Instead of directly pointing you to your error you may find it out by yourself:

Something seems wrong with your f and how Maple understands it,
so look at some of the steps where it is used:

restart; # well ...
N:=4;
a:=-2; b:=2;
f:= cosh(x); # that is what you enter

i:=1; # ... and so on, the body of your proc
h:=(b-a)/N;

Now there is some point where you want f(a+(2*i-1)*h);

What do you get and how you  could comment that?

Instead of directly pointing you to your error you may find it out by yourself:

Something seems wrong with your f and how Maple understands it,
so look at some of the steps where it is used:

restart; # well ...
N:=4;
a:=-2; b:=2;
f:= cosh(x); # that is what you enter

i:=1; # ... and so on, the body of your proc
h:=(b-a)/N;

Now there is some point where you want f(a+(2*i-1)*h);

What do you get and how you  could comment that?

Using 'paper & pencil' here may be better than all the Maple:

Recall that abs(a)*abs(b) = abs(a*b). Now multiplying by I = sqrt(-1) gives
sqrt(5) = abs(1+z*I)*abs(I) = abs( I - z ) which you may remember as the
equation for a circle in the complex plane.

Using 'paper & pencil' here may be better than all the Maple:

Recall that abs(a)*abs(b) = abs(a*b). Now multiplying by I = sqrt(-1) gives
sqrt(5) = abs(1+z*I)*abs(I) = abs( I - z ) which you may remember as the
equation for a circle in the complex plane.

First: thank you for the attempt to provide it for 'usual interfaces'.

I tested it with M 12 Classic interface on Win XP SP 2 using fname := "D:/temp/minttemp.txt";

My test proc is

p:=proc(x)
option hfloat;
y:=exp(t);
#I;
end proc; maplemint(%);

As stated maplemint would warn "These parameters were never used explicitly: x"
and would moan about the variables y and t.

Activating the last statement it would refuse to work (which I find annoying).

Using mint(p) only writes the proc to the file, nothing else.

I also would prefer to read the results of mint in the worksheet, say displaying them,
it would be a bit more comfortable.

PS (for the admin): this thread was not displayed if I enter the board through
http://www.mapleprimes.com/tracker until a first reply was made ... but that
is a bit off the topic and hard to reproduce ...

searching for grrr or grrr* does not give those threads with grrrr (4 times an 'r'), which is not what I would expect from a Google search

thx for reminding the additional parameter in 'factor'

In the example it works, however not for x^2 + 5, except the ugly
factor(x^2+5,complex); identify(%);

Is the some shorthand? I mean a 'handy' one?

thx for reminding the additional parameter in 'factor'

In the example it works, however not for x^2 + 5, except the ugly
factor(x^2+5,complex); identify(%);

Is the some shorthand? I mean a 'handy' one?

My thought was to use your condition solve(diff) ... and now being at home
(and having a beer) it is like this:

diff(prof, w) = 0 means D(f)(e(w))*diff(e(w),w)-L = 0 and your 'solve' is
evaluating that in w=-b/(lambda-1), which is 

                    b        /d      \|
       D(f)(e(- ----------)) |-- e(w)||                 - L = 0
                lambda - 1   \dw     /|          b
                                      |w = - ----------
                                      |      lambda - 1

Abbreviating t = e(w) = ((-b/(lambda-1)-b)/b)^lambda now that writes as 

D(f)(t)*t *const -L, const = -1/b*(lambda-1)*(-lambda/(lambda-1))^(-lambda).

Using dsolve(%, f(t)) gives a solution. Now evaluate the const.

Do not know whether it is sound or gives what you expect. Just a shot ...

My thought was to use your condition solve(diff) ... and now being at home
(and having a beer) it is like this:

diff(prof, w) = 0 means D(f)(e(w))*diff(e(w),w)-L = 0 and your 'solve' is
evaluating that in w=-b/(lambda-1), which is 

                    b        /d      \|
       D(f)(e(- ----------)) |-- e(w)||                 - L = 0
                lambda - 1   \dw     /|          b
                                      |w = - ----------
                                      |      lambda - 1

Abbreviating t = e(w) = ((-b/(lambda-1)-b)/b)^lambda now that writes as 

D(f)(t)*t *const -L, const = -1/b*(lambda-1)*(-lambda/(lambda-1))^(-lambda).

Using dsolve(%, f(t)) gives a solution. Now evaluate the const.

Do not know whether it is sound or gives what you expect. Just a shot ...

The 2nd problem has no maximum, only for fixed g (and that seems to be h = L/2 (or close)),
for h = L/2 it explodes for g ---> 0

The 2nd problem has no maximum, only for fixed g (and that seems to be h = L/2 (or close)),
for h = L/2 it explodes for g ---> 0

upload your problem, use the almost unvisible green arrow which you find at the end of the box, in which you write your comments

it may make things clearer and you can write down what you really want

it is a bit like stumbling in the dark to guess your needs :-)

upload your problem, use the almost unvisible green arrow which you find at the end of the box, in which you write your comments

it may make things clearer and you can write down what you really want

it is a bit like stumbling in the dark to guess your needs :-)

Let us assume, that Maple know the (numerical) answer for all your inputs, which are 3 variables.
Something like x + y + z evaluated in specific points.

What do you want to see as picture for that case?