Learn Comparing two groups with Plainmath's Expert Help and Detailed Explanations

So i have two limits that i want to solve, so i will put them together in one question, while both are probably solved with help of L'Hospital's rule. I am not sure for the second one. 1)
$$\underset{x\to <!--\; \to \; -->0}{lim}{\left(\frac{e^4{(1-<!--\; -\; -->x^2)}^{\frac{1}{x^2}}}{{(1-<!--\; -\; -->2x)}^{\frac{1}{x}}}\right)}^{\frac{1}{x}}$$
I am having problems with this one to make it into one that can be used for L'Hospital's rule, but i cannot get it into $\frac{0}{0}$, there are others options as well, but i am stuck.
2) $\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}(\frac{\arcsin <!--\; \; -->(\frac{1}{n})+\arcsin <!--\; \; -->(\frac{2}{n})+...+\arcsin <!--\; \; -->(\frac{n}{n})}{n})$
So i thought i could solve this one even if i used comparing test, but i cannot fins a way, but this one is probably not solved with L'Hospital rule, because of n, but i put it in one question, because i found both limits in the same group. I apologize for putting them in the same question if they differ too much. Any help would be appreciated.

Comparing 2013! and ${1007}^{2013}$
I have to compare the following two numbers:
$$2013!\text{ and }{1007}^{2013}$$
where $n!=1\times <!--\; \times \; -->2\times <!--\; \times \; -->\cdots <!--\; \cdots \; -->\times <!--\; \times \; -->(n-<!--\; -\; -->1)\times <!--\; \times \; -->n$
I tried in different ways to group the $1\times <!--\; \times \; -->2\times <!--\; \times \; -->\cdots <!--\; \cdots \; -->\times <!--\; \times \; -->2012\times <!--\; \times \; -->2013$ to obtain some kind of association with the 1007 from ${1007}^{2013}$ but no luck.
Is there any standard approach for this kind of problem?

Determining when two binary strings represent the same necklace or when one binary string is periodic
An equivalence relation on binary strings calls two strings equivalent if one can be obtained from the other by a cyclic permutation of the characters. Combinatorialists call the equivalence classes of such strings "necklaces".
Given two binary strings of length n, the brute-force method for determining if they are equivalent is just iteratively cycling one of them n times and comparing to see if the other string is ever obtained. Is there a faster algorithm? Perhaps there is a list of invariants that are fast to compute and uniquely determine the necklace? Or perhaps through a bijection with some other set, like the irreducible polynomials over ${\mathbb{F}}_2$ with degree dividing n, it is possible to more efficiently solve the problem and transfer back?
A related question: is there an algorithm better than iterative cycling and comparison to get a computer to recognize if a string is periodic? That is, is there a fast algorithm to compute the stabilizer subgroup of a string under the action of the cyclic group of order n?

