Maple 2018 Questions and Posts

These are Posts and Questions associated with the product, Maple 2018

In this question at here https://tex.stackexchange.com/questions/503745/how-can-i-get-correct-the-point-a-and-b-automatically-in-this-picture/503838?noredirect=1#comment1272706_503838
The points A and B lie on the circle is intersection of the sphere x^2 + y^2 + (z-3)^2 = 25 and the plane z = 0. 
How can I find coordinates of the points A and B by Maple?

 

unprotect(D); f := proc (x, y) options operator, arrow; (295849/5841396)*x^2-(29441/324522)*y*x+(33995/216348)*y^2-(5989/14751)*x+(3635/4917)*y+1 end proc; 295849 2 29441 33995 2 5989 f := (x, y) -> ------- x - ------ y x + ------ y - ----- x 5841396 324522 216348 14751 3635 + ---- y + 1 4917 coeffs(f(x, y)); -5989 3635 295849 33995 -29441 1, -----, ----, -------, ------, ------ 14751 4917 5841396 216348 324522 A, B, C, D, E, F := %; -5989 295849 33995 -29441 3635 A, B, C, D, E, F := 1, -----, -------, ------, ------, ---- 14751 5841396 216348 324522 4917

Hi,

In statistics a  "degree of freedom" is a strictly positive integer.

The three distributions ChiSquare, StudentT and FRatio from package Statistics have degrees of freedom as parameters. Nevertheless they accept any strictly positive real values for them.
(one can verify that their "Conditions" attribute is of the form [0 < n] instead of [n::posint]).

I think this ought to be corrected in future versions
 

Hello,

I have installed the Physics Package with:

PackageTools:-Install("5137472255164416", version = 329, overwrite)

I have a question about the calculation of the Tensor determinant:

I use two ways to calculate the determinant of the same matrix:

The first with Physics package and the second with LinearAlgebra Package:

It is easy to calculate analytically the result it is : -1

But Physics Package gives : +1 and LinearAlgebra gives: -1

My question is why I found +1 with Physics Package : What I doing wrong ?.

Please find my source hereafter:

**************************************************

restart

with(Physics)

Setup(coordinates = {X = [t, x, y, z]}, dimension = 4, metric = {(1, 1) = 1, (2, 2) = -1, (3, 3) = -1, (4, 4) = -1}, signature = `+---`, mathematicalnotation = true)

Define(H[mu, nu] = Matrix(4, {(1, 1) = 1+h, (1, 2) = h, (2, 2) = -1+h, (3, 3) = -1, (4, 4) = -1}, shape = symmetric))

H[]

H[determinant]

with(LinearAlgebra)

B := Matrix(4, {(1, 1) = 1+h, (1, 2) = h, (2, 2) = -1+h, (3, 3) = -1, (4, 4) = -1}, shape = symmetric)

Determinant(B)

***************************************

Thanks for your answer,

Sincerely yours

Xavier

 

Hi 

I am studying Marine Engineer, and we are gonna work alot with vectors. I am using Maple as my mathematical software and i would like to know if you can do a simple vector addition in Maple?
I have tried to make a simple vector diagram to help understand my problem. 
My vector U, is my reference vector, and my two vectors I1 and I2 are the vectors i would like to plus. I know the length of both I1 and I2, and i know the angle between I1 and U and I2 and U.

If i am using my TI-nspire, i can just type in the length and angle on both I1 and I2, to my reference vector, and then plus them together. Is this possible to do in maple?

For the vectors onmy example the result of vector additon of I1+I2 = 2.42 to an angle of 48.07 degree.

The image is just for representation and is not accurate according to the lengths and angles.

Hope someone out there can help me.

 

Let be given a cylinder with radius of the base circle is equal to 3 and the altitude of the cylinder is equal to 2. How can I draw square ABCD so that A, B and C, D lie on two base circles. How can I get one option of all vetices A, B, C, D ?

I would like to draw an ellipse by orthonal affinity of a circle. Thank you.

i want to solve the systems of diff equation what's the problem   0.mw

I calculate volume of tetrahedron SABC by this way. Now, I want to find one option  of coordinates of all vertices of this tetrahedron. How can I get the coordinates of the points S, A, B, C so that coordinates  center of circumcircle of the triangle ABC is (0,0,0)?

