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Kitonum

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17 years, 182 days
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Adjustment...

Kitonum 21665
September 11 2024
2 0
> 

restart;
> 

 

> 

vars:=[x,y]:
> 

k:=17:
> 

expr:=-2*sqrt(118)*(((-4*x + y + 51/32)*sqrt(k) + (k*x)/4 - (51*y)/4 + 153/32)*sqrt(-4012 + 1003*sqrt(k)) + ((x + 4*y)*sqrt(k) - (85*x)/4 - (17*y)/4)*sqrt(4012 + 1003*sqrt(k)))*k^(1/4)/(17051*(-1 + sqrt(k)));

-2*118^(1/2)*(((-4*x+y+51/32)*17^(1/2)+(17/4)*x-(51/4)*y+153/32)*(-4012+1003*17^(1/2))^(1/2)+((x+4*y)*17^(1/2)-(85/4)*x-(17/4)*y)*(4012+1003*17^(1/2))^(1/2))*17^(1/4)/(-17051+17051*17^(1/2))

 (1)
> 

indets(expr);

{x, y}

 (2)
> 

factor(expr);

(1/64192)*(-4012+1003*17^(1/2))^(1/2)*118^(1/2)*17^(1/4)*(17^(1/2)+5)*(40*x-24*y-3)

 (3)
> 

op(factor(expr));

1/64192, (-4012+1003*17^(1/2))^(1/2), 118^(1/2), 17^(1/4), 17^(1/2)+5, 40*x-24*y-3

 (4)
> 

select(has, [op(factor(expr))], vars);

[40*x-24*y-3]

 (5)
 

 

Download Download_2024-09-11_Has_Select_Question_new.mw

select...

Kitonum 21665
September 09 2024
3 0

I think in Maple the most natural way to solve this particular problem is to use the  select  command:

restart;
L:=combinat:-permute([1,2,3,4]):
select(t->ListTools:-Search(2,t)<ListTools:-Search(3,t), L);

[[1, 2, 3, 4], [1, 2, 4, 3], [1, 4, 2, 3], [2, 1, 3, 4], [2, 1, 4, 3], [2, 3, 1, 4], [2, 3, 4, 1], [2, 4, 1, 3], [2, 4, 3, 1], [4, 1, 2, 3], [4, 2, 1, 3], [4, 2, 3, 1]]

Volume and plot...

Kitonum 21665
September 08 2024
2 3

Here is a simple calculation of the volume of this body and its drawing. Your body is like a torus. First we calculate the volume of this "torus" without a hole, and then simply subtract the volume of the hole. For drawing we use parametric equations of the surface of this body. For clarity, a cutout of 1/4 of the body is made.

restart;
f:=y->y^2-3: g:=y->y-y^2:
int(Pi*(2-f(y))^2, y=-1..3/2)-int(Pi*(2-g(y))^2, y=-1..3/2); # The volume of the solid
evalf(%);
M:=<cos(t),-sin(t); sin(t),cos(t)>:
P1:=plot3d([convert(M.<f(z)-2,0>, list)[1]+2, convert(M.<f(z)-2,0>, list)[2], z], t=-Pi/2..Pi, z=-1..3/2, color="Blue", scaling=constrained, axes=normal):
P2:=plot3d([convert(M.<g(z)-2,0>, list)[1]+2, convert(M.<g(z)-2,0>, list)[2], z], t=-Pi/2..Pi, z=-1..3/2, color="Red", scaling=constrained, axes=normal):
L:=plottools:-line([2,0,-2.9],[2,0,2.9], color=green, linestyle=3, thickness=2):
plots:-display(P1, P2, L, view=[-4.3..8.1,-6.1..6.1,-2.1..3.1]);

                   

 

Inconsistent system...

