Hi,

I'm trying to use the Explore command to examine the effect of two parameters (mu and sigma) on the density function curve. The visualization isn't very optimal, especially with the mu parameter, and it's difficult to add options (range, color, gridlines, etc.). Any suggestions to optimize this idea? Thanks for your insights!

Q_Explore.mw

when we have ode equation we say what is type of equation then  i want solve by this method say the name of method and if possible i want to solve this equation by the method step by step too, maple can do that? also can we plot the solution or any geometricall presentation , also i have error in writing exact form of equation

Download ode-example.mw

i want try all number to my parameter for check the shape of plot there is any way for doing that?

Download control-trajectory.mw

Hi,

just a quick design question regarding my animation. The color "AliceBlue" disappears from my background... Could it be a background option issue?

Q_AlceBlue_background.mw

In the book by W.G. Chinn, N.E. Steenrod "First Concepts of Topology" the another remarkable theorem was proved: any two flat bounded regions can be cut by a single straight line so that each of these regions is divided into two regions of equal area (the second  pancake problem). This is an existence theorem which does not provide any way to find this cut. In this post I made an attempt to find such cut for any 2 convex regions on the plane bounded by a piecewise smooth self-non-intersecting curves.
The Each_Into_2_Equal_Areas procedure returns a list of coordinates of 4 endpoints of the cutting segments. This procedure significantly uses my old procedures  Area  and  Picture , which can be found in detail at the link  https://mapleprimes.com/posts/145922-Perimeter-Area-And-Visualization-Of-A-Plane-Figure-  . The formal arguments of the Each_Into_2_Equal_Areas procedure are the lists  L1  and  L2 specifying the boundaries of the regions to be cut. When specifying  L1  and  L2 , the boundary can be passed clockwise or counterclockwise, but it is necessary that the parameter t (when specifying each link) should go in ascending order. This can always be achieved by replacing  t  with  -t  if necessary. The Pic procedure draws a picture of the source regions and cutting segments. For ease of use, the code for the  Area  and  Picture   procedure is also provided. It is also worth noting that the procedure also works for "not too" non-convex regions (see examples below).

restart;
Area := proc(L) 
local i, var, e, e1, e2, P; 
for i to nops(L) do 
if type(L[i], listlist(algebraic)) then 
P[i] := (1/2)*add(L[i, j, 1]*L[i, j+1, 2]-L[i, j, 2]*L[i, j+1, 1], j = 1 .. nops(L[i])-1) else 
var := lhs(L[i, 2]); 
if type(L[i, 1], algebraic) then e := L[i, 1]; 
if nops(L[i]) = 3 then P[i] := (1/2)*(int(e^2, L[i, 2])) else 
if var = y then P[i] := (1/2)*simplify(int(e-var*(diff(e, var)), L[i, 2])) else 
P[i] := (1/2)*simplify(int(var*(diff(e, var))-e, L[i, 2])) end if end if else e1 := L[i, 1, 1]; e2 := L[i, 1, 2]; 
P[i] := (1/2)*simplify(int(e1*(diff(e2, var))-e2*(diff(e1, var)), L[i, 2])) end if end if end do; 
abs(add(P[i], i = 1 .. nops(L))); 
end proc:

Picture := proc(L, C, N::posint := 100, Boundary::list := [linestyle = 1]) 
local i, var, var1, var2, e, e1, e2, P, Q, h; 
global Border;
uses plottools; 
for i to nops(L) do 
if type(L[i], listlist(algebraic)) then P[i] := op(L[i]) else 
var := lhs(L[i, 2]); var1 := lhs(rhs(L[i, 2])); var2 := rhs(rhs(L[i, 2])); h := (var2-var1)/N; 
if type(L[i, 1], algebraic) then e := L[i, 1]; 
if nops(L[i]) = 3 then P[i] := seq(subs(var = var1+h*i, [e*cos(var), e*sin(var)]), i = 0 .. N) else 
P[i] := seq([var1+h*i, subs(var = var1+h*i, e)], i = 0 .. N) fi else e1 := L[i, 1, 1]; e2 := L[i, 1, 2]; P[i] := seq(subs(var = var1+h*i, [e1, e2]), i = 0 .. N) fi; fi; od; 
Q := [seq(P[i], i = 1 .. nops(L))]; Border := curve([op(Q), Q[1]], op(Boundary)); [polygon(Q, C), Border] 
end proc:

