Maple Questions and Posts

These are Posts and Questions associated with the product, Maple

odetest should be made more robust.

Here is an example where the same exact solution and same exact IC, but when solution is just writtent in a  little different form, odetest no longer verifies it.

Do you consider this a bug? How is the user supposed to know their solution is correct or not now, since it depends on how it is written? What can a user then do to help odetest in this case verify the solution?


 

interface(version);

`Standard Worksheet Interface, Maple 2024.0, Windows 10, March 01 2024 Build ID 1794891`

ode:=diff(y(x), x)*x^2 + cos(2*y(x)) = 1;
ic:=y(infinity)=10/3*Pi;
e1:=2/x+1/3*sqrt(3);
SOL1:=y(x)=arccot(e1) + Pi*3;
odetest(SOL1,[ode,ic]);

(diff(y(x), x))*x^2+cos(2*y(x)) = 1

y(infinity) = (10/3)*Pi

2/x+(1/3)*3^(1/2)

y(x) = arccot(2/x+(1/3)*3^(1/2))+3*Pi

[0, 0]

#now we rewrite the solution a little different. But same solution
e2:=simplify(e1);

(1/3)*(3^(1/2)*x+6)/x

#Now maple no longer verifies the solution

SOL2:=y(x)=arccot(e2) + Pi*3;
odetest(SOL2,[ode,ic])

y(x) = arccot((1/3)*(3^(1/2)*x+6)/x)+3*Pi

[0, -(1/6)*Pi]

 


 

Download same_solution_not_verified_june_13_2024.mw

How can I get Prof. Wilhelm Werner's  FourierSeries  package? (See also "Fourier and Other Orthogonal Function Expansions in Maple" by Dr. Robert Lopez):

How to get logarithm expression using Maple command?

For example, enter log[2](x^2-3x+5)+x^3-1-log[3](x-1)

Output [x^2-3x+5,x-1]

Thank you very much for your help.

Hallo every body 

How to add vector fields to the figure of this example of a three-dimensional differential system.

in maple 18

Porgram_of_corollary_1_in_Maple.mw

NULL

restart

X[j] := x^3*a[j, 0]+x^2*y*a[j, 1]+x^2*z*a[j, 2]+x*y^2*a[j, 3]+x*y*z*a[j, 4]+x*z^2*a[j, 5]+y^3*a[j, 6]+y^2*z*a[j, 7]+y*z^2*a[j, 8]+z^3*a[j, 9]

x^3*a[j, 0]+x^2*y*a[j, 1]+x^2*z*a[j, 2]+x*y^2*a[j, 3]+x*y*z*a[j, 4]+x*z^2*a[j, 5]+y^3*a[j, 6]+y^2*z*a[j, 7]+y*z^2*a[j, 8]+z^3*a[j, 9]

(1)

s := sum(epsilon^j*X[j], j = 0 .. 2)

x^3*a[0, 0]+x^2*y*a[0, 1]+x^2*z*a[0, 2]+x*y^2*a[0, 3]+x*y*z*a[0, 4]+x*z^2*a[0, 5]+y^3*a[0, 6]+y^2*z*a[0, 7]+y*z^2*a[0, 8]+z^3*a[0, 9]+epsilon*(x^3*a[1, 0]+x^2*y*a[1, 1]+x^2*z*a[1, 2]+x*y^2*a[1, 3]+x*y*z*a[1, 4]+x*z^2*a[1, 5]+y^3*a[1, 6]+y^2*z*a[1, 7]+y*z^2*a[1, 8]+z^3*a[1, 9])+epsilon^2*(x^3*a[2, 0]+x^2*y*a[2, 1]+x^2*z*a[2, 2]+x*y^2*a[2, 3]+x*y*z*a[2, 4]+x*z^2*a[2, 5]+y^3*a[2, 6]+y^2*z*a[2, 7]+y*z^2*a[2, 8]+z^3*a[2, 9])

(2)

s1 := subs(a = b, s)

x^3*b[0, 0]+x^2*y*b[0, 1]+x^2*z*b[0, 2]+x*y^2*b[0, 3]+x*y*z*b[0, 4]+x*z^2*b[0, 5]+y^3*b[0, 6]+y^2*z*b[0, 7]+y*z^2*b[0, 8]+z^3*b[0, 9]+epsilon*(x^3*b[1, 0]+x^2*y*b[1, 1]+x^2*z*b[1, 2]+x*y^2*b[1, 3]+x*y*z*b[1, 4]+x*z^2*b[1, 5]+y^3*b[1, 6]+y^2*z*b[1, 7]+y*z^2*b[1, 8]+z^3*b[1, 9])+epsilon^2*(x^3*b[2, 0]+x^2*y*b[2, 1]+x^2*z*b[2, 2]+x*y^2*b[2, 3]+x*y*z*b[2, 4]+x*z^2*b[2, 5]+y^3*b[2, 6]+y^2*z*b[2, 7]+y*z^2*b[2, 8]+z^3*b[2, 9])

