Dear Sir,

I have a question: I have an analytic function depending on a real parameter that is of the form  F_y (z) with y>0 and z is complex.

I search the zeros  of F_y(z), that is the complex z staisfying F_y(z)=0. I used  the Maple function

RootFinding:-Analytic(F, z, re = -5 .. 0, im = -100 .. 100, );

but he displays me an error message  "Error, (in RootFinding:-Analytic) the function, -(100+I*z)^(1/2)*(80-I*z)^2*cosh((1/150)*(100-I*z)^(1/2)*Pi)*sinh((1/150)*(100+I*z)^(1/2)*Pi)+(100-I*z)^(1/2)*(80+I*z)^2*cosh((1/150)*(100+I*z)^(1/2)*Pi)*sinh((1/150)*(100-I*z)^(1/2)*Pi)+(2*I)*y*z^2*cosh((1/150)*(100-I*z)^(1/2)*Pi)*cosh((1/150)*(100+I*z)^(1/2)*Pi), depends on more than one variable: {y, z}. "

Can you help me to resolve my problem?

Best regards,

Zayd.

Is there an option or setting to tell Maple not to return trivial solution as a solution to a PDE? Even though trivial solution is correct, it is not something I wanted, and it is making it hard for me to check if the solution is trivial or not. 

Here is an example:

pde:=I*diff(f(x,t),t)=-diff(f(x,t),x$2)+2*x^2*f(x,t):
bc:=f(-infinity ,t)=0,f(infinity,t)=0:
pdsolve([pde,bc],f(x,t))

Mathematica returns this

If there is no such option in Maple, what would be a good way to check for trivial solution using Maple code?

Just checking if rhs(sol)=0 does not work all the time, as some actual solutions are returned with 0 in rhs, like this example

pde := diff(u(x, y), x) + u(x,y)*diff(u(x, y),y) =0:
sol:=pdsolve(pde,u(x,y)):
sol:=DEtools:-remove_RootOf(sol)

So I need either find a way to tell Maple not to return trivial solution, or good robust way to check if solution returned is the trivial solution so I can reject it.

Any suggestion from the Maple experts how to handle this? I am now just interested in result from pdsolve, not dsolve.

The input is the PDE itself, read from a file. So this has to be done without visual inspection on the screen as the program reads the PDE from a file and process it. So the dependent variable and the independent variables are all included in the PDE itself. To make it more clear, I need a function such as

is_solution_trivial :=proc(pde,sol)
     #decide if sol is trivial solution or not.
     #one possibility is to find from the PDE the dependent variable
     #and the independent variables as in u(x,y,t,....) and then check
     #if sol  has the form u(x,y,t,....) =0 ?
end proc;

The above I can use until I figure how to tell Maple NOT to return trivial solution in first place!

Maple 2019.1

Hi,

I am trying to find p and q from this simultaneous equation as a function of system parameters. I do not know the parameters and I need an expression. But Maple simply just gives p=0 and q=0 as an answer

Eq1:=61*q*L__1^2*C*e*eta/(16*omega__n^2)+5*q*L__1^2*C*e^3*eta^3/(8*omega__n^4)+3*C*p^3*gamma__1*(1/4)+3*q*C*p^2*R__n/(4*omega__n)+q*L__1^2*C*e^4*eta^4/(16*omega__n^5)+145*q*L__1^2*C/(64*omega__n)+3*q^3*C*R__n/(4*omega__n)+3*p*C*q^2*gamma__1*(1/4)+q*R*C/(4*omega__n)+19*q*L__1^2*C*e^2*eta^2/(8*omega__n^3):
Eq2:=-3*C*p^3*R__n/(4*omega__n)-3*p*C*q^2*R__n/(4*omega__n)-p*L__1^2*C*e^4*eta^4/(16*omega__n^5)-5*p*L__1^2*C*e^3*eta^3/(8*omega__n^4)-19*p*L__1^2*C*e^2*eta^2/(8*omega__n^3)-61*p*L__1^2*C*e*eta/(16*omega__n^2)-145*p*L__1^2*C/(64*omega__n)-p*R*C/(4*omega__n)+3*q*C*p^2*gamma__1*(1/4)+3*q^3*C*gamma__1*(1/4):
sys := { Eq1 , Eq2 };solve( sys, {p,q} );

Is there any way to help Maple to try other conditions, I know the only solution should not be just p=0 and q=0.

Thanks,

Baharm31

Hi everyone.I need to know what is "t" in this equation.

(1+0.25cost)×(cos7t)=0.1

I dont need numeric solution.I need step by step solution.

Thanks.

