## plot3d for a complex function ...

I want to plot the argument for a complex function. The input (x,y) represented in polar coordinates (r,phi) by default puts the cut at -I*Pi. Likewise the argument function:

argument(f(x)) plots the range -Pi..Pi.

However the function f(x)=x^2 could typically be plotted with 2 riemann surfaces on top of each other. When phi becomes 2Pi f(x) becomes 4Pi and only then I want to identify the 0 with 4Pi again since the points are equivalent in the preimage.

On the other hand the function f(x)=sqrt(x) never surpasses its own domain. The values always stay within the argument range of (0,2Pi) (in fact it only goes till Pi, or -Pi/2..Pi/2 in maple) when the preimage is taken to be (0,2Pi). Thus when plotting a preimage value of (x,y) with argument phi and 2Pi+phi they will have the same value since phi=2Pi+phi and I see a step in the plot. This step is actually there since the function has a cut at this point.

This step in the plotting image is also shown for f(x)=x^2 (e.g. at phi=+-Pi/2) but it is not of importance since it just comes from the argument function being constrained to -Pi..Pi.

So is it possible to change this behaviour?

## How to achieve its numerical solution？...

I  encountered a non-integrable integral in the process of solving the following process, . How to achieve its numerical solution? Such as in a looping   code：

#######
pa[i] := pa[i-1]-(Int(subs(t = tau, Lpa[i-1]+Na1[i-1]-Na2[i-1]), tau = 0 .. t));

pw[i] := pw[i-1]-(Int(subs(t = tau, Lpw[i-1]+Nw1[i-1]-Nw2[i-1]), tau = 0 .. t)); u[i] := u[i-1]-(Int(subs(t = tau, Lu[i-1]+Nu1[i-1]+Nu2[i-1]), tau = 0 .. t))；

######
Detailed code see annexBC2.mw

## Problem with shooting method...

Dear all,

I would like to ask you to help me with the following problem that I got error.

restart; Shootlib := "C:/Shoot9"; libname := Shootlib, libname; with(Shoot);
with(plots);
N1 := 1; N3 := .5; N2 := 5; Bt := 6; AA := N1*Bt; gamma1 := .2; blt1 := 1;
FNS := {f(eta), fp(eta), fpp(eta), g(eta), gp(eta), h(eta), hp(eta), i(eta), ip(eta), fppp(eta)};
ODE := {diff(f(eta), eta) = fp(eta), diff(fp(eta), eta) = fpp(eta), diff(fpp(eta), eta) = fppp(eta), diff(g(eta), eta) = gp(eta), diff(gp(eta), eta) = N1*(2*g(eta)+eta*gp(eta)+2*g(eta)*fp(eta)-2*f(eta)*gp(eta)+2*N2*N3*(h(eta)*ip(eta)-i(eta)*hp(eta))), diff(h(eta), eta) = hp(eta), diff(hp(eta), eta) = AA*(h(eta)+eta*hp(eta)-2*f(eta)*hp(eta)+2*h(eta)*fp(eta)), diff(i(eta), eta) = ip(eta), diff(ip(eta), eta) = AA*(2*i(eta)+eta*ip(eta)-2*f(eta)*ip(eta)+2*N2*h(eta)*gp(eta)/N3), diff(fppp(eta), eta) = N1*(3*fpp(eta)+(eta-2*f(eta))*fppp(eta)-(2*N2*N2)*(diff(hp(eta), eta)))};
IC := {f(0) = 0, fp(0) = gamma1*fpp(0), g(0) = 1+gamma1*gp(0), gp(0) = beta, h(0) = 0, hp(0) = beta1, i(0) = 0, ip(0) = beta2, fppp(0) = alpha};
BC1 := {f(blt1) = .5, fp(blt1) = gamma1*fpp(blt1), g(blt1) = gamma1*gp(blt1), h(blt1) = 1, i(blt1) = 1};
infolevel[shoot] := 1;
S := shoot(ODE, IC, BC1, FNS, [alpha = .1, beta = .2, beta1 = .3, beta2 = .4], maxfun = 50000);

## implicit animate plot...

Is it possible to animate plot like this parametically:
animate(plot, [[cos(t), sin(t), t = 0 .. A]], A = 0 .. 2*Pi, scaling = constrained, frames = 50)

however given the x and y components as solutions of an implicit equation.

I know I could run RootOf. But it seems that there is a part missing due to a branch cut :-/

## plotting PDE solution with derivative as axis...

I had problems plotting the solutions of my PDE with the derivative as the axis. Anyone knows what went wrong?

p1 := pds:-plot(diff(f(x,y),y), y = 0 .. 10, x = 0, numpoints = 100)

