This is a simplified case from a colleague who could not get has to work.

While this

I/sqrt(a);
has(%, sqrt(a));
                               I   
                             ------
                              (1/2)
                             a     

                             false

is a technically correct output from Maple, false is mathematically an incorrect statement (IMO).

I think new users cannot find the explanation for that on their own because in other instances the output is true

has(I/(1 - sqrt(a)), sqrt(a))
                              

                              true

Are there better ways than

I/sqrt(a);
has(%, sqrt(a))or has(%,1/sqrt(a));
                               I   
                             ------
                              (1/2)
                             a     

                              true

Its a bit complicated to include allways the inverse of an expression to capture all instances of an expression.

Should has do this an give a warning that also the inverse is present?

Hello,

I'm trying to perform a nonlinear curve-fit on a data(named Data in the code) in Maple, but I get an error :

Error, (in Statistics:-NonlinearFit) could not store .476190476190476164*(4.41000000000000014*(4410./[2.08075582127286]^(2/3)+952.380952380952294*[2.08075582127286]^(1/3)+1000.)/[2.08075582127286]^(5/3)-19448.1000000000022/[2.08075582127286]^(7/3)+[1040.37791063642999]-500.-166.666666666666657*(8.82000000000000028/[2.08075582127286]^(2/3)+1.90476190476190466*[2.08075582127286]^(1/3))/[2.08075582127286]-166.666666666666657*(16.8000000000000007/[2.08075582127286]^(1/3)+.907029478458049820*[2.08075582127286]^(2/3))/[2.08075582127286])*[2.08075582127286]-.500000000000000000 in a floating-point rtable

I suspect there's a problem associated with the variable "r". I am trying to find J by solving the equation p__3 = 0. Then J(stored in "r") is used in P__eng1 to perform the curve-fit operation. I have tried using fsolve, and NLPSolve but they ended up in errors. I read the documents but I was unable to solve the problem.

I'm new to Maple and I'd appreciate any comment that helps me solve p__3=0 and to use the J in P__eng1. I have also attached the code.test.mw

Code and data : mapp.zip

Thanks,

I called dsolve with timelimit on this ode.  All of Maple instantly locks up.

I do not mean the worksheet, but everything. I have each worksheet using its own engine,. Not able to open new worksheet, nothing clicks. The whole front end locks up. Can't click on anything. 

Killing every mserver.exe does not even resolve this. I had to terminate all of Maple from the task manager. The strange thing, is that looking at task manager I see mserver.exe doing nothing. zero CPU. Only the front end process is at high CPU (the one with the Java icon).  It looks like the Java frontend is locked up for some reason.

Do others see the same thing? does this happen on the mac also? Please make sure to save all your work before trying to run this on your PC.

Maple 2024.1 on windows 10. This is first order Riccati ode.

Download maple_lock_up_july_17_2024.mw

These are GUI options on my end.

UPDATE 1

in Maple 2023.2 also on windows 10,  there is no lock up at all. same code. The solution to the ode is very large. I tried using both typesetting level  to EXTENDED  and  typesetting level MAPLE STANDARD and no hang.

I wonder if this has something to do with why Maple 2024.1 locks up? But I have not changed anything in my end. I made no change in options or anything else. same PC, same graphics card. 

UPDATE 2

Found the BUG!!  

In Maple 2024.1, if I change the display option typesetting level  to EXTENDED, the front end do not hang.

If I change the display typesetting level MAPLE STANDARD, the ftont end hangs.

But why??

Download maple_lock_up_july_17_2024_2.mw

ps. Reported to Maplesoft support. July 17, 2024. Hopefully this will be fixed in Maple 2024.2

i am looking for all outcome of this function how i can do it

phi := (`\`p_1,p__2`, q__1, q__2, xi) -> p__1*exp(q__1*xi) - p__2*exp(q__2*xi);
for p[1 ] in [0,1,-1,I,-I] do ;
for p[2 ] in [0,1,-1,I,-I] do ;

 for q[1 ] in [0,1,-1,I,-I] do ;
 for q[2 ] in [0,1,-1,I,-I] do ;
result1 := evalf(phi(`\`p_1,p__2`, q__1, q__2, xi));
print('result1);

my code is not correct and not run in fact but it is my try

i am looking for special solution i want give the maple equation and give what answer i want with condition for example i just want thus solution which is A_0,A_1,B_1 not equal to zero and other parameter like (w,lambda,k) are free just this three not equal to zero.

