MaplePrimes Questions

How do I get the susset that contains unknowns on the rhs of the elements?

restart

 

# I need this subset {a=1/sqrt(2+A), b=6*sqrt(4+N),  d=5*H}

 

C:={a=1/sqrt(2+A),b=6*sqrt(4+N) ,c=sqrt(7),d=5*H,,e=-12,f=-96}

{a = 1/(2+A)^(1/2), b = 6*(4+N)^(1/2), c = 7^(1/2), d = 5*H, e = -12, f = -96}

(1)

selectremove(has,indets(rhs~(C)),C)

{}, {A, H, K, N, 1/(2+A)^(1/2), (4+N)^(1/2)}

(2)

selectremove(has,lhs~(C)=indets(rhs~(C)),C)

() = (), {a, b, c, d, e} = {H, K, N, (4+N)^(1/2)}

(3)
 

 

Download 2024-06-05_Q_Select_Remove_indet_elements.mw

Hello, in a Maple script intended for Maple Learn I need to use a slider but I don't know how to get its value.

What should I do to get it?

I have a positive surface that I plot3d for bounded x- and y- value ranges. It reaches 1000 but it's mostly flat except almost at the edges of the x- and y-axes. Therefore, I inlcude view=[default,default,0..10] among the options of plot3d so that I can focus on the features of interest.

For coloring I use 'colorscheme'=["zgradient",["LightGray", "Gray", "Green"]]. However, this applies to the whole surface (up to 1000) and NOT exclusively to the viewed part as I would like to. 

Question: how to scale the zgradient so that it only applies to my "view" portion?

Note that I don't want to do this by using the 'markers' options to readjust the color splits manually, as I don't know the exact proportion and I don't want to randomly play around with different triplets until I achieve a visually satisfying result...

We see that this ode (x + y(x))*(1+diff(y(x),x)) = 0  has 2 solutions, One when (x + y(x))=0 and one when (1+diff(y(x),x))=0. Maple gives 3 solutions. They are correct but why?

Also when changing (1+diff(y(x),x)) to (a+diff(y(x),x)) now it gives only two solution.

Why does this happen? Should it not just return 2 solutions in both cases? and more strange one

(x + y(x))^2 *(1+diff(y(x), x))=0; now it gives 4 solutions. But this is no different. We also have 2 solutions. One when (x + y(x))=0 and one when (1+diff(y(x), x))=0. This time the extra two solutions are complex. 

17168

interface(version);

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

Physics:-Version();

`The "Physics Updates" version in the MapleCloud is 1754 and is the same as the version installed in this computer, created 2024, June 3, 20:39 hours Pacific Time.`

ode:=(x + y(x))*(a+diff(y(x),x)) = 0;
dsolve(ode);
map(X->odetest(X,ode),[%])

(x+y(x))*(a+diff(y(x), x)) = 0

y(x) = -x, y(x) = -a*x+c__1

[0, 0]

ode:=(x + y(x))*(1+diff(y(x),x)) = 0;
dsolve(ode);
map(X->odetest(X,ode),[%])

(x+y(x))*(1+diff(y(x), x)) = 0

y(x) = -x, y(x) = -x-c__1, y(x) = -x+c__1

[0, 0, 0]

ode:= (x + y(x))^2 *(1+diff(y(x), x))=0;
dsolve(ode);
map(X->odetest(X,ode),[%])

 

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

y(x) = -x, y(x) = -x+c__1, y(x) = -(1/2)*c__1-((1/2)*I)*3^(1/2)*c__1-x, y(x) = -(1/2)*c__1+((1/2)*I)*3^(1/2)*c__1-x

[0, 0, 0, 0]

 

 

Download why_extra_solution_from_dsolve_june_5_2024.mw

Any idea why this sometimes happens? odeadvisor says ode is quadrature but when asking it to solve it using quadrature sometimes it works and sometimes not.

Am I doing something wrong?

