-4*a^2*(-10*a^8*m*r*sin(theta)^2+r^10-10*m*r^9+(5*(a^2+8*m^2))*r^8+(40*(-a^2*m-2*m^3))*r^7+(10*(a^4+12*a^2*m^2+8*m^4))*r^6+(4*(-15*a^4*m-40*a^2*m^3-8*m^5))*r^5+(10*(a^6+12*a^4*m^2+8*a^2*m^4))*r^4+(40*(-a^6*m-2*a^4*m^3))*r^3+(5*(a^8+8*a^6*m^2))*r^2+a^10)*r^6*cos(theta)^2+(1/9)*r^8*(a^2-2*m*r+r^2)^5)*m^3*r/((r^2+a^2*cos(theta)^2)^9*(a^2-2*m*r+r^2)^5)

(a^8*(r^10-10*m*r^9+5*(a^2+8*m^2)*r^8-80*m^3*r^7+10*(a^4+12*a^2*m^2+8*m^4)*r^6+4*(-15*a^4*m-40*a^2*m^3-8*m^5)*r^5+10*(a^6+12*a^4*m^2+8*a^2*m^4)*r^4+40*(-a^6*m-2*a^4*m^3)*r^3+5*(a^8+8*a^6*m^2)*r^2-10*a^8*m*r+a^10)*cos(theta)^8-(28/3)*(-(30/7)*sin(theta)^2*a^4*m*r^5+r^10-10*m*r^9+5*(a^2+8*m^2)*r^8+40*(-a^2*m-2*m^3)*r^7+10*(a^4+12*a^2*m^2+8*m^4)*r^6+32*(-5*a^2*m^3-m^5)*r^5+10*(a^6+12*a^4*m^2+8*a^2*m^4)*r^4+40*(-a^6*m-2*a^4*m^3)*r^3+5*(a^8+8*a^6*m^2)*r^2-10*a^8*m*r+a^10)*a^6*r^2*cos(theta)^6+14*(-(260/7)*sin(theta)^2*a^6*m*r^3+r^10-10*m*r^9+5*(a^2+8*m^2)*r^8+40*(-a^2*m-2*m^3)*r^7+10*(a^4+12*a^2*m^2+8*m^4)*r^6+4*(-15*a^4*m-40*a^2*m^3-8*m^5)*r^5+10*(a^6+12*a^4*m^2+8*a^2*m^4)*r^4-80*a^4*m^3*r^3+5*(a^8+8*a^6*m^2)*r^2-10*a^8*m*r+a^10)*a^4*r^4*cos(theta)^4-4*a^2*(-10*a^8*m*r*sin(theta)^2+r^10-10*m*r^9+5*(a^2+8*m^2)*r^8+40*(-a^2*m-2*m^3)*r^7+10*(a^4+12*a^2*m^2+8*m^4)*r^6+4*(-15*a^4*m-40*a^2*m^3-8*m^5)*r^5+10*(a^6+12*a^4*m^2+8*a^2*m^4)*r^4+40*(-a^6*m-2*a^4*m^3)*r^3+5*(a^8+8*a^6*m^2)*r^2+a^10)*r^6*cos(theta)^2+(1/9)*r^8*(a^2-2*m*r+r^2)^5)*m^3*r/((r^2+cos(theta)^2*a^2)^9*(a^2-2*m*r+r^2)^5)

(2)