restart
with(SolveTools);
with(LinearAlgebra);
eq12 := -alpha*k^2*A[0] - alpha*k^2*A[1] - alpha*k^2*B[1] + A[0]^3*beta[4] + (3*A[0]^2)*A[1]*beta[4] + (3*A[0]^2)*B[1]*beta[4] + (3*A[0])*A[1]^2*beta[4] + (6*A[0])*A[1]*B[1]*beta[4] + (3*A[0])*B[1]^2*beta[4] + A[1]^3*beta[4] + (3*A[1]^2)*B[1]*beta[4] + (3*A[1])*B[1]^2*beta[4] + B[1]^3*beta[4] + A[0]^2*beta[3] + (2*A[0])*A[1]*beta[3] + (2*A[0])*B[1]*beta[3] + A[1]^2*beta[3] + (2*A[1])*B[1]*beta[3] + B[1]^2*beta[3] - w*A[0] - w*A[1] - w*B[1] = 0

eq10 := (2*alpha)*k^2*A[1] - (2*alpha)*k^2*B[1] - (8*alpha)*lambda^2*A[1] + (8*alpha)*lambda^2*B[1] - (8*gamma)*lambda^2*A[1] + (8*gamma)*lambda^2*B[1] - (6*A[0]^2)*A[1]*beta[4] + (6*A[0]^2)*B[1]*beta[4] - (12*A[0])*A[1]^2*beta[4] + (12*A[0])*B[1]^2*beta[4] - (6*A[1]^3)*beta[4] - (6*A[1]^2)*B[1]*beta[4] + (6*A[1])*B[1]^2*beta[4] + (6*B[1]^3)*beta[4] - (4*A[0])*A[1]*beta[3] + (4*A[0])*B[1]*beta[3] - (4*A[1]^2)*beta[3] + (4*B[1]^2)*beta[3] + (2*w)*A[1] - (2*w)*B[1] = 0

eq8 := -(3*A[1]^2)*B[1]*beta[4] - (3*A[1])*B[1]^2*beta[4] - (2*A[0])*A[1]*beta[3] - (2*A[0])*B[1]*beta[3] + w*A[1] + w*B[1] - (3*A[0]^2)*B[1]*beta[4] + alpha*k^2*A[1] + alpha*k^2*B[1] - (3*A[0]^2)*A[1]*beta[4] + (3*alpha)*k^2*A[0] + (32*alpha)*lambda^2*A[1] + (32*alpha)*lambda^2*B[1] + (32*gamma)*lambda^2*A[1] + (32*gamma)*lambda^2*B[1] - (3*A[0]^3)*beta[4] + (15*A[0])*A[1]^2*beta[4] - (18*A[0])*A[1]*B[1]*beta[4] + (15*A[0])*B[1]^2*beta[4] + (15*A[1]^3)*beta[4] + (15*B[1]^3)*beta[4] - (3*A[0]^2)*beta[3] + (5*A[1]^2)*beta[3] - (6*A[1])*B[1]*beta[3] + (5*B[1]^2)*beta[3] + (3*w)*A[0] = 0

eq6 := -(4*alpha)*k^2*A[1] + (4*alpha)*k^2*B[1] - (48*alpha)*lambda^2*A[1] + (48*alpha)*lambda^2*B[1] - (48*gamma)*lambda^2*A[1] + (48*gamma)*lambda^2*B[1] + (12*A[0]^2)*A[1]*beta[4] - (12*A[0]^2)*B[1]*beta[4] - (20*A[1]^3)*beta[4] + (12*A[1]^2)*B[1]*beta[4] - (12*A[1])*B[1]^2*beta[4] + (20*B[1]^3)*beta[4] + (8*A[0])*A[1]*beta[3] - (8*A[0])*B[1]*beta[3] - (4*w)*A[1] + (4*w)*B[1] = 0

