## 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).