[Home Squat University Mathematics Magazine] No. 389 issue of Chinese Academy of Sciences 2014-2015-1 calculus Midterm Exam Questions reference

1. Set the $A, b,c$ is a subset of the collection $M $, please prove: $$\bex (C\subset A) \wedge (C\subset b) \lra (C\subset a\cap b). \eex$$

Proof: clearly established.

2. Set the set $X $ meet $\bar{\bar \bbn}\leq \bar{\bar x}$. Please prove: Collection $Y =x\cup\bbn$ meet $\bar{\bar X}=\bar{\bar y}$.

Proof: Apparently $\bar{\bar X}\leq \bar{\bar y}$. On the other hand, by $\bar{\bar N}\leq \bar{\bar x}$ know $X $ by a subset $A $, $$\bex y=x\cup\bbn = (X\bs a) \cup (A\CUP\BBN) \sim (X\bs a) \cup A =x\ra \bar{\bar y}\leq \bar{\bar X}. \eex$$ According to Cantor-bernstein theorem is known $\bar{\bar x}=\bar{\bar y}$.

3. Set $\al>1$, seek limit $$\bex \vlm{n}\frac{n^\al}{\al^n},\quad \vlm{n}\frac{\al^n}{n!},\quad \vlm{n}\frac{n!} {n^n}. \eex$$

Solution: By $$\beex \bea \frac{n^\al}{\al^n}&=\frac{n^\al}{(1+\GM) ^n}\quad\sex{\gm=\al-1>0}\\ &\leq \frac{n^{[\ Al]+1}}{\sex{n\atop [\al]+2}\gamma^{[\al]+2}}\quad\sex{n>[\al]+2} \eea \eeex$$ is known as the first limit $=0$. Also by $$\beex \bea \frac{\al^n}{n!} &\leq \frac{([\al]+1) ^n}{n!} =\frac{[\al]+1}{1}\cdots \frac{[\al]+1}{[\al]+1}\ Cdots \frac{[\al]+1}{n}\quad\sex{n>[\al]+1}\\ &\leq \frac{([\al]+1] ^{[\al]+1}}{([\al]+1)!} \frac{[\al]+1}{n } \eea \eeex$$ know the second limit $=0$. Last by $$\bex \frac{n!} {n^n}=\frac{1\cdots n}{n\cdots N}\leq \frac{1}{n} \eex$$ know the third limit $=0$.

4. Set the sequence $\sed{a_n}$ and $\sed{b_n}$ to meet

(1). $a _n\searrow 0$.

(2). $\dps{\exists\ m>0,\st \sev{\sum_{k=1}^nb_k}\leq m}$, $\forall\ n\in\bbn$. Please prove that the series $\dps{\vsm{n}a_nb_n}$ converge.

Proof: Set $\dps{t_n=\sum_{k=1}^n b_k}$, then $$\beex \bea \sev{\sum_{k=n}^{n+p}a_kb_k} &=\sev{\sum_{k=n}^{n+p}a_k (T_k-T_{ K-1})}\\ &=\sev{\sum_{k=n}^{n+p}a_kt_k-\sum_{k=n-1}^{n+p-1}a_{k+1}t_k}\\ &=\sev{\sum_{k=n}^{n+p} (a_k-a_{ K+1}) T_k +a_{n+p}t_{n+p}-a_nt_{n-1}}\\ &\leq m\sum_{k=n}^{n+p}|a_k-a_{k+1}| +m a_{n+p}+m a_n\\ &\leq M (a_n-a_{n+p+1}) +2ma_n\\ &\leq 3ma_n. \eea \eeex$$ According to the Cauchy convergence criterion, the conclusion is established.

5. Set $n (k) \ (k=1,2,\cdots) $ is a $1$-$1$ mapping from the natural number set $\bbn=\sed{1,2,\cdots}$ to $\bbn$. Make $b _k=a_{n (k)}\ (k=1,2,\cdots) $. We call the series $\dps{\vsm{n}b_n}$ a rearrangement of the progression $\dps{\vsm{n}a_n}$. Proving to any real number $\al$, there is a rearrangement of the series $\dps{\vsm{n}\frac{( -1) ^n}{n}}$, which makes the sum of the re-beat series $\al$.

