Eufisky - The lost book

## 许以超书上一行列式求解

${\Delta _n} = \det \left( {\begin{array}{*{20}{c}}{1 + {x_1}{y_1}}&{1 + {x_1}{y_2}}& \cdots &{1 + {x_1}{y_n}}\\{1 + {x_1}y_1^2}&{1 + {x_1}y_2^2}& \cdots &{1 + {x_1}y_n^2}\\\vdots & \vdots &{}& \vdots \\{1 + {x_1}y_1^n}&{1 + {x_1}y_2^n}& \cdots &{1 + {x_1}y_n^n}\end{array}} \right).$

\begin{align*}&\det A + x\sum\limits_{j,k = 1}^n {{A_{jk}}} = \det \left( {\begin{array}{*{20}{c}}{{a_{11}} + x}& \cdots &{{a_{1n}} + x}\\\vdots &{}& \vdots \\{{a_{n1}} + x}& \cdots &{{a_{nn}} + x}\end{array}} \right)\\=& \det A + x\det \left( {\begin{array}{*{20}{c}}1&1& \cdots &1&1\\{{a_{21}} - {a_{11}}}&{{a_{22}} - {a_{12}}}& \cdots &{{a_{2,n - 1}} - {a_{1,n - 1}}}&{{a_{2n}} - {a_{1n}}}\\{{a_{31}} - {a_{21}}}&{{a_{32}} - {a_{22}}}& \cdots &{{a_{3,n - 1}} - {a_{2,n - 1}}}&{{a_{3n}} - {a_{2n}}}\\\vdots & \vdots &{}& \vdots & \vdots \\{{a_{n1}} - {a_{n - 1,1}}}&{{a_{n2}} - {a_{n - 1,2}}}& \cdots &{{a_{n,n - 1}} - {a_{n - 1,n - 1}}}&{{a_{nn}} - {a_{n - 1,n}}}\end{array}} \right).\end{align*}

\begin{align*}&{\Delta _n} = \det \left( {\begin{array}{*{20}{c}}{1 + {x_1}{y_1}}&{1 + {x_1}{y_2}}& \cdots &{1 + {x_1}{y_n}}\\{1 + {x_1}y_1^2}&{1 + {x_1}y_2^2}& \cdots &{1 + {x_1}y_n^2}\\\vdots & \vdots &{}& \vdots \\{1 + {x_1}y_1^n}&{1 + {x_1}y_2^n}& \cdots &{1 + {x_1}y_n^n}\end{array}} \right)\\= &\det A + \det \left( {\begin{array}{*{20}{c}}1&1& \cdots &1\\{{x_1}\left( {y_1^2 - {y_1}} \right)}&{{x_1}\left( {y_2^2 - {y_2}} \right)}& \cdots &{x_1\left( {y_n^2 - {y_n}} \right)}\\\vdots & \vdots &{}& \vdots \\{{x_1}\left( {y_1^n - y_1^{n - 1}} \right)}&{{x_1}\left( {y_2^n - y_2^{n - 1}} \right)}& \cdots &{{x_1}\left( {y_n^n - y_n^{n - 1}} \right)}\end{array}} \right)\\=& \det A + x_1^{n - 1}\det \left( {\begin{array}{*{20}{c}}1&1& \cdots &1\\{y_1^2 - {y_1}}&{y_2^2 - {y_2}}& \cdots &{y_n^2 - {y_n}}\\\vdots & \vdots &{}& \vdots \\{y_1^n - y_1^{n - 1}}&{y_2^n - y_2^{n - 1}}& \cdots &{y_n^n - y_n^{n - 1}}\end{array}} \right).\end{align*}