(3)

s2 := subs(a = c, s)

x^3*c[0, 0]+x^2*y*c[0, 1]+x^2*z*c[0, 2]+x*y^2*c[0, 3]+x*y*z*c[0, 4]+x*z^2*c[0, 5]+y^3*c[0, 6]+y^2*z*c[0, 7]+y*z^2*c[0, 8]+z^3*c[0, 9]+epsilon*(x^3*c[1, 0]+x^2*y*c[1, 1]+x^2*z*c[1, 2]+x*y^2*c[1, 3]+x*y*z*c[1, 4]+x*z^2*c[1, 5]+y^3*c[1, 6]+y^2*z*c[1, 7]+y*z^2*c[1, 8]+z^3*c[1, 9])+epsilon^2*(x^3*c[2, 0]+x^2*y*c[2, 1]+x^2*z*c[2, 2]+x*y^2*c[2, 3]+x*y*z*c[2, 4]+x*z^2*c[2, 5]+y^3*c[2, 6]+y^2*z*c[2, 7]+y*z^2*c[2, 8]+z^3*c[2, 9])

(4)

Considérons le système suivant:

a[1] := 0; c[1] := 0

a[0, 9] := 0; c[0, 8] := 0; b[0, 7] := 0; a[0, 4] := 0; a[0, 7] := 0; c[0, 3] := 0; c[0, 0] := 0; c[0, 5] := 0; b[0, 4] := 0; a[0, 2] := 0; c[0, 6] := 0; c[0, 1] := 0; c[0, 7] := 0; a[0, 8] := 0; b[0, 5] := 0

b0 := 5; a[4] := 0; c[4] := 0; c[2, 9] := 0; c[2, 2] := 0; c[2, 7] := 0; a[2, 5] := 0; b[2, 8] := 0; a[2, 0] := 0; b[2, 6] := 0; b[2, 1] := 0; a[2, 3] := 0; b[0, 9] := 0

b[1] := 0; b[2] := 0; b[3] := 0; b[4] := 0; a[1, 2] := 0; a[1, 1] := 0; a[1, 4] := 0; a[1, 6] := 0; a[1, 7] := 0; a[1, 8] := 0; a[1, 9] := 0; a[2, 9] := 0; a[2, 8] := 0; a[2, 7] := 0; a[2, 6] := 0; a[2, 4] := 0; a[2, 2] := 0; a[2, 1] := 0; b[1, 0] := 0; b[1, 2] := 0; b[1, 3] := 0; b[1, 4] := 0; b[1, 5] := 0; b[1, 7] := 0; b[1, 9] := 0; b[2, 0] := 0; b[2, 2] := 0; b[2, 3] := 0; b[2, 4] := 0; b[2, 5] := 0; b[2, 7] := 0; b[2, 9] := 0; c[1, 0] := 0; c[1, 1] := 0; c[1, 3] := 0; c[1, 4] := 0; c[1, 5] := 0; c[1, 6] := 0; c[1, 8] := 0; c[2, 0] := 0; c[2, 1] := 0; c[2, 3] := 0; c[2, 4] := 0; c[2, 5] := 0; c[2, 6] := 0; c[2, 8] := 0; b[0, 2] := 0; c[1, 7] := 0

a[1, 0] := 0; a[1, 3] := 0; a[1, 5] := 0; b[1, 1] := 0; b[1, 6] := 0; b[1, 8] := 0; c[1, 2] := 0; c[1, 9] := 0; a[3] := 0; c[3] := 0; a[2] := 1/2; c[2] := 3/2; a[0, 0] := -1/2; a[0, 3] := 5/4; a[0, 1] := 0; a[0, 5] := 0; a[0, 6] := 0; b[0, 6] := -1; b[0, 1] := 3/2; b[0, 0] := 0; b[0, 3] := 0; b[0, 8] := 0; c[0, 2] := 0; c[0, 4] := 0; c[0, 9] := -1/3

eq1 := (epsilon^4*a[4]+epsilon^3*a[3]+epsilon^2*a[2]+epsilon*a[1])*x-(epsilon^4*b[4]+epsilon^3*b[3]+epsilon^2*b[2]+epsilon*b[1]+b0)*y+s