Proof: Set $$\bex S_n=-1-\frac{1}{3}-\cdots-\frac{1}{2n-1},\quad t_n=\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2n}. \eex$$ $S _n$ to $-\infty$, $T _n$ to $+\infty$. Take $n _1$ sufficiently large to make $S _{n_1}<\al$. Then take $n _2>n_1$ make $S _{n_1}+t_{n_2}>\al$. Then take $n _3>n_2$ make $T _{n_2}+s_{n_3}<\al$. have been doing so, get $\sed{n_k}$, make $$\bee\label{389_5_eq} k\ odd\ra s_{n_k}+t_{n_{k+1}}>\al,\quad k\ even\ra T_{n_k}+S_{n_{k +1}}<\al. \eee$$ reflow $\dps{\vsm{n}\frac{( -1) ^n}{n}}$ as follows: $$\bex-1-\cdots-\frac{1}{2n_1+1} +\frac{1}{2}+\cdots+\ frac{1}{2n_2}\\-\frac{1}{2 (n_1+1) +1}-\cdots-\frac{1}{2n_3+1} +\frac{1}{2 (n_2+1)} +\cdots +\frac{1}{2n_4}-\cdots. \eex$$ by \eqref{389_5_eq} The above series converge to $\al$. In fact, the $s _n$ for the above-mentioned series of parts and, then $$\bex \vls{n}s_n\leq \al\leq \vli{n}s_n. \eex$$

6. Using the $\ve-\delta$ language proof function $f (x) =x^2+x\sin x$ is continuous at the $x =1$.

Proof: To $\forall\ \ve>0$, take $$\bex \delta=\min\sed{\frac{\ve}{5},1}>0, \eex$$ when $|x-1|<\delta$, $$\beex \bea |f ( x)-F (1) | &\leq |x^2-1|+|x\sin x-\sin 1|\\ &\leq |x+1|\cdot |x-1| +|x-1|\cdot |\sin x| +|\sin x-\sin 1|\\ &\leq 3|x-1| +|x-1| +|x-1|\\ &<\ve. \eea \eeex$$

7. Define the function $f: \bbr\to \bbr$ as follows: $$\bex f (x) =\sedd{\ba{ll} \frac{1}{n},&x\in\bbq,\ n=\min\sed{q;x=\frac{p}{q};\ p\in\ Bbz,\ q\in\bbn},\\ 0,&X\NOT\IN\BBQ. \ea} \eex$$ analyze the continuity of the $f (x) $ on $\bbr$.

Proof: If $x _0\in\bbq$, then $f (x_0) =1/n$, $N =\min\{q;\ x=p/q,\ p\in\bbz,\ q\in\bbn\}$. Take $\bbr\bs\bbq\ni x_n\to x_0$, then $f (X_n) =0\not\to F (x_0) $. Therefore $f $ in the $x _0$ place is not continuous. If the $x _0\not\in\bbq$, then to $\forall\ \ve>0$, to meet the $1/n\geq \ve$ of the $N $ only a limited number, while in $ ([x_0],[x_0]+1) $ can form into $M/n$\ ($M \in\bbz,\ 1/ N\geq \ve$) has only a finite number of rational numbers. Take $\delta>0$, so that the $U (X_0,\delta) $ does not contain these rational numbers, then $$\bex x\in U (X_0,\delta) \ra |f (x)-F (x_0) |<\ve. \eex$$

8. Prove that the continuous function $f (x) =\sin 1/x$ in $ (0,1) $ is not consistent continuous.

Proof: Take $$\bex x_n=\frac{1}{2n\pi},\quad y_n=\frac{1}{(2N+1/2) \pi}, \eex$$ then $$\bex |x_n-y_n|<\frac{1}{2n\pi}\to 0,\ Quad |f (X_n)-F (y_n) |=1. \eex$$

The topic originates from Shing Teacher's Renren.

[Home Squat University Mathematics Magazine] No. 389 issue of Chinese Academy of Sciences 2014-2015-1 calculus Midterm Exam Questions reference