## general topology – In a locally compact \$ 2 \$ Hausdorff space locally, there is a sequence of compact subsets \$ K_n \$ with \$ K_ {n-1} overs overset∘ {K_n} \$ and \$ bigcup_nK_n = E \$

Leave $$(E, tau)$$ Being a second space of Hausdorff locally compact. I want to show that there is a $$(K_n) _ {n in mathbb N_0} subseteq E$$ such that $$K_n$$ It is compact and $$K_ {n-1} subseteq overset { circ} {K_n}$$ for all $$n en mathbb N$$ Y $$bigcup_ {n in mathbb N_0} K_n = E$$.

The question is basically answered in mathoverflow, but there is a part of the test that I do not understand.

As $$E$$ he is the second accountant, $$tau$$ has an accounting base $$mathcal B$$. It's easy to see that $$mathcal B_c: = left {U in mathcal B: overline U text {is compact} right }$$ is again a basis for $$tau$$.

Now, the construction described in the answer is as follows:
Leave $$U_0 en mathcal B_c$$ Y $$K_0: = overline {U_0}$$. Dice $$(U_0, K_0), ldots, (U_ {n-1}, K_ {n-1})$$, leave $$U_n in mathcal B_c setminus left {U_0, ldots, U_ {n-1} right }$$, $$mathcal C$$ be a finite sublayer of $$K_ {n-1}$$ Y $$K_n: = overline {U_n} cup overline {C}$$.

The problematic part is the cover. $$mathcal C$$ of $$K_ {n-1}$$ (why subcover?). Power $$mathcal C$$ Actually be arbitrary. And if we choose? $$mathcal C = E$$?

## Dynamic programming: calculation of all factors products \$ n-1 when factors \$ n \$ are given

Let's assume we have an operator
$$times: E ^ 2 a E$$
Of which we only know that it is associative. Let's say a multiplication $$e times f$$ It always takes a while $$M$$ for all $$e, f in E$$.

Now we are given $$n$$ elements $$e_1, …, e_n in E$$, and have the task of calculating all $$n$$ products
$$def bigtimes { mathop { vcenter { huge times}}} p_j: = bigtimes_ {i = 1 \ i neq j} ^ n e_i$$

Naively multiplying everything $$n$$ products take $$O (n ^ 2M)$$.

A more sophisticated approach that runs on $$O (n ^ {3/2} M)$$ divisions $$bigtimes_ {i = 1} ^ n e_i$$ in arbitrary positions in $$sqrt n$$ Factors
For each of the $$n$$ products, now we only have to recalculate one factor and multiply the resulting factor, and the other $$k-1$$ factors together.

What is the fastest algorithm for this problem and how does it look?

## What is the asymptotic limit of \$ 1n + 2 (n-1) + 3 (n-2) + … + (n-1) 2 + n \$?

From Ryan's comment, the sum, when you distribute $$i$$ and divide it into more sums comes to
$$sum_ {i = 1} ^ ni (n- (i-1)) = sum_ {i = 1} ^ n (i * n) + sum_ {i = 1} ^ n (- (i ^ 2) -i)$$ which is approximately $$frac {1} {2} n ^ 3 – frac {1} {3} n ^ 3$$ so the asymptotic link is, in fact, $$O (n ^ 3)$$

## Why is \$ frac {X ^ p ^ N-1} {X ^ p ^ {N-1} -1 \$ about \$ mathbb {Q} _p \$ irreducible?

Can anyone tell me why this polynomial is irreducible? Thank you.

## calculation – How do I calculate \$ lim limits_ {x to 8} ( sqrtx-2 / (x-8)) = 1 \$ using \$ x ^ na ^ n = (xa) (x ^ {n-1} + x ^ {n-2} a + x ^ {n-3} a ^ 2 + … + xa ^ {n-2} + a ^ {n-1}) \$?

They ask me to calculate the $$lim limits_ {x to 8} ( sqrtx-2 / (x-8)) = 1$$ using the factoring formula

$$x ^ na ^ n = (xa) (x ^ {n-1} + x ^ {n-2} a + x ^ {n-3} a ^ 2 + … + xa ^ {n-2} + a ^ {n-1})$$

I have rewritten the limit as

$$lim limits_ {x a 8} (x ^ {1/3} -8 ^ {1/3} / (x-8)) = 1$$

I know that $$x-8$$ will be canceled, but I do not know how to connect the values ​​of $$x$$ Y $$a$$ in the formula. I do not know where to stop connecting values, the fractional exponent confuses me.