eq4 := -(3*A[1]^2)*B[1]*beta[4] - (3*A[1])*B[1]^2*beta[4] - (2*A[0])*A[1]*beta[3] - (2*A[0])*B[1]*beta[3] + w*A[1] + w*B[1] - (3*A[0]^2)*B[1]*beta[4] + alpha*k^2*A[1] + alpha*k^2*B[1] - (3*A[0]^2)*A[1]*beta[4] - (3*alpha)*k^2*A[0] + (32*alpha)*lambda^2*A[1] + (32*alpha)*lambda^2*B[1] + (32*gamma)*lambda^2*A[1] + (32*gamma)*lambda^2*B[1] + (3*A[0]^3)*beta[4] - (15*A[0])*A[1]^2*beta[4] + (18*A[0])*A[1]*B[1]*beta[4] - (15*A[0])*B[1]^2*beta[4] + (15*A[1]^3)*beta[4] + (15*B[1]^3)*beta[4] + (3*A[0]^2)*beta[3] - (5*A[1]^2)*beta[3] + (6*A[1])*B[1]*beta[3] - (5*B[1]^2)*beta[3] - (3*w)*A[0] = 0

eq2 := (2*alpha)*k^2*A[1] - (2*alpha)*k^2*B[1] - (8*alpha)*lambda^2*A[1] + (8*alpha)*lambda^2*B[1] - (8*gamma)*lambda^2*A[1] + (8*gamma)*lambda^2*B[1] - (6*A[0]^2)*A[1]*beta[4] + (6*A[0]^2)*B[1]*beta[4] + (12*A[0])*A[1]^2*beta[4] - (12*A[0])*B[1]^2*beta[4] - (6*A[1]^3)*beta[4] - (6*A[1]^2)*B[1]*beta[4] + (6*A[1])*B[1]^2*beta[4] + (6*B[1]^3)*beta[4] - (4*A[0])*A[1]*beta[3] + (4*A[0])*B[1]*beta[3] + (4*A[1]^2)*beta[3] - (4*B[1]^2)*beta[3] + (2*w)*A[1] - (2*w)*B[1] = 0

eq0 := alpha*k^2*A[0] - alpha*k^2*A[1] - alpha*k^2*B[1] - A[0]^3*beta[4] + (3*A[0]^2)*A[1]*beta[4] + (3*A[0]^2)*B[1]*beta[4] - (3*A[0])*A[1]^2*beta[4] - (6*A[0])*A[1]*B[1]*beta[4] - (3*A[0])*B[1]^2*beta[4] + A[1]^3*beta[4] + (3*A[1]^2)*B[1]*beta[4] + (3*A[1])*B[1]^2*beta[4] + B[1]^3*beta[4] - A[0]^2*beta[3] + (2*A[0])*A[1]*beta[3] + (2*A[0])*B[1]*beta[3] - A[1]^2*beta[3] - (2*A[1])*B[1]*beta[3] - B[1]^2*beta[3] + w*A[0] - w*A[1] - w*B[1] = 0

COEFFS := solve({eq0, eq10, eq12, eq2, eq4, eq6, eq8}, {k, lambda, w, A[0], A[1], B[1]})

Good day, all.

Please I want to solve the following delay differential equation:

ODE := diff(y(t), t$2) = (2*(1-y(t-1)^2))*(diff(y(t), t))-y(t)

ics := y(0) = 1, (D(y))(0) = 0

using the following codes but there is an error. Please kindly help to modify the codes.

restart;
Digits:=30:

f:=proc(n)
	2*(1-(y[n-1])^2)*delta[n]+y[n]:
end proc:

g:=proc(n)
	-4*y[n-1]*delta[n-1]+2*(1-(y[n-1])^2)*f(n)-delta[n]:
end proc:


e1:=y[n+2] = -y[n]+2*y[n+1]+(1/120)*h^2*(-3*h*g(n+2)+3*g(n)*h+16*f(n+2)+16*f(n)+88*f(n+1)):
e2:=h*delta[n] = -y[n]+y[n+1]-(1/1680)*h^2*(-128*h*g(n+1)-11*h*g(n+2)+59*g(n)*h+40*f(n+2)+520*f(n)+280*f(n+1)):
e3:=h*delta[n+1] = -y[n]+y[n+1]+(1/1680)*h^2*(-152*h*g(n+1)-10*h*g(n+2)+32*g(n)*h+37*f(n+2)+187*f(n)+616*f(n+1)):
e4:=h*delta[n+2] = -y[n]+y[n+1]+(1/1680)*h^2*(128*h*g(n+1)-101*h*g(n+2)+53*g(n)*h+744*f(n+2)+264*f(n)+1512*f(n+1)):

