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combine plottools shapes using display. How to mak...

Asked by: nm 11173
plotting display
November 09 2024
3  3

I wanted to draw vertical line (as y axis) and add a solid disk at some y location (to indicate some point location). I used plottools:-disk  to make the small disk and used plottools:-line to make the y-axis then used plots:-display to combine them.

All is ok, except the disk is transparent. i.e. I can see the y-axis line below it. Is there a way to make it not transparent? i.e. solid color, so that the y-axis below it do not show?

> 

interface(version);

`Standard Worksheet Interface, Maple 2024.2, Windows 10, October 29 2024 Build ID 1872373`

> 

#plot vertical line
L:=plottools:-line([0,0],[0,3],color="blue"):
c1:=plottools:-disk([0,1],.1,color="red"):
c2:=plottools:-disk([0,2],.1,color="green"):
plots:-display([L,c1,c2],scaling=constrained,axes=none)

> 

 


i.e. 

Download plotools_nov_9_2024.mw

How to setup special type check in a procedure?...

Asked by: Ronan 1336
module type typetools
November 09 2024
1  3

 About 2 years ago I asked this question Application of how to test types in a module? - MaplePrimes on type checking and @CarlLove provided a great answer for inputs to a procedure in that package.

In the package I have a procdure to plot/draw lines. They require sorting out when there is a list of them The procedure recognises the different possible input formats. Could the types be defined in the procdure body. It would make sorting out a lot easier. I could possibly add them to Module load at the start of the package.

> 

restart
> 

 

> 

# Line type
#type _TNL1  expression type {`+`,`*`,`=`, `symbol`,procedure}                         2D line or any expression function equation

> 

#type _TNL2  listlist [[a,b],[c,d]] or [[a,b,c],[d,e,f]]                               2D  line just draws a line between the 2 points
> 

#type _TNL4  listlist [[a,b,c],[d,e,f]]                                                3D line just draws a line between the 2 points
> 

#type _TNL4  list of list & vector 2D [[a,b],<c,d>]  or possibly  [<c,d>,[a,b]]        2D generates line A*x+B*y+C                        
> 

#type _TNL5  list of list & vector 3D [[a,b,c],<d,e,f>] or possibly [<d,e,f>,[a,b,c]]  3D line <a+alpha*d, b+alpha*e, c+alpha*f>

> 

#type _TNL6  vector 3D <a+alpha*d, b+alpha*e, c+alpha*f>                               3D line <a+alpha*d, b+alpha*e, c+alpha*f> checks indets to know
> 

#type _TNL7  list of vector[row] and vector[row]  [<a|b|c>, <d|e|f>]                   projective points are converted to [a/c,b/c] and [d/f,e/f] and line drawn between the points
> 

 

> 

 

> 

 

> 

sorts:=proc(l1::{list,set,Vector[column],`+`,`*`,`=`, `symbol`,procedure},
{range::list:=[-5,5,-4,4]},
{vars::list:=[':-x',':-y']},
{rangep::list:=[-3,3]},
{plopts::list:=[':-colour'=':-blue,magenta']})

description " sort the list of line formats and converts to plotable format  ";

local i,n,Listsystem,subplt, subL, tmp, vv,ll,subpltemp;

      #single line;
if l1::{'Vector[column]',`+`,`*`,`=`, `symbol`,procedure} then
   n:=1;
   subplt:=[{}];
   if l1::'Vector[column]'(3) and indets(l1)={} then
      subL:=[l1[1]*vars[1]+l1[2]*vars[2]+l1[3]];
    else
      subL:=[l1];
   end if;
 elif l1::list and nops(l1)= 2 and not(hastype(l1,set))then
    #for j to 2 do
    #   if  l1[j][1]::listlist or l1[j][2]::listlist  then

                                                        # this section not finished supposed
                                                        # to detect is the list is two line without extra plopts, range, rangexy                                                  
   #   end if;
   # end do;
   n:=1;
   subplt:=[{}];
    #print("here now");
   if hastype(l1,'Vector[column]'(3)) and hastype(l1,list) then  
      #print("in  here");
      vv:=(select(type,l1,('Vector[column]'(3)))[]);
      ll:=select(type,l1,list)[];
      #print("vv " ,vv, "ll",ll);
      subL:=[ll+~alpha*vv];  
    elif hastype(l1,'Vector[column]'(2)) and hastype(l1,list)  then
      vv:=(select(type,l1,('Vector[column]'(2)))[]);
      ll:=select(type,l1,list)[];
      subL:=[ (ll[2] - vars[2])*vv[1] - vv[2]*(ll[1] - vars[1])];    
    else
      subL:=[l1];
  end if;
  
      #single line with own sub-options plopts range rangep bundled as a set
elif l1::set then
   n:=1;
   subpltemp,tmp:=selectremove(has,l1,{':-rangep',':-range',':-plopts',':-rangep'});
   #print("00",subpltemp,tmp);
   subplt:=`if`(has(subpltemp,{':-plopts',':-range',':-rangep'}),[subpltemp],[{}]);
   tmp:=op(tmp);
   #print("TMP ",tmp ,nops(tmp));
   if tmp::{'Vector[column]',`+`,`*`,`=`, `symbol`,procedure} then
  
     if tmp::'Vector[column]'(3) and indets(tmp)={} then
        subL:=[tmp[1]*vars[1]+tmp[2]*vars[2]+tmp[3]];
      elif hastype(tmp,'Vector[column]'(2)) and hastype(tmp,list)  then
        vv:=select(type,tmp,'Vector[column]'(2))[];
        ll:=select(type,tmp,list)[];
        subL:= [(ll[2] - vars[2])*vv[1] - vv[2]*(ll[1] - vars[1])];  
     else
       subL:=[tmp];
   end if;
 elif nops(tmp)= 2  then
   #print("0th nops(tmp)",nops(tmp),tmp);
   # print("here here now");
   if hastype(tmp,'Vector[column]'(3)) and hastype(tmp,list) then  
      #print("in  here here");
      vv:=(select(type,tmp,('Vector[column]'(3)))[]);
      ll:=select(type,tmp,list)[];
      #print("vv " ,vv, "ll",ll);
      subL:=[ll+~alpha*vv];
     elif hastype(tmp,'Vector[column]'(2)) and hastype(tmp,list)  then
      vv:=(select(type,tmp,('Vector[column]'(2)))[]);
      ll:=select(type,tmp,list)[];
      subL:=[ (ll[2] - vars[2])*vv[1] - vv[2]*(ll[1] - vars[1])];         
    else
      subL:=[tmp];
   end if;
   end if;