I seem to be getting a plot of y=0. Any help would be greatly appreciated:))

fyp2.mw

## Problems with Solving PDE with solution of ODE as ...

I have a system of pde as follow,

PDE := [(x*y+1)*(diff(f(x, y), y, y, y))+(x+(3/4)*f(x, y))*(diff(f(x, y), y, y))-(1/2)*(diff(f(x, y), y))^2+T(x, y) = (1/4)*x*(diff(f(x, y), y))*(diff(diff(f(x, y), y), x))-(1/4)*x*(diff(f(x, y), x))*(diff(f(x, y), y, y)), (x*y+1)*(diff(T(x, y), y, y))/(.733)+(x/(.733)+(3/4)*f(x, y))*(diff(T(x, y), y)) = (1/4)*x*(diff(f(x, y), y))*(diff(T(x, y), x))-(1/4)*x*(diff(f(x, y), x))*(diff(T(x, y), y))]

sys_ode := diff(g(y), y, y, y)+(3/4)*g(y)*(diff(g(y), y, y))-(1/2)*(diff(g(y), y))^2+h(y) = 0, (diff(h(y), y, y))/(.733)+(3/4)*g(y)*(diff(h(y), y)) = 0

ics := g(0) = 0, h(0) = 1, (D(g))(10) = 0, g(10) = 0, h(10) = 0

sol2 := dsolve([sys_ode, ics], numeric)

BC := {T(0, y) = h(y), T(x, 0) = 1, T(x, 10) = 0, f(0, y) = g(y), f(x, 0) = 0, f(x, 10) = 0, (D[2](f))(x, 0) = 0}

pds := pdsolve(PDE, BC, numeric)

module() ... end module

pds:-plot(T, y = 0 .. 10, x = 0);

Error, (in pdsolve/numeric/plot) unable to compute solution for x<HFloat(0.0):
solution becomes undefined, problem may be ill posed or method may be ill suited to solution

When I try to use the solution of the Ode as the boundary condition for PDE, by subbing g(y) and h(y) into BC. The plot returns me the error. Anyone knows the reason behind this and how to solve? Any help would be really greatly appreciated. Thanks

## Error, (in solve) cannot solve for an unknown func...

hi.

i have problem with this eq.

## Coefficients extraction to form Equations...

N := 4;
print(`output redirected...`); # input placeholder
4
y := sum(A[2*n].cos(2.*n.x), n = 0 .. N);

eq1 := diff(y, `\$`(x, 2))+(a+2*q*cos(2*x))*y

eq2 := map(combine, eq1, trig)

for i from 0 to 4 do eq4[i] := coeff(eq2, cos(2*n*x)) end do

From these I want to extract the co-ffficients of cos(0x),cos(2x),cos(4x)..

and form a simultaneous linear equation containg A0,A2,A4

The solution is

aA0+qA2=0

2q*A0+(a-4)*A2+q*A4=0

Can anybody tell me how to do it

## cant get result from solve, pls help...

Am here again, pls help me check out this adm code, cant get a result.

below is the attached file

## Getting rid of a frame...

Consider the following code snippet:

`with(DifferentialGeometry):DGsetup([x],M);RemoveFrame(M);`

By itself it is, of course, pretty meaningless, but that is not the point. My problem is that after Maple has executed it, it will annoyingly continue using M> for each new execution group that is inserted. Having removed the frame with the command RemoveFrame, I would have expected that not to be the case. How can the frame be truly removed?

## Maple gets stuck...

The keyboard shortcut sequence for executing an entire worksheet is Alt+e+e+w. Occasionally, I hit Alt+e+w+w by accidence. In that case, Maple sometimes, not always!?, gets completely stuck and can only be shut down through the job list window (Windows Ctrl+Alt+Del). Can anybody else replicate that? And if so, has this erratically odd behaviour been fixed in later versions of Maple?

## collect a product...

S:=...+a^4*b*c*x^2+...

So I know it is possible to collect for example any letter collect(S,x)

But I want to collect in particular for an entire product in this case

e.g. collect(S,a^4*b*c)

This obviously doesnt work but is it possible to get it working somehow?

## Trouble by exporting 3dplot to PS...

There are 4 points which I want to plot as solid spheres in 3d. Here is the code:

Rplot := pointplot3d(`<,>`(r, Transpose(Vector([0, 0, 0]))), style = point, symbolsize = 40, color = [red, grey, grey, green])

plotsetup(ps, plotoutput = `E:\\.../C.eps`, plotoptions = `width=2000, height=2000, noborder`); print(plots:-display(Rplot, axes = boxed, labels = [x, y, z], symbol = solidsphere));

plotsetup(default, plotoptions = `width=2000, height=2000, noborder`); display(Rplot, axes = boxed, labels = [x, y, z], symbol = solidsphere)

In Maple itself everything looks fine, but once I try to export it to ".ps" I see this.

The problem lies in "symbol = solidsphere". When I change it to "diamond", for example, export is successful:

But I need solidsphere. What should I do to fix this error (except manual editing in Adobe Illustrator)?

## Expansion of expression in terms of 'small' functi...

This may be a silly question, but does there exist some simple way of (Taylor) expanding an expression of 'small' functions in terms of these functions.

A simple example: Assume that diff(f(x),x) and g(x) are two functions both with range, say, in [-a,+a], where a << 1, and consider the following expression:

`sqrt(1 + diff(f(x),x)) * (2 + g(x));`

Its expansion to first order in terms of diff(f(x),x) and g(x) should be 2 + diff(f(x),x) + g(x). My problem is that mtaylor does not accept functions as variables to expand on, and I would prefer not to have to substitute back and forth with some 'placeholders'.

## Airy's stress function (convert the results into s...

Can we calculate the following equations in Maple?

Substituting equations (21) and (22) into (17), and then obtain equation (23). How to do that? I have done this, but the results are complex and large. They are not in a sum form, but in an expansion form. The reference and the maple file are attached.