# 数学分析|MATH0004 Analysis 2代写2023

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## Instructions:

We will start with the basic properties of real numbers and use them to prove the main results in elementary differential calculus. We assume familiarity with the properties of real numbers such as completeness, order, and the field axioms.

SEQUENCES:

A sequence is a function $f: \mathbb{N} \rightarrow \mathbb{R}$, where $\mathbb{N}$ is the set of natural numbers. We say that a sequence $(a_n)$ converges to a limit $L$ if for every $\epsilon > 0$, there exists an $N \in \mathbb{N}$ such that $|a_n – L| < \epsilon$ for all $n \geq N$. We write $\lim_{n \rightarrow \infty} a_n = L$.

Find the radius of convergence $R$ of $\sum_0^{\infty} \frac{(2 n) ! x^n}{n !(n+1) !}$.

We can apply the ratio test to determine the radius of convergence of the given series. Let

$a_n=\frac{(2 n) ! x^n}{n !(n+1) !}$

Then, the ratio of successive terms is

\begin{align*} \frac{a_{n+1}}{a_n} &= \frac{(2(n+1))!x^{n+1}}{(n+1)!(n+2)!} \cdot \frac{n!(n+1)!}{(2n)!x^n} \ &= \frac{(2n+2)(2n+1)x}{(n+2)(n+1)} \ &= \frac{4x^2}{(n+2)(n+1)} \cdot \frac{(2n+1)(2n+2)}{4(n+1)} \ &= \frac{4x^2}{(n+2)(n+1)} \cdot \left(1 + \frac{1}{n+1}\right) \cdot \left(1 + \frac{2}{n+1}\right). \end{align*}

Taking the limit as $n \to \infty$, we get

$\lim {n \rightarrow \infty} \frac{a{n+1}}{a_n}=\lim _{n \rightarrow \infty} \frac{4 x^2}{(n+2)(n+1)} \cdot\left(1+\frac{1}{n+1}\right) \cdot\left(1+\frac{2}{n+1}\right)=4 x^2$.

Therefore, the series converges absolutely if $4x^2 < 1$, or equivalently, $|x| < \frac{1}{2}$. On the other hand, the series diverges if $4x^2 > 1$, or equivalently, $|x| > \frac{1}{2}$. Thus, the radius of convergence is $R = \frac{1}{2}$.

Let $h(n)$ be the largest prime factor of the integer $n>1$, and $s(n)$ be the sum of its prime factors, so $h(12)=3, s(12)=7$.

Prove the sequence ${h(n) / s(n)}, n=2,3,4, \ldots$ has $1 / k$ as a cluster point for every positive integer $k$, but no limit.

To show that $1/k$ is a cluster point of the sequence ${h(n)/s(n)}$ for every positive integer $k$, we need to show that for any $\epsilon>0$ and positive integer $k$, there exist infinitely many $n$ such that $|h(n)/s(n)-1/k|<\epsilon$.

Let $\epsilon>0$ and $k$ be given. Choose $p$ to be any prime greater than $k/\epsilon$, and consider the numbers $n=p^2,p^3,\ldots,p^{k+1}$. Note that $h(n)=p$ and $s(n)=1+p+p^2+\cdots+p^k=\frac{p^{k+1}-1}{p-1}$.

For any $n=p^m$ with $2\leq m\leq k+1$, we have

$\left|\frac{h(n)}{s(n)}-\frac{1}{k}\right|=\left|\frac{k p-p^{k+1}}{(p-1)\left(p^{k+1}-1\right)}\right|=\frac{p\left(k-p^k\right)}{(p-1)\left(p^{k+1}-1\right)}<\frac{p(k-p)}{(p-1) p^{k+1}}<\frac{k}{p^k-1}<\epsilon$,

where we used the fact that $p>k/\epsilon$ and $p^{k+1}>kp$. Thus, we have shown that for any $\epsilon>0$ and positive integer $k$, there exist infinitely many $n$ such that $|h(n)/s(n)-1/k|<\epsilon$. Hence, $1/k$ is a cluster point of the sequence ${h(n)/s(n)}$ for every positive integer $k$.

To show that the sequence ${h(n)/s(n)}$ has no limit, we need to find two increasing sequences of integers ${a_n}$ and ${b_n}$ such that $h(a_n)/s(a_n)$ and $h(b_n)/s(b_n)$ have different limits as $n\rightarrow\infty$.

Let ${p_n}$ be the sequence of prime numbers, and define $a_n=p_{2n-1}^2$ and $b_n=p_{2n}^2$. Then we have

$\frac{h\left(a_n\right)}{s\left(a_n\right)}=\frac{p_{2 n-1}}{1+p_{2 n-1}} \rightarrow 1, \quad \frac{h\left(b_n\right)}{s\left(b_n\right)}=\frac{p_{2 n}}{1+p_{2 n}+p_{2 n}^2} \rightarrow 0$,

as $n\rightarrow\infty$, since $p_{2n-1}/(1+p_{2n-1})\rightarrow 1$ and $p_{2n}/(1+p_{2n}+p_{2n}^2)\rightarrow 0$. Therefore, the sequence ${h(n)/s(n)}$ does not have a limit as $n\rightarrow\infty$.

A function $f(x)$ has three distinct zeros $a_0<a_1<a_2$ on an interval $I$, and in addition $f^{\prime}\left(a_2\right)=0$. Assume $f(x)$ has a third derivative $f^{\prime \prime \prime}(x)$ at all points of $I$. Prove there is a point $c \epsilon I$ such that $f^{\prime \prime \prime}(c)=0$.

Since $f(x)$ has three distinct zeros on the interval $I$, we know that $f(x)$ changes sign at each of these zeros. Without loss of generality, assume that $f(x)$ is positive on $(a_0, a_1)$, negative on $(a_1, a_2)$, and positive on $(a_2, b)$.

Since $f(x)$ has a local extremum at $x=a_2$ (because $f^{\prime}(a_2)=0$), the second derivative $f^{\prime \prime}(x)$ must change sign at $x=a_2$. Specifically, $f^{\prime \prime}(x)$ is negative on $(a_1, a_2)$ and positive on $(a_2, b)$.

Now consider the third derivative $f^{\prime \prime \prime}(x)$. Since $f^{\prime \prime}(x)$ changes sign at $x=a_2$, we know that $f^{\prime \prime \prime}(x)$ must have a local extremum at $x=a_2$. Without loss of generality, assume that $f^{\prime \prime \prime}(x)$ is negative on $(a_1, a_2)$ and positive on $(a_2, b)$.

Since $f(x)$ changes sign at each of the zeros $a_0<a_1<a_2$, we know that $f(x)$ must have at least one local extremum (a maximum or minimum) between each pair of zeros. Specifically, $f(x)$ has a local minimum at some point $c_1$ in $(a_0, a_1)$, a local maximum at some point $c_2$ in $(a_1, a_2)$, and a local minimum at some point $c_3$ in $(a_2, b)$.

Since $f^{\prime \prime \prime}(x)$ is negative on $(a_1, a_2)$ and positive on $(a_2, b)$, we know that $f^{\prime \prime \prime}(c_2)>0$ and $f^{\prime \prime \prime}(c_3)<0$. Therefore, by the intermediate value theorem, there must be some point $c$ in $(c_2, c_3)$ such that $f^{\prime \prime \prime}(c)=0$, as desired.