    #a list of lines with possible sub-options if
 else
   n:=nops(l1);
   subplt:=[];
   subL:=[];
   for i to n do
     subpltemp:=select(has,l1[i],{':-plopts',':-range',':-rangep'});
     #print("subpltemp",subpltemp);
     #print("subpltemp",has(subpltemp,{':-plopts',':-range',':-rangep'}));
     if has(subpltemp,{':-plopts',':-range',':-rangep'})then subplt:=  [subplt[],subpltemp]else subplt:=[subplt[],{}] end if;
     tmp:=select(not(has),l1[i],{':-plopts',':-range',':-rangep'});
     if type(tmp,set)then tmp:=op(tmp)end if;
     #print("TMP ",tmp ,nops(tmp));
     #print("1st nops(tmp)",nops(tmp),tmp);
     if tmp::{'Vector[column]',`+`,`*`,`=`, `symbol`,procedure} then
  
     if tmp::'Vector[column]'(3) and indets(tmp)={} then
        subL:=[subL[],tmp[1]*vars[1]+tmp[2]*vars[2]+tmp[3]];
      else
        subL:=[subL[],tmp];
     end if;
 elif nops(tmp)= 2  then
   #print("nops(tmp)",nops(tmp),tmp);
   #print("2nd here here now");
   if hastype(tmp,'Vector[column]'(3)) and hastype(tmp,list) then  
      #print("in  here here");
      vv:=(select(type,tmp,('Vector[column]'(3)))[]);
      ll:=select(type,tmp,list)[];
      #print(tmp,"vv " ,vv, "ll",ll);
      subL:=[subL[],ll+~alpha*vv];
    elif hastype(tmp,'Vector[column]'(2)) and hastype(tmp,list)  then
      vv:=(select(type,tmp,('Vector[column]'(2)))[]);
      ll:=select(type,tmp,list)[];
      #print(tmp,"vv " ,vv, "ll",ll);
      subL:= [subL[],(ll[2] - vars[2])*vv[1] - vv[2]*(ll[1] - vars[1])];           
    else
      subL:=[subL[],tmp];
  end if;
end if;
    
end do;
end if:
return subplt, subL, n
end proc:
> 

 

> 

L1:=2*x+3*y-4;
L1ext:={L1, range=[-5,2,-4,4],plopts=[colour=red ,linestyle=dot] };
L2:=<5,-1,3>:
L2ext:={L2,plopts=[linestyle=dash, thickness=4]}:
L3:=[[2,1],[4,-5]]:
L3ext:={L3,plopts=[color=red,thickness=1]}:
L4:=[[2,1,-3],[4,-5,-2]]:
L4ext:={L4,plopts=[color=green]}:
L5:=[[2,3],<-1,4>]:
L5ext:={L5,plopts=[color=green,thickness=2],range=[-6,6,-6,6]}:
L6:=[[1,2,-3],<-3,1,4>]:
L6ext:={L6,rangep=[-4,1],plopts[linestyle=dash]}:
L7:=<1+3*alpha,4-1*alpha,-1+2*alpha>:
L7ext:={L7,rangep=[-4,1],plopts=[colour=purlpe,thickness=3]}:
L8:=[<2|3|1>,<4|7|2>]:
L8ext:={L8,plopts=[colour=red]}:

2*x+3*y-4

  

{2*x+3*y-4, plopts = [colour = red, linestyle = dot], range = [-5, 2, -4, 4]}

 (1)
> 

 

> 

Lst:=[L1,L1ext,L2,L2ext,L3,L3ext,x^2+2*y^2+3-4*x*y,L4,L4ext,L5,L5ext,L6,L6ext,L7,L7ext,L8,L8ext]:
> 

A,B,c:=sorts(Lst):
> 

for i to c do
print(i,"  A= ",A[i],"    B= ",B[i]);
end do

1, " A= ", {}, " B= ", 2*x+3*y-4

  

2, " A= ", {plopts = [colour = red, linestyle = dot], range = [-5, 2, -4, 4]}, " B= ", 2*x+3*y-4

  

3, " A= ", {}, " B= ", 5*x-y+3

  

4, " A= ", {plopts = [linestyle = dash, thickness = 4]}, " B= ", 5*x-y+3

  

5, " A= ", {}, " B= ", [[2, 1], [4, -5]]

  

6, " A= ", {plopts = [color = red, thickness = 1]}, " B= ", [[2, 1], [4, -5]]

  

7, " A= ", {}, " B= ", x^2-4*x*y+2*y^2+3

  

8, " A= ", {}, " B= ", [[2, 1, -3], [4, -5, -2]]

  

9, " A= ", {plopts = [color = green]}, " B= ", [[2, 1, -3], [4, -5, -2]]

  

10, " A= ", {}, " B= ", -11+y+4*x

  

11, " A= ", {plopts = [color = green, thickness = 2], range = [-6, 6, -6, 6]}, " B= ", -11+y+4*x

  

12, " A= ", {}, " B= ", Vector[column](%id = 36893490129638901388)

  

13, " A= ", {plopts[linestyle = dash], rangep = [-4, 1]}, " B= ", Vector[column](%id = 36893490129638901628)

  

14, " A= ", {}, " B= ", Vector[column](%id = 36893490129638901988)

  

15, " A= ", {plopts = [colour = purlpe, thickness = 3], rangep = [-4, 1]}, " B= ", Vector[column](%id = 36893490129661594308)

  

16, " A= ", {}, " B= ", [Vector[row](%id = 36893490129661594428), Vector[row](%id = 36893490129661594548)]

  

17, " A= ", {plopts = [colour = red]}, " B= ", [Vector[row](%id = 36893490129661594428), Vector[row](%id = 36893490129661594548)]

 (2)
 

 

Download 2024-11-09_A_Line_Type.mw

Adomian Decomposition - Source in the help text?...

Asked by: Alfred_F 295
ode differential-equations adomian-decomposition
November 09 2024
0  2

In the publication
https://www.scirp.org/pdf/apm_2021062513594181.pdf
I read that Maple18 contains the "Adomian Decomposition". In "Help" of "Maple2024" I found no direct reference to this method. I would like to ask you to provide references in the help text to this method for solving differential equations.

Quadratic Solution??...

Asked by: DavidMcClain 5
solve
November 08 2024
1  2

The other night, first time using Maple in more than 20 years, I asked for the solution to this, looking for u1 in terms of u2:

equ := u2 - u1/(-u1^2 + 1)

solve(equ, [u1])

==> u1 = (-1 + sqrt(4*u2^2 + 1))/(2*u2)

But this is sheer nonsense. The correct solution should be:

u1 = (-1 + sqrt(u2^2 + 1))/u2   ;; note the lack of 4x scaling inside the Sqrt, nor the 2x in the denominator.

What the heck??

How collect coefficient of term of ODE equation?...

