# 线性代数代写 Real Analysis|MATH 8810 Boston College Assignment

0

Assignment-daixieTM为您提供波士顿学院Boston College MATH 8810 Real Analysis 线性代数代写代考辅导服务！

## Instructions:

The basic idea of Fourier theory is that any periodic function can be expressed as a sum of sine and cosine functions of different frequencies. These frequencies are called harmonics, and the sum of all harmonics is called the Fourier series.

The Fourier series representation of a periodic function can be used to analyze its properties, such as its frequency content and its response to different inputs. It is also used to design filters and other signal processing techniques.

In addition to the Fourier series, there is also the Fourier transform, which is a mathematical operation that transforms a function from the time domain to the frequency domain. The Fourier transform is widely used in signal processing and image processing to analyze and manipulate signals and images.

Overall, Fourier theory is a powerful tool for understanding and analyzing periodic phenomena, and it has a wide range of applications in many fields.

Let $A$ be a subset of $\mathbb{R}$ which is bounded above, and let $a_0$ be an upper bound for $A$. Prove that $a_0=\sup A$ if and only if for every $\epsilon>0$, there exists $a \in A$ such that $a_0-\epsilon<a$.

To prove that $a_0=\sup A$ if and only if for every $\epsilon>0$, there exists $a \in A$ such that $a_0-\epsilon<a$, we need to prove both directions of the statement:

$(\Rightarrow)$ Assume $a_0 = \sup A$. Let $\epsilon>0$ be given. Since $a_0$ is the least upper bound of $A$, there exists $a\in A$ such that $a_0-\epsilon<a\leq a_0$. This is because if there were no such $a$, then $a_0-\epsilon$ would be a smaller upper bound for $A$ than $a_0$, contradicting the assumption that $a_0=\sup A$.

$(\Leftarrow)$ Assume that for every $\epsilon>0$, there exists $a\in A$ such that $a_0-\epsilon<a$. We need to show that $a_0=\sup A$. Suppose, for the sake of contradiction, that $a_0$ is not the least upper bound of $A$. Then there exists $b>a_0$ such that $b$ is an upper bound for $A$. Let $\epsilon = b-a_0>0$. Since $a_0$ is not the least upper bound of $A$, there exists $a\in A$ such that $a>b$. But then $a_0< a< b = a_0 + \epsilon$, which contradicts the assumption that for every $\epsilon>0$, there exists $a\in A$ such that $a_0-\epsilon<a$. Therefore, $a_0$ is the least upper bound of $A$, i.e., $a_0=\sup A$.

Hence, we have proven that $a_0=\sup A$ if and only if for every $\epsilon>0$, there exists $a \in A$ such that $a_0-\epsilon<a$.

Prove that
$$\lim _{n \rightarrow \infty} \frac{1}{20 n^2+20 n+2020}=0 .$$

We can prove this limit using the squeeze theorem. First, note that for any positive $₫ \mathcal{P}$ integer \$n \$, we have:
$$20 n^2+20 n+2020 \geq 20 n^2$$
This is because \$20 n+2020 \$ is always positive for positive \$n \$, so adding it to \$20 n^{\wedge} 2 \$only increases the value. Using this inequality, we can write:
$$0 \leq \frac{1}{20 n^2+20 n+2020} \leq \frac{1}{20 n^2}$$
Now, taking the limit as n approaches infinity of both sides of this inequality, we get:
$$\lim {n \rightarrow \infty} 0 \leq \lim {n \rightarrow \infty} \frac{1}{20 n^2+20 n+2020} \leq \lim {n \rightarrow \infty} \frac{1}{20 n^2}$$ The left-hand side simplifies to 0 , and the right-hand side simplifies to 0 as well, since the denominator grows faster than the numerator as $\$ \mathrm{n} \$$approaches infinity. Therefore, by the squeeze theorem, we have:$$ \lim {n \rightarrow \infty} \frac{1}{20 n^2+20 n+2020}=0
$$as desired. 问题 3. Let \left{x_n\right} be a bounded sequence of real numbers. Prove$$
\lim {n \rightarrow \infty} x_n=0, $$if and only if$$ \limsup {n \rightarrow \infty}\left|x_n\right|=0 .


First, we will prove that if $\lim_{n \rightarrow \infty} x_n = 0$, then $\limsup_{n \rightarrow \infty} |x_n| = 0$.

Suppose $\lim_{n \rightarrow \infty} x_n = 0$. Let $\epsilon > 0$ be arbitrary. Since $\lim_{n \rightarrow \infty} x_n = 0$, there exists $N$ such that for all $n \geq N$, we have $|x_n| < \epsilon$.

Thus, for any $k \in \mathbb{N}$, there are only finitely many $n$ such that $|x_n| \geq \frac{1}{k}$. Therefore, $\limsup_{n \rightarrow \infty} |x_n| = 0$.

Next, we will prove that if $\limsup_{n \rightarrow \infty} |x_n| = 0$, then $\lim_{n \rightarrow \infty} x_n = 0$.

Suppose $\limsup_{n \rightarrow \infty} |x_n| = 0$. Then, for any $\epsilon > 0$, there exists $N$ such that for all $n \geq N$, we have $|x_n| < \epsilon$.

Since ${x_n}$ is bounded, there exists a positive number $M$ such that $|x_n| \leq M$ for all $n$. Therefore, for any $\epsilon > 0$, we have

$\left|x_n\right| \leq M<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$

for all $n \geq N$. This implies that $\lim_{n \rightarrow \infty} x_n = 0$.

Thus, we have shown both directions of the equivalence and conclude that

$\lim {n \rightarrow \infty} x_n=0 \quad$ if and only if $\quad \limsup {n \rightarrow \infty}\left|x_n\right|=0$.