The two expressions are the same because:

simplify(m^3*r*(cos(theta)^6*a^6-9*cos(theta)^4*a^4*r^2+11*cos(theta)^2*a^2*r^4-(1/3)*r^6)*(a^2*cos(theta)^2-(1/3)*r^2)/(r^2+a^2*cos(theta)^2)^9-(a^8*(r^10-10*m*r^9+5*(a^2+8*m^2)*r^8-80*m^3*r^7+10*(a^4+12*a^2*m^2+8*m^4)*r^6+4*(-15*a^4*m-40*a^2*m^3-8*m^5)*r^5+10*(a^6+12*a^4*m^2+8*a^2*m^4)*r^4+40*(-a^6*m-2*a^4*m^3)*r^3+5*(a^8+8*a^6*m^2)*r^2-10*a^8*m*r+a^10)*cos(theta)^8-(28/3)*(-(30/7)*sin(theta)^2*a^4*m*r^5+r^10-10*m*r^9+5*(a^2+8*m^2)*r^8+40*(-a^2*m-2*m^3)*r^7+10*(a^4+12*a^2*m^2+8*m^4)*r^6+32*(-5*a^2*m^3-m^5)*r^5+10*(a^6+12*a^4*m^2+8*a^2*m^4)*r^4+40*(-a^6*m-2*a^4*m^3)*r^3+5*(a^8+8*a^6*m^2)*r^2-10*a^8*m*r+a^10)*a^6*r^2*cos(theta)^6+14*(-(260/7)*sin(theta)^2*a^6*m*r^3+r^10-10*m*r^9+5*(a^2+8*m^2)*r^8+40*(-a^2*m-2*m^3)*r^7+10*(a^4+12*a^2*m^2+8*m^4)*r^6+4*(-15*a^4*m-40*a^2*m^3-8*m^5)*r^5+10*(a^6+12*a^4*m^2+8*a^2*m^4)*r^4-80*a^4*m^3*r^3+5*(a^8+8*a^6*m^2)*r^2-10*a^8*m*r+a^10)*a^4*r^4*cos(theta)^4-4*a^2*(-10*a^8*m*r*sin(theta)^2+r^10-10*m*r^9+5*(a^2+8*m^2)*r^8+40*(-a^2*m-2*m^3)*r^7+10*(a^4+12*a^2*m^2+8*m^4)*r^6+4*(-15*a^4*m-40*a^2*m^3-8*m^5)*r^5+10*(a^6+12*a^4*m^2+8*a^2*m^4)*r^4+40*(-a^6*m-2*a^4*m^3)*r^3+5*(a^8+8*a^6*m^2)*r^2+a^10)*r^6*cos(theta)^2+(1/9)*r^8*(a^2-2*m*r+r^2)^5)*m^3*r/((r^2+a^2*cos(theta)^2)^9*(a^2-2*m*r+r^2)^5))

0

(3)

But I don't get the expression (1) from (2) with the help of simplify() command:

simplify((a^8*(r^10-10*m*r^9+5*(a^2+8*m^2)*r^8-80*m^3*r^7+10*(a^4+12*a^2*m^2+8*m^4)*r^6+4*(-15*a^4*m-40*a^2*m^3-8*m^5)*r^5+10*(a^6+12*a^4*m^2+8*a^2*m^4)*r^4+40*(-a^6*m-2*a^4*m^3)*r^3+5*(a^8+8*a^6*m^2)*r^2-10*a^8*m*r+a^10)*cos(theta)^8-(28/3)*(-(30/7)*sin(theta)^2*a^4*m*r^5+r^10-10*m*r^9+5*(a^2+8*m^2)*r^8+40*(-a^2*m-2*m^3)*r^7+10*(a^4+12*a^2*m^2+8*m^4)*r^6+32*(-5*a^2*m^3-m^5)*r^5+10*(a^6+12*a^4*m^2+8*a^2*m^4)*r^4+40*(-a^6*m-2*a^4*m^3)*r^3+5*(a^8+8*a^6*m^2)*r^2-10*a^8*m*r+a^10)*a^6*r^2*cos(theta)^6+14*(-(260/7)*sin(theta)^2*a^6*m*r^3+r^10-10*m*r^9+5*(a^2+8*m^2)*r^8+40*(-a^2*m-2*m^3)*r^7+10*(a^4+12*a^2*m^2+8*m^4)*r^6+4*(-15*a^4*m-40*a^2*m^3-8*m^5)*r^5+10*(a^6+12*a^4*m^2+8*a^2*m^4)*r^4-80*a^4*m^3*r^3+5*(a^8+8*a^6*m^2)*r^2-10*a^8*m*r+a^10)*a^4*r^4*cos(theta)^4-4*a^2*(-10*a^8*m*r*sin(theta)^2+r^10-10*m*r^9+5*(a^2+8*m^2)*r^8+40*(-a^2*m-2*m^3)*r^7+10*(a^4+12*a^2*m^2+8*m^4)*r^6+4*(-15*a^4*m-40*a^2*m^3-8*m^5)*r^5+10*(a^6+12*a^4*m^2+8*a^2*m^4)*r^4+40*(-a^6*m-2*a^4*m^3)*r^3+5*(a^8+8*a^6*m^2)*r^2+a^10)*r^6*cos(theta)^2+(1/9)*r^8*(a^2-2*m*r+r^2)^5)*m^3*r/((r^2+a^2*cos(theta)^2)^9*(a^2-2*m*r+r^2)^5))