Asked by: salim-barzani 1390
differential-equations coefficients power
November 08 2024
0  10

I want collect 1/G(xi) & G'(xi)/G(xi) also 1/G(xi)*G(xi)/G(xi)  when they have power ,and give me what i looking for, i can do by hand but it take time any one can do this by maple code? like this picture below, and if possible find some arbitrary parameter

collect.mw

Diff displayed incorrectly ...

Asked by: matmxhu 85
derivative syntax 2dmath
November 07 2024
2  3

Hey all Maple experts.I could really use some help with  diff,D,Diff

> 

restart
> 

interface(version)

`Standard Worksheet Interface, Maple 2024.2, Windows 10, October 29 2024 Build ID 1872373`

 (1)
> 

with(Physics[Vectors])
> 

NULL
> 

CompactDisplay(A_(x, y, z, t), `&varphi;`(x, y, z, t), v_(x, y, z, t), F_(x, y, z, t), v__x(x, y, z, t), v__y(x, y, z, t), v__z(x, y, z, t), A__x(x, y, z, t), A__y(x, y, z, t), A__z(x, y, z, t), quiet)
> 

macro(Av = A_(x, y, z, t), `&vartheta;` = `&varphi;`(x, y, z, t), Vv = v_(x, y, z, t), Fv = F_(x, y, z, t))

show, ON, OFF, kd_, ep_, Av, vartheta, Vv, Fv

 (2)
> 

Fv = q*('-VectorCalculus[Nabla](`&vartheta;`)'-(diff(Av, t))+`&x`(Vv, `&x`(VectorCalculus[Nabla], Av)))

F_(x, y, z, t) = q*(-Physics:-Vectors:-Nabla(varphi(x, y, z, t))-(diff(A_(x, y, z, t), t))+Physics:-Vectors:-`&x`(v_(x, y, z, t), Physics:-Vectors:-Curl(A_(x, y, z, t))))

 (3)
> 

Av = A__x(x, y, z, t)*_i+A__y(x, y, z, t)*_j+A__z(x, y, z, t)*_k, Vv = v__x(x, y, z, t)*_i+v__y(x, y, z, t)*_j+v__z(x, y, z, t)*_k, F_(x, y, z, t) = F__x*_i+F__y*_j+F__z*_k

A_(x, y, z, t) = A__x(x, y, z, t)*_i+A__y(x, y, z, t)*_j+A__z(x, y, z, t)*_k, v_(x, y, z, t) = v__x(x, y, z, t)*_i+v__y(x, y, z, t)*_j+v__z(x, y, z, t)*_k, F_(x, y, z, t) = F__x*_i+F__y*_j+F__z*_k

 (4)
> 

subs[eval](A_(x, y, z, t) = A__x(x, y, z, t)*_i+A__y(x, y, z, t)*_j+A__z(x, y, z, t)*_k, v_(x, y, z, t) = v__x(x, y, z, t)*_i+v__y(x, y, z, t)*_j+v__z(x, y, z, t)*_k, F_(x, y, z, t) = F__x*_i+F__y*_j+F__z*_k, F_(x, y, z, t) = q*(-Physics[Vectors][Nabla](varphi(x, y, z, t))-(diff(A_(x, y, z, t), t))+Physics[Vectors][`&x`](v_(x, y, z, t), Physics[Vectors][Curl](A_(x, y, z, t)))))

F__x*_i+F__y*_j+F__z*_k = q*(-(diff(varphi(x, y, z, t), x))*_i-(diff(varphi(x, y, z, t), y))*_j-(diff(varphi(x, y, z, t), z))*_k-(diff(A__x(x, y, z, t), t))*_i-(diff(A__y(x, y, z, t), t))*_j-(diff(A__z(x, y, z, t), t))*_k+(-v__y(x, y, z, t)*(diff(A__x(x, y, z, t), y))+v__y(x, y, z, t)*(diff(A__y(x, y, z, t), x))+v__z(x, y, z, t)*(diff(A__z(x, y, z, t), x))-v__z(x, y, z, t)*(diff(A__x(x, y, z, t), z)))*_i+(-v__z(x, y, z, t)*(diff(A__y(x, y, z, t), z))+v__z(x, y, z, t)*(diff(A__z(x, y, z, t), y))+v__x(x, y, z, t)*(diff(A__x(x, y, z, t), y))-v__x(x, y, z, t)*(diff(A__y(x, y, z, t), x)))*_j+(v__y(x, y, z, t)*(diff(A__y(x, y, z, t), z))-v__y(x, y, z, t)*(diff(A__z(x, y, z, t), y))-v__x(x, y, z, t)*(diff(A__z(x, y, z, t), x))+v__x(x, y, z, t)*(diff(A__x(x, y, z, t), z)))*_k)

 (5)
> 

map(Component, F__x*_i+F__y*_j+F__z*_k = q*(-(diff(varphi(x, y, z, t), x))*_i-(diff(varphi(x, y, z, t), y))*_j-(diff(varphi(x, y, z, t), z))*_k-(diff(A__x(x, y, z, t), t))*_i-(diff(A__y(x, y, z, t), t))*_j-(diff(A__z(x, y, z, t), t))*_k+(-v__y(x, y, z, t)*(diff(A__x(x, y, z, t), y))+v__y(x, y, z, t)*(diff(A__y(x, y, z, t), x))+v__z(x, y, z, t)*(diff(A__z(x, y, z, t), x))-v__z(x, y, z, t)*(diff(A__x(x, y, z, t), z)))*_i+(-v__z(x, y, z, t)*(diff(A__y(x, y, z, t), z))+v__z(x, y, z, t)*(diff(A__z(x, y, z, t), y))+v__x(x, y, z, t)*(diff(A__x(x, y, z, t), y))-v__x(x, y, z, t)*(diff(A__y(x, y, z, t), x)))*_j+(v__y(x, y, z, t)*(diff(A__y(x, y, z, t), z))-v__y(x, y, z, t)*(diff(A__z(x, y, z, t), y))-v__x(x, y, z, t)*(diff(A__z(x, y, z, t), x))+v__x(x, y, z, t)*(diff(A__x(x, y, z, t), z)))*_k), 1)

F__x = -v__y(x, y, z, t)*(diff(A__x(x, y, z, t), y))*q+v__y(x, y, z, t)*(diff(A__y(x, y, z, t), x))*q+v__z(x, y, z, t)*(diff(A__z(x, y, z, t), x))*q-v__z(x, y, z, t)*(diff(A__x(x, y, z, t), z))*q-(diff(varphi(x, y, z, t), x))*q-(diff(A__x(x, y, z, t), t))*q

 (6)
> 

collect(F__x = -v__y(x, y, z, t)*(diff(A__x(x, y, z, t), y))*q+v__y(x, y, z, t)*(diff(A__y(x, y, z, t), x))*q+v__z(x, y, z, t)*(diff(A__z(x, y, z, t), x))*q-v__z(x, y, z, t)*(diff(A__x(x, y, z, t), z))*q-(diff(varphi(x, y, z, t), x))*q-(diff(A__x(x, y, z, t), t))*q, [q, v__x(x, y, z, t), v__y(x, y, z, t), v__z(x, y, z, t)])