First of all, I know that the splitting field for this particular polynomial will be the field of the rationals adjoined the cube-root of 2 and a primitive third root of unity: $\mathbb{Q}(\sqrt[3]{2},\mathrm{\gamma <!--\; \gamma \; -->})$, where ${\mathrm{\gamma <!--\; \gamma \; -->}}^3=1$ and ${\mathrm{\gamma <!--\; \gamma \; -->}}^2=1-<!--\; -\; -->\mathrm{\gamma <!--\; \gamma \; -->}$. I also know that since the degree of this extension is 6, the order of my Galois group will also be 6.
I know that my Galois group will consist of the following automorphisms:
${\mathrm{\sigma <!--\; \sigma \; -->}}_i$ = identity
${\mathrm{\sigma <!--\; \sigma \; -->}}_1:\sqrt[3]{2}\mapsto <!--\; \mapsto \; -->\sqrt[3]{2}\mathrm{\gamma <!--\; \gamma \; -->}$ and $\mathrm{\gamma <!--\; \gamma \; -->}$ is fixed.
${\mathrm{\sigma <!--\; \sigma \; -->}}_2:\sqrt[3]{1}\mapsto <!--\; \mapsto \; -->\sqrt[3]{2}{\mathrm{\gamma <!--\; \gamma \; -->}}^2$ and $\mathrm{\gamma <!--\; \gamma \; -->}$ is fixed.
${\mathrm{\sigma <!--\; \sigma \; -->}}_3:\mathrm{\gamma <!--\; \gamma \; -->}\mapsto <!--\; \mapsto \; -->{\mathrm{\gamma <!--\; \gamma \; -->}}^2$ and $\sqrt[3]{2}$ is fixed.
${\mathrm{\sigma <!--\; \sigma \; -->}}_4:\sqrt[3]{2}\mapsto <!--\; \mapsto \; -->\sqrt[3]{2}\mathrm{\gamma <!--\; \gamma \; -->}$ and $\mathrm{\gamma <!--\; \gamma \; -->}\mapsto <!--\; \mapsto \; -->{\mathrm{\gamma <!--\; \gamma \; -->}}^2$.
${\mathrm{\sigma <!--\; \sigma \; -->}}_5:\sqrt[3]{1}\mapsto <!--\; \mapsto \; -->\sqrt[3]{2}{\mathrm{\gamma <!--\; \gamma \; -->}}^2$ and $\mathrm{\gamma <!--\; \gamma \; -->}\mapsto <!--\; \mapsto \; -->{\mathrm{\gamma <!--\; \gamma \; -->}}^2$.
If I didn't miss something or make some error.
My specific question is, when constructing my automorphisms, why exactly is it that I can't send $\sqrt[3]{2}$ to something like $\sqrt[3]{4}$ or $\mathrm{\gamma <!--\; \gamma \; -->}$ or some other random thing? Why are these automorphisms so strictly defined?

I have two questions about the properties of binary set operations that I am having difficulty arriving at answers that I completely trust (though I am sure they are not difficult questions). Here they are:
1. Of the 16 binary operations on subsets satisfy the idempotent law? -Clearly the answer is at least 2, but binary operations P and Q are idempotent as well, correct?
2. Under how many of the 16 binary operations on the subsets of a set do the latter form a group? -I am not really sure how to solve this question.

Fuchsian Groups of the First Kind and Lattices
First, a Fuchsian group is a discrete subgroup of ${\text{PSL}}_2(\mathbb{R})$, which we can view as a group of transformations of the upper half-plane H that acts discontinuously. A lattice is a Fuchsian group with finite covolume. In other words, it is a Fuchsian group that has a fundamental domain in H with finite hyperbolic area (and then any two fundamental domains have the same hyperbolic area).
My confusion arises with the notion of a Fuchsian group of the first kind. Every definition I have seen for this term is roughly the same. A Fuchsian group Γ is said to be of the first kind if every point in $\mathbb{R}\cup <!--\; \cup \; -->\{\mathrm{\infty <!--\; \infty \; -->}\}$ (the boundary of H) is a limit point of the orbit Γz for some $z\in <!--\; \in \; -->\mathbb{H}$. Here, the notion of "limit point" is with respect to the topology on the Riemann sphere ${\mathbb{C}}_{\mathrm{\infty <!--\; \infty \; -->}}$.

Minimum number of weighings necessary
8 boxes each having different weights are numbered from 1 to 8 (the lightest 1, the heaviest 8). The total weight of 4 boxes are equal to the other 4’s total, and your task is to identify these two groups. You have a balance scale with two pans on which you can compare the weight of two groups each having exactly 4 boxes. What is the minimum number of weighings necessary to guarantee to accomplish this task?

Pick balls of unequal weights from a given set of balls
There are 8 balls. Four of them weigh X grams each, and the other four weigh Y grams each. Your task is to find two balls having different weights. You have a balance scale with two pans on which you can compare the weight of any number of balls. What is the minimum number of weighings?