(1/2)*epsilon^2*x-5*y-(1/2)*x^3+(5/4)*x*y^2

(5)

eq2 := (epsilon^4*b[4]+epsilon^3*b[3]+epsilon^2*b[2]+epsilon*b[1]+b0)*x+(epsilon^4*a[4]+epsilon^3*a[3]+epsilon^2*a[2]+epsilon*a[1])*y+s1

5*x+(1/2)*epsilon^2*y+(3/2)*x^2*y-y^3

(6)

eq3 := (epsilon^4*c[4]+epsilon^3*c[3]+epsilon^2*c[2]+epsilon*c[1])*z+s2

(3/2)*epsilon^2*z-(1/3)*z^3

(7)

Faisons le changement (x,y,z)=(εX,εY,εZ)

 

x := epsilon*X; y := epsilon*Y; z := epsilon*Z

epsilon*X

 

epsilon*Y

 

epsilon*Z

(8)

Xpoint := collect(eq1/epsilon, epsilon)

((1/2)*X-(1/2)*X^3+(5/4)*X*Y^2)*epsilon^2-5*Y

(9)

Ypoint := collect(eq2/epsilon, epsilon)

((1/2)*Y+(3/2)*X^2*Y-Y^3)*epsilon^2+5*X

(10)

Zpoint := collect(eq3/epsilon, epsilon)

((3/2)*Z-(1/3)*Z^3)*epsilon^2

(11)

Faisons le changement (X, Y, Z) = (`ϱ`*cos(theta), `ϱ`*sin(theta), eta)

 

X := `ϱ`*cos(theta); Y := `ϱ`*sin(theta); Z := eta

`ϱ`*cos(theta)

 

`ϱ`*sin(theta)

 

eta

(12)

`ϱt` := collect(simplify((X*Xpoint+Y*Ypoint)/`ϱ`), epsilon)

-(1/4)*`ϱ`*epsilon^2*(17*`ϱ`^2*cos(theta)^4-19*cos(theta)^2*`ϱ`^2+4*`ϱ`^2-2)

(13)

`θt` := collect(simplify((X*Ypoint-Xpoint*Y)/`ϱ`^2), epsilon)

5+((17/4)*`ϱ`^2*cos(theta)^3*sin(theta)-(9/4)*`ϱ`^2*sin(theta)*cos(theta))*epsilon^2

(14)

`ηt` := collect(Zpoint, epsilon)

((3/2)*eta-(1/3)*eta^3)*epsilon^2

(15)

Utilisons le développpement de taylor

p := series(`ϱt`/`θt`, epsilon, 5)

series(-((1/20)*`ϱ`*(17*`ϱ`^2*cos(theta)^4-19*cos(theta)^2*`ϱ`^2+4*`ϱ`^2-2))*epsilon^2+((1/100)*`ϱ`*(17*`ϱ`^2*cos(theta)^4-19*cos(theta)^2*`ϱ`^2+4*`ϱ`^2-2)*((17/4)*`ϱ`^2*cos(theta)^3*sin(theta)-(9/4)*`ϱ`^2*sin(theta)*cos(theta)))*epsilon^4+O(epsilon^6),epsilon,6)

(16)

q := series(`ηt`/`θt`, epsilon, 5)

series(((3/10)*eta-(1/15)*eta^3)*epsilon^2+((1/5)*(-(3/10)*eta+(1/15)*eta^3)*((17/4)*`ϱ`^2*cos(theta)^3*sin(theta)-(9/4)*`ϱ`^2*sin(theta)*cos(theta)))*epsilon^4+O(epsilon^6),epsilon,6)

(17)

NULL

Averaging d'ordre 1

Les fonctions F11 et F21 sont données comme suit:

NULL

F11 := coeff(p, epsilon)

0

(18)

F21 := coeff(q, epsilon)

0

(19)

NULL

Calculons les fonctions moyennées f11et f12

f11 := (int(F11, theta = 0 .. 2*Pi))/(2*Pi)

0

(20)

f12 := (int(F21, theta = 0 .. 2*Pi))/(2*Pi)

0

(21)

solve({f11 = 0, f12 = 0}, {eta, `ϱ`})

{eta = eta, `ϱ` = `ϱ`}

(22)

NULL

Averaging d'ordre 2

NULL

F12 := simplify(coeff(p, epsilon^2))

-(1/20)*`ϱ`*(17*`ϱ`^2*cos(theta)^4-19*cos(theta)^2*`ϱ`^2+4*`ϱ`^2-2)

(23)

F22 := simplify(coeff(q, epsilon^2))

(3/10)*eta-(1/15)*eta^3

(24)

NULL

Calculons les fonctions moyennées "f21 "et "f22"

f21 := simplify((int(F12, theta = 0 .. 2*Pi))/(2*Pi))

