@Kitonum 

Yes, it would be interesting to look at a beginner's face seeing:

 [seq(Array(`..`~(1, numelems~([L]))[], ()-> `?[]`~([L], `[]`~([args]))))]

Of course I understand that Carl's intention was to simply use the procedure, not to understand it.
(And the procedure is really nice as a substitute for CartesianProduct!)

@Carl Love 

I think that the default limit for real arguments is standard and generally accepted.

Note that for limit(f(x),x=c, right), where c is complex, Maple computes limit(f(c+t), t=0, right).

@minhthien2016 

The real option in limit(f(x), x=a, real)  makes sense only when a=infinity and means that limit(f(x), x=infinity) and limit(f(x), x=-infinity) both exist and are equal.

For example,
limit(sqrt(1-x^2)/abs(x), x=infinity, real);  # the limit is complex!
        I
 

@Rouben Rostamian  

Nice problem.
The exact answers are:
Q1:

P:= n -> add( binomial(n-5*k,k)/6^(6*k)*(-1)^(k-1), k=1..floor(n/6))

Q2:
N = 32341

(P(32340)<1/2, P(32341)>1/2)

@Rouben Rostamian  

It's only a (~good) approximation for small n.

E.g. P(12) = 7/6^6 - 1/6^12.

Probably the answer for Q2 is almost correct.

@Joe Riel 

It would be interesting to see the Maple answer in the form
Y := n -> [1,-1,0][n mod 3 +1]

@JAMET 

I don't see any problem.

Download FC.mw

@JAMET 

u runs in the set {0, 1/n, ..., a/b, ..., 1}  constructed just above it.

Please explain for this simpler expression.

f:=diff(v(x, y), x, y) * diff(u(x, y), y, y)^2;

What are you expecting for int(f, x=0..1, y=0..1); ?

@Carl Love 

Are you saying that the global variables are saved in a .m file without being explicitely present in the arguments of save?

I have not seen this behavior. Probably you mean something else.

@acer 

Thank you, good to know. Probably I should have guessed it. Actually, if B is not definite, some eigenvalues could be undefined (even if symmetric and commuting A,B implies real eigenvalues -- but their number could be < n).

@Carl Love 

You are right, I have assumed (without checking) that A,B commute (A.B=B.A).

Edit. Even when A,B commute and have hfloats, the eigenvalues still have imaginary 0s.
So, the algorithm in this case ignores the symmetry.

@Carl Love 

Yes, M is symmetric, but it seems that this is ignored by Eigenvalues, because 0 imaginary parts still appear. Probably the symmetry is taken into account only for hfloats.

@tomleslie

Download JK.mw

@tomleslie 

The OP does not want to reassign anything; he wants some computations with U.