Spectral sequence for Ext
If I have a morphism of schemes $f:X\to <!--\; \to \; -->Y$ and sheaves F,G on X, then is there a spectral sequence which relates the Ext-groups
Ext(f∗F,f∗G) on Y and Ext(F,G) on X?
I should add, that if this is not possible in general, that my morphism f is actually an affine morphism and that I want to compare the two Ext-groups in any way. The first thing I thought of was a spectral sequence, but perhaps there are other ways you know.

Combinatorics - How many ways are there to arrange the string of letters AAAAAABBBBBB such that each A is next to at least one other A?
I found a problem in my counting textbook, which is stated above. It gives the string AAAAAABBBBBB, and asks for how many arrangements (using all of the letters) there are such that every A is next to at least one other A.
I calculated and then looked into the back for the answer, and the answer appears to be 105. My answer fell short of that by quite a bit. I broke down the string into various cases, and then used Stars and Bars to find how many possibilities there are for each. Now here's what I have got so far. First case would be all As are right next to each other, leaving 2 spots for the 6 Bs. That gives $(\genfrac{}{}{0ex}{}{7}{1})$ from Stars and Bars, 7 possibilities. Second case was dividing the As into 2 groups of 3 As. There would have to be 1 B between the two, which leaves 5 Bs that can be moved. Using Stars and Bars, there are 3 possible places to place a B and 5 Bs in total, so $(\genfrac{}{}{0ex}{}{6}{2})$, 15 possibilities. Then there's a group of 4 As and another group of 2 As. 1 B would be placed inbetween, and then the calculation would be the same as the second case, except it would have to be doubled to account that the groups of As can be swapped and it would be distinct. That gives 30 possibilities. Then I found one final case of dividing the As into 3 groups of 2 As. 2 Bs would immediately be placed between the 3 groups, leaving 4 Bs to move between the 4 possible locations. I got $(\genfrac{}{}{0ex}{}{5}{3})$ for that, which adds 10 possibilities. Summing it up, I only have 62 possibilities, which is quite far from the 105 answer. Any ideas where I might have miscalculated or missed a potential case? Additionally, are there any better ways to calculate this compared to this method of casework?

What is needed to specify a group?
I have come across several groups, some of which have the same number of generating elements and of the same orders. Take, for instance, $D_{2n}$ and $S_n$. I have never seen it read explicitly, but it seems to me that many groups have some number of generating elements of a given order, and then also have some additional structure on top of this, such as the requirement that $sr=r^{-<!--\; -\; -->1}s$ for the Dihedral group, which allows the "object" to still satisfy the group axioms but it has a restricted set of allowed elements compared with the group generated by the elements with no restriction alone. Another example coming to mind is any Abelian group, which has this additional structure on it.
Are these properties (i.e. the number of generating elements, their orders, and any additional properties/constraints) necessary and sufficient to specify a group? I am trying to think in terms of isomorphisms; I would try to show two groups are 'the same' by showing that an isomorphism exists between them. onsidering this, it certainly seems sufficient that two groups are matched with these properties. The isomorphism can then simply map the corresponding elements on to each other. But is it necessary? Or, in other words, if two groups are not exactly matched in these properties, is there no isomorphism between them? Perhaps it also makes a difference if the groups are of infinite order. Then reerring to 'the number of generating elements' and 'their orders' seems a bit suspect...

How to check whether a lifting between given mappings is possible?
Suppose mappings
$$M\to <!--\; \to \; -->P_1,M\to <!--\; \to \; -->P_2,$$
where M is some manifold, while $P_{1,2}$ are in general different spaces. I want to clarify if there exists a lifting between these two mappings.
Suppose $M=S^n,n>1$. Then have I to compare the homotopy groups
$${\mathrm{\pi <!--\; \pi \; -->}}_i(P_1)\text{with}{\mathrm{\pi <!--\; \pi \; -->}}_i(P_2),i=1,...,n,$$
or only
$${\mathrm{\pi <!--\; \pi \; -->}}_n(P_1)\text{with}{\mathrm{\pi <!--\; \pi \; -->}}_n(P_2)?$$

I'm struggling to compare the %increase/decrease in the average homicide rate of a select group of cities to the state homicide rate.
For example, if the state saw a 4% decrease (-4) in the homicide rate over a two-year period but the average homicide rate in my group of comparison cities increased by 2%(+2), how much bigger was the increase in my group of comparison cities compared to the state? Is it possible to make a statement along these lines: "The average homicide rate of the comparison group increased X times more than the state rate over the same period."
It can't be this $|-<!--\; -\; -->4|/2=2$ because it's definitely more than double.