F__x = (v__y(x, y, z, t)*(diff(A__y(x, y, z, t), x)-(diff(A__x(x, y, z, t), y)))+(diff(A__z(x, y, z, t), x)-(diff(A__x(x, y, z, t), z)))*v__z(x, y, z, t)-(diff(varphi(x, y, z, t), x))-(diff(A__x(x, y, z, t), t)))*q

 (7)
> 

convert(F__x = (v__y(x, y, z, t)*(diff(A__y(x, y, z, t), x)-(diff(A__x(x, y, z, t), y)))+(diff(A__z(x, y, z, t), x)-(diff(A__x(x, y, z, t), z)))*v__z(x, y, z, t)-(diff(varphi(x, y, z, t), x))-(diff(A__x(x, y, z, t), t)))*q, Diff)

F__x = (v__y(x, y, z, t)*(Diff(A__y(x, y, z, t), x)-(Diff(A__x(x, y, z, t), y)))+(Diff(A__z(x, y, z, t), x)-(Diff(A__x(x, y, z, t), z)))*v__z(x, y, z, t)-(Diff(varphi(x, y, z, t), x))-(Diff(A__x(x, y, z, t), t)))*q

 (8)
> 

convert(F__x = (v__y(x, y, z, t)*(diff(A__y(x, y, z, t), x)-(diff(A__x(x, y, z, t), y)))+(diff(A__z(x, y, z, t), x)-(diff(A__x(x, y, z, t), z)))*v__z(x, y, z, t)-(diff(varphi(x, y, z, t), x))-(diff(A__x(x, y, z, t), t)))*q, D)

F__x = (v__y(x, y, z, t)*((D[1](A__y))(x, y, z, t)-(D[2](A__x))(x, y, z, t))+((D[1](A__z))(x, y, z, t)-(D[3](A__x))(x, y, z, t))*v__z(x, y, z, t)-(D[1](varphi))(x, y, z, t)-(D[4](A__x))(x, y, z, t))*q

 (9)

 
Hello everyone, in the result of this command execution process, it appears that the symbols for partial derivatives and derivatives in equation (8) are displayed incorrectly. What should I do?

Download error_display.mw

Nice recursion to solve...

Asked by: Alfred_F 295
sum recurrence limit
November 05 2024
0  2

After exertion with ordinary differential equations now relaxation:

Determine the formation law, limit and sum limit for
u_n+3=(13/12)*u_n+2 - (3/8)*u_n+1 + (1/24)*u_n .
Starting values ​​u_1=0, u_2=1, u_3=1.

How calculate Adomian polynomial by different algo...

Asked by: salim-barzani 1390
ode pde differential-equations
November 04 2024
0  22

there is four formula for calculate them which i know them by name of author the first one is adomian second one is (BiazarShafiofAdomian) which one member of mableprimes write code for me,but i don't know how use for all kind function maybe in future i upload this program for fix this issue, the third one is by zhao which is i think i easy for calculate just  i need someone one to wite the program and do some test for some example i  upload some picture in case for getting algorithm to writting and have some example for testing  so  lets see who can do this algorithm is very usfule when we solve ODE or PDE by LDM, also last method is by taking integral have a good method, in this question this algorithm is zhao which is usfull one

Exploring Oscillators: A Simple Pendulum Worksheet...

by: Paras31 230 Maple 2024
ode education differential-equations
November 03 2024
1  0

Over the last few days, I’ve been creating worksheets on oscillators to support my class’s understanding of these fundamental physics concepts. I wanted to share one of these worksheets that I found particularly useful for illustrating energy exchange and motion dynamics.

A simple pendulum is a classic physics example that exhibits periodic motion. It consists of a mass m (called a bob) attached to a string or rod of length L, which swings back and forth under the influence of gravity. When the bob is displaced from its equilibrium position and released, it swings back and forth under the influence of gravity.
To derive the equation of motion, we can examine the forces acting on the pendulum bob and use Newton’s second law.

Period of a Pendulum:

• 

Frequency (f) "-" the number of cycles the pendulum completes in one second. Measued in hertz ("Hz)."

f = 1/T

• 

Period ("T) -" the time it takes the pendulum to complete one cycle. Measued in seconds (s).

T = 2*Pi*sqrt(L/g)

This period depends only on the length Land gravitational acceleration "g,"meaning it is independent of the amplitude for small oscillations.

What is the period and the frequency of a single pendulum that is 70 cm long on the earth and on the moon?

> 

L := .7; g__earth := 9.8; g__moon := 1.6

.7

  

9.8

  

1.6

 (1)
> 

T__earth := 2*Pi*sqrt(L/g__earth); f__earth := L/T__earth

1.679251909

  

.4168522878

 (2)
> 

T__moon := 2*Pi*sqrt(L/g__moon); f__moon := L/T__moon

4.155936442

  

.1684337597

 (3)

The above image is taken from https://www.researchgate.net/publication/365297210_Scientific_counterfactuals_as_make-believe

1. 

Forces on the Pendulum Bob:

The main forces acting on the bob are:

• 

The gravitational force"`f__g`=mg, "acting vertically downward.
• 

The tension Tauin the string, acting along the string toward the pivot point.
2. 

Components of the Gravitational Force:

Since the pendulum swings in an arc, it’s helpful to resolve the gravitational force into two components:

• 

Radial Component (along the string): This component, "`f__y`=mgcostheta ," is countered by the tension in the string and does not contribute to the pendulum’s motion.
• 

Tangential Component (perpendicular to the string): This component, f__z = -`mgsin&theta;`(restoring force), acts along the arc of the pendulum’s swing and is responsible for its motion.
3. 

Applying Newton's Second Law
Since the tangential component of the gravitational force causes the pendulum’s motion, we can apply Newton's second law in the tangential direction:``

f__X = ma__x

Substituting for f__x and the tangential acceleration a__xNULL

m*(diff(s(t), t, t)) = -`mgsin&theta;`

where diff(s(t), t, t) = a__x and a__x = diff(x, t, t)

Now, we want to write everything in terms of θ

s = `L&theta;`

we obtain:

diff(theta(t), t, t) = -g*`sin&theta;`/L

Small-Angle Approximation

For small angles (typically) theta  , the approximation

`&approx;`(sin(theta), theta)(where theta is in radians) can be used. This simplifies the equation:

diff(theta(t), t, t) = -g*theta/L

This equation is now in the form of a simple harmonic oscillator

diff(theta(t), t, t) = -omega^2*theta

where omega = sqrt(g/L)is the angular frequency of the pendulum.