inx:=0:
ind:=0:
iny:=1:
h:=1/2:
n:=1:
omega:=10:
u:=omega*h:
N:=solve(h*p = 10, p):

err := Vector(round(N)):
exy_lst := Vector(round(N)):
numerical_y1:=Vector(round(N)):

c:=1:
for j from 0 to 2 do
	t[j]:=inx+j*h:
end do:

vars:=y[n+1],y[n+2],delta[n+1],delta[n+2]:

step := [seq](eval(x, x=c*h), c=1..N):
printf("%6s%45s%45s\n", 
	"h","Num.y","Num.z");
#eval(<vars>, solve({e||(1..4)},{vars}));


st := time():
for k from 1 to N/2 do

	par1:=x[0]=t[0],x[1]=t[1],x[2]=t[2]:
	par2:=y[n]=iny,delta[n]=ind:
	res:=eval(<vars>, fsolve(eval({e||(1..4)},[par1,par2]), {vars}));

	for i from 1 to 2 do
		printf("%6.5f%45.30f%45.30f\n", 
		h*c,res[i],res[i+2]):
		
		numerical_y1[c] := res[i]:
		
		c:=c+1:
	end do:
	iny:=res[2]:
	ind:=res[4]:
	inx:=t[2]:
	for j from 0 to 2 do
		t[j]:=inx + j*h:
	end do:
end do:
v:=time() - st;
v/4;
printf("Maximum error is %.13g\n", max(err));
NFE=evalf((N/4*3)+1);
#get array of numerical and exact solutions for y1
numerical_array_y1 := [seq(numerical_y1[i], i = 1 .. N)]:
#exact_array_y1 := [seq(exy[i], i = 1 .. N)]:

#get array of time steps
time_t := [seq](step[i], i = 1 .. N):

#display graphs for y1
with(plots):
numerical_plot_y1 := plot(time_t, numerical_array_y1, style = [point], symbol = [asterisk],
				color = [blue,blue],symbolsize = 20, legend = ["TFIBF"]);

 Thank you, and best regards.

I can't isolate a variable from the constraint and move it to one side of the inequality. For example, it's like having theta <= all other terms. I've tried using the isolate and eval functions, but they aren't producing any results or simplifying the expression. What am I doing wrong?

Attaching sheet below (issue marked yellow in background): N_1.mw 

Dear all,

I'd like to explore graphically a polynomial surface depending on two parameters a and b.

The problem is that, as soon as I start playing with the sliders, Maple freezes and I have to 'force quit'.

Can you please tell me if you have the same problem with this example?

Thanks.

Download Explore.mw

Hello Dear Professional users,

I have a question regarding the "fsolve" command and also the "assign" command in Maple.

In my previous codes, I just used one time from "fsolve" and then "assign" command.

Today, I want to use "fsolve" and "assign" in a for loop. But, I can not get the results correctly.

For example, previously I reach a system of algebraic equations and the solve my problem easily:

N:=8:
y:=sum(a[n]*t^n,n=0..N):
y:=unapply(y,t):
*****some calculations with a[n] coefficients *****
A:={a set of nonlinear algebraic equations in terms of a[n]}%The number of equations is N+1 (same as the number of a[n] for n=0..N)
sol:=fsolve(A):
assign(sol)
plot(y(t),t=0..1)

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

Today, my problem is:

N:=8:
M:=4:
for i from 1 to M do
y[i]:=sum(a[i,n]*t^n,n=0..N):
y[i]:=unapply(y[i],t):
end do
for i from 1 to M do
*****some calculations with a[n] coefficients *****
@A[i]:={a set of nonlinear algebraic equations in terms of a[i,n]}%The number of equations is N+1 (same as the number of a[n] for n=0..N)
@sol[i]:=fsolve(A[i]):
@assign(sol[i])
@plot(y[i](t),t=0..1)
end do
****************************************************************************************************
****************************************************************************************************
****************************************************************************************************

What I should write instead of "A[i]", "sol[i]", and "assign(sol[i])" the lines started with
@A[i],....
@sol[i],...
@assign(sol[i]),....

Thanks for your attention in advance

With kind regards,
Emran

Hello

I am trying to understand how to use Maple to solve a PDE.  Below it is a problem (Henon-Heilles system) where the answer is known.  

with(PDEtools);

infolevel[pdsolve]:=3:

declare(Hamil(x,y,u,v));

PDEHamil := u*diff(Hamil(x, y, u, v), x) + v*diff(Hamil(x, y, u, v), y) + (-2*x*y - x)*diff(Hamil(x, y, u, v), u) + (-x^2 + y^2 - y)*diff(Hamil(x, y, u, v), v) = 0;

pdsolve(PDEHamil)

Maple returns no solution, but one solution is:

sH:=1/2*(u^2+v^2)+1/2*(x^2+y^2-2/3*y^3)+x^2*y;

simplify(eval(subs(Hamil(x,y,u,v)=sH,PDEHamil)));
    0=0

What am I missing?

Many thanks.

I am trying to install Syrup in my home computer (I have it installed in my work computer).  I followed the instructions in the Readme file: 

From Standard Maple:
    Open the file Syrup-Installer.mla.
    To do so, use File -> Open, choose file type 
    "Maple Library Archive (.mla)", select the file, and
    click "Open".

Everything seemed to work and the help page opened up:  But, it is not the syrup help page.  furthermore, when I type ?Syrup, it doesn't open it either.

When I try to run a worksheet that uses Syrup (that works on my work computer), I get these errors:

I"m going to reboot now and try again.  

Jorge

I was working with a Dataframe when I wanted to change the datatype of multiple columns at the same time as this is quite a large dataframe. I found in the helpfile that I can change datatype by the following command: 

SubsDatatype(Data, plts, float) which then change the datatype of "plts" into float. I had hoped that using multiple columns in the command would work in this way: SubsDatatype(Data, [plts, act], float)  but apparently not. Is there a way to do this or do I have to do it column by column?

Additionally I have another question about dataframes. I would like to replace "0" in the dataframe by a "blank" as you can do in excel. How do you do this in a dataframe?

Thanks in advance for any help given!

restart;
with(DEtools);
ode := diff(y(x), x) = epsilon - y(x)^2;
                       d                       2
               ode := --- y(x) = epsilon - y(x) 
                       dx                       

sol := dsolve(ode);
                  /           (1/2)            (1/2)\        (1/2)
sol := y(x) = tanh\_C1 epsilon      + x epsilon     / epsilon     

P := particularsol(ode);
                          (1/2)                 (1/2)  
       P := y(x) = epsilon     , y(x) = -epsilon     , 

                /    y(x)    \            (1/2)          
         arctanh|------------| - x epsilon      + _C1 = 0
                |       (1/2)|                           
                \epsilon     /                           

i am looking for finding all solution of this equation like this picture below

Dear all,

I would like to find how I can calculate covariant derivative of Einstein tensor for an arbitrary metric ds_2=-A(r)*dt^2+B(r)*dr^2+dtheta^2+sin(theta)^2*dphi^2

with best

Dear all,

I have this polynomial function

G(x, y) := (-0.14*y^3 + 1.20000000000000*y^2 - 1.26000000000000*y + 0.200000000000000)*x^3 + (1.20*y^3 - 10.0800000000000*y^2 + 10.0800000000000*y - 1.20000000000000)*x^2 + (-8.82*y + 10.08*y^2 - 1.26*y^3)*x + 1. - 1.2*y^2 + 0.2*y^3

I don't understand why the command

minimize(G(x,y),x=0..1,y=0..1);

produces the error

Error, (in RootOf/RootOf:-algnum_in_range) invalid input: RootOf/RootOf:-rootof_in_range expects its 1st argument, rt, to be of type ('RootOf')(polynom(rational,_Z),identical(index) = posint), but received RootOf(7*_Z^3-93*_Z^2+327*_Z-187)

Thanks for your advices, Nicola