## Show that there is no computability reduction HP \$ le \$ \$ Sigma \$ *

Let's try more: yes $$L$$ it is a language and $$L not in mathsf {R}$$, so $$L not le_ mathrm {T} L & # 39;$$ For any $$L & # 39; in mathsf {R}$$. (Here, $$le_ mathrm {T}$$ indicates a reduction of Turing; this is synonymous with his notion of a reduction of "computability". In other words, $$mathsf {R}$$ It is closed under the Turing reductions.

Suppose, in the face of a contradiction, that such a reduction exists and is calculated by a Turing machine. $$M$$. Also, let's go $$M & # 39;$$ be a decision maker for $$L & # 39;$$. Then there is a Turing machine that solves $$L$$, that is, simulating directly $$M$$, answering all your oracle queries by simulating $$M & # 39;$$ about them, and, in the end, responding to what $$M$$ make. This contradicts $$L not in textsf {R}$$.

The case of the problem of detention is deduced from not being in $$mathsf {R}$$ (since he is not even in $$mathsf {RE}$$) and, of course, $$Sigma ^ ast in mathsf {R}$$.

A side note: if you find the idea of ​​closed classes interesting under the Turing reductions, I suggest you look for the theoretical notion of calculating the degree of Turing. $$mathsf {R}$$ it would be the "base case" (ie zero degree) in that context.

## It is \$ k (| a_1 | + | a_2 | + … + | a_n |) le | b_1 | + | b_2 | + … + | b_n | + k | S | \$ Right?

Is inequality as true?

Leave $$k> 0, a_i> 0$$ for $$1 le i le n$$ and let $$S: = a_1 + a_2 + …. + a_n$$ Suppose that $$b_i: = S-ka_i text {for} 1 le i le n.$$
So

$$k (| a_1 | + | a_2 | + … + | a_n |) le | b_1 | + | b_2 | + … + | b_n | + k | S |$$

Equality if only if $$S = 0$$

## probability: why is the event \$ {X_ {(j)} le x_i } \$ equivalent to the event \$ {Y_i ge j } \$?

From the statistical inference of Casella and Berger:

Leave $$X_1, points X_n$$ Be a random sample of a discrete distribution.
with $$f_X (x_i) = p_i$$, where $$x_1 lt x_2 lt dots$$ they are possible
values ​​of $$X$$ in ascending order. Leave $$X _ {(1)}, dots, X _ {(n)}$$
denotes the order statistics of the sample. Define $$Y_i$$ as the number of $$X_j$$ that are less than or equal to
$$x_i$$. Leave $$P_0 = 0, P_1 = p_1, dots, P_i = p_1 + p_2 + dots + p_i$$.

Yes $${X_j le x_i }$$ it's a "success" and $${X_j gt x_i }$$ it is a "failure", then $$Y_i$$ it is binomial with parameters $$(n, P_i)$$.

Then the event $${X _ {(j)} le x_i }$$ is equivalent to the event $${Y_i ge j }$$

Can anyone explain why these two are equivalent?

$${X_ {(j)} le x_i } = {s in text {sun} (X _ {(j)}): X _ {(j)} (s) le x_i }$$

$${Y_i ge j } = {s & # 39; in text {sun} (Y_i): Y_i (s & # 39;) ge j }$$

I have trouble understanding how these functions of random variables show this equivalence.

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## integration – Proof that \$ int_ { pi / 6} ^ { pi / 2} frac {x} { sin {x}} le frac { pi ^ 2} {6} \$

Proof that $$int _ { pi / 6} ^ { pi / 2} frac {x} { sin {x}} le frac { pi ^ 2} {6}$$

After some calculations I get it if I take $$frac {3} {2} x$$ then after the integral I get $$frac {3} {4} x ^ 2 + C$$ Y
$$int _ { pi / 6} ^ { pi / 2} frac {3} {4} x ^ 2 + C = pi ^ 2/6$$ so I should show that
$$frac {x} { sin {x}} le frac {3x} {2}$$

But the last inequality is not true …

## real analysis – For \$ g (x) = int_0 ^ infty frac {1} {x + y} f (y) , dy \$, Show \$ m {x in (0, infty): g (x)> lambda } le 1 / lambda cdot lVert f rVert_ {L ^ 1} \$

Q: for $$x in (0, infty)$$ leave:
begin {align *} g (x) & = int_0 ^ infty frac {1} {x + y} f (y) , dy \ end {align *}
Show that to $$f in L ^ 1 (0, infty)$$:
begin {align *} m {x in (0, infty): g (x)> lambda } & le frac { lVert f rVert_ {L ^ 1}} { lambda} \ end {align *}

My job:

begin {align *} E_ lambda & = {x in (0, infty): g (x)> lambda } \ end {align *}

By tchebychev

begin {align *} m (E_ lambda) & le frac {1} { lambda} int_0 ^ infty g (x) , dx \ & = frac {1} { lambda} int_0 ^ infty int_0 ^ infty frac {1} {x + y} f (y) , dy , dx \ end {align *}

I'm a little stuck here. Is there any integration technique to solve or simplify this integral?

## nt.number theory – About the determinants \$ det left[(ipm j)left(frac{ipm j}pright)right]_ {1 le i, j le (p-1) / 2} \$

Leave $$p$$ be an odd cousin and define
$$D_p ^ +: = det left[(i+j)left(frac{i+j}pright)right]_ {1 le, j le (p-1) / 2}$$
Y $$D_p ^ {-}: = det left[(i-j)left(frac{i-j}pright)right]_ {1 le i, j le (p-1) / 2},$$
where $$( frac { cdot} p)$$ It is the symbol of Legendre.

QUESTION. Is my next conjecture true (formulated in 2013)? How to solve it?

Guess. For any cousin $$p> 5$$ with $$p equiv1 pmod4$$, we have
$$left ( frac {D_p ^ +} p right) = 1 = left ( frac {D_p ^ -} p right).$$

Observation. (i) I have verified the conjecture of all the cousins $$5 with $$p equiv1 pmod4$$.

(ii) For any prime $$p equiv1 pmod4$$, clearly $$D_p ^ –$$ it is a determinant of symmetry of bias and, therefore, it is a whole square by a Cayley result. But I can not show that $$p nmid D_p ^ –$$ for all the cousins $$p> 5$$ with $$p equiv 1 pmod4$$.

In my opinion, the conjecture does not seem so difficult. Your comments towards your solution are welcome!

## Show that \$ Lambda (t) le- frac C2 Lambda & # 39; (t) \$ if and only if \$ e ^ {2t / C} Lambda (t) \$ does not increase

Leave $$Lambda in C ^ 1 ([0infty))[0infty))[0infty))[0infty))$$ Y $$C> 0$$. Why $$Lambda (t) le- frac C2 Lambda & # 39; (t) ; ; ; text {for all} t> 0$$ keep if and only if $$e ^ {2t / C} Lambda (t)$$ is not increasing in $$t$$?

Is this just an application of Grönwall's inequality?

## 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.