$\det \left( {\begin{array}{*{20}{c}}1&1& \cdots &1\\{y_1^2 - {y_1}}&{y_2^2 - {y_2}}& \cdots &{y_n^2 - {y_n}}\\\vdots & \vdots &{}& \vdots \\{y_1^n - y_1^{n - 1}}&{y_2^n - y_2^{n - 1}}& \cdots &{y_n^n - y_n^{n - 1}}\end{array}} \right) = \det \left( {\begin{array}{*{20}{c}}1&{{y_1}}&{{y_2}}& \cdots &{{y_n}}\\0&1&1& \cdots &1\\0&{y_1^2 - {y_1}}&{y_2^2 - {y_2}}& \cdots &{y_n^2 - {y_n}}\\\vdots & \vdots & \vdots &{}& \vdots \\0&{y_1^n - y_1^{n - 1}}&{y_2^n - y_2^{n - 1}}& \cdots &{y_n^n - y_n^{n - 1}}\end{array}} \right).$

\begin{align*}\det \left( {\begin{array}{*{20}{c}}1&{{y_1}}&{{y_2}}& \cdots &{{y_n}}\\0&1&1& \cdots &1\\1&{y_1^2}&{y_2^2}& \cdots &{y_n^2}\\\vdots & \vdots & \vdots &{}& \vdots \\1&{y_1^n}&{y_2^n}& \cdots &{y_n^n}\end{array}} \right) &= \det \left( {\begin{array}{*{20}{c}}1&{{y_1}}&{{y_2}}& \cdots &{{y_n}}\\1&1&1& \cdots &1\\1&{y_1^2}&{y_2^2}& \cdots &{y_n^2}\\\vdots & \vdots & \vdots &{}& \vdots \\1&{y_1^n}&{y_2^n}& \cdots &{y_n^n}\end{array}} \right) + \det \left( {\begin{array}{*{20}{c}}1&{{y_1}}&{{y_2}}& \cdots &{{y_n}}\\{ - 1}&0&0& \cdots &0\\1&{y_1^2}&{y_2^2}& \cdots &{y_n^2}\\\vdots & \vdots & \vdots &{}& \vdots \\1&{y_1^n}&{y_2^n}& \cdots &{y_n^n}\end{array}} \right)\\& = \left( { - 1} \right) \cdot \prod\limits_{k = 1}^n {\left( {{y_k} - 1} \right)} \cdot \prod\limits_{1 \le i < j \le n} {\left( {{y_j} - {y_i}} \right)} + \prod\limits_{k = 1}^n {{y_k}} \cdot \prod\limits_{1 \le i < j \le n} {\left( {{y_j} - {y_i}} \right)} \\& = \prod\limits_{1 \le i < j \le n} {\left( {{y_j} - {y_i}} \right)} \cdot \left[ { - \prod\limits_{k = 1}^n {\left( {{y_k} - 1} \right)} + \prod\limits_{k = 1}^n {{y_k}} } \right].\end{align*}

\begin{align*}{\Delta _n} &= \det A + x_1^{n - 1}\prod\limits_{1 \le i < j \le n} {\left( {{y_j} - {y_i}} \right)} \cdot \left[ { - \prod\limits_{k = 1}^n {\left( {{y_k} - 1} \right)} + \prod\limits_{k = 1}^n {{y_k}} } \right]\\& = x_1^n \cdot \prod\limits_{k = 1}^n {{y_k}} \cdot \prod\limits_{1 \le i < j \le n} {\left( {{y_j} - {y_i}} \right)} + x_1^{n - 1}\prod\limits_{1 \le i < j \le n} {\left( {{y_j} - {y_i}} \right)} \cdot \left[ { - \prod\limits_{k = 1}^n {\left( {{y_k} - 1} \right)} + \prod\limits_{k = 1}^n {{y_k}} } \right]\\& = x_1^{n - 1}\prod\limits_{1 \le i < j \le n} {\left( {{y_j} - {y_i}} \right)} \cdot \left[ { - \prod\limits_{k = 1}^n {\left( {{y_k} - 1} \right)} + \left( {{x_1} + 1} \right)\prod\limits_{k = 1}^n {{y_k}} } \right].\end{align*}

## 与$\sin n^2$类似的一些问题

1.证明: $\sum_{k=1}^n\sin k^2$无界.

