Good afternoon.

I have a differential equation of non-integer degree and would like to know if it is possible to express a solution in terms of elementary or special-functions for certain values of the exponent, n>0.

For this equation, Maple provides an analytical solution for the exponent values n=0 and n=1, otherwise, there is no solution returned. I am particularly interested in the cases where n=1/2, 3/2, 2, 5/2, and 3

I am hoping that someone can help me resolve this - if a closed-form solution is not possible, then a numerical solution would also be welcome.

I have provided the details in the attached worksheet.

Thanks for reading!

MaplePrimes_Dec_19.mw

Given the center x1 of a circle in R^2 with radius d12, and a point p2 on the circle, so that d12=||p2-x1||, denote the points on the line segment from x1 to p2 as x1(t) = x1+t*v12, with t=0..d12, and v12 =( p2-x1)/d12.  I want to animate the points x1(t) moving along the line segment from x1 to p2 and the corresponding circles of decreasing radius, with center x(t) and radius d12-t, so that p2 remains on the circle.

I can animate the points along the line segment from x1 to p2 using ‘style=point, symbol=solidcircle’.

I would like to use plottools-circle, to plot the circles. I have also tried the following type commands for the circles of decreasing radius.

Plot([x1(1)+t*v12(1)+(d12-t)*cos(theta)*v12(1)+ (d12-t)*sin(theta)*u12(1), x1(2)+ +t*v12(2)+(d12-t)*d12*cos(theta)*v12(2)+(d12-t)*sin(theta)*u12(2), theta=0..2*PI]

where u12 is a unit vector orthogonal to v12.

I have not been able to combine the two plots into an animation. Thank you

On considère une ellipse x^2/a^2+y^2/b^2-1=0 et 2 sommets de cette ellipse A(a,0) et B(0,b). On imagine une hyperbole équilatère variable passant par les points O, A et B. Cette courbe rencontre l'ellipse en 2 autres points A1 et B1. Montrer que la droite A1B1 passe par un point fixe. Même avec l'intelligence artificielle, je ne parviens pas à résoudre ce problème. Pourriez-vous d'aider. Merci.

Machine translation by moderator:

We consider an ellipse x^2/a^2+y^2/b^2-1=0 and 2 vertices of this ellipse A(a,0) and B(0,b). We imagine a variable equilateral hyperbola passing through the points O, A and B. This curve meets the ellipse at 2 other points A1 and B1. Show that the line A1B1 passes through a fixed point. Even with artificial intelligence, I can't solve this problem. Could you help. Thank you.

I didn't put it in the title, but of course this is a post about Advent of Code, in particular Days 16 and 18 which feature a perenial favorite type of problem: finding shortest paths in mazes.

Your input for these is always a maze given as an ascii map.  Like so:

###############
#.......#....E#
#.#.###.#.###.#
#.....#.#...#.#
#.###.#####.#.#
#.#....#....#.#
#.#.#####.###.#
#...........#.#
###.#.#####.#.#
#...#.....#.#.#
#.#.#.###.#.#.#
#.....#...#.#.#
#.###.#.#.#.#.#
#S......#.#...#
###############

There's lots of ways to import one of these into Maple and then solve the maze, but I am to highlight how to do it with GraphTheory.  I am going to start with a GridGraph and then remove the walls in order to leave a just the vertices that represent the paths:

with(StringTools): with(GraphTheory):
maze:=
"###############
#.......#....E#
#.#.###.#.###.#
#.....#.#...#.#
#.###.#####.#.#
#.#....#....#.#
#.#.#####.###.#
#...........#.#
###.#.#####.#.#
#...#.....#.#.#
#.#.#.###.#.#.#
#.....#...#.#.#
#.###.#.#.#.#.#
#S......#.#...#
###############
":
mazelines := (Split(Trim(maze), "\n")):
sgrid := ListTools:-Reverse((map(Explode, mazelines)) ):
m,n := nops(sgrid), nops(sgrid[1]);
tgrid := table([seq(seq([i,j]=sgrid[i,j],i=1..m),j=1..n)]):
start := lhs(select(e->rhs(e)="S", [entries(tgrid,'pairs')])[]);
finish := lhs(select(e->rhs(e)="E", [entries(tgrid,'pairs')])[]);

Now the maze is stored in the table tgrid, and it is easy to find the walls and paths.  In a GridGraph the vertices are labeled with their coordinates as "x,y" and so we rewrite our list of paths in that form, so we can create the induced subgraph of the Grid that includes only those vertices.

