That LN:= ln@abs is an anti-derivative in the real (!) case

  diff(g(x),x); eval(%, g=Ln); 
  convert(%, piecewise); # that assumes x to be real

                     { undefined          x = 0
                     {
                     {    1/x           otherwise

but that is not a solution for complex variables, not even in
the right half plane (which would avoid branch cuts):

Define the real and imaginary parts of a function f: IR^2 ---> IR^2

  f(x+I*y);
  #eval(%, f=exp);     # a good example 
  #eval(%, f=abs);     # to see, that abs is not good
  #eval(%, f=ln@abs);  # for the case discussed here
  u:=unapply(evalc(Re(%)), x,y);
  v:=unapply(evalc(Im(%%)), x,y);

Then the Cauchy Riemann DE say: f is analytic iff the following holds:

  ['D[1](u)(x,y) = D[2](v)(x,y)', 'D[2](u)(x,y) = -D[1](v)(x,y)'];
  evalc(%);

Now activate the discussed case and you get

                          x            y
                      [------- = 0, ------- = 0]
                        2    2       2    2
                       x  + y       x  + y

On the other hand 'one knows', that 1/z has an anti-derivative in
the analytic functions and it is ln, so there no 'choice', if one
wants to cover that (I am unprecise and vague here).

Note that most of integration only holds in the complex case, if the
functions are complex analytic (and not only harmonic) while taking
the univariate case (what usually is understood by Maple).

For example M would tell, that D(LN)(z) has no meaning, if taken
in a non-real point z:

  diff(Ln(z),z); eval(%, z=x+I*y); evalc(%); 
  subs(x=0, y=1, %); 
  eval(%);
                              abs(1, z)
                              ---------
                                | z |

                              abs(1, I)

  Error, (in simpl/abs) abs is not differentiable at non-real arguments


But it is only a rephrasing, what is meant in the link given by acer:
ln@abs is false, if working beyond the Reals. Which Maple is designed for.

'limit( x/sqrt(1-cos(x)), x=0,right)': '%'=%;
                                   x            1/2
                     lim    ---------------- = 2
                   x -> 0+  sqrt(1 - cos(x))

'limit( x/sqrt(1-cos(x)), x=0,left)': '%'=%;

                                  x             1/2
                    lim    ---------------- = -2
                  x -> 0-  sqrt(1 - cos(x))

Also never hesitate to plot your task in case of doubts and troubles:

plot( x/sqrt(1-cos(x)), x=-0.1 .. 0.1);

I probably would use xj.

Else `xi ` (with a blank!).

For me it would be more convenient to write common factors in front,
so would write 1/4 * newMatrix (where newMatrix := 4 * yourMatrix ).

A matter of taste ... but I forgot, how to find the factor, which
would make all the coefficients to live in integers.


Others may prefer to have it with |det| = 1:

  M:=Matrix([[1/4, 1/2], [1/2, 1/4]]);
  v:=LinearAlgebra[Determinant](M);
  newM:= sqrt(1/abs(v))*M; # has |det| = 1

and then one way is

  sqrt(3)*newM;

Under 'some' not so terrible assumptions that means to solve
0 = y*xi^(t+1)+xi^t*(y-1)+(y-s)*xi-1+y-s for xi,
where y,s,t are given.

That neither can hope for a closed form - I would say - nor
does it have a reason for unique solutions (not even in the Reals).

For considering your network as data transformation (what ever
that should mean) it should be enough to see, that it is a function,
"ie assigning unique results to inputs", no?

Edited to give the reduction:

I have to upload that: that ugly new board again does not accept
'greather than' etc ... what a mess. Was it ever tested seriously?

reference: http://www.mapleprimes.com/questions/89281-Problem-With-Solution-Due-To-RootOf

Writing t = (b+i*d)/(a+i*c) and x = (a+i*c) the equation becomes
y= g/(1+exp(x)) + h/(1+exp(t*x)) and knowing x then gives the i.

Deviding by g, again naming the lhs y and 'pulling out t' (0 < x)
we get y= 1/(1+xi) + s/(1+xi^t), now xi = exp(x) and solve for xi.

y= 1/(1+xi) + s/(1+xi^t);
normal(%):
% * denom(rhs(%)):
0=lhs(%) - rhs(%):
simplify(%, symbolic); # 0 < xi, t real
f:=rhs(%);

t (1 + t) t
f := y + y xi + y xi + y xi - 1 - xi - s - s xi


u:=coeftayl(f,xi=0, 0);
v:=coeftayl(f,xi=0, 1) assuming 0 < t;

u := -1 + y - s
v := y - s

f-u-v*xi: simplify(%, size):
'f = % + v*xi + u'; is(%);

(1 + t) t
f = y xi + xi (y - 1) + v xi + u

true

So one wants a root xi of f.

Download reduction_for_expone.txt

I think "deformation" is the magic word you are searching for,

i e a path phi: [0, ... 1] ---> (some space) with phi(0) standing for x^2, phi(1) standing for x^3

(1-t)*x^2 + t*x^3 ?

But that is in no way unique and heavily depends on your "desire" (i e what is meaningful in your working context)

Here one can do it by the Abel-Plana formula as well
(though it is much slower):

Digits:=14;

Sum(ln(x)/((x)^(2)+3*x+2),x=1..infinity);
changevar( x=n+1,%,n): factor(%);

                       infinity
                        -----
                         \         ln(n + 1)
                          )     ---------------
                         /      (n + 3) (n + 2)
                        -----
                        n = 0

f:= n -> ln(n+1)/(n+3)/(n+2);

'Sum(f(n),n=0..infinity)'= 
   'int(f(x),x = 0 .. infinity) + 1/2*f(0)'+ 
   I*'Int( (f(I*x)-f(-I*x))/(exp(2*Pi*x) - 1), x=0..infinity)';
 %:
 ``=rhs(evalf(%));

  infinity           infinity
   -----            /
    \              |
     )     f(n) =  |          f(x) dx + 1/2 f(0)
    /              |
   -----          /
   n = 0            0

              infinity
             /
            |          f(x I) - f(-I x)
         +  |          ---------------- dx I
            |          exp(2 Pi x) - 1
           /
             0


                                                     -17
             = 0.60178245837480 - 0.20538903175872 10    I

'Obviously' the imaginary is just spurious and can be deleted.

> for i from 1 to 7 do
> if i=2 or i=6 then
>   next
> end if:
> print(i);
> end do:

                                  1

                                  3

                                  4

                                  5

                                  7

Can one define 1 as the identity and then use composition of functions, telling the system that one talks about additive functions (ok, linear, but there seem no scalars in the question) ?

You want to pernute the 3 rows, there are 3! = 6 of them and to that is taking one of them, i.e. a number in [1, ... , 6], and RandomTools:-Generate(integer(range=1...6)) does that

Besides the answers so far: what do you intend, what is your actual desire?

What you want is to expand the expression, not to simplify it.

Also you may try to post your code through an upload and ask for a review,
some times you may get really helpful suggestions for improvements.

(x^3)^(2/3): [%, simplify(%, symbolic)];
eval(%, x=I): evalf(%); 

                              3 (2/3)   2
                           [(x )     , x ]

                 [0.5000000000 - 0.8660254038 I, -1.]

I do not have the link handy (it is in Maple's Application Center I think), but this one
is very good and fast (it is almost as fast as if one would compile it, I once got some
code from him for that test).

The only thing is that those things meanwhile are only available as standard sheets
 

And - of course - I do not mind at all slight typos or grammatical errors (may be that
I would not even discover half of them; I am not aware of tools working in Maple and
typing outside and copying into it is a mess).