## $\forall a > 0$ $\sum_{n=1}^{\infty} f(na)$ is convergent. Prove that $\int_{0}^{\infty}f(x) dx$ is convergent.

Hi can you help me solve this exercise? Thanks. Let $$f: [0;+\infty) \to \mathbb{R}$$ be nonnegative and continuous function. Suppose $$\forall a > 0$$ $$\sum_{n=1}^{\infty} f(na)$$ is convergent. Prove that $$\int_{0}^{\infty}f(x) dx$$ is convergent. I tried to solve it by using the Riemann sum, but for fixed a it doesn’t work. I have no other ideas.

## If $\space$ $\forall$ $x \in \Bbb R$, $\space$ $f(f(x))=x^2-x+1$. Find the value of $f(0)$.

If $$\space$$ $$\forall$$ $$x \in \Bbb R$$, $$\space$$ $$f(f(x))=x^2-x+1$$. Find the value of $$f(0)$$.

I thought that making $$f(x)=0$$ implies that $$f(0)= 0^2 – 0 + 1 = 1$$, but i think that this isn’t correct, because the $$x$$ in $$f(f(x))$$ isn’t equal to $$f(x)$$ .

Any hints?

## If $\abs(z_{0}) = 1, z_{0} \in \mathbb{C}$ prove that then $\forall z\in \mathbb{C} , z \neq z_{0}$

If $$\abs(z_{0}) = 1, z_{0} \in \mathbb{C}$$ prove that then $$\forall z\in \mathbb{C} , z \neq z_{0}$$

$$\abs \frac{z-z_{0}}{1- \bar{z}z_{0}} = 1$$.

I tried by multiplying given fraction with $$\frac{\bar{z_{0}}{\bar{z_{0}}$$ but I got nowhere.

## Does “$\forall x\in L, \sigma(\neg x)=\neg \sigma(x)$” hold given that $\sigma(F)\equiv F$ for a CNF formula $F$ built on a set $L$ of literals?

Suppose we have a CNF formula $$F$$ built on the set of literals $$L=\{x_1,\neg x_1,\cdots,x_n,\neg x_n\}$$ where each variable is used in at least one clause of $$F$$. Consider a permutation $$\sigma$$ of $$L$$ such that $$\sigma(F)$$ is logically equivalent to $$F$$ i.e. $$\sigma(F)\equiv F$$.

Does it hold that $$\forall x\in L, \sigma(\neg x)=\neg \sigma(x)$$ ?

I tried to find a counter example without success.

## Prove that $\forall \ A_k\subset [3n]$, with $|A_k|=2n$, $\exists\ a_i,\ a_j\in A_k\$ s.t. $\ a_i-a_j=n-1$

Let $$n\geq2$$ be an integer. Prove that every subset $$A_k\subset [3n]$$, with $$|A_k|=2n$$, contains two elements $$a_i,\ a_j\$$ s.t. $$\ a_i-a_j=n-1$$, and prove that $$\nexists\ a_i,\ a_j \in A_k\ \forall k\$$ s.t. $$\ a_i – a_j =n$$.

It seems obvious that I will have to apply the Pigeonhole Principle, but I have no clear idea about how to start. Could you give me some hints? Thanks in advance!

## Proove that $\forall m \in \mathbb{N}^{*},\exists n \in \mathbb{N},\forall k \geq n, p_{k+1}^m I need to proove : $$\forall m \in \mathbb{N}^{*},\exists n \in \mathbb{N},\forall k \geq n, p_{k+1}^m<\prod_{i=1}^{k}p_i$$ I can proove this assertion using Prime number theoreme : For fixed $$m$$ we have : $$\displaystyle m \log(p_{k+1}) < \log\left( \prod_{i=1}^{k}p_i \right)$$ And $$\log\left(\displaystyle \prod_{i=1}^{k}p_i \right) \sim p_k$$ and $$\log(p_{k+1}) \sim \log(k+1)$$ give the result. But i need An other proof not asymptotic. ## Showing$\forall x(\exists i \in I(x = x_i) \rightarrow x \in A) = \forall i \in I (x_i \in A)$A little background information, I’m reviewing Example 2.3.1(2) from the book How To Prove It. Basically it says that $$\forall x(\exists i \in I(x = x_i) \rightarrow x \in A)$$ is the same as $$\forall i \in I (x_i \in A)$$. What I’m not sure of is the $$(\rightarrow)$$ part of the proof. Given: $$\forall x (\exists i \in I(x = x_i) \rightarrow x \in A)$$ (1) Goal: $$\forall i \in I (x_i \in A)$$. Scratch work We first let i be an arbitrary element of I. Our goal thus becomes $$x_i \in A$$. Since (1) starts with $$\forall x$$, we apply universal instantiation and plug in $$x_i$$ to get $$\exists i \in I(x_i = x_i) \rightarrow x_i \in A)$$. Since $$i \in I$$ and $$x_i$$ is obviously equal to $$x_i$$, by modus ponens we get $$x_i \in A$$? Is this how it is done? Thanks in advance. ## If$(\alpha,\beta)$is the factor pair congruences of algebra$\mathbb{A},$ia$(\forall \gamma\in ConA)\alpha\circ\gamma=\gamma\circ\beta?$Let $$\mathbb{A}$$ be an algebra such that $$ConA$$ is the distributive lattice. If $$(\alpha,\beta)$$ is the factor pair congruences of algebra $$\mathbb{A},$$ prove that $$(\forall \gamma\in ConA)\alpha\circ\gamma=\gamma\circ\beta.$$ ##$H \triangleleft G$,$h \in H$.$C_G(h)g \cap H \ne \emptyset, \forall g \in G$implies$\lbrace C_G(h)g \cap H, g \in G \rbrace$partition of$H$? Let $$G$$ be a group and $$H \triangleleft G$$. I know that $$G$$ fixes the conjugacy classes of $$H$$ under conjugation if and only if $$\forall h \in H, C_G(h)g \cap H \ne \emptyset, \forall g \in G$$. Now, $$\forall h \in H$$, call $$R_g:=C_G(h)g \cap H$$. I’m arguing that, perhaps under stronger conditions that above, $$R:=\lbrace R_g, g \in G \rbrace$$ is a different partition of $$H$$ than the one induced by conjugacy (namely $$O=\lbrace O_h, h \in H \rbrace$$, $$O_h$$ being the orbits). Is that it? ## If$\forall s\in\mathbb R^d, \forall F \in \mathcal F\ E[e^{i}I_F]=E[e^{i}]P[F]$then$X$is independen of$\mathcal F\$

The claim in the title seems very plausible since the characteristic function “characterizes” or determines the distribution of $$X$$, but I don’t know how to derive it. There is a similar result for the characteristic functions of two Random variables, eg here, but I’m not sure if it could be deduced from that.

Any help would be appreciated!