2.证明$\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = 1}^n {\sin \sqrt k } = 0.$

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TeX数学公式网页预览版

## 多元里的两道问题

“数学是你们的选择，你们随时都可以放弃。但当数学仍是你们的选择时，就必须为此负责。”
——S.Lang对他学生上课前说的话

## i的i次方等于多少?

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$，这是我们从初中就开始熟悉的概念。

$i= (-1)^{1/2} = ((-1)(-1)(-1))^{1/2} =(-1)^{1/2}(-1)^{1/2}(-1)^{1/2}=iii=-i$，

$-1=(-1)^1 = ((-1)^2)^{1/2}=1^{1/2}=1$。

$i^i= \exp(i Ln i) = \exp(i (i\pi/2 + i2k\pi)) = \exp(-\pi/2 - 2k\pi)=\exp(-\pi/2) \exp(-2k\pi)$

$(-2)^{1/3}, \hspace{5 mm}, (-1)^i$

（把你们的解答放在评论里，我的答案后天附在下面）

【答案】

$(-2)^{1/3} = \exp(Ln(-2)/3) = \{-\sqrt[3]{2}, \;\;\; \sqrt[3]{2}(1/2 -i\sqrt{3}/2), \;\;\; \sqrt[3]{2}(1/2 + i\sqrt{3}/2) \}$

$(-1)^i = \exp(iLn(-1))= e^{-\pi+2k\pi} = 0.043213918 \times 535.4916555^k$

$x_0 = -\sqrt[5]{2}$

$x_1 = -\sqrt[5]{2}((\sqrt{5}-1)/4+ i \sqrt{10+2\sqrt{5}}/4), \;\;\; x_2 =-\sqrt[5]{2}(-(\sqrt{5}+1)/4 + i \sqrt{10-2\sqrt{5}}/4)$

$x_3 = -\sqrt[5]{2}(-(\sqrt{5}+1)/4 - i \sqrt{10-2\sqrt{5}}/4),\;\;\; x_4 = -\sqrt[5]{2}((\sqrt{5}-1)/4- i \sqrt{10+2\sqrt{5}}/4),$

$x_n = -\sqrt[5]{2}(\cos(2n\pi/5) +i\sin(2n\pi/5)), \;\;\; n=0,1,2,3,4$

$\cos(2\pi/5)=(\sqrt{5}-1)/4= 0.309016994...$，

$\sin(2\pi/5)=\sqrt{10+2\sqrt{5}}/4= 0.951056516...$，

$\cos(\pi/5)=(\sqrt{5}+1)/4= 0.809016994…$

$\sin(\pi/5)=\sqrt{10-2\sqrt{5}}/4= 0.587785252...$

## 谢之题解16.2级数求和计算篇

1.设已知$\sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^{n - 1}}{a_n}} = A,\sum\limits_{n = 1}^\infty {{a_{2n - 1}}} = B$,证明: $\sum\limits_{n = 1}^\infty {{a_n}}$收敛并求其和.

$\sum\limits_{n = 1}^\infty {{a_n}} = 2\sum\limits_{n = 1}^\infty {{a_{2n - 1}}} - \sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^{n - 1}}{a_n}} = 2B - A.$

2.设$P(x)=a_0+a_1x+\cdots+a_mx^m$为$m$次多项式,求级数$\sum\limits_{n = 0}^\infty {\frac{{P\left( n \right)}}{{n!}}}$的和.

\begin{align*}{b_k} &= \sum\limits_{n = 0}^\infty {\frac{{{n^k}}}{{n!}}} = \sum\limits_{n = 1}^\infty {\frac{{{n^{k - 1}}}}{{\left( {n - 1} \right)!}}} = \sum\limits_{n = 0}^\infty {\frac{{{{\left( {n + 1} \right)}^{k - 1}}}}{{n!}}} \\&= {b_{k - 1}} + C_{k - 1}^1{b_{k - 2}} + \cdots + C_{k - 1}^{k - 2}{b_1} + {b_0},\end{align*}

$B\left( x \right) = \sum\limits_{n = 0}^\infty {\frac{{B\left( n \right)}}{{n!}}{x^n}} = {e^{{e^x} - 1}}.$

3.求$1 - \frac{{{2^3}}}{{1!}} + \frac{{{3^3}}}{{2!}} - \frac{{{4^3}}}{{3!}} + \cdots$的和.

