It would be useful to know how exactly simplify acts when applied to an inert function, because this does not seem to be documented (only diff, expand and print are mentioned).

simplify(%sin(x+Pi));
                          %sin(x + Pi)
simplify(%sin(Pi));
                               0

%MeijerG() does not appear to be very inert!

@Mariusz Iwaniuk 

Unfortunately Maxima cannot expand tanh(x+1) at 0.

diff(f, $n) is very powerful in Maple, so convert/FPS can be easily made to work in these cases.

BTW, comparing the two expansions, the following identity appears:

which seems difficult to prove directly.

@asa12 

You should plot the series (t2) for a not too large interval; it will be much faster.

No, you did not. And a "thanks" is missing too.

Why don't you post the worksheet?

Download tanh.mw

@asa12 

For f(x)=tanh(x+1) , f(0)  <> 0  sou you must add f(0) (or let n >= 0, n = 0 .. infinity).

@asa12 

Is "a" computed? AFAIK diff(f, x$n) exists in Maple 12.

assuming n>=1 is not necessary but the result is simpler, and n=0 is not useful as tanh(0)=0.

@Rouben Rostamian  

My point was that in our case the problem can be formulated and solved easily without differential inclusions.
(I do not reject the subject, I know something about it, I work in Convex Analysis).
Note also that a similar question may be asked for differential inclusions whenever the solution is not unique.

The solution of the ode has exactly x(0)=A, x'(0)=0. We are looking for x() having the stationary points isolated, so x'(0)=0 can be replaced by limit(x'(t), t=0) = 0. It is not unusual for an ode defined in an open interval to have conditions at the endpoints.

@Rouben Rostamian  

But in this particular case, the classical ode has a C^1 solution which is C^2 except a finite set of points (in a bounded interval).
This solution obviously satisfies the differential inclusion.

@Preben Alsholm 

The formula

sol:=(A-n*C)*cos(omega*t)+(-1)^n*C/2;

is valid only for t < tmax = A*Pi*sqrt(k/m)/(2*mu*g)

because after that   A - nC  becomes <0.
That is why the graph obtained by dsolve is different from sol for t>tmax.

Another interesting thing. sol is only an approximation! A rather good one.
It is possible to compute the exact solution but this will be of the form

A(t)*cos(omega*t + alpha(t)) + S(t)  where A(t), alpha(t) , S(t) are locally (piecewise) constant.

I like this problem!   :-)

Edit. I was intrigued and I have computed by hand the solution. sol is correct (i.e. alpha(t)=0), it is dsolve which makes the difference (for t<tmax of course).

@Preben Alsholm 

The following graph (yours) includes (-1)^n in red. It is obvious now that the ode refers to signum(x') not signum(x'');

@tomleslie 

It depends on how the ode is understood. A natural assumption is that x''() exists and is nonzero except on a discrete set A.
In this case x() cannot be continuous.
Otherwise any piecewise linear function is a solution because x''() = 0 except on a discrete set, so we don't have any condition for x(). So, it can even oscillate:

@Preben Alsholm 

The ode is given but the problem is that x() cannot be continuous.
So, the desired shape of the solution is out of the question.

This direct consequence of Green's formula is used:

     ∫∫_D f(x,y) dx dy = ∫_C Q(x,y) dy,  where Q := ∫ f(x,y) dx

It remains to compute the curvilinear integral.
We must find a positively oriented parametrization (by hand!):

x=X(t), y=Y(t), t ∈ [a,b] of the (closed) curve C.

Then the curvilinear integral reduces to the usual Riemann integral

       ∫_a^b Q(X(t),Y(t))·Y^′(t) dt

The boring (sometimes difficult) part is to write the parametrization (usually piecewise).

@Kitonum 

I have used a compiled procedure.
BTW, I have corrected the comment which was wrong.
X(10^9)  needed several minutes but X(100000) only 65ms.