> 

restart; with(plots); with(DEtools)
> 

L := 1; m := .2; g := 9.8

1

  

.2

  

9.8

 (4)
> 

T := 2*Pi*sqrt(L/g)

2.007089924

 (5)
> 

omega := sqrt(g/L)

3.130495168

 (6)
> 

ODE__1 := diff(theta(t), t, t)+omega^2*theta = 0; IC := theta(0) = A, (D(theta))(0) = 0

diff(diff(theta(t), t), t)+9.799999997*theta(t) = 0

  

theta(0) = A, (D(theta))(0) = 0

 (7)
> 

sol := dsolve({IC, ODE__1}, theta(t))

theta(t) = A*cos((1/100000)*97999999970^(1/2)*t)

 (8)
> 

plot_1 := subs(A = 0.873e-1, sol); plotsresult := plot([rhs(plot_1)], t = 0 .. 2, color = [red])

 
> 

`&theta;_t` := rhs(subs(A = 0.873e-1, sol)); v_t := diff(`&theta;_t`, t)

0.873e-1*cos((1/100000)*97999999970^(1/2)*t)

  

-0.8730000000e-6*97999999970^(1/2)*sin((1/100000)*97999999970^(1/2)*t)

 (9)
> 

T := (1/2)*m*L^2*v_t^2; V := m*g*L*(1-cos(`&theta;_t`)); H := T+V

0.7468864200e-2*sin((1/100000)*97999999970^(1/2)*t)^2

  

1.96-1.96*cos(0.873e-1*cos((1/100000)*97999999970^(1/2)*t))

  

0.7468864200e-2*sin((1/100000)*97999999970^(1/2)*t)^2+1.96-1.96*cos(0.873e-1*cos((1/100000)*97999999970^(1/2)*t))

 (10)
> 

energy_plot := plot([eval(T), eval(V), eval(H)], t = 0 .. 5, color = [red, blue, green], legend = ["Kinetic Energy", "Potential Energy", "Total Energy"], title = "Energy Exchange in Simple Pendulum", labels = ["Time (s)", "Energy (Joules)"])

 
> 

directionfield := DEplot([diff(theta(t), t) = v(t), diff(v(t), t) = -omega^2*theta(t)], [theta(t), v(t)], t = 0 .. 20, theta = -20 .. 20, v = -40 .. 40, arrows = medium, title = "Direction Field for Simple Harmonic Oscillator", axes = boxed, color = navy)
> 

sol1 := dsolve({ODE__1, theta(0) = 3, (D(theta))(0) = 0}, theta(t)); sol2 := dsolve({ODE__1, theta(0) = 6.5, (D(theta))(0) = 0}, theta(t)); sol3 := dsolve({ODE__1, theta(0) = -8, (D(theta))(0) = 0}, theta(t)); sol4 := dsolve({ODE__1, theta(0) = 9.7, (D(theta))(0) = 2.5}, theta(t))

theta(t) = 3*cos((1/100000)*97999999970^(1/2)*t)

  

theta(t) = (13/2)*cos((1/100000)*97999999970^(1/2)*t)

  

theta(t) = -8*cos((1/100000)*97999999970^(1/2)*t)

  

theta(t) = (25000/9799999997)*97999999970^(1/2)*sin((1/100000)*97999999970^(1/2)*t)+(97/10)*cos((1/100000)*97999999970^(1/2)*t)

 (11)
> 

theta1 := rhs(sol1); theta2 := rhs(sol2); theta3 := rhs(sol3); theta4 := rhs(sol4)

3*cos((1/100000)*97999999970^(1/2)*t)

  

(13/2)*cos((1/100000)*97999999970^(1/2)*t)

  

-8*cos((1/100000)*97999999970^(1/2)*t)

  

(25000/9799999997)*97999999970^(1/2)*sin((1/100000)*97999999970^(1/2)*t)+(97/10)*cos((1/100000)*97999999970^(1/2)*t)

 (12)
> 

v1 := diff(theta1, t); v2 := diff(theta2, t); v3 := diff(theta3, t); v4 := diff(theta4, t)

-(3/100000)*97999999970^(1/2)*sin((1/100000)*97999999970^(1/2)*t)

  

-(13/200000)*97999999970^(1/2)*sin((1/100000)*97999999970^(1/2)*t)

  

(1/12500)*97999999970^(1/2)*sin((1/100000)*97999999970^(1/2)*t)

  

(5/2)*cos((1/100000)*97999999970^(1/2)*t)-(97/1000000)*97999999970^(1/2)*sin((1/100000)*97999999970^(1/2)*t)

 (13)
> 

phase_plot := plot([[eval(theta1, t = tval), eval(v1, t = tval), tval = 0 .. 20], [eval(theta2, t = tval), eval(v2, t = tval), tval = 0 .. 20], [eval(theta3, t = tval), eval(v3, t = tval), tval = 0 .. 20], [eval(theta4, t = tval), eval(v4, t = tval), tval = 0 .. 20]], style = line, title = "Phase Portrait for Simple Harmonic Oscillator", labels = ["x (Displacement)", "v (Velocity)"], color = ["red", "blue", "green", "orange"], axes = boxed)
> 

display(directionfield, phase_plot)

 
> 

theta_at_1_sec := subs(t = 1, A = 0.873e-1, rhs(sol)); evalf(theta_at_1_sec)

0.873e-1*cos((1/100000)*97999999970^(1/2))

  

-0.8729462437e-1

 (14)
 

NULL

Download Simple_Pendulum.mw

How collect series solution of PDE?...

Asked by: salim-barzani 1390
pde differential-equations laplace
November 02 2024
1  31

this example is easiest one for getting solution but i can't collect each part and do like elite i can do each part seperatly but it take to much time i want collect solution and do by easier way if possible this is a laplace adomian decomposition methd which contain adomian polynomial too i want upgrade the code, can any one help the  process for get better vision of this topic 
i do upload some picture and my mw. for more undrestanding

and please can any one explan why when i take laplace why is write D[2](u)(x, 0) must be D[1]?