(a^8*(r^10-10*m*r^9+5*(a^2+8*m^2)*r^8-80*m^3*r^7+10*(a^4+12*a^2*m^2+8*m^4)*r^6+4*(-15*a^4*m-40*a^2*m^3-8*m^5)*r^5+10*(a^6+12*a^4*m^2+8*a^2*m^4)*r^4+40*(-a^6*m-2*a^4*m^3)*r^3+5*(a^8+8*a^6*m^2)*r^2-10*a^8*m*r+a^10)*cos(theta)^8-(28/3)*(-(30/7)*sin(theta)^2*a^4*m*r^5+r^10-10*m*r^9+5*(a^2+8*m^2)*r^8+40*(-a^2*m-2*m^3)*r^7+10*(a^4+12*a^2*m^2+8*m^4)*r^6+32*(-5*a^2*m^3-m^5)*r^5+10*(a^6+12*a^4*m^2+8*a^2*m^4)*r^4+40*(-a^6*m-2*a^4*m^3)*r^3+5*(a^8+8*a^6*m^2)*r^2-10*a^8*m*r+a^10)*a^6*r^2*cos(theta)^6+14*(-(260/7)*sin(theta)^2*a^6*m*r^3+r^10-10*m*r^9+5*(a^2+8*m^2)*r^8+40*(-a^2*m-2*m^3)*r^7+10*(a^4+12*a^2*m^2+8*m^4)*r^6+4*(-15*a^4*m-40*a^2*m^3-8*m^5)*r^5+10*(a^6+12*a^4*m^2+8*a^2*m^4)*r^4-80*a^4*m^3*r^3+5*(a^8+8*a^6*m^2)*r^2-10*a^8*m*r+a^10)*a^4*r^4*cos(theta)^4-4*a^2*(-10*a^8*m*r*sin(theta)^2+r^10-10*m*r^9+5*(a^2+8*m^2)*r^8+40*(-a^2*m-2*m^3)*r^7+10*(a^4+12*a^2*m^2+8*m^4)*r^6+4*(-15*a^4*m-40*a^2*m^3-8*m^5)*r^5+10*(a^6+12*a^4*m^2+8*a^2*m^4)*r^4+40*(-a^6*m-2*a^4*m^3)*r^3+5*(a^8+8*a^6*m^2)*r^2+a^10)*r^6*cos(theta)^2+(1/9)*r^8*(a^2-2*m*r+r^2)^5)*m^3*r/((r^2+cos(theta)^2*a^2)^9*(a^2-2*m*r+r^2)^5)

(4)

Why not? Is there any solution for this?