22020

interface(version);

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

Physics:-Version();

`The "Physics Updates" version in the MapleCloud is 1754 and is the same as the version installed in this computer, created 2024, June 3, 20:39 hours Pacific Time.`

restart;

18368

Example 1 that does not work

 

ode:=diff(y(x),x)=-1;
DEtools:-odeadvisor(ode);

diff(y(x), x) = -1

[_quadrature]

sol:=dsolve(ode,y(x),['quadrature']);

"sol := "

Example 2 that works

 

ode:=diff(y(x),x)=x;
DEtools:-odeadvisor(ode);

diff(y(x), x) = x

[_quadrature]

sol:=dsolve(ode,y(x),['quadrature']);

y(x) = (1/2)*x^2+c__1

 

 

Download sometimes_dsolve_works_on_quadrature_june_5_2024.mw

Dear all

I have a data, how can I study this data using N-soft set : Normalized the data, membership function, analyse the risk, ..... compute the risk, interpret the results

Cancer_patient.xlsx

N_soft_set.mw

Thank you for your help

Maybe someone get the code working ?
 

with(plots):
with(VectorCalculus):

# Example 1: Vector Field and Visualization
V := [x, y, z]:
print("Vector Field V:", V):
fieldplot3d([V[1], V[2], V[3]], x = -2..2, y = -2..2, z = -2..2, arrows = slim, title = "Vector Field in 3D"):

# Example 2: Tangent Vector to a Curve
curve := [cos(t), sin(t), t]:
print("Curve:", curve):
tangent := diff(curve, t):
print("Tangent Vector:", tangent):
plot3d([cos(t), sin(t), t], t = 0..2*Pi, labels = [x, y, z], title = "Curve in 3D"):

# Example 3: Curvature of a Surface
u := 'u': v := 'v':
surface := [u, v, u^2 - v^2]:
print("Surface:", surface):

# Compute the first fundamental form
ru := [diff(surface[1], u), diff(surface[2], u), diff(surface[3], u)]:
rv := [diff(surface[1], v), diff(surface[2], v), diff(surface[3], v)]:
E := ru[1]^2 + ru[2]^2 + ru[3]^2:
F := ru[1]*rv[1] + ru[2]*rv[2] + ru[3]*rv[3]:
G := rv[1]^2 + rv[2]^2 + rv[3]^2:
firstFundamentalForm := Matrix([[E, F], [F, G]]):
print("First Fundamental Form:", firstFundamentalForm):

# Compute the second fundamental form
ruu := [diff(surface[1], u, u), diff(surface[2], u, u), diff(surface[3], u, u)]:
ruv := [diff(surface[1], u, v), diff(surface[2], u, v), diff(surface[3], u, v)]:
rvv := [diff(surface[1], v, v), diff(surface[2], v, v), diff(surface[3], v, v)]:
normal := CrossProduct(ru, rv):
normal := eval(normal / sqrt(normal[1]^2 + normal[2]^2 + normal[3]^2)):
L := ruu[1]*normal[1] + ruu[2]*normal[2] + ruu[3]*normal[3]:
M := ruv[1]*normal[1] + ruv[2]*normal[2] + ruv[3]*normal[3]:
N := rvv[1]*normal[1] + rvv[2]*normal[2] + rvv[3]*normal[3]:
secondFundamentalForm := Matrix([[L, M], [M, N]]):
print("Second Fundamental Form:", secondFundamentalForm):

# Compute the Christoffel symbols
# Ensure DifferentialGeometry package is loaded
with(DifferentialGeometry):
DGsetup([u, v], N):
Gamma := Christoffel(firstFundamentalForm):
print("Christoffel Symbols:", Gamma):

# Visualize the surface
plot3d([u, v, u^2 - v^2], u = -2..2, v = -2..2, labels = [u, v, z], title = "Saddle Surface in 3D"):

 

I was about to convert -1 to exp(I*Pi).