> 

restart
> 

with(inttrans)
> 

with(PDEtools)
> 

with(LinearAlgebra)
> 

NULL
> 

with(SolveTools)
> 

undeclare(prime)

`There is no more prime differentiation variable; all derivatives will be displayed as indexed functions`

 (1)
> 

with(PDEtools)
> 

declare()

`Declared :`

  

u(x, t)*`to be displayed as`*u

  

`The prime differentiation variable has not been declared yet`

  

`Displayed derivatives and declared functions will be copied and pasted "as they were entered"`

 (2)
> 

declare(u(x, t))

u(x, t)*`will now be displayed as`*u

 (3)
> 

eq := diff(u(x, t), `$`(t, 2))+u(x, t)^2-(diff(u(x, t), x))^2 = 0

diff(diff(u(x, t), t), t)+u(x, t)^2-(diff(u(x, t), x))^2 = 0

 (4)
> 

NULL
> 

eqs := laplace(eq, t, s)

s^2*laplace(u(x, t), t, s)-(D[2](u))(x, 0)-s*u(x, 0)+laplace(u(x, t)^2, t, s)-laplace((diff(u(x, t), x))^2, t, s) = 0

 (5)
> 

solve({eqs}, {laplace(u(x, t), t, s)})

{laplace(u(x, t), t, s) = (s*u(x, 0)+(D[2](u))(x, 0)-laplace(u(x, t)^2, t, s)+laplace((diff(u(x, t), x))^2, t, s))/s^2}

 (6)
> 

subs({u(x, 0) = 0, (D[2](u))(x, 0) = exp(x)}, %)

{laplace(u(x, t), t, s) = (exp(x)-laplace(u(x, t)^2, t, s)+laplace((diff(u(x, t), x))^2, t, s))/s^2}

 (7)
> 

eq3 := invlaplace(%, s, t)

{u(x, t) = exp(x)*t-(int(u(x, _U1)^2*(t-_U1), _U1 = 0 .. t))+int((diff(u(x, _U1), x))^2*(t-_U1), _U1 = 0 .. t)}

 (8)
> 

NULL
> 

NULL
> 

NULL
> 

"u[0](x,t):=exp(x)*t"

proc (x, t) options operator, arrow, function_assign; exp(x)*t end proc

 (9)
> 

n := N

N

 (10)
> 

k := K

K

 (11)
> 

f := proc (u) options operator, arrow; u^2 end proc

proc (u) options operator, arrow; u^2 end proc

 (12)
> 

for j from 0 to 3 do A[j] := subs(lambda = 0, (diff(f(seq(sum(lambda^i*u[i](x, t), i = 0 .. 20), m = 1 .. 2)), [`$`(lambda, j)]))/factorial(j)) end do

(exp(x))^2*t^2

  

2*exp(x)*t*u[1](x, t)

  

u[1](x, t)^2+2*exp(x)*t*u[2](x, t)

  

2*u[1](x, t)*u[2](x, t)+2*exp(x)*t*u[3](x, t)

 (13)
> 

NULL
> 

NULL
> 

n := N

N

 (14)
> 

k := K

K

 (15)
> 

f := proc (u) options operator, arrow; (diff(u, x))^2 end proc

proc (u) options operator, arrow; (diff(u, x))^2 end proc

 (16)
> 

for j from 0 to 3 do B[j] := subs(lambda = 0, (diff(f(seq(sum(lambda^i*u[i](x, t), i = 0 .. 20), m = 1 .. 2)), [`$`(lambda, j)]))/factorial(j)) end do

(exp(x))^2*t^2

  

2*exp(x)*t*(diff(u[1](x, t), x))

  

(diff(u[1](x, t), x))^2+2*exp(x)*t*(diff(u[2](x, t), x))

  

2*(diff(u[1](x, t), x))*(diff(u[2](x, t), x))+2*exp(x)*t*(diff(u[3](x, t), x))

 (17)
> 

"#` know we need find all term of u[0]=exp(x)*t` #` u`[1]=-invlaplace(1/(s^(2))(`A__0`+`B__0`)) u[2]=-invlaplace(1/(s^(2))(`A__1`+`B__1`)) ans so on ... at end i want collect all of them and find final result even if is aproximate and want do test of pde too "
> 

NULL

Download explananing_of_get_solution.mw

Typesetting internal error when Physics update is ...

Asked by: nm 11173
libname physics latex
November 02 2024
0  3

[moderator: The Physics update Library fixes this bug with the same error generated and reported by the same Mapleprimes member on another ODESteps example.]

I have removed Physics update from libname path. 

Now I find I get error calling latex command. When Physics update is on libname, no error.

The error is 

         Error, (in Typesetting:-Parse) too many levels of recursion

I am using worksheet with typesetting extended. But also when I change it to typesetting standard, same error. 

Does this mean one must keep Physics update on libname path for Maple to work OK?

Is this error expected if Physics update is not on libname?

Worksheet below that shows this problem

> 

restart;
> 

interface(version);

`Standard Worksheet Interface, Maple 2024.2, Windows 10, October 29 2024 Build ID 1872373`

CASE 1. With Physics update in libname path, no error

 
> 

restart;

> 

libname;

"C:\Users\Owner\maple\toolbox\2024\Physics Updates\lib", "C:\Program Files\Maple 2024\lib"

> 

ode:=[diff(x__1(t),t)=2*x__1(t)+x__2(t),diff(x__2(t),t)=2*x__1(t)+3*x__2(t)];

[diff(x__1(t), t) = 2*x__1(t)+x__2(t), diff(x__2(t), t) = 2*x__1(t)+3*x__2(t)]

> 

the_output:=Student:-ODEs:-ODESteps(ode,output=typeset):

> 

latex(the_output,'output'=string):

> 

 

CASE 2.  Removing Physics from libname path, gives internal error

 
> 

restart;

> 

libname:=libname[2];

"C:\Program Files\Maple 2024\lib"

> 

ode:=[diff(x__1(t),t)=2*x__1(t)+x__2(t),diff(x__2(t),t)=2*x__1(t)+3*x__2(t)];

[diff(x__1(t), t) = 2*x__1(t)+x__2(t), diff(x__2(t), t) = 2*x__1(t)+3*x__2(t)]

> 

the_output:=Student:-ODEs:-ODESteps(ode,output=typeset):

> 

latex(the_output,'output'=string):

Error, (in Typesetting:-Parse) too many levels of recursion

> 

 

 

 

Download internal_error_from_latex_when_libname_changed_nov_2_2024.mw

why order of odetest makes difference in how it wo...

Asked by: nm 11173
ode remember differential-equations
October 31 2024
3  8

I think I just found one of the most serious problems in Maple I've seen (other than timelimit hanging).

This is using Maple 2024.2 on windows 10.

I'll explain in words the problem, then give worksheet below to reproduce this. I can reproduce this all the time.

I have implicit solution in y(x) to an ode.

If I first solve for y(x) from the solution, so solution is now explicit, then call odetest to check if this explicit solution is correct, and if I use assumptions on the odetest call, and then after that call odetest on the original implicit solution, then odetest fails to verify the implicit solution.

But, if I change the order, and first call odetest to verify the implicit solution first, it verifies it OK !  

So the problem ONLY happens if I change the order of calling odetest and if I use assumptions on the odetest call that was used before on the explicit solution.

This tells me that Maple remembers something from earlier call. Does it remember the assumptions used? If so, this is very risky. As some part of program might call odetest with some assumptions, and another part of the program can use no assumptions.  I thought assuming is only applied to the call it is used at only and will not affect future calls.