 

NULL


 

Download QuestionSimplifyMultivariateRationalFunction.mw

Dear Maple users

I have a question which may seem very simple, but for some reason I have not been able to find a solution to it in Maple:

If plotting a function in Maple by specifying an x-range, but not a y-range, Maple will somehow find a reasonable y-range itself. That's basic stuff. But how do I ensure that this y-range contains the value 0, and Maple still decide the min and max by itself? I could set it myself, but my worksheet require that it is done automatically independent of the given function at hand. 

Erik

 Hello,

  Doing simple calculations I found Maple 2024 and 2023 don't recognize Kronecker delta in the expression

Full worksheet attached. Is there any way to let Maple handle this?

Remaining questions:
1. Doing calculations using Int and value left the result unevaluated eq. (6) and eq. (7) while using int gives 0. Why?
2. Is there possibility to substitute indexed (i.e.) atomic variables like p__m? For example changing by command p__m to p__k.

Thank you in advance,

delta_kron-problem.mw

Hi. I want to plot an sphere animating equator on the sphere(earth). please help me. Tnx.

I have a repetative set of parameter names used in procedures in a package. I settled on making all optional inputs in the format {name::type:=something}. Originally I wanted to use non capitalised names for the optional inputs. But the names clash with Maple commands. I have all sorts of quicky abbreviations like prnt etc. These are both messy and tacky. For many of my choices I would seem to have to use capitalised naming to give a meaningful name. The problem here is these names are used in other Maple packages. So sooner or later there is going to be a clash.

Is there a way a of handling this? Like can I defiine package parameter names? Or should I just stick with my abbreviations?

I read somewhere about this problem in the help years ago. I think Basis was the example used in different packages GroebinerBasis and Polynomial package and using PackageName:-Basis to avoid the clash with both loaded. But that is more a command level handling.

restart

 

illegal:=proc(x,y,{Point:=[symbol=solidcircle,colour=purple]},{Line:=[colour=green,thickness=2]},{Colour:="Blue"},{Scale:=5},{Print:="y"})

print("1 Point= ",Point);  #I currently use points
print("2 Line= ",Line);   #                 linetype
print("3 Colour=  ",Colour); #              clr     this has nothing to do with plotting colour
print("4 Scale= ",Scale);   #               scl   
print("5 Print= ",Print);   #               prnt
Scale*x/y
end proc

 

proc (x, y, { Colour := "Blue", Line := [colour = green, thickness = 2], Point := [symbol = solidcircle, colour = purple], Print := "y", Scale := 5 }) print("1 Point= ", Point); print("2 Line= ", Line); print("3 Colour=  ", Colour); print("4 Scale= ", Scale); print("5 Print= ", Print); Scale*x/y end proc

(1)

illegal(3,7)

"1 Point= ", [symbol = solidcircle, colour = purple]

 

"2 Line= ", [colour = green, thickness = 2]

 

"3 Colour=  ", "Blue"

 

"4 Scale= ", 5

 

"5 Print= ", "y"

 

15/7

(2)

illegal(3,7,line=[linestyle=dash,colour=black,thickness=4])

"1 Point= ", [symbol = solidcircle, colour = purple]

 

"2 Line= ", [colour = green, thickness = 2]

 

"3 Colour=  ", "Blue"

 

"4 Scale= ", 5

 

"5 Print= ", "y"

 

15/7

(3)
 

 

Download 2024-03-09_Illegal_or_Not.mw

I need admin's help
I use evalf(3*21/100,3)=0.630
and evalf(3*89/100,3)=2.67
Is there a way for me to get 2 decimal places
so evalf(3*21/100,3)=0.63?

My apologies for such a basic question. But when I do: my_function:=int((t/τ)*exp(-t/τ),t=0..infinity), I get: lim_{t-->∞} (exp(-t/τ))*τ-(t*exp(-t/τ)) + τ, which with simplify I can get to: lim_{t-->∞} (-t + τ)*(exp(-t/τ))+τ

But why doesn't Maple automatically solve this limit to τ? What do I need to do more?