First try by intuition

convert(-1, exp);
                               -1

False intuition. Next try by right click on -1 and consulting the context pannel: -> nothing that rings a bell

Entering a complex number 0.5000000001 + 0.8660254037*I to see if that changes something

-> something but no exponential form
                          

Next: AI assitant

->   OK, I know this. -> refine query

Polar, of course!

convert(-1, polar);
                          polar(1, Pi)

Not what I want. Maybe

convert(convert(-1, polar), exp);
                               -1

Next -> Ask for help in MaplePrimes whether there is a command

In relation to my comment below:
https://www.mapleprimes.com/questions/238277-Robustness-Of-Plotevalsomething#comment301920 (whose my worksheet rho-analysis_mmcdara_Gammapositive_MaPal.mw builds on the Sturm's analysis of @mmcdara)

I would like to better understand how to isolate/pin down the 4 real roots whose existence was confirmed by Sturm's analysis. To do so, @acer suggested to look into RootFinding:-Parametric: rho-analysis_acc(1).mw
Questions:

  1. Why the plot doesn't change if I also include -1<rho and rho<1 among the equations to solve? I understood that regions 1, 4 and 5 have 0 solutions anyway but if I add the constraints on rho I should expect 2 regions instead of 5, right?

  2. I don't understand SampleSolutions(m,2)=0.56 and SampleSolutions(m,3)=0.51 (even after reading help page). How are these numeric values found?

  3. Looking at the big picture, how to reconcile this CellPlot with A) my plot3d of Eq in https://www.mapleprimes.com/questions/238277-Robustness-Of-Plotevalsomething#comment301920 and B) @mmcdara Sturm's analysis. I am having a hard time putting all together.

    Thank you. 

    EDIT: I formulated this question by branching out from my comment 
    https://www.mapleprimes.com/questions/238277-Robustness-Of-Plotevalsomething#comment301920 in the corresponding thread. If inappropriate, please help me migrate this question as appropriate. 

I tried to solve a hogher order ODE system to reduce first order ODE sytem. In this case I have assume a initial conditions as a variable. (sse the attachment, Conver_to_first_order.mw).

I got an error in specification of initial value...

How can I solve this problem in RKF45 method with shooting technique.

Here, I have assume f=x[1], f'=x[2]. f''=x[3], f'''=x[3]' and theta=x[4], theta'=x[5], theta''=x[5]'

I try to dot product 2 vector that contain time derivative

the answer should be v . u_rho = rhodot

but Maple doesn't give me a proper answer. What should i do?

kind regard

restart;

with(LinearAlgebra);

with(Typesetting);
Settings(typesetprime = true);
Settings(typesetdot = true);

v := <diff(rho(t), t)*cos(theta(t)) - rho(t)*diff(theta(t), t)*sin(theta(t)), diff(rho(t), t)*sin(theta(t)) + rho(t)*diff(theta(t), t)*cos(theta(t)), diff(z(t), t)>

u_rho := <cos(theta(t)), sin(theta(t)), 0>

v . u_rho

Any one could find why Maple 2024 gives Error, (in evalf/int) invalid arguments on this ode? Why is it even calling evalf in first place as this is all symbolic.

Also reported to Maplesof just in case.

25504

restart;

25504

interface(version);

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

Physics:-Version();

`The "Physics Updates" version in the MapleCloud is 1752 and is the same as the version installed in this computer, created 2024, May 31, 18:17 hours Pacific Time.`

ode:=diff(y(x), x)^3 = (x - 2)^2;
ic:=y(2)=1;
Student:-ODEs:-ODESteps([ode,ic]);

(diff(y(x), x))^3 = (x-2)^2

y(2) = 1

Error, (in evalf/int) invalid arguments

dsolve([ode,ic])

y(x) = 1+(1/5)*(3*x-6)*RootOf(_Z^3-x^2+4*x-4)

DEtools:-odeadvisor(ode)

[_quadrature]

 

 

Download odetest_internal_error_evalf_int_june_3_2024.mw

Probably I am missing the obvious. In the attached worksheet the for statement gives me all results as desired but if I try to put this in a procedure I only get the last value. Could someone help me out. Thank you  ProcQuestion.mw

I wanted to do 

map(X->odetest(X,ode) assuming x>0,[sol]) 

where sol are the solutions I wanted to test. The above does not work as is. After some trial and errors, I found this works

map(X->[odetest(X,ode) assuming x>0],[sol]) 

I am not sure why. But it works. Is this the above the correct way to add assuming to a command inside Map? Or is there a different way or better way to do this?