-(1/160)*`ϱ`*(7*`ϱ`^2-16)

(25)

f22 := simplify((int(F22, theta = 0 .. 2*Pi))/(2*Pi))

-(1/30)*eta*(2*eta^2-9)

(26)

solve({f21 = 0, f22 = 0}, {eta, `ϱ`})

{eta = 0, `ϱ` = 0}, {eta = 3*RootOf(2*_Z^2-1), `ϱ` = 0}, {eta = 0, `ϱ` = 4*RootOf(7*_Z^2-1)}, {eta = 3*RootOf(2*_Z^2-1), `ϱ` = 4*RootOf(7*_Z^2-1)}

(27)

allvalues({eta = 0, `ϱ` = 4*RootOf(7*_Z^2-1)})

{eta = 0, `ϱ` = (4/7)*7^(1/2)}, {eta = 0, `ϱ` = -(4/7)*7^(1/2)}

(28)

allvalues({eta = 3*RootOf(2*_Z^2-1), `ϱ` = 4*RootOf(7*_Z^2-1)})

{eta = (3/2)*2^(1/2), `ϱ` = (4/7)*7^(1/2)}, {eta = -(3/2)*2^(1/2), `ϱ` = (4/7)*7^(1/2)}, {eta = (3/2)*2^(1/2), `ϱ` = -(4/7)*7^(1/2)}, {eta = -(3/2)*2^(1/2), `ϱ` = -(4/7)*7^(1/2)}

(29)

NULL

with(VectorCalculus)

M, d := Jacobian([f21, f22], [`ϱ`, eta] = [(4/7)*sqrt(7), 0], 'determinant')

Matrix(%id = 18446744074358842782), -3/50

(30)

factor(d)

-3/50

(31)

M1, d1 := Jacobian([f21, f22], [`ϱ`, eta] = [(4/7)*sqrt(7), (3/2)*sqrt(2)], 'determinant')

Matrix(%id = 18446744074358843142), 3/25

(32)

d1 := factor(d1)

3/25

(33)

M2, d2 := Jacobian([f21, f22], [`ϱ`, eta] = [(4/7)*sqrt(7), -(3/2)*sqrt(2)], 'determinant')

Matrix(%id = 18446744074358843382), 3/25

(34)

factor(d2)

3/25

(35)

restart

with(DEtools):

epsilon := 10^(-2)

1/100

(36)

eq1 := diff(x(t), t) = (1/2)*epsilon^2*x(t)-5*y(t)-(1/2)*x(t)^3+(5/4)*x(t)*y(t)^2

diff(x(t), t) = (1/20000)*x(t)-5*y(t)-(1/2)*x(t)^3+(5/4)*x(t)*y(t)^2

(37)

eq2 := diff(y(t), t) = 5*x(t)+(1/2)*epsilon^2*y(t)+(3/2)*x(t)^2*y(t)-y(t)^3

diff(y(t), t) = 5*x(t)+(1/20000)*y(t)+(3/2)*x(t)^2*y(t)-y(t)^3

(38)

eq3 := diff(z(t), t) = (3/2)*epsilon^2*z(t)-(1/3)*z(t)^3

diff(z(t), t) = (3/20000)*z(t)-(1/3)*z(t)^3

(39)

DEplot3d([eq1, eq2, eq3], [x(t), y(t), z(t)], t = -10 .. 10, [[x(0) = 0.1511857892e-1, y(0) = 0, z(0) = 0], [x(0) = 0.1511857892e-1, y(0) = 0, z(0) = 0.2121320343e-1], [x(0) = 0.1511857892e-1, y(0) = 0, z(0) = -0.2121320343e-1]], linecolor = [blue, red, black], stepsize = 0.1e-1)

 

Download Porgram_of_corollary_1_in_Maple.mw

Hi,

I am trying to produce two statistical tables representing a weighted series in a horizontal manner (Tab1) and in a vertical manner (Tab0). The xi and ni in Tab1 do not display well... Perhaps, there is a simpler way to do this?S5S4BoxPlotPondération.mw

Hello, 

do you have an idea how could be filtered several values from dataframe? 
I have a dataframe called "TestData". I need to select rows from data frame which are equal to the list called "SelectionList". 

Thank you for a comment. 
 