 

restart:
SA := 3: 
SB := 5: 
SC := 7:
AB := 3:
BC := 4: 
AC := 5:
V := (1/12)*sqrt(SA^2*BC^2*(AB^2+AC^2-BC^2-SA^2+SB^2+SC^2)+SB^2*AC^2*(AB^2-AC^2+BC^2+SA^2-SB^2+SC^2)+SC^2*AB^2*(-AB^2+AC^2+BC^2+SA^2+SB^2-SC^2)-SA^2*AB^2*SB^2-SB^2*BC^2*SC^2-SA^2*AC^2*SC^2-AB^2*BC^2*AC^2);

From here https://www.mapleprimes.com/questions/227573-How-To-Get-Coordinates-Of-Triangle-ABC
I find coordinates of the point A, B, C. Now I tried
 

with(geom3d):
point(A, -3/2, -2, 0):
point(B, 3/2, -2, 0):
point(C, -3/2, 2, 0):
point(S, x, y, z):
solve([distance(S, A) = 3, distance(S, B) = 5, distance(S, C) = 7], [x, y, z], explicit);
with(geom3d):
point(A, -3/2, -2, 0):
point(B, 3/2, -2, 0):
point(C, -3/2, 2, 0):
point(S, x, y, z):
solve([distance(S, A) = 5, distance(S, B) = 3, distance(S, C) = 7], [x, y, z], explicit);

Use the above code, I obtain the result.

Lang_2_output_as_tswitch_varies.mw

^ in this worksheet I have made a graph of a variable in a simple ODE against time (shown below), at t=150 a switch condition, in the worksheet called tswitch, changes the rates of change of the ODE.

I am thinking about afunction that maps tswitch->solution of the ODE and would like to visualise it as a 3d surface, but couldn't work it out in Maple

 

How to get correct result this equation
restart; with(Student:-MultivariateCalculus);
A := [0, 0, 0];
B := [c, 0, 0];
S := [x, 0, z];
solve([Distance(S, A) = a, Distance(S, B) = b], [x, z]) assuming and(a > 0, b > 0, c > 0);

I have just made a bar graph in Visualising_numerous_derivatives_of_the_L2_model.mw

and i would like to set it to be logscaled, but could not find an option in the documentation, or a way of using a logplot to get similar functionality

This post is closely related to the previous one  https://www.mapleprimes.com/posts/210930-Numbrix-Puzzle-By-The-Branch-And-Bound-Method  which presents the procedure  NumbrixPuzzle   that allows you to effectively solve these puzzles (the text of this procedure is also available in the worksheet below).  
This post is about generating these puzzles. To do this, we need the procedure  SerpentinePaths  (see below) , which allows us to generate a large number of serpentine paths in a matrix of a specified size, starting with a specified matrix element. Note that for a square matrix of the order  n , the number of such paths starting from [1,1] - position is the sequence  https://oeis.org/search?q=1%2C2%2C8%2C52%2C824&language=english&go=Search .

The required parameter of  SerpentinePaths procedure is the list  S , which defines the dimensions of the matrix. The optional parameter is the list  P  - this is the position of the number 1 (by default P=[1,1] ).
As an example below, we generate 20 puzzles of size 6 by 6. In exactly the same way, we can generate the desired number of puzzles for matrices of other sizes.


 

restart;

SerpentinePaths:=proc(S::list, P::list:=[1,1])
local OneStep, A, m, F, B, T, a;

OneStep:=proc(A::listlist)
local s, L, B, T, k, l;

s:=max[index](A);
L:=[[s[1]-1,s[2]],[s[1]+1,s[2]],[s[1],s[2]-1],[s[1],s[2]+1]];
T:=table(); k:=0;
for l in L do
if l[1]>=1 and l[1]<=S[1] and l[2]>=1 and l[2]<=S[2] and A[op(l)]=0 then k:=k+1; B:=subsop(l=a+1,A);
T[k]:=B fi;
od;
convert(T, list);
end proc;
A:=convert(Matrix(S[], {(P[])=1}), listlist);
m:=S[1]*S[2]-1;
B:=[$ 1..m];
F:=LM->ListTools:-FlattenOnce(map(OneStep, `if`(nops(LM)<=30000,LM,LM[-30000..-1])));
T:=[A];
for a in B do
T:=F(T);
od;
map(convert, T, Matrix);

end proc:
 

