vv

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7 years, 139 days

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These are replies submitted by vv

You have no answers because the problem has not much sense. If you really need it, try to explain exactly what you want.

@Zeineb Yes, it was a mistake, the limit of (a_n)^(1/n) if obviously oo for this subsequence.
Actually Maple computes these (though not necessary):

a := n -> (n - sqrt(n*(n + 7))*signum(sin(n)) + 9)^n:
simplify((abs(a(n))^(1/n))) assuming sin(n)>0:
L1:=limit(%, n=infinity):
simplify((abs(a(n))^(1/n))) assuming sin(n)<0:
L2:=limit(%, n=infinity):
R:=min(1/L1,1/L2);

       R = 0

The two subsequences cannot be given explicitly. But we know that the sets

N_1 = {n in N: sin(n)>0},  N_2 = {n in N: sin(n)<0},  

are infinite, so the subsequences are ( |a_n|^(1/n) )_(n in N_1) ,  ( |a_n|^(1/n) )_(n in N_2).

The fact that N_1, N_2 are infinite follows from Kronecker's Approximation Theorem
because sin(n) = sin(n+2*k*PI) , Pi being irrational (k,n in N)

@stud_8013  You may assign the sequence to a variable using s:= %;  or  s:=seq(...);
then  s[k]   returns the k-th polynomial in the sequence.

@stud_8013 You have changed the conditions. Then:

restart;
A:={seq(0..2)}:  A0:=A minus {0}:
a:=Iterator:-CartesianProduct(A0,A$3):
seq(add(v[i]*x^(i-1)+x^4,i=1..4), v=a);

 

@ecterrab Mathematics has also fundamental problems which wait to be solved, for example the Riemann  hypothesis. Unfortunately, in Maple, the Zeta function cannot be computed efficiently with high precision for large arguments. Maybe in the future, such problems will be supported by Maple too.

@Axel Vogt  It was just a guess, after converting the general term into factorials (no GAMMA with noninteger arguments).

@dharr Actually, a tail recursion is essentially a loop, which we can write directly.

It is interesting that the compiler accepts option remember. This is not documented!
The main restriction is that a compiled procedure cannot call a user procedure (nested or not, compiled or not).
So the compiler can be used only for very small pieces of code.

Edit. Unfortunalely, the option remember is accepted but ignored by the compiler!

@Carl Love If C_n denotes the cyclic group of order n, then for C_2 x C_2 (direct product, isomorphic with Klein's group),   the product of elements equals 1.

I don't understand your "the" unique nonidentity (...) ; a group may have many elements of order 2; e.g. in (C_2)^n, all the elements except "1" have order 2.

@Zeineb For any nonabelian group (of even or odd order), there is an ordering of the elements for which the product is <> 1; but you need a procedure to find this ordering.
For an abelian group, the product is 1 if the order is odd, but if the order is even, the product (of the elements) may be <>1. 

The problem is not correctly posed. The order of the terms does matter, so, which order dou you want?
For your group there are orders for which the product is 1.

@Carl Love My objection is that the omitted userinfo's in the second call could contain essential information about the algorithm (which I suppose the OP wants to know).

@nm I don't know. You could redefine userinfo, but that seems useless becuse you problably don't know the source code details, in order to filter the output. After all, the facility is supposed to be used interactively.

@Ronan You have a more subtle mistake, corrected but not mentioned by the responders, such that just replacing 0 with NULL is not enough. Try to find and explain it!

This reminds me that for a polynomial with random coefficients, its roots tend to lie near the unit circle.

p:=randpoly(x, degree=200, dense):
plots:-complexplot([fsolve(p, complex)], style=point, scaling=constrained)

Maybe LaTeX should accept empty matrices (even if they do not make much sense).

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