I want to stick to the map command as above and nothing else such as ~ or some other ones. I find map more clear to use for me.

Here is worksheet.

9592

restart;

9592

ode:=x = diff(y(x), x)*sqrt(1 + diff(y(x), x)^2);

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

sol:=dsolve(ode);
map(X->[odetest(X,ode) assuming x>0],[sol])

y(x) = ((1/32)*I)*2^(1/2)*(-(256/3)*Pi^(1/2)*2^(1/2)*x^3*cosh((3/2)*arcsinh(2*x))-8*Pi^(1/2)*2^(1/2)*(-(64/3)*x^4-(8/3)*x^2+2/3)*sinh((3/2)*arcsinh(2*x))/(4*x^2+1)^(1/2))/Pi^(1/2)+c__1, y(x) = -((1/32)*I)*2^(1/2)*(-(256/3)*Pi^(1/2)*2^(1/2)*x^3*cosh((3/2)*arcsinh(2*x))-8*Pi^(1/2)*2^(1/2)*(-(64/3)*x^4-(8/3)*x^2+2/3)*sinh((3/2)*arcsinh(2*x))/(4*x^2+1)^(1/2))/Pi^(1/2)+c__1, y(x) = Int(-(1/2)*(-2+2*(4*x^2+1)^(1/2))^(1/2), x)+c__1, y(x) = Int((1/2)*(-2+2*(4*x^2+1)^(1/2))^(1/2), x)+c__1

[[(1/2)*(2*x*(4*x^2+1)^(1/2)+(-8*x^2*cosh((3/2)*arcsinh(2*x))*(4*x^2+1)^(1/2)+16*x^3*sinh((3/2)*arcsinh(2*x))-cosh((3/2)*arcsinh(2*x))*(4*x^2+1)^(1/2)+4*sinh((3/2)*arcsinh(2*x))*x)*(-2+2*(4*x^2+1)^(1/2))^(1/2))/(4*x^2+1)^(1/2)], [-(1/2)*(-2*x*(4*x^2+1)^(1/2)+(-8*x^2*cosh((3/2)*arcsinh(2*x))*(4*x^2+1)^(1/2)+16*x^3*sinh((3/2)*arcsinh(2*x))-cosh((3/2)*arcsinh(2*x))*(4*x^2+1)^(1/2)+4*sinh((3/2)*arcsinh(2*x))*x)*(-2+2*(4*x^2+1)^(1/2))^(1/2))/(4*x^2+1)^(1/2)], [2*x], [0]]

for item in [sol] do
    odetest(item,ode) assuming x>0;
od;

(1/2)*(2*x*(4*x^2+1)^(1/2)+(-8*x^2*cosh((3/2)*arcsinh(2*x))*(4*x^2+1)^(1/2)+16*x^3*sinh((3/2)*arcsinh(2*x))-cosh((3/2)*arcsinh(2*x))*(4*x^2+1)^(1/2)+4*sinh((3/2)*arcsinh(2*x))*x)*(-2+2*(4*x^2+1)^(1/2))^(1/2))/(4*x^2+1)^(1/2)

-(1/2)*(-2*x*(4*x^2+1)^(1/2)+(-8*x^2*cosh((3/2)*arcsinh(2*x))*(4*x^2+1)^(1/2)+16*x^3*sinh((3/2)*arcsinh(2*x))-cosh((3/2)*arcsinh(2*x))*(4*x^2+1)^(1/2)+4*sinh((3/2)*arcsinh(2*x))*x)*(-2+2*(4*x^2+1)^(1/2))^(1/2))/(4*x^2+1)^(1/2)

2*x

0

 

 

Download how_to_use_map_with_assuming.mw

These  are good example(s)  why I think Maple's simplify needs to be improved. 

Example 1

Given an expression which is zero for nonnegative x, one would expect simplify to simplify it to zero when told that x is positive. 

But nothing I tried with simplify worked. combine  figured it out.

But why? Is this not the job of simplify? 

