Kitonum

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16 years, 218 days

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These are answers submitted by Kitonum

data[2 .. 3];

 

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); 

                               

 

 

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);

       

 


 

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

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

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(M);

 

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.

         

 

 

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.

The  has(a, b)  command returns true if  b  is an operand of  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);

                       

In the help for the  Algebraic:-Remainder(a, b, x, options) command it is written: "The arguments a and b must be univariate polynomials in the variable x, which is typically a name. If other names or non-algebraic sub-expressions are included and they cannot be evaluated to algebraic numbers, an "unable to divide multivariate polynomials" error will be returned". So use he standard command  rem :

restart;
rem(a*x^3 + b*x^2 + c*x + d, 3*a*x^2 + 2*b*x + c, x);

 

Use the  empty symbol  ``  for this.

Example:

restart;
plot(sqrt(x), x = 0 .. 4, tickmarks = [[seq(i=``,i=0..4)], [seq(i=``,i=0..2)]], scaling=constrained);

                               

restart;
expr:=factor(x^10/36 - 4/25*y^24*z^8);
r:=op(expr);
d:=r[1];
sort(``(sqrt(d)*r[2])*``(sqrt(d)*r[3]), x);

                    

 

 

It's very simple. You must write your code as a procedure with the parameter  xP , and then use the  plots:-animate command, specifying a range for the parameter  xP :

restart;
Proc:=proc(xP)
local xA, yA, xB, yB, yP;
uses plots,geometry;
_EnvHorizontalName := 'x';
_EnvVerticalName := 'y';
xA := 5;
yA := 0;
point(A, xA, yA);
xB := 5;
yB := -7;
point(B, xB, yB);
midpoint(C, A, B);
segment(sg1, A, B);

yP := 0;
point(P, xP, yP);
PerpenBisector(L, C, P);
line(YYp, y = yB);
line(XXp, y = 0);
intersection(M, L, YYp);
line(PM, [P, M]);
projection(H, C, PM);
triangle(CMP, [C, M, P]);
triangle(ABH, [A, B, H]);
distance(B, H);
circle(cir, [B, 7]);
display(textplot([[coordinates(A)[], "A"], [coordinates(B)[], "B "], [coordinates(C)[], "C"], [coordinates(M)[], "M"], [coordinates(H)[], "H"], [coordinates(P)[], "P"]], align = {"above", 'right'}),
draw([YYp(color = red), XXp(color = black), PM(color = green), L(color = green), sg1(color = black), cir(color = magenta), P(color = black, symbol = solidcircle, symbolsize = 10), M(color = black, symbol = solidcircle, symbolsize = 10), H(color = black, symbol = solidcircle, symbolsize = 10), A(color = blue, symbol = solidcircle, symbolsize = 10), B(color = blue, symbol = solidcircle, symbolsize = 10), CMP(color = blue, filled = true, transparency = 0.8), ABH(color = red, filled = true, transparency = 0.8), C(color = blue, symbol = solidcircle, symbolsize = 10)]),
axes = none, view = [-15 .. 14, -15 .. 3]);
end proc:

plots:-animate(Proc, [a], a=-12..12, frames=90);

     

Also we can use the Explore command:

Explore(Proc(xP), xP=-12..12);

     

 

When specifying a function using the arrow, you must specify the function name. Your function names will be y1 and y2, not y1(x) and y2(x).
Also, your expression is somewhat better simplified with the  combine  command rather than the  simplify.

y1 := proc (x) options operator, arrow; 2*sin(x)-sin(2*x)+cos(2*x) end proc; y2 := proc (x) options operator, arrow; 4*sin(x)+sin(2*x)-cos(2*x) end proc; diff(y1(x), x); Expr := 1/2*((diff(y1(x), x))^2+(diff(y2(x), x))^2)+1/2*(3*y1(x)^2-y1(x)*y2(x)+y2(x)^2); simplify(Expr); combine(Expr)

33/2+(3/2)*sin(4*x)+(5/2)*sin(3*x)+(5/2)*cos(3*x)+(3/2)*cos(x)+(3/2)*sin(x)

(1)

NULL

Download simplify_expression_new.mw

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