Kitonum 21665
September 06 2024
0 0


 

> 

restart; _local(D); sys := {F*b+L*c+Q*d+V*d2+a^2 = 1, G*b+M*c+R*d+W*d2+a*b = 0, H*b+N*c+S*d+X*d2+a*c = 0, B*a+H2*b+O*c+T*d+Y*d2 = 0, B*f+H2*g+O*h+T*h2+Y*k = 0, B*l+H2*m+O*n+T*o+Y*p = 0, B*q+H2*r+O*s+T*t+Y*u = 1, B*v+H2*w+O*x+T*y+Y*z = 0, C*a+K*b+P*c+U*d+Z*d2 = 0, C*f+K*g+P*h+U*h2+Z*k = 0, C*l+K*m+P*n+U*o+Z*p = 0, C*q+K*r+P*s+U*t+Z*u = 0, C*v+K*w+P*x+U*y+Z*z = 1, F*g+L*h+Q*h2+V*k+a*f = 0, F*m+L*n+Q*o+V*p+a*l = 0, F*r+L*s+Q*t+V*u+a*q = 0, F*w+L*x+Q*y+V*z+a*v = 0, G*g+M*h+R*h2+W*k+b*f = 1, G*m+M*n+R*o+W*p+b*l = 0, G*r+M*s+R*t+W*u+b*q = 0, G*w+M*x+R*y+W*z+b*v = 0, H*g+N*h+S*h2+X*k+c*f = 0, H*m+N*n+S*o+X*p+c*l = 1, H*r+N*s+S*t+X*u+c*q = 0, H*w+N*x+S*y+X*z+c*v = 0}; LL := {A, B, C, D, D2, F, G, H, H2, K, L, M, N, O, P, Q, R, S, T, U, V, W, X, Y, Z}; MM := LinearAlgebra:-GenerateMatrix(sys, LL); LinearAlgebra:-LinearSolve(MM)

Error, (in LinearAlgebra:-LinearSolve) inconsistent system

  
> 

``


 

Download Count_operations_new.mw

Dihedral angle of two halfplanes...

Kitonum 21665
September 04 2024
1 4

To solve this, we can follow the plan:
1. Take a point M inside the pyramid - we can take any linear convex combination of the pyramid's vertices.
2. Find the projections of this point M1 and M2 onto the desired planes.
3. Find the angle alpha between the vectors MM1 and MM2.
4. The required angle is Pi - alpha.

restart;
local D:
with(Student:-MultivariateCalculus):
A, B, C, D, S :=  [0,0,0], [1,0,0], [1,1,0], [0,1,0], [0,0,1]:
p1, p2 := Plane(S,B,C), Plane(S,C,D):
M:=(A+B+C+D+S)/5;
M1:=Projection(M, p1);
M2:=Projection(M, p2);
v1:=convert(M1-M,Vector);
v2:=convert(M2-M,Vector);
Pi-Angle(v1,v2);  # The answer

                               

               

Another way...

Kitonum 21665
September 04 2024
1 0 
data[2 .. 3];

 

Area...

Kitonum 21665
August 29 2024
1 0

This figure is not bounded and its angles go to infinity in three directions. Therefore, to accurately calculate the area, the integration limits should be changed:

restart;
y0 := -ln(1 - exp(-x)):
y1 := -ln(-1 + exp(-x)):
y2 := -ln(1 + exp(-x)):
Area := int(y1-y2, x=-infinity..0) + int(y0-y2, x=0..infinity);
evalf(Area);

                               

 

 

Solution...

Kitonum 21665
August 24 2024
2 0 
restart;
Square:=proc(t, r)
local A, B, M, N, v, R, P, c, s, T;
uses plots, plottools;
A := [-r, 0];
M := <r*cos(t), r*sin(t)>;
v:=M-convert(A,Vector);
R:=<cos(Pi/2),-sin(Pi/2); sin(Pi/2),cos(Pi/2)>;
N:=M + R.v;
P:=N - v;
M, N, P := convert~([M, N, P], list)[];
c:=circle(r, color = blue);
s:=curve([A,M,N,P,A], color=red);
T:=textplot([[A[],"A"],[r,0,"B", align=right],[M[],"M"],[N[],"N"],[P[],"P"]], font=[times,16], align={left,above});
display(c, s, T, scaling=constrained, thickness=2);
end proc:

plots:-animate(Square, [t,1], t=0..2*Pi, frames=90);

       

 

`if`...

