学委真不负责
微积分
$1^\infty$ 的处理
令 $$\lim_{x\to x_o} u(x)=1,\ \lim_{x\to x_o} v(x)=\infty$$
则 $$\begin{eqnarray} \lim_{x\to x_o} u(x)^{v(x)}&=&\lim_{x\to x_o} [1+(u-1)]^{\frac{1}{u-1}\cdot (u-1)\cdot v}\\&=&\exp\Big\{\lim_{x\to x_o}(u-1)\cdot v\Big\} \end{eqnarray}$$
例题:$$ \begin{eqnarray} & &\lim_{x\to \infty}(\sin \frac{1}{x}+\cos\frac{1}{x})^x\\ &= &\exp\Big\{\lim_{x\to\infty}x\cdot(\sin\frac{1}{x}+\cos\frac{1}{x}-1)\Big\}\\&=&\exp\Big\{\lim_{x\to\infty}x\cdot (\frac{1}{x}+1-1+o(\frac{1}{x}))\Big\}\\&=&\exp(1)=e\end{eqnarray}$$
还有习题课 函数连续性 第二题 秒出
(可以试一下作业第八题)
各种小量运算
大家都会泰勒展开那我就不说了
$\sin x\sim x-\frac{x^3}{6}+o(x^3)$
$\cos x\sim 1-\frac{x^2}{2}+o(x^2)$
$\tan x\sim x+\frac{x^3}{3}+o(x^3)$
$e^x\sim 1+x+\frac{x^2}{2}+\frac{x^3}{6}+o(x^3)$
$\ln(x+1)\sim x-\frac{x^2}{2}+\frac{x^3}{3}+o(x^3)$
$(1+x)^\alpha\sim 1+\alpha x+\frac{\alpha^2-\alpha}{2}x^2+o(x^2)$
当 $x\to 0$ 时可以直接替代,否则不能用
Stolz定理的应用
我们知道Stolz定理就是离散洛必达,所以洛必达能干的事情Stolz应该也能干,比如 $0\cdot \infty$ 型求极限
方法差不多长这样:
$n\cdot x_n=\frac{n}{\frac{1}{x_n}}\sim \frac{n+1-n}{\frac{1}{x_{n+1}}-\frac{1}{x_n}}=\frac{1}{\frac{1}{x_{n+1}}-\frac{1}{x_n}}$
于是就能把n给消掉了。多用于 $x_{n+1}=f(x_n)$ 型的数列。
例题:设 $x_n\in (0,1), x_{n+1}=x_n(1-x_n),$ 试证明 $\lim_{n\to \infty} nx_n=1$
直接按照上面的来就好了。
练习:设$x_n\in (0,1), x_{n+1}=\sin(x_n),$ 试证明 $\lim_{n\to \infty}\sqrt{\frac{n}{3}}x_n=1$
习题课上的收获
只要是“证明xx收敛”,而没有求值的题目,多半是用柯西;否则只要把要证的东西通过放缩/迭代/奇技淫巧,变成题目里的东西就好
好题推荐
试证明$$\lim_{n\to\infty} n\sin(2\pi e n!)=2\pi$$
线代
“加边法”
线代第二次作业P36/11(1)有无脑做法。对4班那次推送忍了很久了,因此点了几次右下角的投诉,理由是“常识性谣言”。
就是加一行一列不改变行列式:$$D=\begin{vmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{vmatrix}=\begin{vmatrix}1&bla&bla\\0&a_{11}&a_{12}\\0&a_{21}&a_{22}\end{vmatrix}$$
可以自行选取 bla 的值,然后通过行列变换将原来矩阵里面的东西消掉,比如11(1)
$$\begin{eqnarray} & &\begin{vmatrix}1+a_1+b_1&a_1+b_2&\cdots&a_1+b_n\\a_2+b_1&1+a_2+b_2&\cdots&a_2+b_n\\\vdots&\vdots&\ddots&\vdots\\a_n+b_1&a_n+b_2&\cdots&1+a_n+b_n\end{vmatrix}\\ &=&\begin{vmatrix}1&-b_1&-b_2&\cdots&-b_n\\0&1+a_1+b_1&a_1+b_2&\cdots&a_1+b_n\\0&a_2+b_1&1+a_2+b_2&\cdots&a_2+b_n\\\vdots&\vdots&\vdots&\ddots&\vdots\\0&a_n+b_1&a_n+b_2&\cdots&a_n+b_n\end{vmatrix}\\ &=&\begin{vmatrix}1&-b_1&-b_2&\cdots&-b_n\\1&1+a_1&a_1&\cdots&a_1\\1&a_2&1+a_2&\cdots&a_2\\\vdots&\vdots&\vdots&\ddots&\vdots\\1&a_n&a_n&\cdots&1+a_n\end{vmatrix}\\ &=&\begin{vmatrix}1&0&0&0&\cdots&0\\0&1&-b_1&-b_2&\cdots&-b_n\\-a_1&1&1+a_1&a_1&\cdots&a_1\\\vdots&\vdots&\vdots&\vdots&\ddots&\vdots\\-a_n&1&a_n&a_n&\cdots&1+a_n\end{vmatrix}\\ &=&\begin{vmatrix}1&0&1&1&\cdots&1\\0&1&-b_1&-b_2&\cdots&-b_n\\-a_1&1&1&0&\cdots&0\\-a_2&1&0&1&\cdots&0\\\vdots&\vdots&\vdots&\vdots&\ddots&\vdots\\-a_n&1&0&0&\cdots&1\end{vmatrix}\\ &=&\begin{vmatrix}1+\sum a_i&0&1&1&\cdots&1\\-\sum a_ib_i&1&-b_1&-b_2&\cdots&-b_n\\0&1&1&0&\cdots&0\\\vdots&\vdots&\vdots&\vdots&\ddots&\vdots\\0&1&0&0&\cdots&1\end{vmatrix}\\ &=&\begin{vmatrix}1+\sum a_i&-n&1&\cdots&1\\-\sum a_ib_i&1+\sum b_i&-b_1&\cdots&-b_n\\0&0&1&\cdots&0\\\vdots&\vdots&\vdots&\ddots&\vdots\\0&0&0&\cdots&1\end{vmatrix}\\ &=&\Big(1+\sum_{i=1}^n a_i\Big)\Big(1+\sum_{i=1}^n b_i\Big)-n\sum_{i=1}^n a_ib_i \end{eqnarray}$$
程设
要讲些啥呢?
大家把平时困惑的问题,哪怕是很小很小的问题,哪怕你最后自己解决了,发到公众号里面。我们才能根据大家的需求编写推送哦~
#include <cstdio> char a[20],len; int check(int x){ for(len=0;x;x/=10)a[++len]=x%10; for(int i=1;i<=len;i++,len--) if(a[i]!=a[len])return 0; return 1; } int n; int main(){ scanf("%d",&n); for(int i=1;i*i*i<=n;i++) if(check(i)&&check(i*i)&&check(i*i*i)) printf("%d ",i); return 0; }
我会说上面的代码是我刚睡醒在手机上直接敲的吗233
#include <cstdio> int n,a[60]={1}; int main(){ scanf("%d",&n); for(int i=1;i<n;i++) for(int j=i;j;j--) a[j]+=a[j-1]; for(int i=0;i<n;i++) printf("%d ",a[i]); puts(""); return 0; }
听说周二选李宛洲的课的同学要考试?害怕
思修习题十答案
D;C;ABC;ABCD;
2022年10月03日 16:51
How to Apply for the SBI Credit Card Online? · Visit Finserv MARKETS. Visit us online · Fill in your details. How to Apply for the SBI Credit Card Online? · Visit Finserv MARKETS. Visit us online · Fill in your details. Help us with a few of your details. SBI Credit Card Eligibility. In order to apply for SBI Credit card offers one needs to meet the following eligibility criterias, sbi bank credit card as listed below: Income. Apply online for an SBI Credit Card. Choose from multiple variants to suit your lifestyle. Get offers Shopping. Help us with a few of your details. SBI Credit Card Eligibility. In order to apply for SBI Credit card offers one needs to meet the following eligibility criterias, as listed below: Income. Apply online for an SBI Credit Card. Choose from multiple variants to suit your lifestyle. Get offers Shopping.