(walls,paths) := selectremove(e->rhs(e)="#", [entries(tgrid, 'pairs')]):
paths := map(s->sprintf("%d,%d",lhs(s)[]), paths):
H := SpecialGraphs:-GridGraph(m,n);
G := InducedSubgraph(H, paths);

We can use StyleVertex to highlight the start and finish.

StyleVertex(G, sprintf("%d,%d",start[]), color="LimeGreen");
StyleVertex(G, sprintf("%d,%d",finish[]), color="Red");

plots:-display(<
DrawGraph(H, stylesheet=[vertexshape="square", vertexborder=false, vertexcolor="Black"], showlabels=false) | 
DrawGraph(G, stylesheet=[vertexshape="square", vertexborder=false, vertexcolor="Black"], showlabels=false)>);

(I omitted a step where I set the vertex locations of the maze grid, you can see that in the attached worksheet)

Now finding a path through the maze is as easy as calling GraphTheory:-ShortestPath

sp := ShortestPath(G, sprintf("%d,%d",start[]), sprintf("%d,%d",finish[]) ):

StyleVertex(G, sp[2..-2], color="Orange");
StyleEdge(G, [seq({sp[i],sp[i+1]}, i=1..nops(sp)-1)], color="Orange");
DrawGraph(G, stylesheet=[vertexshape="square", vertexpadding=10, vertexborder=false,
             vertexcolor="Black"],  showlabels=false, size=[800,800]);

Now, Advent of Code seldom gives you a completely simple maze like this, often these is a twist like having to calculate the costs of turns seperately from the cost of steps, or each direction or position has a seperate cost associated with it.

For example, Day 16 has us starting facing east, and then turns cost 1000, while moving forward costs 1. That sort of problem is no longer exactly a maze, instead of the vertices being representing an "x,y" position, instead you increase the number of vertices by a factor of 4, so that you have a vertex for every position and orientation "x,y,o" with edges of weight 1 between adjacent vertices with the same orientation and edges of wieght 1000 to connect "x,y,N" to "x,y,E" and "x,y,W" e.g.  In that sort of weighted graph, we can use GraphTheory:-DijkstrasAlgorithm to find the shortest path and it's weighted cost.

In this code, we expand our list of maze locations with directions, and the use the grid table to generate a list of weighted edges:

dtable := table([0=[0,1], 1=[1,0], 2=[0,-1], 3=[-1,0]]):
dname := table([0="N",1="E",2="S",3="W"]):
dpaths := map(s->local d;seq(cat(s,",",d), d in ["N","E","S","W"]), paths):

edges := NULL:
for i from 1 to m do for j from 1 to n do
    if tgrid[[i,j]] = "#" then next; end if;
    for d from 0 to 3 do
        dir := dtable[d];
        if tgrid[[i,j]+dir] <> "#" then
            edges := edges, [{cat("",i,",",j,",",dname[d]), cat("",i+dir[1],",",j+dir[2],",",dname[d])},1];
        end if;
        edges := edges, [{cat("",i,",",j,",",dname[d]), cat("",i,",",j,",",dname[d+1 mod 4])}, 1000],
                 [{cat("",i,",",j,",",dname[d]), cat("",i,",",j,",",dname[d-1 mod 4])}, 1000];
    end do;
end do; end do:

Gd := Graph(dpaths,weighted,{edges});

Once that is done, it's a simple matter of calling Dijkstra's Algorithm on the graph, but notice that we can reach the finsh while traveling north or east, so we need to find the sortest path to both (you can pass a list of vertices to Dijkstra, and it will efficiently calculate paths to all of them), and select the smaller of the two:

spds := DijkstrasAlgorithm(Gd, cat("",start[1],",",start[2],",E"), 
    [cat("",finish[1],",",finish[2],",N"), cat("",finish[1],",",finish[2],",E")] , 
    distance):
i := min[index](map2(op,2,spds)):
spd := spds[i];

spd := [["2,2,E", "3,2,E", "4,2,E", "4,2,N", "4,3,N", "4,4,N", "4,5,N", "4,6,N", "4,6,E", "5,6,E",
 "6,6,E", "7,6,E", "8,6,E", "8,6,N", "8,7,N", "8,8,N", "8,9,N", "8,10,N", "8,11,N", "8,12,N", 
"8,12,W", "7,12,W", "6,12,W", "5,12,W", "4,12,W", "3,12,W", "2,12,W", "2,12,N", "2,13,N", 
"2,14,N", "2,14,E", "3,14,E", "4,14,E", "5,14,E", "6,14,E", "7,14,E", "8,14,E", "9,14,E", 
"10,14,E", "11,14,E", "12,14,E", "13,14,E", "14,14,E"], 6036]

