## Solution verification of \$ lim_ {n to infty} left ( frac {n ^ 3} {1 cdot 3 + 3 cdot5 + cdots + (2n-1) (2n + 1)} right) \$

I resolved this limit and got the solution $$frac {3} {4}$$. I tried to check WolframAlpha, but when generating the presentation the expression shows $$lim {n to n}$$ instead of $$lim {n to infty}$$ and tells me that the limit diverges. So I am not sure if it diverges due to misunderstanding of the problem or if the limit is really divergent.

$$lim_ {n to infty} left ( frac {n ^ 3} {1 cdot 3 + 3 cdot5 + cdots + (2n-1) (2n + 1)} right)$$

I applied the Stolz-Cesaro theorem and eventually (after the initial steps) I obtained

$$lim_ {n to infty} frac {(n + 1) ^ 3-n ^ 3} {1 cdot 3 + 3 cdot5 + cdots + (2n-1) (2n + 1) – (1 cdot 3 + 3 cdot5 + cdots + (2n-1) (2n + 1) + (2n + 1) (2n + 3))} = lim_ {n to infty} frac {3n ^ 2 + 3n +1} {(2n + 1) (2n + 3)} = frac {3} {4}$$

Is the result correct?

P.S.
Should I eliminate these types of questions if the answer is a simple yes, since they do not provide much information and may not be very useful for anyone but me?

## For which positive integers \$ n \$ does \$ x ^ n – 1 \$ factor like \$ (1 + x) (1 + x + cdots + x ^ {n-1}) \$ over \$ mathbb {F} _2[x]\$

For positive integers $$n$$ make $$x ^ n – 1$$ factor as $$(1 + x) (1 + x + cdots + x ^ n-1)$$ finished $$mathbb {F} _2 (x)$$

I think the first terms are 1,3,5,11,13, …

## co.combinatorics – Important combinatorial and algebraic interpretations of the coefficients in the polynomial \$[n]! _q = (1 + q) (1 + q + q ^ 2) ldots (1 + q + cdots + q ^ n-1) \$

What are some important combinatorial and algebraic interpretations of the coefficients in the polynomial?

$$(n)! _ q = (1 + q) (1 + q + q ^ 2) ldots (1 + q + cdots + q ^ n-1})?$$

As motivation, I will give three interpretations, ask for a fourth one and ask a related question about unimodality. I would be particularly interested in the answers using RSK correspondence or subspaces of $$mathbb {F} _q ^ n$$.

1. Given a permutation $$sigma in mathrm {Sym} _n$$, leave $$mathrm {inv} ( sigma)$$ denote the number of investments from $$sigma$$; that is, peers $$(x, y)$$ with $$x Y $$sigma (x)> sigma (y)$$. Then $$(n)! _ q = sum _ { sigma in mathrm {Sym} _n} q ^ { mathrm {inv} ( sigma)}$$.

2. An element $$x in {1, ldots, n-1 }$$ is a offspring from $$sigma in mathrm {Sym} _n$$ Yes $$sigma (x)> sigma (x + 1)$$. the major index $$mathrm {maj} ( sigma)$$ is the sum of the declines of $$sigma$$. Then $$(n)! _ q = sum _ { sigma in mathrm {Sym} _n} q ^ { mathrm {maj} ( sigma)}$$. I think this is due to MacMahon.

3. In the version & # 39; from the inside out & # 39; of Fisher-Yates shuffles in a $$n$$-bar of cards, in step $$j-1$$card $$j-1$$ from the top it is exchanged with one of the cards in positions $$0, 1, ldots, j-1$$ from above, chosen uniformly at random. These options are listed by $$1 + q + cdots + q ^ {j-1}$$. After $$n$$ steps (starting with $$j = 1$$), each permutation has the same probability. (This is essentially a coset enumeration in the symmetric group by the string $$mathrm {Sym} _1 le mathrm {Sym} _2 le ldots le mathrm {Sym} _n$$.) Thus $$(n) _q!$$ list the permutations according to the sum of the positions chosen at each stage.

Does the normal Fisher-Yates mix have a similar combinatorial interpretation? Is there a more natural interpretation of the $$q$$-power, are you still using the Fisher-Yates shuffle?

Finally, (1) makes it easy to see that $$(n) _q!$$ it is symmetric, that is, the coefficients of $$q ^ m$$ Y $$q ^ n-m$$ they are the same: use the Coxeter involution, thinking about $$(n) _q!$$ as the Poincaré series of the Coxeter group $$mathrm {Sym} _n$$. This can also be seen similarly from (2). But it doesn't seem obvious from (3).

What interpretation is the best way to show that the coefficients in $$(n)! _ q$$ Are they unimodal, that is, they increase first and then decrease?