Data:=<"LC1", "LC2", "LC3", "LC4", "LC5", "LC6", "LC7", "LC8", "LC9", "LC10", "LC11", "LC12", "LC13", "LC14", "LC15", "LC16", "LC17", "LC18", "LC19", "LC20">;
LoadValue2:=<10,15,100,82,18,89,25,84,46,18,79,12,0,28,147,15,86,444,18,65>;

TestData:=DataFrame(<Data|LoadValue2>,columns=[Case,Load]);

SelectionList:={"LC3", "LC4", "LC5", "LC6", "LC7", "LC8", "LC9"};

I want to place a grid on this drawing 16x7=112 small squares. Thank you.
with(plots);
with(geometry);
_EnvHorizontalName := 'x';
_EnvVerticalName := 'y';
unprotect(D);
point(A, 0, 0);
point(B, 8, 0);
point(C, 8, 1);
point(D, 16, 2);
point(E, 15, 5);
line(AD, [A, D]);
line(AE, [A, E]);
line(DE, [D, E]);
line(AB, [A, B]);
segment(s1, B, C);
segment(s2, B, A);
triangle(Tr, [A, D, E]);
alpha := FindAngle(AB, AD);
beta := FindAngle(AD, AE);
is(alpha + beta = arctan(1/3));
pl := plot(gridlines = true, title = "Dessin pour montrer que arctan(1/3)=arctan(1/5)+arctan(1/8)", titlefont = ["ROMAN", 20]);
display(textplot([[coordinates(A)[], "A"], [coordinates(B)[], "B"], [coordinates(C)[], "C"], [coordinates(D)[], "D"], [coordinates(E)[], "E"]], align = {"above", 'right'}), draw([A(color = black, symbol = solidcircle, symbolsize = 16), D(color = black, symbol = solidcircle, symbolsize = 16), E(color = black, symbol = solidcircle, symbolsize = 16), s1(color = blue), s2(color = blue), Tr(color = orange, filled = true, transparency = 0.9), Tr(color = blue)]), pl, gridlines = true, axes = none);

Dears, 

Can you look the code bellow and send me my error please? I used Maple 18.

restart;
with(plots);
theta(t) = 19.592+1.2697*cos(.5240*t+4.3391)-.6343*cos((2*.5240)*t-.6963);
omicron(t) = 99.4876+89.8581*cos(.5232*t+15.4500)+19.1069*sin((2*.5232)*t)-8.5891*cos((3*.5232)*t+3.7723)+6.4660*sin((5*.5232)*t);
`&varphi;`(theta):=0.000203*theta*(theta - 11.7)*sqrt(42.3-theta);
mu[v](theta,omicron):=0.0886*exp(((0.01*omicron +1.01*theta  -21.211)/(14.852))^(4));
p[0](theta):=(-0.153* theta*theta + 8.61*theta - 97)/(mu[v](theta,omicron)):
p[2](omicron):=(4*0.25)/(2500)*omicron*( 50 -omicron);
p[3](omicron):=(4*0.75)/(2500)*omicron*( 50 -omicron);
p[2](theta):=exp (0.06737 - 0.00554*theta);
theta[EA](theta):=1/(-0.00094*theta*theta + 0.049*theta - 0.552);
L[v](theta,omicron):=(3.375*(4*omicron*(50-omicron))^(3)*exp(0.0054*theta+0.6737))/(50^(6)*(2+(0.00554*theta-0.06737)^(-1)));
eta(theta,omicron):=(p[0](theta)*p[1](omicron)*p[2](omicron)*p[3](omicron)*p[2](theta))/(theta[EA](theta));
lambda[v] := beta[v]*`&varphi;`(theta)*i[v](t)/n[h](t);
lambda[h] := (beta[h]*`&varphi;`(theta)*i[h](t)+beta[h]*`&varphi;`(theta)*omega*r[h](t))/n[h](t);
n[v](t):=s[v](t)+i[v](t);
beta[h] := 0.9e-1; beta[v] := 0.2e-1; Lambda[h] := .50; sigma[1] := 0.15e-1; sigma[2] := 0.71e-1; Omega[h] := .50; mu[h] := 0.128e-1; delta[h] := .45; k[v] := .66; omega := .3; mu[d] := 0.14e-2;
sys := {diff(i[h](t), t) = Omega[h]+sigma[2]*r[h](t)+lambda[v]*s[h](t)-(delta[h]+mu[d]+mu[h])*i[h](t), diff(i[v](t), t) = lambda[h]*s[v](t)-mu[v](theta, omicron)*i[v](t), diff(j[v](t), t) = L[v](theta, omicron)*(1-j[v](t)/k[v])*n[v](t)-(eta(theta, omicron)+mu[j](theta, omicron))*j[v](t), diff(r[h](t), t) = delta[h]*i[h](t)-(sigma[1]+sigma[2]+mu[h])*r[h](t), diff(s[h](t), t) = Lambda[h]+sigma[1]*r[h](t)-(lambda[v]+mu[h])*s[h](t), diff(s[v](t), t) = eta(theta, omicron)*j[v](t)-(lambda[h]+mu[v](theta, omicron))*s[v](t), i[h](0) = 100, i[v](0) = 100, j[v](0) = 200, r[h](0) = 0, s[h](0) = 10000, s[v](0) = 5000};
p1 := dsolve(sys, numeric, method = rkf45, output = procedurelist);
Error, (in dsolve/numeric/process_input) input system must be an ODE system, found {mu[j](theta, omicron), mu[v](theta, omicron)}
p1o := odeplot(p1, [theta, omicron, i[h](t)], 0 .. 10, numpoints = 100, labels = ["Time (Days)", " infectious population"], labeldirections = [horizontal, vertical], style = line, color = red, axes = boxed, legend = [front, rear, ideal]);

