@Lali_miani 

You already got the explanation: sqrt(2*x+1)/(2*x+1) is automatically simplified to 1/sqrt(2*x+1). The automatic simplifications cannot be prevented, e.g. you cannot see 2/10 instead of 1/5. There are some tricks, but I think that it is much more important to focus on maths!

@Carl Love 

This is probably the most efficient method.
But I think that for a beginner, a more useful construction (to be often used) is

MyMatrix := Matrix( 2,3, (i,j) -> elem[i,j](t) );

@dharr 

So, I'd formulate:

When x is a sparse Matrix, Vector or rtable, for efficiency reasons, only the nonzero entries are scanned and the zero entries are then appended.

@Kitonum 

evalc does not work well if Zeta is present.

@Al86 

It is easy to compute by hand if you know elementary Calculus. E.g. the limit of the numerator is:
exp(-t) - exp(-infinity)  = exp(-t).

@Al86 You can try (e.g. for t=2)

Student[Calculus1][LimitTutor]();

but AFAIK it works for (very) simple limits.

@digerdiga 

In the future it would be nice to mention the method(s) you know about for your problem.
Otherwise, the answers seem to be useless because you knew them.

@Carl Love 

Nice and professional (as always :-)).

@digerdiga 

asympt cannot switch branches. You can obtain the same result without puiseux.

restart;
sol:=RootOf((8*n-8)*_Z^6+(n^4+36*n^2-68*n+56)*_Z^5+(n^5+10*n^4+80*n^3-200*n^2+224*n-152)*_Z^4+(n^6+28*n^5+69*n^4-268*n^3+468*n^2-356*n+200)*_Z^3+(3*n^7+32*n^6+7*n^5-204*n^4+380*n^3-544*n^2+272*n-128)*_Z^2+(3*n^8+14*n^7-20*n^6-32*n^5+252*n^4-240*n^3+304*n^2-80*n+32)*_Z-n^9-12*n^8-44*n^7-40*n^6-4*n^5-128*n^4+48*n^3-64*n^2):
f:=eval(op(sol), [_Z=z,n=1/x]):
g:=numer(eval(f, z=a/x+b+c*x)):
S:=solve({seq(coeff(g,x,k),k=0..2)}):
subs(S[],x=1/n, a/x+b+c*x);

@Kitonum 

I know. SymFun just calls solve/identity which could (and should) be improved a lot (you are free to send a SCR for it, and if successful, SymFun will work for your examples). Actually, if we are interested only in an odd/even result, this is easy to obtain.
BTW, the "hack" part was included just to answer (c). SymFun can be of course improved, but I prefer a capable solve/identity.

@Carl Love 

To find the initial point of a p-cycle simply do:

f:=y-> 4*y*(1-y):
p:=2;
solve((f@@p)(x)=x); # or fsolve

@tomleslie 

No need for DirectSearch. The min of expr^2 is of course 0 (attained for all variables 0).

@mamehrashi 

Write your problem in the text area, not in the title.
Maybe a smaller example will help.
If   P = 2*x*y + 3*x*z + 4*y*t + 5*t^2   is the polynomial in the variables x,y,z,t,
what result do you expect?

You just posted a polynomial in 16 variables, homogeneous, total degree 4,  with float coefficients.
What do you want to do with it?

@acer 

You have used the parametric version

plot3d([1,u,v],u=0..2*Pi-2*v,v=0..Pi, coords=spherical)

instead of the explicite

plot3d(1, u=0..2*Pi-2*v,v=0..Pi, coords=spherical)

It seems that you knew that the latter does not work correctly. Is it documented, or just a bug?