We can then plot to compare this to the unweighted shortest path:

dsp := ListTools:-MakeUnique( map(s->s[1..-3], spd[1]) );
StyleVertex(G, dsp[2..-2], color="DarkBlue");
StyleEdge(G, [seq({dsp[i],dsp[i+1]}, i=1..nops(dsp)-1)], color="DarkBlue");

DrawGraph(G, stylesheet=[vertexshape="square", vertexpadding=10,
             vertexborder=false, vertexcolor="Black"],  showlabels=false,
          size=[800,800]);

And you can see it's a path that requires more steps, but definitely uses fewer turns if we start facing east/right (6 vs. 9):

I hope this has given you a little bit of a flavor of how to use GraphTheory commands to solve path finding problems.  Like with the second part here, usually the biggest challenge is figuring out how to encode and construct a graph that represents your problem.  Then the actual commands to solve it, are easy. You can see all the code, and a couple steps I left out from above in this worksheet: Mazeblog.mw

And just for fun, here's a Maple workbook that imports a maze from an image and solves it: MazeFromImage.maple

I have this system

eq1 := diff(x(t), t)-(1/6)*(6*x(t)^3*y(t)+(2*y(t)^2-2)*x(t)^2+3*y(t)*(z(t)-2)*x(t)-2*y(t)^2+2)*sqrt(3) = 0;
                         
eq2 := diff(y(t), t)-(1/6)*(y(t)-1)*sqrt(3)*(y(t)+1)*(6*x(t)^2+2*y(t)*x(t)+3*z(t)-2) = 0;
                                    
eq3 := diff(z(t), t)-(1/3)*z(t)*sqrt(3)*(6*y(t)*x(t)^2+2*x(t)*y(t)^2+3*z(t)*y(t)-2*x(t)-3*y(t)) = 0;

I solved it numerically using these ics

ics := x(0) = -0.01, y(0) = .99, z(0) = 0.01

sol1 := dsolve({ics, op(syst)}, {x(t), y(t), z(t)}, type = numeric, output = listprocedure)

I need to use the x(t), y(t), z(t) as follows
  X :=  eval(x(t), sol1)
  Y :=  eval(y(t), sol1)

Z :=  eval(z(t), sol1)

to solve the following system for P(t), Q(t), R(t) numerically 
eq4 := diff(R(t), t)-P(t)*Z-(-2*(-Y^2+2)*X/sqrt(3)+sqrt(3)*(-2*X^2+Z+4/3)*Y)*R(t) = 0;
eq5 := diff(Q(t), t)-(2/3)*R(t)+2*((1/3)*Y+X)*P(t)/sqrt(3)-(-2*(-Y^2+2)*X/sqrt(3)+2*sqrt(3)*(X^2-(1/2)*Z-2/3)*X)*Q(t) = 0;
eq6 := diff(P(t), t)+(1/2)*R(t)+2*sqrt(3)*X*Q(t)+(2*(-Y^2+2)*X/sqrt(3)+sqrt(3)*(-2*X^2+Z+1)*Y)*P(t) = 0;

Any help please? 

Hey Guys, 

I have to solve multiple system of equations under some restrictions given as inequalities. Sometimes solve is not able to find the result in houres so I tryd to break the problem in half. So in the first step I just want to solve my 8 polynomial equations with 8 variables and in a second step I want so take the solutions, bring them together with the set of inequalities and solve it again. Since also some sets of equations are to hard for the simple solve command I got the advice from people of this plattform to try PolynomialSystem with the diffrent engines. However I have the feeling they make misstakes and now Im not sure If I can trust my results. 