Maple gives same solution for two different equations.

eq1 := 1/5*sqrt(-20*y + 1) - 1/5*ln(1 + sqrt(-20*y + 1)) = x + 2;
eq2 := -1/5*sqrt(-20*y + 1) - 1/5*ln(1 - sqrt(-20*y + 1)) = x + 2;

Solving these for y, gives same exact solution. But this is not correct. As this worksheet shows.

Is this a bug? How could two different equations give same solution?
 

15172

interface(version);

`Standard Worksheet Interface, Maple 2024.0, Windows 10, March 01 2024 Build ID 1794891`

Case 1. Solve first then plugin x value in solution

 

eq1:=(sqrt(a^2 - 4*b*y) - a*ln(a + sqrt(a^2 - 4*b*y)))/b=x+c__1;
eq2:=(-sqrt(a^2 - 4*b*y) - a*ln(a - sqrt(a^2 - 4*b*y)))/b=x+c__1;

eq1:=eval(eq1,[a=1,b=5,c__1=2]);
eq2:=eval(eq2,[a=1,b=5,c__1=2]);

((a^2-4*b*y)^(1/2)-a*ln(a+(a^2-4*b*y)^(1/2)))/b = x+c__1

(-(a^2-4*b*y)^(1/2)-a*ln(a-(a^2-4*b*y)^(1/2)))/b = x+c__1

(1/5)*(-20*y+1)^(1/2)-(1/5)*ln(1+(-20*y+1)^(1/2)) = x+2

-(1/5)*(-20*y+1)^(1/2)-(1/5)*ln(1-(-20*y+1)^(1/2)) = x+2

sol1:=simplify(solve(eq1,y));

-(1/20)*LambertW(-exp(-11-5*x))*(LambertW(-exp(-11-5*x))+2)

sol2:=simplify(solve(eq2,y));

-(1/20)*LambertW(-exp(-11-5*x))*(LambertW(-exp(-11-5*x))+2)

eval(sol1,x=10.);

0.3221340286e-27

eval(sol2,x=10.);

0.3221340286e-27

Case 2. Plugin in same x value in equation and then solve, we get different answers

 

eq1:=(sqrt(a^2 - 4*b*y) - a*ln(a + sqrt(a^2 - 4*b*y)))/b=x+c__1;
eq2:=(-sqrt(a^2 - 4*b*y) - a*ln(a - sqrt(a^2 - 4*b*y)))/b=x+c__1;

eq1:=eval(eq1,[a=1,b=5,c__1=2,x=10]);
eq2:=eval(eq2,[a=1,b=5,c__1=2,x=10]);

((a^2-4*b*y)^(1/2)-a*ln(a+(a^2-4*b*y)^(1/2)))/b = x+c__1

(-(a^2-4*b*y)^(1/2)-a*ln(a-(a^2-4*b*y)^(1/2)))/b = x+c__1

(1/5)*(-20*y+1)^(1/2)-(1/5)*ln(1+(-20*y+1)^(1/2)) = 12

-(1/5)*(-20*y+1)^(1/2)-(1/5)*ln(1-(-20*y+1)^(1/2)) = 12

sol1:=evalf(solve(eq1,y));

-205.8850616

sol2:=evalf(solve(eq2,y));

0.3221340286e-27


 

Download different_equations_give_same_solution_june_12_2024.mw

 

I am using intersectplot  to make projective coordinate plots. Everything intersects the plane z=1. I find the plot quality poor, i.e. dotty dashy lines and circle. This seem to be the best linestyle=solid can do here. gridrefine can't be applied here. 
Any suggestions to improve quality here?
Maybe intersectplot is not the best aprroach here but so far it is all if have figured out.


restart

 

 

with(plottools)