## Algorithms – Solving T (n) = 3T (n-1) + 2

I am trying to improve the resolution of recurring relationships, so I am creating my own simple relationships and try to solve them. I have made the following recurrence.

``````T (n) = 3T (n-1) + 2 and we say that T (1) = 1

T (n) = 3 [3T(n-2) + 2] + 2
T (n) = 3 [3[3T(n-3) + 2] + 2]+ 2
T (n) = 3 [3[3[3T(n-4) + 2] + 2]+ 2]+ 2 = T (n) = 81 T (n-4) + 8
.
.
.
We say that k = n-1.
T (n) = 3 ^ k T (n-k) + 2 * k
T (n) = 3 ^ k T (1) + 2 * k
``````

I'm stuck on how to finish this and how to find out what is the complexity of time in Big O.

## turing machines: complexity of space and time of \$ L = {a ^ nb ^ {n ^ 2} mid n≥1 } \$

Consider the following language:
$$L = {a ^ nb ^ {n ^ 2} mid n≥1 } ,$$

When it comes to determining the time and space complexity of a multi-tape TM, we can use two memory tapes, the first to count $$n$$, and the second to repeat. $$n$$ times the count of $$n$$. Therefore, due to the way we are using the second tape, you should have a $$Theta (n ^ 2)$$ The complexity of space, and would say the same with respect to time one. I thought it was correct, but the solution is $$TM (x) = | x | + n + 2$$, where, $$x$$ is, supposedly, the length of the chain, therefore $$Theta (| x |)$$. Sounds right, then, is my reasoning completely wrong, or just a different way of expressing it?

Could we have reasoned differently, and say, for example, for each $$a$$ we write a symbol on the first tape, and then we count the $$b$$By scanning the symbols from one side to the other $$n$$ times? This time, the complexity of space should be $$Theta (n)$$, while the complexity of time should remain unchanged. What would change if we had a single TM tape?

## Probability: anti-concentration: upper limit for \$ P ( sup_ {a in mathbb S_ {n-1}} sum_ {i = 1} ^ na_i ^ 2Z_i ^ 2 ge epsilon) \$

Leave $$mathbb S_ {n-1}$$ be the unitary sphere in $$mathbb R ^ n$$ Y $$z_1, ldots, z_n$$ be a sample i.i.d of $$mathcal N (0, 1)$$.

Dice $$epsilon> 0$$ (It can be assumed that it is very small), which is reasonable upper limit for the tail probability $$P ( sup_ {a in mathbb S_ {n-1}} sum_ {i = 1} ^ na_i ^ 2z_i ^ 2 ge epsilon)$$ ?

• Using ideas from this other answer (MO link), you can set the not uniform Limit of anti-concentration: $$P ( sum_ {i = 1} ^ na_i ^ 2z_i ^ 2 le epsilon) le sqrt {e epsilon}$$ for all $$a in mathbb S_ {n-1}$$.

• The uniform analog is another story. Can coverage numbers be used?

## Additional explanation of the steps of an equation that proves that \$ sum ^ {n} _ {k = 0} k cdot binom {n} {k} = n cdot2 ^ {n-1} \$

So, this is one of the questions in my textbook, which seems to be quite common: $$sum ^ {n} _ {k = 0} k cdot binom {n} {k} = n cdot2 ^ {n-1}$$

The same book provides the following solution:
$$sum_ {k = 0} ^ {n} k cdot binom {n} {k} = sum ^ {n} _ {k = 0} n cdot binom {n-1} {k- 1} = n sum ^ {n-1} _ {k = 0} binom {n-1} {k} = n cdot2 ^ {n-1}$$

What is not clear to me is (1) how do I get from $$sum ^ {n} _ {k-0} n cdot binom {n-1} {k-1}$$ to$$n sum ^ {n-1} _ {k = 0} binom {n-1} {k}$$ and (2) of $$n sum ^ {n-1} _ {k = 0} binom {n-1} {k}$$ to $$n cdot2 ^ {n-1}$$ respectively.