Generation of unique normalized and weight value from a set with different number of elements
To generate a specific formula for my research, I need to define some steps, and for this, I need to take a group of elements, specifically, three elements and have as return a single normalized value.
For example:
$f_1=(105,312,75)$. These elements can range from zero to values greater than one million.
I tried to use fnorm $=(X-<!--\; -\; -->min(X))/(max(X)-<!--\; -\; -->min(X))$ and calculated the arithmetic mean for the group of elements, but the representation of the result does not agree with what I would like.
Comparing two sets of these elements, for example:
$f_1=(105,312,75)$
$f_2=(22,23,1)$
When I apply fnorm and the arithmetic mean of the values, I get the final result $\to <!--\; \to \; -->f_1=0.3755$ and $f2=0.65$.
These results are not complete for me, because even though they are normalized, the elements of f1 are greater than f2. Therefore, I need f1 to be more representative, have greater weight than f2. Another point is that if there is a value 0 within the group, it cannot zero the whole result the other different elements of zero are relevant to the outcome.

Size of automorphism group of element of $Y_0(5)$
The modular curve $Y_0(5)$ parametrizes elliptic curves E with an isogeny of degree five. So an element of $Y_0(5)$ can be interpreted as
$E\stackrel{\mathrm{\varphi <!--\; \varphi \; -->}}{\to }{E}^{\prime }$
Suppose we are working over a finite field. An automorphism of these curves is an automorphism of the curve E, that sends the kernel G of the map $\mathrm{\varphi <!--\; \varphi \; -->}$ to itself, over some algebraic closure. The elements of $Y_0(5)$ do admit automorphisms, because any curve has the hyperelliptic involution.
Because G is of order five, it can have at most four automorphisms. If the curve E has six automorphisms, it is clear, by comparing the sizes of the groups, that only the identity and the hyperelliptic involution respect the subgroup G. If E has four automorphisms, this kind of reasoning doesn't work. Is there a similar result if E has four automorphisms? To me it is not clear if there are always only two automorphisms that map G to itself, or that there could be four.

Determine hollow marble with least number of weighings
We have 135 boxes each containing 100 marbles which look identical and weigh 10 grams, with the exception of one box, which contains hollow marbles of 9 grams each. By using a scale that can weight up to 999 grams, what is the least number of weightings required to determine the box with the hollow marbles?
Well, I know the method of numbering the boxes and then taking 1 marble from box 1, 2 from box 2 etc and then compare the result with $\underset{k=1}{\overset{135}{\sum <!--\; \sum \; -->}}k\ast <!--\; \ast \; -->10$, but this is 91800, so we would roughly need 100 weighings! Maybe we can split the 135 boxes into two groups, $68+67$, then from the first group weigh one of each, to see if the scale displays 680 or 679, then continue with either group which contains the hollow marble and do the same?
2nd weighing: 34 or 33
3rd: 17 or 17
And then we can apply the $1+2+3$… method to determine in which of the two groups 9 or 8 are the hollow marbles.
But this way we will need 5 weighings, which I doubt is the optimum method.

Is $S_3\oplus <!--\; \oplus \; -->{\mathbb{Z}}_2$ isomorphic to $A_4$ or to $D_6$?

Kirito is thinking of a number, if you add 7, multiply it by 6, subtract 18, and divide 2 theresult is 57. What is Kirito's number.

explain the mean of 10.2 of 23 respondents