Thank you for your patient help! Mischa

I very welcome this new feature Thumbs up - Free signs icons but here I am stuck

"restart; f(x):=( sin(x))/(x) :  plot(f(x),title=f(x))"

 

solve(f(x) = 0, x, allsolutions); about(_Z1)

Originally _Z1, renamed _Z1~:
  is assumed to be: integer
 

 

SolveTools:-DisplaySolutions(%)

%PIECEWISE([2*Pi*_Z1, ``])

(1)

Questions:
Q1: How to get also the uneven multiples of Pi?
Q2: Why is zero not excluded?
Q3: How to get the desirable output "{  Pi n        n in `&Zopf;` , n<>0"

 

DisplaySolutions.mw

 

 

How to interpret the output to limit()?

restart;
local gamma;

gamma

(1)

A := -sigma__v^2*(((-2*gamma^2*sigma__d^4*sigma__e^4 - 16)*sigma__v^6 + (-6*gamma^2*sigma__d^4*sigma__e^6 - 4*gamma^2*sigma__d^2*sigma__e^4 - 48*sigma__e^2)*sigma__v^4 - sigma__e^4*(gamma^4*sigma__d^6*sigma__e^6 + 4*gamma^2*sigma__d^4*sigma__e^4 + 8*gamma^2*sigma__d^2*sigma__e^2 + 48)*sigma__v^2 - 4*gamma^2*sigma__d^2*sigma__e^8 - 16*sigma__e^6)*sqrt(gamma^2*sigma__d^2*sigma__e^4 + 4*sigma__e^2 + 4*sigma__v^2) + (2*sigma__d^2*sigma__v^8 + (12*sigma__d^2*sigma__e^2 + 8)*sigma__v^6 + 2*(12 + gamma^2*sigma__d^4*sigma__e^4 + sigma__d^2*(gamma^2 + 13)*sigma__e^2)*sigma__e^2*sigma__v^4 + 8*(3 + gamma^2*sigma__d^4*sigma__e^4 + sigma__d^2*(gamma^2 + 6)*sigma__e^2/2)*sigma__e^4*sigma__v^2 + sigma__e^6*(gamma^2*sigma__d^2*sigma__e^2 + 4)*(gamma^2*sigma__d^4*sigma__e^4 + 2*sigma__d^2*sigma__e^2 + 2))*sigma__v^2*gamma*sigma__d)*sigma__d/(4*(sigma__e^2 + sigma__v^2)^2*(gamma^2*sigma__d^2*sigma__e^4 + 4*sigma__e^2 + 4*sigma__v^2)^2);

-(1/4)*sigma__v^2*(((-2*gamma^2*sigma__d^4*sigma__e^4-16)*sigma__v^6+(-6*gamma^2*sigma__d^4*sigma__e^6-4*gamma^2*sigma__d^2*sigma__e^4-48*sigma__e^2)*sigma__v^4-sigma__e^4*(gamma^4*sigma__d^6*sigma__e^6+4*gamma^2*sigma__d^4*sigma__e^4+8*gamma^2*sigma__d^2*sigma__e^2+48)*sigma__v^2-4*gamma^2*sigma__d^2*sigma__e^8-16*sigma__e^6)*(gamma^2*sigma__d^2*sigma__e^4+4*sigma__e^2+4*sigma__v^2)^(1/2)+(2*sigma__d^2*sigma__v^8+(12*sigma__d^2*sigma__e^2+8)*sigma__v^6+2*(12+gamma^2*sigma__d^4*sigma__e^4+sigma__d^2*(gamma^2+13)*sigma__e^2)*sigma__e^2*sigma__v^4+8*(3+gamma^2*sigma__d^4*sigma__e^4+(1/2)*sigma__d^2*(gamma^2+6)*sigma__e^2)*sigma__e^4*sigma__v^2+sigma__e^6*(gamma^2*sigma__d^2*sigma__e^2+4)*(gamma^2*sigma__d^4*sigma__e^4+2*sigma__d^2*sigma__e^2+2))*sigma__v^2*gamma*sigma__d)*sigma__d/((sigma__e^2+sigma__v^2)^2*(gamma^2*sigma__d^2*sigma__e^4+4*sigma__e^2+4*sigma__v^2)^2)