[annulus, arc, arrow, circle, colorbar, cone, cuboid, curve, cutin, cutout, cylinder, disk, dodecahedron, ellipse, ellipticArc, exportplot, extrude, getdata, hemisphere, hexahedron, homothety, hyperbola, icosahedron, importplot, line, octahedron, parallelepiped, pieslice, point, polygon, polygonbyname, prism, project, pyramid, rectangle, reflect, rotate, scale, sector, semitorus, sphere, stellate, tetrahedron, torus, transform, translate, triangulate]

(1)

with(plots)

[animate, animate3d, animatecurve, arrow, changecoords, complexplot, complexplot3d, conformal, conformal3d, contourplot, contourplot3d, coordplot, coordplot3d, densityplot, display, dualaxisplot, fieldplot, fieldplot3d, gradplot, gradplot3d, implicitplot, implicitplot3d, inequal, interactive, interactiveparams, intersectplot, listcontplot, listcontplot3d, listdensityplot, listplot, listplot3d, loglogplot, logplot, matrixplot, multiple, odeplot, pareto, plotcompare, pointplot, pointplot3d, polarplot, polygonplot, polygonplot3d, polyhedra_supported, polyhedraplot, rootlocus, semilogplot, setcolors, setoptions, setoptions3d, shadebetween, spacecurve, sparsematrixplot, surfdata, textplot, textplot3d, tubeplot]

(2)

 

 

DistCircle:=x^2+y^2=1

x^2+y^2 = 1

(3)

pt1:=[1/4,3/4]

[1/4, 3/4]

(4)

pt2:=[7/8,-1/3]

[7/8, -1/3]

(5)

pt3:=[-3/2,3/7]

[-3/2, 3/7]

(6)

pt4:=[3/5,-4/5]

[3/5, -4/5]

(7)

pt5:=[-1/10,-3/2]

[-1/10, -3/2]

(8)

 

L12:=-(13*x)/12 - (5*y)/8 + 71/96; #LnPeqns(pt1,pt2);

-(13/12)*x-(5/8)*y+71/96

(9)

L13:=-(9*x)/28 + (7*y)/4 - 69/56; #LnPeqns(pt1,pt3);

-(9/28)*x+(7/4)*y-69/56

(10)

L23:=(16*x)/21 + (19*y)/8 + 1/8; #LnPeqns(pt2,pt3);

(16/21)*x+(19/8)*y+1/8

(11)

L35:=(27*x)/14 + (7*y)/5 + 321/140; #LnPeqns(pt5,pt3)

(27/14)*x+(7/5)*y+321/140

(12)

nullline:=3/5*x-4/5*y-1

(3/5)*x-(4/5)*y-1

(13)

ptplt:=point([pt1,pt2,pt3,pt4,pt5],color="Green",symbol=solidcircle,symbolsize=10):
txtplt:=textplot([pt4[],typeset("pt4")],align={below,right}):

plt1:=display(txtplt,implicitplot([DistCircle,L12,L13,L23,L35,nullline],x=-2..2,y=-1.5...1.5,color=[red,blue,blue,blue,blue,cyan]),ptplt,scaling=constrained)

 

 

# Projective Geometry Version

DistCirclez:=x^2+y^2-z^2;  #a Cone

 

x^2+y^2-z^2

(14)

pt1p:=[pt1[],1];
pt2p:=[pt2[],1];
pt3p:=[pt3[],1];
pt4p:=[pt4[],1];
pt5p:=[pt5[],1];

[1/4, 3/4, 1]

 

[7/8, -1/3, 1]

 

[-3/2, 3/7, 1]

 

[3/5, -4/5, 1]

 

[-1/10, -3/2, 1]

(15)

 

 

 

L12p:=(13*x)/12 + (5*y)/8 - (71*z)/96;#LnPeqns([pt1p,pt2p,[0,0,0]]);

(13/12)*x+(5/8)*y-(71/96)*z

(16)

L13p:=(13*x)/12 + (5*y)/8 - (71*z)/96;#LnPeqns([pt1p,pt3p,[0,0,0]]);

(13/12)*x+(5/8)*y-(71/96)*z

(17)

L23p:=(9*x)/28 - (7*y)/4 + (69*z)/56;#LnPeqns([pt2p,pt3p,[0,0,0]]);

(9/28)*x-(7/4)*y+(69/56)*z

(18)

L35p:=(27*x)/14 + (7*y)/5 + (321*z)/140;#LnPeqns([pt3p,pt5p,[0,0,0]]);

(27/14)*x+(7/5)*y+(321/140)*z

(19)

L04p:=3/5*x-4/5*y-1*z;