## coding theory – \$ (1,1, cdots, 1) en C Leftrightarrow X-1 nmid g (X) \$, cyclic code \$ C \$

$$C$$ is a $$q$$– conical cyclic $$(n, k)$$-code where $$(n, q) = 1$$, with polynomial generator $$g (X)$$. Show that $$(1,1, cdots, 1) en C Leftrightarrow X-1 nmid g (X)$$.

I don't know if I'm right:
$$x ^ n -1 = (x – 1) (1 + x + x2 cdots + x ^ n-1)$$.
As $$g (x)$$ it is a generator polynomial so it divides $$x ^ n −1$$ and he has not $$x – 1$$ as a factor, you must divide $$1 + x + x 2 cdots + x ^ n-1$$. In other words, the length n word consisting of all $$1$$& # 39; s is a code word

## Show that \$ {a_1b_1, a_2b_2 cdots, a_ {100} b_ {100} } \$ reduces the waste system mod 101

Dice $${a_1, …, a_ {100} }, {b_1, …, b_ {100} }$$ they are reduced waste system modules $$101$$. Test it $${a_1b_1, a_2b_2 cdots, a_ {100} b_ {100} }$$ the waste system module is reduced $$101$$

I have no idea of ​​this problem. Help me..

## A limt with sum \$ S_n = 1 + frac {n-1} {n + 2} + frac {n-1} {n + 2} cdot frac {n-2} {n + 3} + cdots + frac {n-1} {n + 2} cdot frac {n-2} {n + 3} cdots frac {1} {2n} \$

$$S_n = 1 + frac {n-1} {n + 2} + frac {n-1} {n + 2} cdot frac {n-2} {n + 3} + cdots + frac {n-1} {n + 2} cdot frac {n-2} {n + 3} cdots frac {1} {2n}$$ So $$S_n / sqrt {n}$$ tends to $$frac { pi} {2}$$.

How to show then? the general term in $$S_n$$ is $$frac {n-1} {n + 2} frac {n-2} {n + 3} cdots frac {nk} {n + k + 1} = frac {C_ {n + 1} ^ {k + 2}} {C_ {n + k + 1} ^ {k + 2}}$$ where $$C_n ^ k = frac {n!} {K! (N-k)!}$$. So, how to do?

## nt.number theory – How many \$ mathbb {Q} \$ – bases of \$\$ can be constructed from the vector set \$ log (1), cdots, log (n) \$?

How many $$mathbb {Q}$$-base  It can be constructed from the set of vectors. $$log (1), cdots, log (n)$$?

Data for $$n = 1,2,3 cdots$$ computed with Sage:

$$1, 1, 1, 2, 2, 5, 5, 7, 11, 25, 25, 38, 38, 84, 150, 178, 178, 235, 235$$

Context:
The real numbers $$log (p_1), cdots, log (p_r)$$ where $$p_i$$ Is it known that the i-th prime are linearly independent over the rational ones? $$mathbb {Q}$$.

Example:

``````[{}]    -> For n = 1, we have 1 = a (1) bases; We count {} as the basis for V_0 = {0}
[{2}] -> For n = 2, we have 1 = a base (2), which is {2};
[{2, 3}] -> for n = 3, we have 1 = a base (3), which is {2,3};
[{2, 3}, {3, 4}] -> for n = 4 we have 2 = a (4) bases, which are {2,3}, {3,4}
[{2, 3, 5}, {3, 4, 5}] -> a (5) = 2;
[{2, 3, 5}, {2, 5, 6}, {3, 4, 5}, {3, 5, 6}, {4, 5, 6}] -> a (6) = 5;
[{2, 3, 5, 7}, {2, 5, 6, 7}, {3, 4, 5, 7}, {3, 5, 6, 7}, {4, 5, 6, 7}] -> a (7) = 5.
``````

## Actual analysis: is there \$ alpha> 0, beta in (0,1) \$ such that \$ dfrac { sum_ {k = 1} ^ n a_k} {n} le alpha (a_1 cdots) a_n) ^ {1 / n} + beta max_i (a_i) \$ has?

Leave $$a_1 ge a_2 ge cdots ge a_n ge 0$$ You will be given non-negative numbers. My question is this:

There are some $$beta in (0,1), alpha> 0$$, such that $$dfrac {a_1 + cdots + a_n} {n} le alpha (a_1 cdots a_n) ^ {1 / n} + beta a_1?$$

This article derives the inequalities from above, with $$alpha = 1$$ Y $$beta = 1 / n$$, but with $$sum_ {1 le i instead of $$a_1$$, multiplied with $$beta$$. Also, your method does not seem applicable to address my question. Is there any known result about this or is it an open question? Any ideas regarding how to proceed? Thanks in advance.