The expression is 

e:=x-1/4*(-2+2*(4*x^2+1)^(1/2))^(1/2)*(2+2*(4*x^2+1)^(1/2))^(1/2);

Here is worksheet

``

interface(version);

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

restart;

28000

e:=x-1/4*(-2+2*(4*x^2+1)^(1/2))^(1/2)*(2+2*(4*x^2+1)^(1/2))^(1/2);

x-(1/4)*(-2+2*(4*x^2+1)^(1/2))^(1/2)*(2+2*(4*x^2+1)^(1/2))^(1/2)

coulditbe(e=0);

true

#we see it is zero for x>=0
plot(e,x=-3..3)

simplify(e) assuming x>=0;

x-(1/4)*(-2+2*(4*x^2+1)^(1/2))^(1/2)*(2+2*(4*x^2+1)^(1/2))^(1/2)

simplify(e,size) assuming x>=0;

x-(1/4)*(-2+2*(4*x^2+1)^(1/2))^(1/2)*(2+2*(4*x^2+1)^(1/2))^(1/2)

expand(e) assuming x>=0;

x-(1/4)*(-2+2*(4*x^2+1)^(1/2))^(1/2)*(2+2*(4*x^2+1)^(1/2))^(1/2)

simplify(e,sqrt) assuming x>=0;

x-(1/4)*(-2+2*(4*x^2+1)^(1/2))^(1/2)*(2+2*(4*x^2+1)^(1/2))^(1/2)

simplify(e,symbolic) assuming x>=0;

x-(1/4)*(-2+2*(4*x^2+1)^(1/2))^(1/2)*(2+2*(4*x^2+1)^(1/2))^(1/2)

combine(e) assuming x>=0;

0


Compare some other software on this same problem

It should be as simple as the above in Maple.

A user should not have to try 100 different commands in Maple to find which works.

simplify should have done it in first place. What Am I overlooking here?

Maple 2024

Download why_simplify_do_not_work_example.mw

EXAMPLE 2

In this example the expression is zero for x<0. Here nothing worked for me. This is challenge for Maple experts here to find the command to simplify this to zero. 

I know it is zero for x<0.

e := -1/2*(-2*x*sqrt(4*x^2 + 1) + (16*x^3*sinh(3/2*arcsinh(2*x)) - 8*x^2*cosh(3/2*arcsinh(2*x))*sqrt(4*x^2 + 1) + 4*sinh(3/2*arcsinh(2*x))*x - cosh(3/2*arcsinh(2*x))*sqrt(4*x^2 + 1))*sqrt(-2 + 2*sqrt(4*x^2 + 1)))/sqrt(4*x^2 + 1)


 

24064

interface(version);

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

Physics:-Version();

`The "Physics Updates" version in the MapleCloud is 1752 and is the same as the version installed in this computer, created 2024, May 31, 18:17 hours Pacific Time.`

e:=-1/2*(-2*x*sqrt(4*x^2 + 1) + (16*x^3*sinh(3/2*arcsinh(2*x)) - 8*x^2*cosh(3/2*arcsinh(2*x))*sqrt(4*x^2 + 1) + 4*sinh(3/2*arcsinh(2*x))*x - cosh(3/2*arcsinh(2*x))*sqrt(4*x^2 + 1))*sqrt(-2 + 2*sqrt(4*x^2 + 1)))/sqrt(4*x^2 + 1);

-(1/2)*(-2*x*(4*x^2+1)^(1/2)+(16*x^3*sinh((3/2)*arcsinh(2*x))-8*x^2*cosh((3/2)*arcsinh(2*x))*(4*x^2+1)^(1/2)+4*sinh((3/2)*arcsinh(2*x))*x-cosh((3/2)*arcsinh(2*x))*(4*x^2+1)^(1/2))*(-2+2*(4*x^2+1)^(1/2))^(1/2))/(4*x^2+1)^(1/2)

plot(e,x=-10..3);

coulditbe(e=0)

true

simplify(e) assuming x<0;