NumbrixPuzzle:=proc(A::Matrix)
local A1, L, N, S, MS, OneStepLeft, OneStepRight, F1, F2, m, L1, p, q, a, b, T, k, s1, s, H, n, L2, i, j, i1, j1, R;
uses ListTools;
S:=upperbound(A); N:=nops(op(A)[3]); MS:=`*`(S);
A1:=convert(A, listlist);
for i from 1 to S[1] do
for j from 1 to S[2] do
for i1 from i to S[1] do
for j1 from 1 to S[2] do
if A1[i,j]<>0 and A1[i1,j1]<>0 and abs(A1[i,j]-A1[i1,j1])<abs(i-i1)+abs(j-j1) then return `no solutions` fi;
od; od; od; od;
L:=sort(select(e->e<>0, Flatten(A1)));
L1:=[`if`(L[1]>1,seq(L[1]-k, k=0..L[1]-2),NULL)];
L2:=[seq(seq(`if`(L[i+1]-L[i]>1,L[i]+k,NULL),k=0..L[i+1]-L[i]-2), i=1..nops(L)-1), `if`(L[-1]<MS,seq(L[-1]+k,k=0..MS-L[-1]-1),NULL)];
OneStepLeft:=proc(A1::listlist)
local s, M, m, k, T;
uses ListTools;
s:=Search(a, Matrix(A1));   
M:=[[s[1]-1,s[2]],[s[1]+1,s[2]],[s[1],s[2]-1],[s[1],s[2]+1]];
T:=table(); k:=0;
for m in M do
if m[1]>=1 and m[1]<=S[1] and m[2]>=1 and m[2]<=S[2] and A1[op(m)]=0 then k:=k+1; T[k]:=subsop(m=a-1,A1);
fi;
od;
convert(T, list);
end proc;
OneStepRight:=proc(A1::listlist)
local s, M, m, k, T, s1;
uses ListTools;
s:=Search(a, Matrix(A1));  s1:=Search(a+2, Matrix(A1));  
M:=[[s[1]-1,s[2]],[s[1]+1,s[2]],[s[1],s[2]-1],[s[1],s[2]+1]];
T:=table(); k:=0;
for m in M do
if m[1]>=1 and m[1]<=S[1] and m[2]>=1 and m[2]<=S[2] and A1[op(m)]=0 and `if`(a+2 in L, `if`(is(abs(s1[1]-m[1])+abs(s1[2]-m[2])>1),false,true),true) then k:=k+1; T[k]:=subsop(m=a+1,A1);
fi;
od;
convert(T, list);   
end proc;
F1:=LM->ListTools:-FlattenOnce(map(OneStepLeft, LM));
F2:=LM->ListTools:-FlattenOnce(map(OneStepRight, LM));
T:=[A1];
for a in L1 do
T:=F1(T);
od;
for a in L2 do
T:=F2(T);
od;
R:=map(t->convert(t,Matrix), T);
if nops(R)=0 then return `no solutions` else R fi;
end proc:


Simple examples

SerpentinePaths([3,3]);  # All the serpentine paths for the matrix  3x3, starting with [1,1]-position
SerpentinePaths([3,3],[1,2]);  # No solutions if the start with [1,2]-position
SerpentinePaths([4,4]):  # All the serpentine paths for the matrix  4x4, starting with [1,1]-position
nops(%);
nops(SerpentinePaths([4,4],[1,2]));  # The number of all the serpentine paths for the matrix  4x4, starting with [1,2]-position
nops(SerpentinePaths([4,4],[2,2]));  # The number of all the serpentine paths for the matrix  4x4, starting with [2,2]-position