(2)

# Limits

A__0 := limit(A, gamma = 0);
A__inf_wo_assumptions := limit(A, gamma = infinity);
A__inf_with_assumptions := limit(A, gamma = infinity) assuming 0 < sigma__e, 0 < sigma__v, 0 < sigma__d;

(1/2)*sigma__v^2*sigma__d/(sigma__e^2+sigma__v^2)^(1/2)

 

signum(sigma__d^3*sigma__e^2*sigma__v^4*(-sigma__d*sigma__e^2+(sigma__d^2*sigma__e^4)^(1/2))/(sigma__e^2+sigma__v^2)^2)*infinity

 

0

(3)

Download limits_signum.mw

Was experimenting with methods to handle the representation of 3D lines and plotting them. Where I normally use a point and a direction vector to dascribe the line.

With a bit of experimenting I see the element wise operation `+`~  or  `-`~  using prefix notation saves a lot of time converting vector to lists and vice a versa.  Would be interested to know if there are better techniques.

On the plotting side using plot3d Can the colour of the lines be changed individually? Or should I use a different plotting command?

restart

with(plottools):

l:=([2,-3,1],<3,7/9,6>);   # 3d line point + vector

P:=[7,-8,9]

l := [2, `&ndash;`(3), 1], Vector(3, {(1) = 3, (2) = 7/9, (3) = 6})

 

[7, -8, 9]

(1)

pl:=`+`~(lambda*l[2],l[1]); #3d line as vector eqn

 

Vector(3, {(1) = 3*lambda+2, (2) = (7/9)*lambda-3, (3) = 6*lambda+1})

(2)

vnl:=`-`~(pl,P) ; #vector from Point P to 3D line

 

Vector(3, {(1) = 3*lambda-5, (2) = (7/9)*lambda+5, (3) = 6*lambda-8})

(3)

vnl.l[2] assuming `real` ; #dot product of vectors= 0 when perpendicular

 

(3694/81)*lambda-532/9

(4)

sol:=solve( { (4) }, [lambda] )[];

[lambda = 2394/1847]

(5)

intP:=eval(pl,sol)  #intersection point

Vector(3, {(1) = 10876/1847, (2) = -3679/1847, (3) = 16211/1847})

(6)

l2:=P,eval(vnl,sol) ;  #perpendicular 3D line through P

l2 := [7, `&ndash;`(8), 9], Vector(3, {(1) = -2053/1847, (2) = 11097/1847, (3) = -412/1847})

(7)

pl2:=`+`~(lambda*l2[2],l2[1]); #3D line as vector eqn

Vector(3, {(1) = -(2053/1847)*lambda+7, (2) = (11097/1847)*lambda-8, (3) = -(412/1847)*lambda+9})

(8)

plots:-display(plot3d([pl,pl2],lambda=-.5..1.8,thickness=0,colour=[orange,purple],axes=normal,scaling=constrained),
                point(P,colour=blue ,symbolsize=15,symbol=solidsphere),
                point(l[1],colour=green ,symbolsize=15,symbol=solidsphere),
                point(eval(pl,sol),colour=red ,symbolsize=15,symbol=solidsphere),
                arrow(l,0.2, 0.4, 0.1,colour=green),
                arrow(l2,0.2, 0.4, 0.1,colour=blue));

 

  

 


 

Download Perpendicular_3D_lines.mw

I see a number of MaplePrimes questions answered with good examples using prefix notation.  Is there good reference material somewhere that can show me how to make use of prefix notation when using Maple?  It's not always obvious (to me) how to construct maple commands using prefix operators so I typically just use the standard commands.  Understanding Maples prefix notation would help in understanding some of the answers given to questions in this forum.  Thanks.