Kitonum 21665
August 24 2024
1 0


 

> 

restart;
> 

v := 1: a := 0: b := 1: t := 0.1e-2: dt := 0.1e-3: N := 5: h := (b-a)/(N-1):
> 

for i to N do x[i] := a+(i-1)*h end do:
p := x->product(x-x[k], k = 1 .. N);
f := expand(p(x));

proc (x) options operator, arrow; product(x-x[k], k = 1 .. N) end proc

  

x^5-(5/2)*x^4+(35/16)*x^3-(25/32)*x^2+(3/32)*x

 (1)
> 

PRime := diff(f, x);

5*x^4-10*x^3+(105/16)*x^2-(25/16)*x+3/32

 (2)
> 

for i to N do
for j to N do
 if i <> j then a[i, j] := (eval(PRime, x = x[i]))/((x[i]-x[j])*(eval(PRime, x  = x[j]))) else
 a[i, j] := -add(`if`(k<>i, a[i, k], undefined), k = 1 .. N) end if;  
 end do:
 end do:
> 

seq(seq(a[i,j], i=1..N), j=1..N);

undefined, -1, 1/3, -1/3, 1, 16, undefined, -8/3, 2, -16/3, -12, 6, undefined, -6, 12, 16/3, -2, 8/3, undefined, -16, -1, 1/3, -1/3, 1, undefined

 (3)
> 

 


 

Download burgers_type_method_new.mw

Circle passing through points C, D, K, H...

Kitonum 21665
August 13 2024
0 0

A circle does not necessarily pass through 4 points, even if the points lie in the same plane. Therefore, we first find the center of the circle passing through points C, D, H, and then check that point K also lies on the same circle:


 

> 

restart;
_local(D, O);
with(Student:-MultivariateCalculus):
A := [0, 0, 0];
B := [a, 0, 0];
C := [a, b, 0];
D := [0, b, 0];
S := [0, 0, h];
O := [x, y, z];
lineSC := Line(S, C);
lineSD := Line(S, D);
H := Projection(A, lineSC);
K := Projection(A, lineSD);
OH := H - O;
OK := K - O;
OC := C - O;
M := Matrix([OH, OK, OC]);
E:=(C+D)/2;
F:=(C+H)/2;
p1:=Plane(E,convert(C-D,Vector));
p2:=Plane(F,convert(C-H,Vector));
Eq1:=GetRepresentation(p1);
Eq2:=GetRepresentation(p2);
Md:=LinearAlgebra:-Determinant(M);
solve({Eq1,Eq2,Md}, {x,y,z});
O := eval(O, %);
d:=simplify(Distance(O, H));
d1:=simplify(Distance(O, K));
is(d=d1);

A := [0, 0, 0]

  

B := [a, 0, 0]

  

C := [a, b, 0]

  

D := [0, b, 0]

  

S := [0, 0, h]

  

O := [x, y, z]

  

Student:-MultivariateCalculus:-Line([0, 0, h], Vector(3, {(1) = a, (2) = b, (3) = -h}), variables = [x, y, z], parameter = t, id = 1)

  

Student:-MultivariateCalculus:-Line([0, 0, h], Vector(3, {(1) = 0, (2) = b, (3) = -h}), variables = [x, y, z], parameter = t, id = 2)

  

H := [h^2*a/(a^2+b^2+h^2), h^2*b/(a^2+b^2+h^2), h*(a^2+b^2)/(a^2+b^2+h^2)]

  

K := [0, h^2*b/(b^2+h^2), h*b^2/(b^2+h^2)]

  

OH := [-x+h^2*a/(a^2+b^2+h^2), -y+h^2*b/(a^2+b^2+h^2), -z+h*(a^2+b^2)/(a^2+b^2+h^2)]

  

OK := [-x, -y+h^2*b/(b^2+h^2), -z+h*b^2/(b^2+h^2)]