-8*((cosh((3/2)*arcsinh(2*x))*(-(1/2)*x^2-1/16)*(4*x^2+1)^(1/2)+sinh((3/2)*arcsinh(2*x))*x*(x^2+1/4))*(-2+2*(4*x^2+1)^(1/2))^(1/2)-(1/8)*x*(4*x^2+1)^(1/2))/(4*x^2+1)^(1/2)

simplify(e,symbolic) assuming x<0;

-8*((cosh((3/2)*arcsinh(2*x))*(-(1/2)*x^2-1/16)*(4*x^2+1)^(1/2)+sinh((3/2)*arcsinh(2*x))*x*(x^2+1/4))*(-2+2*(4*x^2+1)^(1/2))^(1/2)-(1/8)*x*(4*x^2+1)^(1/2))/(4*x^2+1)^(1/2)

combine(e) assuming x<0;

-(1/2)*(-2*x*(4*x^2+1)^(1/2)+(16*x^3*sinh((3/2)*arcsinh(2*x))-8*x^2*cosh((3/2)*arcsinh(2*x))*(4*x^2+1)^(1/2)+4*sinh((3/2)*arcsinh(2*x))*x-cosh((3/2)*arcsinh(2*x))*(4*x^2+1)^(1/2))*(-2+2*(4*x^2+1)^(1/2))^(1/2))/(4*x^2+1)^(1/2)

simplify(e,arctrig) assuming x<0;

-8*((cosh((3/2)*arcsinh(2*x))*(-(1/2)*x^2-1/16)*(4*x^2+1)^(1/2)+sinh((3/2)*arcsinh(2*x))*x*(x^2+1/4))*(-2+2*(4*x^2+1)^(1/2))^(1/2)-(1/8)*x*(4*x^2+1)^(1/2))/(4*x^2+1)^(1/2)

simplify(e,sqrt) assuming x<0;

-8*((cosh((3/2)*arcsinh(2*x))*(-(1/2)*x^2-1/16)*(4*x^2+1)^(1/2)+sinh((3/2)*arcsinh(2*x))*x*(x^2+1/4))*(-2+2*(4*x^2+1)^(1/2))^(1/2)-(1/8)*x*(4*x^2+1)^(1/2))/(4*x^2+1)^(1/2)

simplify(normal(e),sqrt) assuming x<0;

-8*((cosh((3/2)*arcsinh(2*x))*(-(1/2)*x^2-1/16)*(4*x^2+1)^(1/2)+sinh((3/2)*arcsinh(2*x))*x*(x^2+1/4))*(-2+2*(4*x^2+1)^(1/2))^(1/2)-(1/8)*x*(4*x^2+1)^(1/2))/(4*x^2+1)^(1/2)

simplify(normal(e, expanded)) assuming x<0;

-8*((cosh((3/2)*arcsinh(2*x))*(-(1/2)*x^2-1/16)*(4*x^2+1)^(1/2)+sinh((3/2)*arcsinh(2*x))*x*(x^2+1/4))*(-2+2*(4*x^2+1)^(1/2))^(1/2)-(1/8)*x*(4*x^2+1)^(1/2))/(4*x^2+1)^(1/2)

simplify(radnormal(e)) assuming x<0;

(1/8)*((4*x^2+(4*x^2+1)^(1/2)+1)*(-2+2*(4*x^2+1)^(1/2))-64*x*((1/2)*cosh((3/2)*arcsinh(2*x))*(-1/8-x^2)*(4*x^2+1)^(1/2)+sinh((3/2)*arcsinh(2*x))*x*(x^2+1/4))*(-2+2*(4*x^2+1)^(1/2))^(1/2))/((4*x^2+1)^(1/2)*x)

simplify(evala(e)) assuming x<0;

-8*((cosh((3/2)*arcsinh(2*x))*(-(1/2)*x^2-1/16)*(4*x^2+1)^(1/2)+sinh((3/2)*arcsinh(2*x))*x*(x^2+1/4))*(-2+2*(4*x^2+1)^(1/2))^(1/2)-(1/8)*x*(4*x^2+1)^(1/2))/(4*x^2+1)^(1/2)

 


Using some other software gives

I'd like to find command to do the same in Maple. i.e. simplify it to zero. What else to try?

 

Download why_simplify_do_not_work_example_2.mw

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