Question is attached:

I describe in words the problem I want to solve with Maple. I'll need to work with random variables.

I want to compute Var[A+B+C] where A, B, and C are not independent of each other. In particular, I don't know how to compute Cov[A,B], Cov[A,C], and Cov[B,C]. The model specifications follow.

Let:

A = X__1*(-lambda__1*X__1-lambda__1*delta__1+nu__1-nu__01);
B = X__2*(-lambda__2*X__2-lambda__2*delta__2+nu__2-nu__02);
A = X__3*(-lambda__3*X__3-lambda__3*delta__3+nu__1+nu__2-nu__01-nu__02);

A = X__1*(-X__1*lambda__1-delta__1*lambda__1-nu__01+nu__1)

 

B = X__2*(-X__2*lambda__2-delta__2*lambda__2-nu__02+nu__2)

 

A = X__3*(-X__3*lambda__3-delta__3*lambda__3-nu__01-nu__02+nu__1+nu__2)

(1)

where lambda__1, lambda__2, and lambda__3 are constants. Moreover, nu__01 is the mean of nu__1~N(nu__01,sigma__nu^2)
and nu__02 is the mean of nu__2~N(nu__02,sigma__nu^2). Note that nu__1 and nu__2 have the same variance and are independent of each other.

In addition:

X__1 = beta__1*(nu__1+nu__2-nu__01-nu__02)+alpha__1*delta__1+alpha__2s*delta__2;
X__2 = beta__2*(nu__1+nu__2-nu__01-nu__02)+alpha__2*delta__2+alpha__1s*delta__1;
X__3 = beta__3*(nu__1+nu__2-nu__01-nu__02)+alpha__3*delta__3;

X__1 = beta__1*(nu__1+nu__2-nu__01-nu__02)+alpha__1*delta__1+alpha__2s*delta__2

 

X__2 = beta__2*(nu__1+nu__2-nu__01-nu__02)+alpha__2*delta__2+alpha__1s*delta__1

 

X__3 = beta__3*(nu__1+nu__2-nu__01-nu__02)+alpha__3*delta__3

(2)

where beta__1, beta__2, beta__3, alpha__1, alpha__2, alpha__3, alpha__1s, alpha__2s are constants. Moreover, delta__1~N(0,sigma__d^2), delta__2~N(0,sigma__d^2), and delta__3~N(0,sigma__d3^2) (note the different variance for delta__3). The variables delta__1, delta__2, and delta__3 are independent of each other. Moreover, nu__1 and nu__2 are independent of delta__1, delta__2, and delta__3.

Now, A, B, C are all products of the form W*Q. In general, Var[W*Q] can be found by applying a formula*** which here reduces to Var[W*Q] = sigma__W^2*sigma__Q^2+(Cov[W,Q])^2, where Cov[W,Q] is simply E[W*Q] since E[W]=0 and E[Q]=0 in my three cases. In short, it's relatively straightforward to find Var[A], Var[B], and Var[C]. However, I don't know about the covariance terms. How to tackle the covariance terms, i.e., Cov[A,B], Cov[A,C], and Cov[B,C]?


***See @whuber's comment in Prof. Dilip Sarwate's answer here https://stats.stackexchange.com/questions/15978/variance-of-product-of-dependent-variables


Perhaps it would be useful to automate the computation of E[A+B+C] as well. However, I managed to compute the expectation by hand, with pen and paper. It would be nice to double check with a script.

Download variance_of_sum_of_products.mw

Want to make a rectangular graph (in the sense of Graph Theory) of 4 copies of the path A-B-C-D i.e. four rows of this path. Once the rectangular or matrix like graph is made its ediges are to be weighted as the matrix position of vertex stating from (1,1) at the upper left vertex i.e. "A". Help please.

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