Comp
2022非数学类
- \(f(x)=\begin{cases}(1+x)^{1/x}&x\ne0\\e&x=0\end{cases}\) 在 0 处的三次泰勒多项式
- \((1+x)^{1/x}=e^{\frac {ln(1+x)}x} = e^{\frac {\sum\limits_{i=1}^4\frac {(-1)^{i-1}}ix^i+o(x^4)}x} = e^{\sum\limits_{i=1}^4\frac {(-1)^{i-1}}ix^{i-1}+o(x^3)} = e\cdot e^{\sum\limits_{i=2}^4\frac {(-1)^{i-1}}ix^{i-1}+o(x^3)} = e\cdot (\sum\limits_{j=0}^3\frac 1{j!}a^j+o(x^3))\)
- a 记为 \(\sum\limits_{i=2}^4\frac {(-1)^{i-1}}ix^{i-1}+o(x^3)\)
2021数学类
斯托尔兹定理(storz)：对于 \(x_n,y_n\)，其中 \(x_n\) 递增并趋于 \(+∞\)，若 \(\lim\limits_{n\to∞}=\frac {y_n-y_{n-1}}{x_n-x_{n-1}}=a\) （\(a\in \mathbb R\) 或 a 趋于无穷大），则 \(\lim\limits_{n\to ∞}\frac {y_n}{x_n}=\lim\limits_{n\to ∞}\frac {y_n-y_{n-1}}{x_n-x_{n-1}}\)
泰勒公式，绝对值三角不等式
莱布尼茨定理
- \(\displaystyle \lim\limits_{n\to+∞}(\frac 23)^{\frac1{2^{n-1}}}(\frac 47)^{\frac1{2^{n-2}}}\dots(\frac {2^{n-1}}{2^n-1})^{\frac12}\)
- 记 \(a_n = \prod\limits_{i=1}^{n-1}(\frac{2^i}{2^{i+1}-1})^{\frac1{2^{n-i}}} = (\frac 23)^{\frac1{2^{n-1}}}(\frac 47)^{\frac1{2^{n-2}}}\dots(\frac {2^{n-1}}{2^n-1})^{\frac12}\)
- \(\ln a_n = \ln [\prod\limits_{i=1}^{n-1}(\frac{2^i}{2^{i+1}-1})^{\frac1{2^{n-i}}}] = \sum\limits_{i=1}^{n-1}\frac 1{2^{n-i}}\ln\frac{2^i}{2^{i+1}-1} = \frac 1{2^n} \sum\limits_{i=1}^{n-1}{2^i}\ln\frac{2^i}{2^{i+1}-1}\) （此时式子分离为两项 \(y_n\)，\(x_n\)）
- \(\lim\limits_{n\to+∞}\ln a_n = \lim\limits_{n\to+∞}\frac {y_n}{x_n} = \lim\limits_{n\to+∞}\frac {y_n-y_{n-1}}{x_n-x_{n-1}} = \lim\limits_{n\to+∞}\frac {2^{n-1}\ln \frac {2^{n-1}}{2^{n}-1}}{2^n-2^{n-1}} = \lim\limits_{n\to+∞}\ln\frac {2^{n-1}}{2^{n}-1} = \lim\limits_{n\to+∞} \ln \frac 1{2-2^{n-1}} = \ln\frac12\)
- \(\lim\limits_{n\to+∞}a_n = \lim\limits_{n\to+∞}e^{\ln a_n} = \frac12\)
- f 在 \([0,2]\) 上有二阶导数，对于所有 \(x\in[0,2]\) 有 \(f(x)\le 1,~|f''(x)|\le1\)，证明：\(|f'(x)|\le 2\)
- 计算 \(I=\iint\limits_D|xy-\frac14|dxdy,~D=[0,1]\times[0,1]\)
- \(\sum\limits_{n=1}^∞a_n=\frac 1{1^p}-\frac1{2^q}+\frac1{3^p}+\dots+\frac1{(2n-1)^p}-\frac1{(2n)^q}+\dots(p>0,q>0)\)