Attached you can find a file with an example. In the beginning I solve equations and restrictions together and there is a solutions. Then I tryd to solve only the equations with PolynomialSystem and the the four known engines and the eniges traditional and backsolve dont find the solution which as we saw before exist. When a soultions holds under restrictions it should always appear if I omit the restirctions. When I use the enige triade and groebner then the right solution is there. 
However in some other cases it feels the other way round.
So to me it looks like no matter which engine I take, I can never 100% trust my results. Did I something wrong? Whats the reason for those mistakes? Furthermore backsolve gives me 7 solutions, but solutions 2 and 7 are the same. I also recognized, that there is a diffrence between putting in the variable vars as a list or a set. What happens, if I dont specify which engine should be used?

I am happy about any advice. Thank you in advance.

Regards

Felix

Can_I_trust_the_diffrent_eniges_of_Polynomial_Systems.mw

I have a simple question, but don't find a neat solution for it. I have 4 3d Vectors that result in the resulting Vector V5. I now want the unit vector of V5. How do I do it the most efficient way? 

My attempt is in the file attached. I get the error message: "Error, (in rtable/Power) exponentiation operation not defined for Vectors"

Thanks a lot for your help.3d_vector.mw

Assuming I have a sumpro function written in Maple 2016. How can I implement it in C# and what is the process?

Please help me.

sumpro := proc (i) local a, b;

        a := (rand(1 .. 10))(); b := (rand(1 .. 10))();

        print("Sum of ", a, " and ", b, " is ", a+b)

end proc;

save sumpro, "D://Sumpro.m"

Download mapleprime_sumpro_to_c.mw

Hi

If possible, please help me write the steps to solve the following equation.

By setting the coefficients of the same power (Yⁱ) on both sides of equation equal, we solution get

I have some cubic and quartic equations with complex cofficients. Maple 2015 is able to solve these and returns the roots as labelled sets, so I can do things like "plot S[1]". I want to vary some parameters in the coefficients, and see what happens to the roots.
My problem is when I log out and then rerun the code, the labels 1,2,3,(4) are frequently attached to different roots than they were the first time. This is both unexpected and inconvenient. Is there any way to ensure that the same roots are always given the same labels?

[moderator: see also this Question from 2023]

l45 := (x + 7)(x - 1) = (1 + x)^2;
                             "(->)"

Error, (in solve) cannot solve for an unknown function with other operations in its arguments

This is my first time working with plotting data from a matrix. However, with the help of a friends on MaplePrimes, I learned how to plot the data in both Maple and MATLAB. Despite this, I am having trouble with visualization. When I change the delta value, my function experiences vibrations or noise, which is clearly visible in the plot. But when I change delta, I encounter errors with my matrix data. How can I fix this problem? and there is any way for get better visualization by Explore ? also How show this vibration or noise in 2D?

Download data-analysis.mw

A parallelogram is given in the Cartesian coordinate system. If the corner points of the parallelogram are connected to the midpoints of adjacent sides using lines, then the eight connecting lines form an octagon.
It must be proven that its area is one sixth of the parallelogram's area.

is(abs(x)=max(x,-x)) assuming real;

#  FAIL

I wonder if this will work in newer versions of Maple?

These two issues probably came up before, but I can't find where and when searching Maple primes.

So I thought to summarize the issues I see with odetest in one post, in the hope to get clarification on current status on these from the powers who know.

The first issue

The order in which odetest returns the answer. When odetest is called to check the ode and IC, as in 

the_status := odetest(sol,[ode,IC])

One would expect the_status to be a list, where the first entry tells if sol verifies the ode, and the second entry tells if sol verifies IC.

i.e. the order is the same as in the input. right? Since ode is first and IC is second in the input list.

But Maple sometimes mixes the order. See example 1 below. This makes it impossible to determine if the solution verifies the ode or IC,  when one of the entries in the_status is zero and the other is not, since order can be reversed sometimes.

Second issue:

When the solution is implicit, Maple gives invalid odetest result on the IC, unless one rewrites the solution using (lhs-rhs)(sol)=0.

i.e. move everything to the left side of the equation with RHS zero. This happens sometimes and when the solution is implicit.

I have thought this was fixed in current Maple, but it is not.  I remember this came up before, but can't find when and where.

Example 2 below shows an example.

Will these two issues hopefully be fixed in Maple 2025? Sometimes one forgets to rewrite the solution using (lhs-rhs)(sol)=0 and this results in false negative. 

Please see worksheet below. ps. I hope forum manager does not delete this question.

Download issues_with_odetest_dec_16_2024.mw