Hi,

I do not really understand the difference between annrow operator and unapply.
From the help pages it seems that unapply "creates" an arrow operator and thus that they could be two different ways to do the same thing.


restart:

f := x[1]+y[1]:

a := indets(f):                  # just because f can be more complex than the f above
g := (op(a)) -> f;              # generates an error, "operators not of a symbol type"
h := unapply(f, (op(a)))   # ok, but with a strange output
     h := (x__1, y__1) -> x__1+y__1

So it seems that Maple has transform by itself the indexed x[1] and y[1] into symbols x__1 and y__1.

Could you explain me what happened exactly ?

TIA

Assume we have an expression in several variables, x,y,z,..., where all of them are function of one parameter, t, for an example consider the following simple expression;

f := 2*y(t)*(diff(x(t), t))^2+3*(diff(x(t), t$3))-3*x(t)*(diff(y(t), t));

Is there any command or a way to ask Maple to give the highest order of derivation of x or y with respect to t in the expression? For example in the above example, the answer for x is 3 and for y is 1. If we remove the second term, then the answer for x should be 1.

Hello,

I'm confused about Maple's adjoint function (in the DEtools package). When I take the adjoint of the derivative operator:

I get back simply "Dx". However, doing the calculation by hand and integrating by parts seems to indicate that this should return the negative of Dx. The inner product I'm using is . Is Maple perhaps using a different inner product? Or is this a generalization that I'm unaware of? Or is it perhaps just a bug?

Thanks!

Hi,

I have an expression and want to find its derivative respect to T[1] but not T[2], for example: 

u[1](T[1], T[2], T[3]):=exp(T[1]*I)

If I find the 

diff(u[1](T[1], T[2], T[3]), T[1], T[2])

I would like to get something like this:

D__1*diff(u[1](T[1], T[2], T[3]), T[1])

instead of the derivative respect to T[2], which makes the result to be zero.

I need an expression that I can do other things with it.

You might say why do you just remove the diff of T[2], well I want to automate my code and this is part of the expression so I cannot write something else.

Regards,

Bahar

Hi,

I want to find the real and Imaginary part of this equation, without losing the exponential forms. 

(I*(1/2))*(I*R*C*C1*omega__n-L__1*C*exp(I*eta*T[2])*e^2*eta^2*C2-5*L__1*C*exp(I*eta*T[2])*e*eta*omega__n*C2+(3*I)*R__n*C*C1^2*C2*omega__n-3*gamma__1*C*C1^2*omega__n^2*C2-13*L__1*C*exp(I*eta*T[2])*omega__n^2*C2*(1/2))/omega__n

If I use 

evalc(Im(D1C1))

then the exp changes to sin or cos forms. How can I prevent it?

Regards,

Bahar

I have the sequence f(n) satify the conditions

f(n) = 4*f(n-1)-2*f(n-2), f(1) = 1, f(2) = 28.

I find the formula of f(n)

rsolve({f(1) = 1, f(2) = 28, f(n) = 4*f(n-1)-2*f(n-2)}, {f})

I got

f(n) = (-25*sqrt(2)*(1/4)-6)*(2-sqrt(2))^n+(25*sqrt(2)*(1/4)-6)*(2+sqrt(2))^n

Is there an integeral number k satisfy the equation f(n) = k^2, (k <> 0) ?

I have a 3D plot as follows:

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

with(plots);

plot3d(sin(x*y), x = -2 .. 2, y = -2 .. 2, axes = framed, colorscheme = ["zgradient", ["Blue", "Green", "Yellow", "Red"]]);

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

I want to plot its 2d contour (i.e. graph x-y plane).

How do I plot the contour of the graph in which the color of the contour lines have the corresponding colorscheme as the 3d plot?  Thank you. 

(P.S. This question has been updated)

i want to reverse this list using seq : :L := [2, 1, -7, 8];

when i try this way it works: seq(L[i], i = nops(L) .. 1, -1) but this way doesn't work :seq(L[nops(L)-i], i = 1 .. nops(L)), can anyone explain to me why ?

Hi everyone,

i want to to skip some code lines if a certain condition is/isn't given. I would prefer using a if..then expression. Is this possible.

Thanks for your help in advance.

I combined two animations as follow, but the dash line is turned into solid line. How can I solve this?

with(plots);

A := animate([X, X+t, X = -10 .. 10], t = -10 .. 10, color = red, linestyle = dash);

B := animate([X, t^2+X, X = -10 .. 10], t = -10 .. 10, color = blue, linestyle = solid);

display([A, B]);

I have two equations 

(i) x=t^2;

(ii) y=t^3;

I want to plot these two equations into a single graph [i.e. combine these two graphs: graph of x versus t (solid line, bule color), and graph of y versus t (dash line, red color)].

How do I customise the location of the legend? I want the legend located inside the blank space of graph (say, at top-right corner ) with vertical alignment. Thank you.

That is something simililar to the following graph: