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

Using ScatterPlot (or ErrorPlot), one can add error bars to a 2d point plot of data. However, the bars are single lines. I wish to create a plot with H-type error bars in both the horizontal and vertical directions.  Below is an example showing how the bars should appear. (This image is taken from a previous question about adding error bars.) 

I do not need to reproduce this figure exactly. The location of the data points and the size of the error bars are irrelevant. The closest I have seen is using BoxPlot.

Has this question been asked and answered? If so, I cannot find it. 

there is any way for define conformable fractional derivative in partial differential equation

restart;
with(PDEtools);
pde := a*diff(psi(x, t), x $ 2) + (b*abs(psi(x, t))^(-2*n) + c*abs(psi(x, t))^(-n) + d*abs(psi(x, t))^n + f*abs(psi(x, t))^(2*n))*psi(x, t) = 0;
pde + i*diff(u(x, t), [t $ beta]) = 0;

how define a  fractional derivative in sense of conformable derivative

I don't understand where the the csgn(L) comes from in the solution below:

all the variables are defined as real:

coupled_network_vbat.mw

Thanks in advance for any help.

Jorge