[Matrix(3, 3, {(1, 1) = 1, (1, 2) = 6, (1, 3) = 7, (2, 1) = 2, (2, 2) = 5, (2, 3) = 8, (3, 1) = 3, (3, 2) = 4, (3, 3) = 9}), Matrix(3, 3, {(1, 1) = 1, (1, 2) = 8, (1, 3) = 7, (2, 1) = 2, (2, 2) = 9, (2, 3) = 6, (3, 1) = 3, (3, 2) = 4, (3, 3) = 5}), Matrix(3, 3, {(1, 1) = 1, (1, 2) = 8, (1, 3) = 9, (2, 1) = 2, (2, 2) = 7, (2, 3) = 6, (3, 1) = 3, (3, 2) = 4, (3, 3) = 5}), Matrix(3, 3, {(1, 1) = 1, (1, 2) = 4, (1, 3) = 5, (2, 1) = 2, (2, 2) = 3, (2, 3) = 6, (3, 1) = 9, (3, 2) = 8, (3, 3) = 7}), Matrix(3, 3, {(1, 1) = 1, (1, 2) = 2, (1, 3) = 9, (2, 1) = 4, (2, 2) = 3, (2, 3) = 8, (3, 1) = 5, (3, 2) = 6, (3, 3) = 7}), Matrix(3, 3, {(1, 1) = 1, (1, 2) = 2, (1, 3) = 3, (2, 1) = 8, (2, 2) = 7, (2, 3) = 4, (3, 1) = 9, (3, 2) = 6, (3, 3) = 5}), Matrix(3, 3, {(1, 1) = 1, (1, 2) = 2, (1, 3) = 3, (2, 1) = 8, (2, 2) = 9, (2, 3) = 4, (3, 1) = 7, (3, 2) = 6, (3, 3) = 5}), Matrix(3, 3, {(1, 1) = 1, (1, 2) = 2, (1, 3) = 3, (2, 1) = 6, (2, 2) = 5, (2, 3) = 4, (3, 1) = 7, (3, 2) = 8, (3, 3) = 9})]

 

[]

 

52

 

25

 

36

(1)


Below we find 12,440 serpentine paths in the matrix  6x6 starting from various positions (the set  L )

k:=0:  n:=6:
for i from 1 to n do
for j from i to n do
k:=k+1; S[k]:=SerpentinePaths([n,n],[i,j])[];
od: od:
L1:={seq(S[i][], i=1..k)}:
L2:=map(A->A^%T, L1):
L:=L1 union L2:
nops(L);

12440

(2)


Further, using the list  L, we generate 20 examples of Numbrix puzzles with the unique solutions

T:='T':
N:=20:
M:=[seq(L[i], i=combinat:-randcomb(nops(L),N))]:
for i from 1 to N do
for k from floor(n^2/4) do
T[i]:=Matrix(n,{seq(op(M[i])[3][j], j=combinat:-randcomb(n^2,k))});
if nops(NumbrixPuzzle(T[i]))=1 then break; fi;
od:  od:
T:=convert(T,list):
T1:=[seq([seq(T[i+j],i=1..5)],j=[0,5,10,15])]:
DocumentTools:-Tabulate(Matrix(4,5, (i,j)->T1[i,j]), fillcolor = "LightYellow", width=95):


The solutions of these puzzles

DocumentTools:-Tabulate(Matrix(4,5, (i,j)->NumbrixPuzzle(T1[i,j])[]), fillcolor = "LightYellow", width=95):

 


For some reason, these 20 examples and their solutions did not load here.

 Edit. I separately inserted these generated 20 puzzles as a picture:

 

Download SerpPathsinMatrix.mw

 

Hello,

My question is mathematical in nature, so it might be a little out of place but I though I would give it a shot. 

You have a series of chebyshev coefficients in two connecting subdomains lets say S1 = [0,0.5] and S2=[0.5,1]. So far you are still in the spectral space. If you want to compute the solution in real space you can sum the coefficients with the Chebyshev polynomials. 

Now imagine you change the interval to S1 = [0,0.6] and S2 = [0.6,1]. Is there a way to manipulate the Chebyshev coefficients from both initial subdomains to create a new set of Chebyshev coefficients that fit the solution in the new subdomains. 

The brute force method would be to create the real solution of Chebyshev polynomials and then use that to form a new set of Chebyshev coefficients. Or you can use Clenshaw to compute the solution at several points, and then use the points to create new Chebyshev coefficients.

But what if we can stay in spectral space and create the new chebyshev coefficients. Is that possible? If so, how?

P; Q; lma; [3 -42] [-, ---] [2 25 ] [22649 -2304] [-----, -----] [5523 1841 ] 1.627112258 with(geometry); point(Pp, P[1], P[2]); point(Qp, Q[1], Q[2]); ellipse(p, ['foci' = [Pp, Qp], 'MajorAxis' = lma]); Error, (in geometry:-ellipse) the given polynomial/equation is not an algebraic representation of a ellipse; I can't understand this error
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