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These are answers submitted by vv

You must export in a directory with write permission. By default the current directory is Maple's directory (without such permission).
So, you must change the directory, e.g. use "c:/temp/Test.jpg", (if you have such a directory),
or, execute first

Your second equation Y, equivalent to Y = 0,  reduces to a = 0, and does not depend on m. So, you have nothing to eliminate.

For higher-order functionals, use variables to represent derivatives. E.g.
L1:=eval(L, [diff(x(t),t,t)=diff(x1(t),t), diff(theta(t),t,t)=diff(theta1(t),t)]):
L2:=L1 + A*(diff(x(t),t) - x1(t))^2 + B*(diff(theta(t),t) - theta1(t))^2:
EulerLagrange(L2, t, [x(t), theta(t), x1(t), theta1(t)]);



z is on the tangent thru the origin to the circle abs(z-6*I) = 3.

So,  arg(z) equals


a := arccos(3/6);



z = 3*tan(a) * exp(I*a); evalc(%);

z = 3*3^(1/2)*(1/2+((1/2)*I)*3^(1/2))


z = (3/2)*3^(1/2)+(9/2)*I



Such an assignment is not possible due to the standard evaluation in any CAS: psi(t) would be evalaluated to psi(t)(t)(t)... (i.e. a nonsense).
It is possible to use macro (for input only)  or alias (for both input and output). You must be aware that in both cases psi(a) will be evaluated to psi(t)(a).






(diff(psi, t))*t+psi










(diff(psi(t), t))*t+psi(t)







# x,y,z = the unknowns, top to bottom
# h1,h2,h3  = halves of the heights 
# a,b,c,d,e = the bases of the triangles
[260=a*h2, 247=b*h2, 287=d*h3, 209=e*h3,
x=a*h1, y=c*h2, z=(b+c)*h3,
h1/(h1+h2)=a/(b+c), h1/(h1+h2+h3)=a/(d+e)]:
xy:=select( u -> (eval(x,u)::posint and eval(y,u)::posint), XY1):
for v in xy do
  zz:={solve( eval(XY[2],v) )};
  for z1 in zz do  if z1::posint then print(v[],'z'=z1, eval(elim,[v[],'z'=z1])) fi od;

So, two solutions found.
I don't see any relation with the star (100). Maybe I miss something.

It often works.

r := x*(diff(theta(t), t))^2+y*(diff(varphi(t), t))^2+z*(diff(theta(t), t$2))+w*diff(varphi(t), t$2):
g := (4*(f+T))*(diff(theta(t), t))^2+u*(diff(varphi(t), t))^2+(f+9)*(diff(theta(t), t$2))+4*s*diff(varphi(t), t$2):

eval(r-g, [theta=(t->t^3), varphi=(t->t^7)]): # or similar
solve([coeffs(%, t)], [x,y,z,w]);
simplify(eval(r-g, %[1]));   #check
#          [[x = 4*f+4*T, y = u, z = f+9, w = 4*s]]
#                               0



A brute-force search shows that the problem has no solution with at a least one tuple (xi,yi,zi) from the set {-N,...,N}, N = 40.
It shoud not be too hard to prove this in general (mathematically).

It's only a display decision when interface(typesetting=extended) .
For interface(typesetting=standard)  ==>  a0.5

`@W`  is a name (symbol)  enclosed in back quotes. It could be used as a generic procedure, or maybe a procedure `@W` is defined somewhere. Such a name can be used in 2022 too!

See here the correct formula and its proof (it's very simple):

Proof of the Euler product formula for the Riemann zeta function - Wikipedia

It cannot be proved in Maple for obvious reasons.

It is easy to formulate the problem in Maple:

f:=x^(1/(x + 1/x)) + y^(1/(y + 1/y)) - exp(1);
F:=piecewise(f>0, 1, 0);
int(F, x=0..infinity, y=0..infinity); # No chance!

Even the approximation of the integral is not trivial, but it works.

evalf(Int(F, [x=2..7, y=2..7], method = _CubaCuhre, epsilon=1e-6));
#                        13.7592321343801

For a symbolic result we may try to exploit the symmetry, or to find a parametrization of the curve, but this is maths... 

eval([Pi/2<=argument(2*I*z), argument(2*I*z)<=Pi, 0<abs(4*I*z), abs(4*I*z)<=1], z=x+I*y); 
plots:-inequal(evalc(%), x = -1 .. 1, y = -1 .. 1, optionsopen = [linestyle = dash, color = red, thickness = 2]);

Remark.  Here argument(2*I*z)<=Pi  is superfluous. Note however that some authors/books use argument(z) in [0,2*Pi)  instead of (-Pi,Pi]. In such situations we have to adapt the maple code!

This is actually a visual representation of a Riemann surface for Ln (= multi-valued log)

col:=[green, cyan, red, maroon, grey]:
plots:-display( seq( plot3d([ r*cos(u), r*sin(u), u],
u=-Pi+2*k*Pi .. Pi+2*k*Pi, r=0..1,  labels=[Rez,Imz, 'Im(Ln(z))'], grid=[120,30], color=col[k+3], lightmodel=none), k=-2..2));


# I have considered mu a function of xi

`diff/Y` := (u, xi) -> mu(u)*(1-Y(u)^2) * diff(u,xi);

proc (u, xi) options operator, arrow; mu(u)*(1-Y(u)^2)*(diff(u, xi)) end proc


### That's all. Examples. ###

diff(u(Y(xi)), xi);



diff(u(Y(xi)), xi$2);

((D@@2)(u))(Y(xi))*mu(xi)^2*(1-Y(xi)^2)^2+(D(u))(Y(xi))*(diff(mu(xi), xi))*(1-Y(xi)^2)-2*(D(u))(Y(xi))*mu(xi)^2*Y(xi)*(1-Y(xi)^2)


diff(u(Y(xi)), xi$3);

((D@@3)(u))(Y(xi))*mu(xi)^3*(1-Y(xi)^2)^3+3*((D@@2)(u))(Y(xi))*mu(xi)*(1-Y(xi)^2)^2*(diff(mu(xi), xi))-6*((D@@2)(u))(Y(xi))*mu(xi)^3*(1-Y(xi)^2)^2*Y(xi)+(D(u))(Y(xi))*(diff(diff(mu(xi), xi), xi))*(1-Y(xi)^2)-6*(D(u))(Y(xi))*(diff(mu(xi), xi))*Y(xi)*mu(xi)*(1-Y(xi)^2)-2*(D(u))(Y(xi))*mu(xi)^3*(1-Y(xi)^2)^2+4*(D(u))(Y(xi))*mu(xi)^3*Y(xi)^2*(1-Y(xi)^2)


diff(u(Y(xi^7)), xi$2);






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