## functional analysis – \$ conv (A) = { sum_ {i = 1} ^ n lambda _i x_i: sum_ {i = 1} ^ n lambda_i = 1, lambda_i ge 0, x_i in A, i = 1, cdots, n } \$

Leave $$X$$ be a normed vector space and $$A$$ be a subset of $$X$$. $$conv (A)$$
It is called the intersection of all convex subsets of $$X$$ that contain
$$A$$

a) Show that conv (A) is a convex set

b) show that

$$conv (A) = { sum_ {i = 1} ^ n lambda _i x_i: sum_ {i = 1} ^ n lambda_i = 1, lambda_i ge 0, x_i in A, i = 1, cdots, n }$$

c) yes $$A$$ it is compact then it is $$conv (A)$$ compact?

d) Show that if $$A subseteq mathbb {R} ^ n$$ it is compact then $$conv (A)$$ is
compact

a) Take two elements $$a$$ Y $$b$$ at the intersection of all convex subsets of $$X$$ that contain $$A$$. Now take $$lambda a + (1- lambda) b$$. It is contained in each of the subsets of $$X$$ that contain $$A$$, therefore it is contained in the intersection. Q.E.D.

second)

$$leftarrow$$:

Suppose by induction that $$sum_ {i = 1} ^ n lambda_i x_i$$ It belongs to conv (A). We must prove that $$sum_ {i = 1} ^ {k + 1} lambda_i x_i$$, with $$sum_ {i = 1} ^ {k + 1} = 1$$ It also belongs.

$$sum_ {i = 1} ^ {k + 1} lambda_i x_i = sum_ {i = 1} ^ {k} lambda_i x_i + lambda_ {k + 1} x_ {k + 1} = sum_ {i = 1} ^ {k} lambda_i x_i + (1- sum_ {i = 1} ^ n lambda_i) x_ {k + 1}$$

Now choose $$delta$$ such that $$delta sum_ {i = 1} ^ {k} lambda_i = 1$$, so

$$frac { delta} { delta} left ( sum_ {i = 1} ^ {k} lambda_i x_i + (1- sum_ {i = 1} ^ n lambda_i) x_ {k + 1 } right = = left ( frac {1} { delta} sum_ {i = 1} ^ {k} ( delta lambda_i) x_i + frac {1} { delta} ( delta-1) x_ {k + 1} right) = \ frac {1} { delta} x + left (1- frac {1} { delta} right) x_ {k + 1}$$

which is a collection of elements of $$A$$ that adds up to $$1$$

$$rightarrow$$:

I can only say that $$x in conv (A)$$, so $$x = 1x + 0 cdot all$$ Therefore, it is a combination that adds to $$1$$ of elements of $$A$$?

by do), give me a clue: show that conv (A union B) is the image through a continuous function of the compact $${( alpha, beta; alpha, beta ge 0, alpha + beta = 1) } times A times B$$. I do not understand how to show this.

re) Some clue?

## logic – Should \$ x \$ and \$ y \$ be different elements in a declaration of the form \$ for all x for all and cdots \$?

For any formula $$varphi$$ in which $$x$$ Y $$y$$ they are free variables, $$for all x > for all and > varphi$$ quantifies on all possible values ​​for $$x$$ Y $$y$$. As a result, $$x$$ Y $$y$$ may be different objects or them may be equal; for the declaration to maintain, $$varphi$$ must keep in both of them cases.

Following this reasoning, a valid interpretation for its formula would be "an Italian is happy no matter who wins the World Cup" (assuming "WC" means "World Cup"), with "Italian" for $$x$$ and "who" for $$y$$. A) Yes, $$x$$ is happy if they win the World Cup ($$x = y$$) but also happy if someone else does ($$x neq and$$); therefore "independently".

The other statement can be expressed in the predicate logic as follows:
$$forall x ( text {italian} (x) to ( text {winWC} (x) to text {happy} (x)))$$

In fact, this is evaluated with the same truth value as the first formula precisely when you select the same values ​​for $$x$$ Y $$y$$.

Assuming the most reasonable domain (ie people in the world) and interpretation in this context, I could argue that there is an Italian who is happy only if the Italian team wins the World Cup and, otherwise, is not happy . Choose this person as your value to $$x$$ and an arbitrary non-Italian person (for example, German) to $$y$$; then the second formula is evaluated as true under this variable assignment (since $$text {italiano} (x)$$ it's true and $$x$$ he is happy if $$text {winWC} (x)$$, that is, the Italian national team wins), but the first one evaluates to false (since $$x$$ it is unhappy if $$y$$team, that is, the German national team wins).