Is there a way then to clear all assumptions used on earlier call to Maple command before using the command again? Or to tell Maple not to remember assumptions used on a call?

This is a big problem. It took me 14 hrs of debuging to find it. Order of calls to odetest should not make it behave different.

I hope someone could find solution to this, since now I have no idea what to do as I need to use odetest on explicit and implicit solutions and I do not want the order of calling Maple command to make difference in results.

This worksheet shows the problem. 3 cases are given. Notice that when using assumptions on earlier call to odetest, how it fails to verify the implicit solution in later call.

> 

restart;
> 

interface(version);

`Standard Worksheet Interface, Maple 2024.2, Windows 10, October 29 2024 Build ID 1872373`

> 

Physics:-Version();

`The "Physics Updates" version in the MapleCloud is 1824 and is the same as the version installed in this computer, created 2024, October 31, 14:22 hours Pacific Time.`

> 

kernelopts('assertlevel'=2):

CASE 1. Calling odetest on explicit first with assumptions, make odetest hang when calling on implicit after

 
> 

restart;

> 

IMPLICIT_SOL:=ln(10*x^2 + (10*y(x) + 12)*x + 5*y(x)^2 + 8*y(x) + 4)/5 - (4*arctan((5*y(x) + 5*x + 4)/(5*x + 2)))/5 = (3*ln(2))/5 + (4*arctan(3))/5;
ode:=2*x+3*y(x)+2+(y(x)-x)*diff(y(x),x) = 0;
IC:=y(0) = -2;

(1/5)*ln(10*x^2+(10*y(x)+12)*x+5*y(x)^2+8*y(x)+4)-(4/5)*arctan((5*y(x)+5*x+4)/(5*x+2)) = (3/5)*ln(2)+(4/5)*arctan(3)

2*x+3*y(x)+2+(y(x)-x)*(diff(y(x), x)) = 0

y(0) = -2

> 

#find explicit solution first
EXPLICIT_SOL:=solve(IMPLICIT_SOL,y(x)):

> 

#NOw check the explicit solution. Using assumptions to see the problem
timelimit(30,  ( odetest(y(x)=EXPLICIT_SOL,[ode,IC]) assuming positive, y(x)::positive) );

Error, (in evalr) time expired

> 

#Now odetest hangs on the implicit solution

> 

timelimit(30,  odetest(IMPLICIT_SOL,[ode,IC]) );

Error, (in is/internal/rename) time expired

> 

 

> 

 

(1/5)*ln(10*x^2+(10*y(x)+12)*x+5*y(x)^2+8*y(x)+4)-(4/5)*arctan((5*y(x)+5*x+4)/(5*x+2)) = (3/5)*ln(2)+(4/5)*arctan(3)

y(x)

2*x+3*y(x)+2+(y(x)-x)*(diff(y(x), x)) = 0

y(0) = -2

CASE 2. Calling odetest on implicit solution first, then it DOES NOT hang !!

 
> 

restart;

> 

IMPLICIT_SOL:=ln(10*x^2 + (10*y(x) + 12)*x + 5*y(x)^2 + 8*y(x) + 4)/5 - (4*arctan((5*y(x) + 5*x + 4)/(5*x + 2)))/5 = (3*ln(2))/5 + (4*arctan(3))/5;
ode:=2*x+3*y(x)+2+(y(x)-x)*diff(y(x),x) = 0;
IC:=y(0) = -2;

(1/5)*ln(10*x^2+(10*y(x)+12)*x+5*y(x)^2+8*y(x)+4)-(4/5)*arctan((5*y(x)+5*x+4)/(5*x+2)) = (3/5)*ln(2)+(4/5)*arctan(3)

2*x+3*y(x)+2+(y(x)-x)*(diff(y(x), x)) = 0

y(0) = -2

> 

#notice, no hang now, since called before
timelimit(30,  odetest(IMPLICIT_SOL,[ode,IC]) );

[0, 0]

> 

#NOw check the explicit solution. This will timeout which is OK
EXPLICIT_SOL:=solve(IMPLICIT_SOL,y(x)):
timelimit(30,  ( odetest(y(x)=EXPLICIT_SOL,[ode,IC]) assuming positive, y(x)::positive) );

Error, (in type/complex) time expired

> 

#check again that odetest still verifies the implicit solution OK

> 

timelimit(30,  odetest(IMPLICIT_SOL,[ode,IC]) );

[0, 0]

> 

 

CASE 3. Calling odetest with no assumptions on explicit solution, then it also does not hang

 
> 

restart;

> 

IMPLICIT_SOL:=ln(10*x^2 + (10*y(x) + 12)*x + 5*y(x)^2 + 8*y(x) + 4)/5 - (4*arctan((5*y(x) + 5*x + 4)/(5*x + 2)))/5 = (3*ln(2))/5 + (4*arctan(3))/5;
ode:=2*x+3*y(x)+2+(y(x)-x)*diff(y(x),x) = 0;
IC:=y(0) = -2;

(1/5)*ln(10*x^2+(10*y(x)+12)*x+5*y(x)^2+8*y(x)+4)-(4/5)*arctan((5*y(x)+5*x+4)/(5*x+2)) = (3/5)*ln(2)+(4/5)*arctan(3)

2*x+3*y(x)+2+(y(x)-x)*(diff(y(x), x)) = 0

y(0) = -2

> 

#Now check the explicit solution. but DO NOT use assumptions
EXPLICIT_SOL:=solve(IMPLICIT_SOL,y(x)):
timelimit(30, odetest(y(x)=EXPLICIT_SOL,[ode,IC]));

Error, (in factor/remember) time expired

> 

#check again that odetest still verifies the implicit solution OK

> 

timelimit(30,  odetest(IMPLICIT_SOL,[ode,IC]) );

[0, 0]

> 

 


 

Download order_of_ode_test_makes_difference_oct_31_2024.mw

 

update NOV 2, 2024 6 AM

I found the cause.

Removing PHYSICS from libname, then the problem goes away !

So this is caused by PHYSICS package. For some reason, having Physics package in the libname causes odetest to hang/fail compared to when the physics package is not in the libname path. Worksheet below.

> 

restart;
> 

interface(version);

`Standard Worksheet Interface, Maple 2024.2, Windows 10, October 29 2024 Build ID 1872373`

> 

Physics:-Version();

`The "Physics Updates" version in the MapleCloud is 1824 and is the same as the version installed in this computer, created 2024, October 31, 14:22 hours Pacific Time.`

> 

libname;

"C:\Users\Owner\maple\toolbox\2024\Physics Updates\lib", "C:\Program Files\Maple 2024\lib"

CASE 1. without PHYSICS on libname, it works !!