Each_Into_2_Equal_Areas:=proc(L1::list, L2::list)
local D, n, m, L10, L20, S1,S2, f, L11, L21, i, j, k, s, A, B, C , sol;

f:=(X,Y)->expand((y-X[2])*(Y[1]-X[1])-(x-X[1])*(Y[2]-X[2]));
L10:=map(p->`if`(type(p,listlist),[[p[1,1]+t*(p[2]-p[1])[1],p[1,2]+t*(p[2]-p[1])[2]],t=0..1],p), L1);
L20:=map(p->`if`(type(p,listlist),[[p[1,1]+t*(p[2]-p[1])[1],p[1,2]+t*(p[2]-p[1])[2]],t=0..1],p), L2);
S1:=Area(L1); S2:=Area(L2);  
n:=nops(L1); m:=nops(L2);

for i from 1 to n do
for j from i to n do

for k from 1 to m do
for s from k to m do

if not ((nops({i,j})=1 and type(L1[i],listlist)) or (nops({k,s})=1 and type(L2[k],listlist))) then

A:=eval(L10[i,1],t=t1): 
B:=eval(L10[j,1],t=t2):
C:=eval(L20[k,1],t=t3): 
D:=eval(L20[s,1],t=t4):

L11:=`if`(j=i,[subsop([2,2]=t1..t2,L10[i]),[B,A]],`if`(j=i+1,[subsop([2,2]=t1..op([2,2,2],L10[i]),L10[i]),subsop([2,2]=op([2,2,1],L10[j])..t2,L10[j]),[B,A]], [subsop([2,2]=t1..op([2,2,2],L10[i]),L10[i]),op(L10[i+1..j-1]),subsop([2,2]=op([2,2,1],L10[j])..t2,L10[j]),[B,A]])):

L21:=`if`(s=k,[subsop([2,2]=t3..t4,L20[k]),[D,C]],`if`(s=k+1,[subsop([2,2]=t3..op([2,2,2],L20[k]),L20[k]),subsop([2,2]=op([2,2,1],L20[s])..t4,L20[s]),[D,C]], [subsop([2,2]=t3..op([2,2,2],L20[k]),L20[k]),op(L20[k+1..s-1]),subsop([2,2]=op([2,2,1],L20[s])..t4,L20[s]),[D,C]])):

sol:=fsolve(simplify({Area(L11)-S1/2,Area(L21)-S2/2,eval(f(A,B),[x=C[1],y=C[2]]),eval(f(A,B),[x=D[1],y=D[2]])}),{t1=op([2,2,1],L10[i])..op([2,2,2],L10[i]),t2=op([2,2,1],L10[j])..op([2,2,2],L10[j]),t3=op([2,2,1],L20[k])..op([2,2,2],L20[k]),t4=op([2,2,1],L20[s])..op([2,2,2],L20[s])});

if type(sol,set(`=`)) then  return eval([A,B,C,D],sol) fi;

fi;
od: od: od: od:
end proc:

Pic := proc(L1,L2,col1,col2,Size:=[800,400])
local P1, P2, P3, T, P;
uses plots, plottools;
P1, P2 := Picture(L1, color=col1), Picture(L2, color=col2):
P3 := line(Sol[1..2][],color=red,thickness=3), line(Sol[3..4][],color=red,thickness=3), line(Sol[1],Sol[4],linestyle=2,thickness=3,color=red):
T:=textplot([[Sol[1][],"A"],[Sol[2][],"B"],[Sol[3][],"C"],[Sol[4][],"D"]], font=[times,18], align=[left,above]);
P:=pointplot(Sol, symbol=solidcircle, color=red, symbolsize=10);
display(P1,P2,P3,T,P, scaling=constrained, size=Size, axes=none);
end proc: 

Examples of use.

local D:
L1:=[[[8,0],[6,7]],[[6,7],[2,5]],[[2,5],[0,2]],[[0,2],[0,0]],[[0,0],[8,0]]]:
L2:=[[[5*cos(t)+16,5*sin(t)],t=Pi/2..Pi],[[5*cos(t)+16,5*sin(t)/2],t=Pi..2*Pi],[[21,0],[16,5]]]:
Sol:=Each_Into_2_Equal_Areas(L1,L2): Points:=[A,B,C,D]:
seq(Points[i]=Sol[i], i=1..4);
Pic(L1,L2,"Yellow","LightBlue",[900,400]);