(3/5)*x-(4/5)*y-z

(20)

ptpltp:=point([pt1p,pt2p,pt3p,pt4p,pt5p],symbol=solidsphere, symbolsize=8,color="green"):
intp1:=intersectplot(DistCirclez,z=1,x=-2.5..2.5,y=-2.5..2.5,z=0..1,linestyle=solid):#unit circle at z=1
intp12p:=intersectplot(L12p,z=1,x=-2.5..2.5,y=-2.5..2.5,z=0..1,color=blue):
intp13p:=intersectplot(L13p,z=1,x=-2.5..2.5,y=-2.5..2.5,z=0..1,color=blue):
intp23p:=intersectplot(L23p,z=1,x=-2.5..2.5,y=-2.5..2.5,z=0..1,color=blue):
intp35p:=intersectplot(L35p,z=1,x=-2.5..2.5,y=-2.5..2.5,z=0..1,color=blue):
intp04p:=intersectplot(L04p,z=1,x=-2.5..2.5,y=-2.5..2.5,z=0..1,color=cyan):

 

display(ptpltp,intp1,intp12p,intp13p,intp23p,intp35p,intp04p,scaling=constrained,caption="Projective Co-ords on plane z=1",axes=normal,axis[3]=[tickmarks=[1]])

 

 


Download 2024-06-10_Q_Intersectplot_quality.mw

restart:with(Physics):

Setup(dimension=3,coordinates=(X=[x,y,z])):

Define(Tp3[~alpha,~beta,~gamma],B[~lambda,mu],T3[~rho,~epsilon,~sigma]):

Tp3[~alpha,~beta,~gamma]=B[~alpa,rho]B[~beta,epsilon]B[~gamma,sigma]T3[~rho,~epsilon,~sigma]

SumOverRepeatedIndices of the expression does not do anything. Why?

Thanks for the answer.

Dear sir,

In the given problem, eta = 0 to 20, I want the table value of eta for a step size of 1000 (0 to 20 in thousand parts).

i have calculated only for one value, zero

Download Demo_paper_work.mw

Can I solve the given biquadratic equation in terms of sigma. I only need real positive root, if any

6125*_Z^4 + 68644*_Z^3*sigma - 219625*_Z^3 + 255712*_Z^2*sigma - 959250*_Z^2 + 238144*_Z*sigma - 1113500*_Z - 245000

I tried with solve for variable but it does not work. 

These are two examples of challenging ode solutions to show they satisfy the ode.

I tried many things myself but can't do it. Feel free to use any method or trick you want. The goal is simply to show that the solution is correct. The solutions are correct as far as I know, but hard to show by back substitution since the solutions are given in form of integrals and RootOf in them.

Extra credit points will be awarded for those who manage to do both.

28148

interface(version);

`Standard Worksheet Interface, Maple 2024.0, Windows 10, March 01 2024 Build ID 1794891`

Example 1

 

_EnvTry:='hard';
ode:=y(x) = arcsin(diff(y(x), x)) + ln(1 + diff(y(x), x)^2);
sol:=dsolve(ode);
r:=odetest(sol,ode);
coulditbe(r=0);

hard

y(x) = arcsin(diff(y(x), x))+ln(1+(diff(y(x), x))^2)

x-Intat(1/sin(RootOf(-_a+_Z+ln(sin(_Z)^2+1))), _a = y(x))-c__1 = 0

-arcsin(sin(RootOf(-y(x)+_Z+ln(3/2-(1/2)*cos(2*_Z)))))+RootOf(-y(x)+_Z+ln(3/2-(1/2)*cos(2*_Z)))

FAIL

Example 2

 

ode:=(1 + diff(y(x), x)^2)*(arctan(diff(y(x), x)) + a*x) + diff(y(x), x) = 0;
sol:=dsolve(ode);
r:=odetest(sol,ode);
coulditbe(r=0)

(1+(diff(y(x), x))^2)*(arctan(diff(y(x), x))+a*x)+diff(y(x), x) = 0

y(x) = Int(tan(RootOf(a*x*tan(_Z)^2+tan(_Z)^2*_Z+a*x+tan(_Z)+_Z)), x)+c__1

(-arctan(tan(RootOf(2*a*x+sin(2*_Z)+2*_Z)))+RootOf(2*a*x+sin(2*_Z)+2*_Z))*tan(RootOf(2*a*x+sin(2*_Z)+2*_Z))/(a*x+RootOf(2*a*x+sin(2*_Z)+2*_Z))

FAIL

 

 

Download showing_solution_satisfies_ode.mw

Hi,

I am exploring the boxplot, and I see that I do not have the option to integrate 2 lists: One for observations and one for frequencies. The BoxPlot command only accepts one list (List A in my example). Is there a way to create the BoxPlot using the 'Obs' and 'Eff' lists? Thank you for your insight

QBoxPlot.mw

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