 
> 

restart;

> 

kernelopts('assertlevel'=2):

> 

libname;

"C:\Users\Owner\maple\toolbox\2024\Physics Updates\lib", "C:\Program Files\Maple 2024\lib"

> 

libname:=libname[2]; #remove PHYSICS

"C:\Program Files\Maple 2024\lib"

> 

libname;

"C:\Program Files\Maple 2024\lib"

> 

IMPLICIT_SOL:=ln(10*x^2 + (10*y(x) + 12)*x + 5*y(x)^2 + 8*y(x) + 4)/5 - (4*arctan((5*y(x) + 5*x + 4)/(5*x + 2)))/5 = (3*ln(2))/5 + (4*arctan(3))/5;
ode:=2*x+3*y(x)+2+(y(x)-x)*diff(y(x),x) = 0;
IC:=y(0) = -2;

(1/5)*ln(10*x^2+(10*y(x)+12)*x+5*y(x)^2+8*y(x)+4)-(4/5)*arctan((5*y(x)+5*x+4)/(5*x+2)) = (3/5)*ln(2)+(4/5)*arctan(3)

2*x+3*y(x)+2+(y(x)-x)*(diff(y(x), x)) = 0

y(0) = -2

> 

#find explicit solution first
EXPLICIT_SOL:=solve(IMPLICIT_SOL,y(x)):

> 

#NOw check the explicit solution. Using assumptions to see the problem
timelimit(30,  ( odetest(y(x)=EXPLICIT_SOL,[ode,IC]) assuming positive, y(x)::positive) );

Error, (in evalr/ln) time expired

> 

#Now try odetest  on the implicit solution

> 

infolevel[odetest]:=5;

5

> 

timelimit(30,  odetest(IMPLICIT_SOL,[ode,IC]) );

odetest: Performing an implicit solution test

odetest: Performing an explicit (try soft) solution test

odetest: Performing an implicit solution (II) test

[0, 0]

> 

 

> 

 

CASE 2. With Physics on libname, it fails !

 
> 

restart;

> 

kernelopts('assertlevel'=2):

> 

libname;

"C:\Users\Owner\maple\toolbox\2024\Physics Updates\lib", "C:\Program Files\Maple 2024\lib"

> 

IMPLICIT_SOL:=ln(10*x^2 + (10*y(x) + 12)*x + 5*y(x)^2 + 8*y(x) + 4)/5 - (4*arctan((5*y(x) + 5*x + 4)/(5*x + 2)))/5 = (3*ln(2))/5 + (4*arctan(3))/5;
ode:=2*x+3*y(x)+2+(y(x)-x)*diff(y(x),x) = 0;
IC:=y(0) = -2;

(1/5)*ln(10*x^2+(10*y(x)+12)*x+5*y(x)^2+8*y(x)+4)-(4/5)*arctan((5*y(x)+5*x+4)/(5*x+2)) = (3/5)*ln(2)+(4/5)*arctan(3)

2*x+3*y(x)+2+(y(x)-x)*(diff(y(x), x)) = 0

y(0) = -2

> 

#find explicit solution first
EXPLICIT_SOL:=solve(IMPLICIT_SOL,y(x)):

> 

timelimit(30,  ( odetest(y(x)=EXPLICIT_SOL,[ode,IC]) assuming positive, y(x)::positive) );

Error, (in evalr/shake) time expired

> 

#Now try odetest  on the implicit solution

> 

infolevel[odetest]:=5;

5

> 

timelimit(30,  odetest(IMPLICIT_SOL,[ode,IC]) );

odetest: Performing an implicit solution test

odetest: Performing an explicit (try soft) solution test

odetest: Performing an implicit solution (II) test

Error, (in is/internal/rename) time expired

 

 

Download order_of_ode_test_makes_difference_NOV_2_2024.mw

my question is: Is it safe to permanently remove Physics package from libname? Why is having physics package in libname (which is by default) causes this problem?

I do not use Physics package explicitly in my code. i.e. I do not do Physics:- calls. 

Any one knows what effect not having Physics on libname cause? Will Maple still work OK for everything if one is not calling Physics package explicitly?

Maple and MapleSim 2024.2 updates

by: TechnicalSupport 750 Maple 2024 , MapleSim 2024
October 31 2024
3  0

We have just released updates to Maple and MapleSim.

Maple 2024.2 includes ability to tear away tabs into new windows, improvements to scrollable matrices, corrections to PDF export, small improvements throughout the math engine, and more.  We recommend that all Maple 2024 users install this update.

This update also include a fix to the problem with the simplify extension mechanism, as first reported on MaplePrimes. Thanks, as always, for helping us make Maple better.

This update is available through Tools>Check for Updates in Maple, and is also available from the Maple 2024.2 download page, where you can find more details.

At the same time, we have also released an update to MapleSim, which contains a variety of improvements to MapleSim and its add-ons. You can find more information on the MapleSim 2024.2 download page.

Which function is more credible?...

Asked by: sursumCorda 1229
polynomial realdomain
October 30 2024
1  1

Here are two systems over the reals:

sys__1:=And(r*(387*r+52)+2<r*(226*q+121*s)+9*q*(q*(2*q-5)-3*s+2)+6*s,4*q^3+r*(27*r+4)+s^2=q*(q+18*r),q>=0,r>=0):
sys__2:=And((392-1739*q)*r+4*(2-9*q)**2+2151*r**2<75*r*s,4*q**3+r*(27*r+4)+s**2=q*(q+18*r),q>=0,r>=0):

The following results indicate that both and are satisfiable 

QuantifierElimination:-QuantifierEliminate(exists([s,q,r],sys__1));
                              true

QuantifierElimination:-QuantifierEliminate(exists([s,q,r],sys__1));
                              true

but RealDomain:-solve simply returns an empty list (that is, no solution exists) in both cases

RealDomain:-solve(sys__1,[q,s,r]); # ⟹ sys1 cannot be satisfied
                               []

RealDomain:-solve(sys__2,[q,s,r]); # ⟹ sys2 cannot be satisfied
                               []

As discussed in the previous problem, in contrast to using QuantifierElimination:-QuantifierEliminate, the use of RealDomain:-solve is unsafe. Nevertheless, the above output suggests that even the much-more-sophisticated QuantifierElimination:-QuantifierEliminate is still not always reliable (since the correct returnedvalue appears to be in lieu of ). So, what is the right command to handle polynomial systems over real domains in Maple? 

How to simplify the following term?...

Asked by: Alfred_F 295
step-by-step radical simplification
October 30 2024
1  4

Since the puzzle task "A circle is to be disturbed ..." makes no fun, here is a Maple task:
The term to be simplified step by step:
(2+10/(3*sqrt(3)))^(1/3)+(2-10/(3*sqrt(3)))^(1/3)

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