The specified regions may overlap:

L1:=[[[8,0],[6,7]],[[6,7],[2,5]],[[2,5],[0,2]],[[0,2],[0,0]],[[0,0],[8,0]]]:
L2:=[[[5*cos(t)+9,5*sin(t)],t=Pi/2..Pi],[[5*cos(t)+9,5*sin(t)/2],t=Pi..2*Pi],[[14,0],[9,5]]]:
Sol:=Each_Into_2_Equal_Areas(L1,L2):  Points:=[A,B,C,D]:
seq(Points[i]=Sol[i], i=1..4);
Pic(L1,L2,"Yellow","LightBlue");

If there is a solution for which the cutting segments intersect the boundary of each of the regions at 2 points, then the procedure also works for such non-convex regions:

L1:=[[[cos(t),sin(t)],t=Pi/3..2*Pi-Pi/3],[[cos(-t)+1,sin(-t)],t=-Pi-Pi/3..-Pi+Pi/3]]:
L2:=[[[cos(t)+2,sin(t)],t=-Pi/6..Pi+Pi/6],[[cos(-t)+2,sin(-t)-1],t=-5*Pi/6..-Pi/6]]:
Sol:=Each_Into_2_Equal_Areas(L1,L2): Points:=[A,B,C,D]:
seq(Points[i]=Sol[i], i=1..4);
Pic(L1,L2,"Yellow","LightBlue");

A number of other interesting examples can be found in the attached file.

Each_Into_2_Equal_Areas1.mw

THis IC for Abel ode is not valid and should result in no solution. But instead of returning NULL, dsolve throws internal error called Error, (in dsolve) invalid limiting point

Download internal_error_instead_of_no_solution.mw

Is there an equivalent of currentdir() which instead of returning current working directory, returns the directory of the file being read. So assume I have an ".mm" or ".mpl" file saved in some location and there is another file with a location fixed relative to this file, but not fixed in absolute location on the computer and not fixed relative with current working directory. One natural thing is to have a line in the first file that takes its current location (not current working location of the user!) and then use the fixed relative path info, and then gives the location of the second file to the user. Using currendir won't help here because current working directory is not necessarily the same as the location of the file you are reading, the first file. One may say the user is reading the first file so he knows already its location, right? Well, if the user is also the writer of the file, sure, not a problem, he can manually edit the lines of the file and put that location inside the file instead of using currentdir etc. However, if the user is not the writer and also does not know how to edit or whatever else reason, then the file should be able to inform Maple of its own location, so that's why something like currentdir() but not for the working directory, instead for currently being read file's directory is helpful. I checked FileTools package quickly and couldn't notice anything like what I want. Anyone have any clue on name of such command if exists or any other trick that does what I want (except expecting the user to do something ^_^ so all from the writer's side please ^_^).

Hi, I want to generate a matrix of variables, say, something like, 

n:=4:

Matrix(n, n, [[a_11, a_12, a_13, a_14], [a_12, a_22, a_23, a_24], [a_13, a_23, a_33, a_34], [a_14, a_24, a_34, a_44]]) :

How can I do it automatically for arbitrary values of n?

How can I convert sinh(x)/cosh(x) to tanh(x) ?

One of the most difficult/nice tasks from my collection of old problems:
Calculate Int(from 0 to 1)Int(from 0 to 1)[1/(1-x*y)]dxdy .

Hi,

I am looking to design a learning sequence on the applications of integral calculus (arc length, area under the curve, solid of revolution). The animation of the solid is not optimal because I want to display the revolution over [1,2] while maintaining a global view of the curve over [-3,3]. Thank you for your insights.

S7_Arc_Intégrale_Animation.mw

i try find some part of solution of this kind of pde but i can't get results my openion is maybe this pde is wronge when i defined 

pde.mw

Let eps be a real number greater than zero. Calculate the limit lim(x-->oo)[x^(1-eps)*Integral(from x to x+1) sin(t^2)dt] .

my solution is a little bit long which when i click on pdetest command i wait at least more than a hour but still is runing and i don't get any result and not give me error , which i don't know my result is true or not so How i can find that my pde by this solution it will be zero or not ?

Download test.mw

Hi,

I'm looking to animate the Riemann sum only on the interval [0,2], while keeping a global view of the curve (G). I tried the background option, but without success. Any ideas? Thanks!

S7_Riemann_Animation_Aire.mw