Wang Gaoteng, thx. May the very person to be contacted is in the link,
giving Name, email adress and phone number

Edited: deleted the explicit data here given before

Anyway: he does it systematically with that provider, another one
may be the original manufacturer have even more interest in that
than we can imagine

Yes, as I guessed they provide servers, thx.

I do not believe it is a competitor: Maple is not that strong (sorry)
and I have minor reasons to doubt MapleSim is a strongly used
component - it would be much too risky for a competitor (which in
this case is Matlab) to go that way:

S/he would become vulnerable for extorsion attempts (internally)
and no manager would do - not even unofficially ...

What I think is: it is just advertising for plagiaristic products (but
in a quite silly way, may that's why you think about a competitor).
And that those technicians are aware of the weakness of the most
common boards.

Edited: s/he did it at other sites as well, link or here

PS: no need to excuse for China, it is related to the economical
situation of those peoples. The usual attacks are through mails
(at work ~ 98% are killed by our admin's filters and meanwhile
he rejects any mail from CN servers ... but not only from there,
there are lists which can be checked and are updated regularly)

Wang Gaoteng, it is not intended to insult Chinese people

Go into one of the post, like http://www.mapleprimes.com/posts/103581-Hermes-Bags
pick the linked http://www.hermesbags-outlet.org/ and send a ping gives the above IP
and http://www.tiffanyand-cooutlet.com/ gives 95.154.210.219 as another one, which
lead to the Chinese site - however this is only the provider.

But all the pages are Chinese. And refer to www.ptidc.com - providing servers (as far
as I understand a crude translation of that site).

And through whois you can find out more, here

Edited:

www.tiffanyand-cooutlet.com is registered to a person in Shanghai, link
and www.hermesbags-outlet.org to the same person

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?