  

OC := [-x+a, -y+b, -z]

  

Matrix(3, 3, {(1, 1) = -x+h^2*a/(a^2+b^2+h^2), (1, 2) = -y+h^2*b/(a^2+b^2+h^2), (1, 3) = -z+h*(a^2+b^2)/(a^2+b^2+h^2), (2, 1) = -x, (2, 2) = -y+h^2*b/(b^2+h^2), (2, 3) = -z+h*b^2/(b^2+h^2), (3, 1) = -x+a, (3, 2) = -y+b, (3, 3) = -z})

  

E := [(1/2)*a, b, 0]

  

F := [h^2*a/(2*(a^2+b^2+h^2))+(1/2)*a, h^2*b/(2*(a^2+b^2+h^2))+(1/2)*b, h*(a^2+b^2)/(2*(a^2+b^2+h^2))]

  

Student:-MultivariateCalculus:-Plane(Vector(3, {(1) = a, (2) = 0, (3) = 0}), [(1/2)*a, b, 0], variables = [x, y, z], id = 1)

  

_m2776775710912

  

x = (1/2)*a

  

(-h^2*a/(a^2+b^2+h^2)+a)*x+(-h^2*b/(a^2+b^2+h^2)+b)*y-h*(a^2+b^2)*z/(a^2+b^2+h^2) = (-h^2*a/(a^2+b^2+h^2)+a)*((1/2)*h^2*a/(a^2+b^2+h^2)+(1/2)*a)+(-h^2*b/(a^2+b^2+h^2)+b)*((1/2)*h^2*b/(a^2+b^2+h^2)+(1/2)*b)-(1/2)*h^2*(a^2+b^2)^2/(a^2+b^2+h^2)^2

  

-h^2*a*(a^2*b*h-a^2*b*z-a^2*h*y+b^3*h-b^3*z-b^2*h*y)/((a^2+b^2+h^2)*(b^2+h^2))

  

{x = (1/2)*a, y = (1/2)*b*(b^2+2*h^2)/(b^2+h^2), z = (1/2)*h*b^2/(b^2+h^2)}

  

[(1/2)*a, (1/2)*b*(b^2+2*h^2)/(b^2+h^2), (1/2)*h*b^2/(b^2+h^2)]

  

(1/2)*(((b^2+h^2)*a^2+b^4)/(b^2+h^2))^(1/2)

  

(1/2)*(((b^2+h^2)*a^2+b^4)/(b^2+h^2))^(1/2)

  

true

 (1)
> 

 


 

Download center_circle.mw

Empty symbol...

Kitonum 21665
August 08 2024
2 1

In this example, it's easier to do it manually, using the empty symbol  ``   to prevent automatic bracket expansion:

restart;
p:=9*csc(theta)^2+9;
9*``(p/9);

 

evala...

Kitonum 21665
August 02 2024
0 2 
evala(M);

 

Solution...

Kitonum 21665
August 02 2024
1 0

You can use a special program for this. See  https://mapleprimes.com/posts/200677-Partition-Of-An-Integer-With-Restrictions

Examples of use:

Partition(10, 4, 1..4);  # Your problem.
``;
Partition(10, 4, {1,3,5,7,9});  # Partitions of length 4 consisting only of odd numbers.
``;
Partition(10, 1..10, {1,3,5,7,9});  # Any partitions of odd numbers.

         

 

 

Easier...

Kitonum 21665
July 19 2024
2 4 
A:=int(x^(1/2),x);
sqrt(A^2);

                                     

We used the fact that for any non-negative real number  A = sqrt(A^2) . This technique is often useful.

op...

Kitonum 21665
July 17 2024
1 0

The  has(a, b)  command returns true if  b  is an operand of  a  of some level (in this case we say that  b  is a subexpression of  a ) . The operands are returned by the op command. See

restart;
A:=I/sqrt(a);
B:=op(A);
op(B[2]);
has(A,a^(-1/2));
has(A,-1/2);

                       

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