Ok, why not considering it as not to bad ...

I do not understand, why the password can not be altered, just as if some user would have requested that. But I have not checked whether a new and old password works in parallel.

BTW the new flood lets one guess, that it is from China, since the posted links resolve to this IP (which is Chinese) and there one finds www.ptidc.com where one can rent servers.

If that board would allow to sort/select according to first posting or last reply date (for 'recent contibution') it would help, beyond the current mess. But it still is not possible ...

As from http://www.mapleprimes.com/posts/103122-Handling-SPAM the SW seems to have a security hole: locked useres still can post.

For checking multiples: I doubt the SWallows - it obviously is not able to allow easy things like a sound sort ...

And even if so: operating time is so unexpensive in some countries, that it would not matter being recognized once.

BTW: though the server is hosted in US and is a dedicated one I doubt the spamer sits there ... it is just the provider (even tracing the subscription would not give much)

Without solving the security problem it is likely to be continued ...

Working with *.mws I have not recognized that. And usually I write a short note or docu in the header of a file. Done.

Doing it for a *.mw I can find it - filled it in and tried to read it from outside, using the explorer, within a directory. Did not work. But so what ...

At the job however it may be quite convenient to do such for office documents. At least it was. But meanwhile one can use 'speaking file names' instead of 8 characters. And any reasonable kind of process documentation requires more than the tiny fields intend.

But I do not consider that as something important - that's why I replied, of course :-)

evalf is to give 1 numerical value - for that you have to provide some concrete n

1st: Do not panic. 2nd: You are trapped by the Standard interface.

f:= x -> (sin(x-2)+x^3)*(1/(x+1)-2*x)/(sin(x)^2-exp(x));
# compare with your input here         ^^^^^^^^

Now define the derivative - since you have only 1 variable it is easy:

df := D(f);

You should check the help and handbook for that theme.

plot([f(x),df(x)],x=0..10);

For MS Word you may wish to look up 'Exporting', through the menu
you should be able to use RTF format, MS can read it.

This homework certainly expects, that you learn how to write that in Maple's way.

Start with it first.

I think it is 2 times the following, where I leave it to you to simplify to a desired form and for k=0 or k=1 one would take the limit.

1/beta^2*Pi+I*arctanh(tan(1/2*Pi*k)*(I-1/2*beta/Pi+beta*Q)/(-(I-1/2*beta/Pi-beta*Q)*(I-1/2*beta/Pi+beta*Q))^(1/2))/beta^2/k/(-(I-1/2*beta/Pi-beta*Q)*(I-1/2*beta/Pi+beta*Q))^(1/2)-I*arctan(tan(1/2*Pi*k)*(I+1/2*beta/Pi-beta*Q)/((I+1/2*beta/Pi+beta*Q)*(I+1/2*beta/Pi-beta*Q))^(1/2))/beta^2/k/((I+1/2*beta/Pi+beta*Q)*(I+1/2*beta/Pi-beta*Q))^(1/2)

                              Pi k  /    beta         \
                          tan(----) |I - ---- + beta Q|
                               2    \    2 Pi         /
          arctanh(---------------------------------------------) I
                  / /    beta         \ /    beta         \\1/2
                  |-|I - ---- - beta Q| |I - ---- + beta Q||
   Pi             \ \    2 Pi         / \    2 Pi         //
  ----- + --------------------------------------------------------
      2        2   / /    beta         \ /    beta         \\1/2
  beta     beta  k |-|I - ---- - beta Q| |I - ---- + beta Q||
                   \ \    2 Pi         / \    2 Pi         //

                             Pi k  /    beta         \
                         tan(----) |I + ---- - beta Q|
                              2    \    2 Pi         /
           arctan(--------------------------------------------) I
                  //    beta         \ /    beta         \\1/2
                  ||I + ---- + beta Q| |I + ---- - beta Q||
                  \\    2 Pi         / \    2 Pi         //
         - ------------------------------------------------------
                2   //    beta         \ /    beta         \\1/2
            beta  k ||I + ---- + beta Q| |I + ---- - beta Q||
                    \\    2 Pi         / \    2 Pi         //

average = sum all and then divide by number of summands, just do it.

Out of a mood:

with(ArrayTools):
A:=RandomArray(20, 10);
Size(A): k:=%[1]*%[2];
Reshape(A, 1.. k); 
AddAlongDimension(%)/k;

But you should look for your problem ... google ...

I guess, you have to tell something about Q as well.

Note that the integrand is symmetric w.r.t. phi, so 0 ... phi is enough

Your k means, that you integrate to Pi at most, kind of scaling, so it
only has an effect on boundaries.

May be Maple hangs inbetween to find discontinuities.

Anyway the result may be 'long' and not compact.

Here a shot by using

y=1/(2*Pi) + Q* cos(k*phi); isolate(%, phi); # now plug it in without boundaries:

Int((1/(2*Pi) + Q* cos(k*phi))^2/ (1+ beta^2*((1/(2*Pi)) + Q* cos(k*phi) )^2),phi);
changevar(phi = arccos(1/2*(2*y*Pi-1)/Q/Pi)/k, %, y):
value(%):
simplify(%, symbolic):
simplify(%, size);

You have the recursion

  GAMMA(n+1)/GAMMA(n): '%'= simplify(%);

                           GAMMA(n + 1)
                           ------------ = n
                             GAMMA(n)
  isolate(%, GAMMA(n+1));

                      GAMMA(n + 1) = GAMMA(n) n


Now define G:= n -> GAMMA(1/3 + n) and see what it does (no, it does
not satisfy your recursion, it looks different):

  'G(n+1)/G(n)': '%'= simplify(%);

                          G(n + 1)
                          -------- = 1/3 + n
                            G(n)
  isolate(%, G(n+1));

              GAMMA(4/3 + n) = (1/3 + n) GAMMA(1/3 + n)

I avoid to use 'alias' but it says: G(n+1) = (1/3+n)*G(n)

Now give it a go:

  rsolve( g(n+1) = (1/3+n)*g(n), g);

That will show you a solution, which needs g(0), so provide it:

  rsolve( {g(n+1) = (1/3+n)*g(n), g(0) = GAMMA(1/3)}, g);

                            GAMMA(1/3 + n)

Well, then you expect (under some additional assumptions) that
Maple is able to quickly find global extrema in case R^n->R^m,
m=1. I would say that is hopeless in general even for m=1.

A brute way - as suggested by hirnyk - is like 'visual inspection'
and if that is enough for you ...

For the partial question 'is it a supremum/infimum or an achieved
value' my guessing is, that you have to use properties of the domain
and the function itself to say something about is image, like 'open,
closed or proper mapping'.

But I think you are too optimistic.

BTW: for x^2 the open interval maps to the the semi-open interval,
but what do you get for z^2 applied to a circle, reading R^2 as C^1?

Have you tried to find maximum and minimum?

http://www.mapleprimes.com/questions/43542-Interpolation-Or-Approximation-With-Two-Variables

http://www.mapleprimes.com/questions/37579-Extrapolation-Of-3D-Data

?ArrayInterpolation

I would like to hear about a Math answer as well ... ggogled for 0,1 matrix but nothinh illuminating.

BTW: the following slight modification in Joe Riel's code is faster (but I lost patients for n=6)

#        eigs := LinearAlgebra:-Eigenvalues(evalf(A));
#        if andmap(`>`, eigs, 0) then
         if LinearAlgebra:-IsDefinite(A) then  # = positive definite

plot3d( 'dphidx'(x,y,10,5,1/10), x= 0..1, y=0..1, axes=boxed);

tastes like some coding error - ?