\begin{align*}{b_k} &= \sum\limits_{n = 0}^\infty {{{\left( { - 1} \right)}^n}\frac{{{n^k}}}{{n!}}} = \sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^n}\frac{{{n^{k - 1}}}}{{\left( {n - 1} \right)!}}} = \sum\limits_{n = 0}^\infty {{{\left( { - 1} \right)}^{n+1}}\frac{{{{\left( {n + 1} \right)}^{k - 1}}}}{{n!}}} \\& =- {b_{k - 1}} - C_{k - 1}^1{b_{k - 2}} - \cdots - C_{k - 1}^{k - 2}{b_1} - {b_0},\end{align*}

\begin{align*}& 1 - \frac{{{2^3}}}{{1!}} + \frac{{{3^3}}}{{2!}} - \frac{{{4^3}}}{{3!}} + \cdots = \sum\limits_{n = 0}^\infty {{{\left( { - 1} \right)}^n}\frac{{{{\left( {n + 1} \right)}^3}}}{{n!}}} \\=& {b_3} + 3{b_2} + 3{b_1} + {b_0} = - \frac{1}{e}.\end{align*}

4.求下列级数的和:(1) $\sum\limits_{n = 1}^\infty {\arctan \frac{1}{{2{n^2}}}}$; (2) $\sum\limits_{n = 1}^\infty {\arctan \frac{2}{{{n^2}}}}$.

$\sum\limits_{n = 1}^\infty {\arctan \frac{1}{{2{n^2}}}} = \sum\limits_{n = 1}^\infty {\left( {\arctan \frac{1}{{2n - 1}} - \arctan \frac{1}{{2n + 1}}} \right)} = \frac{\pi }{4}.$

$\sum\limits_{n = 1}^\infty {\arctan \frac{2}{{{n^2}}}} = \sum\limits_{n = 1}^\infty {\left( {\arctan \frac{1}{{n - 1}} - \arctan \frac{1}{{n + 1}}} \right)} = \frac{\pi }{2} + \frac{\pi }{4} = \frac{{3\pi }}{4}.$

5.设$a>1$,求$\sum\limits_{n = 0}^\infty {\frac{{{2^n}}}{{{a^{{2^n}}} + 1}}}$的和.

\begin{align*}\sum\limits_{n = 0}^\infty {\frac{{{2^n}}}{{{a^{{2^n}}} + 1}}} &= \frac{1}{{a + 1}} + \sum\limits_{n = 1}^\infty {\frac{{{2^n}}}{{{a^{{2^n}}} + 1}}} = \frac{1}{{a + 1}} - \frac{1}{{a - 1}} + \frac{1}{{a + 1}} + \sum\limits_{n = 1}^\infty {\frac{{{2^n}}}{{{a^{{2^n}}} + 1}}} \\&= \frac{1}{{a + 1}} - \frac{2}{{{a^2} - 1}} + \sum\limits_{n = 1}^\infty {\frac{{{2^n}}}{{{a^{{2^n}}} + 1}}} = \frac{1}{{a + 1}} - \frac{{{2^2}}}{{{a^{{2^2}}} - 1}} + \sum\limits_{n = 2}^\infty {\frac{{{2^n}}}{{{a^{{2^n}}} + 1}}} \\&= \frac{1}{{a + 1}} - \mathop {\lim }\limits_{n \to \infty } \frac{{{2^{n + 1}}}}{{{a^{{2^{n + 1}}}} - 1}} = \frac{1}{{a + 1}}.\end{align*}

6.求$1 + \frac{1}{3} - \frac{1}{5} - \frac{1}{7} + \frac{1}{9} + \frac{1}{{11}} - \cdots$的和.

\begin{align*}&\sum\limits_{n = 1}^\infty {\left( {\frac{1}{{8n - 7}} + \frac{1}{{8n - 5}} - \frac{1}{{8n - 3}} - \frac{1}{{8n - 1}}} \right)} = \sum\limits_{n = 1}^\infty {\int_0^1 {\left( {{x^{8n - 8}} + {x^{8n - 6}} - {x^{8n - 4}} - {x^{8n - 2}}} \right)} } \\=& \int_0^1 {\sum\limits_{n = 1}^\infty {\left( {{x^{8n - 8}} + {x^{8n - 6}} - {x^{8n - 4}} - {x^{8n - 2}}} \right)} dx} = \int_0^1 {\frac{{1 + {x^2} - {x^4} - {x^6}}}{{1 - {x^8}}}dx} \\= &\left. {\frac{{\arctan \left( {1 + \sqrt 2 x} \right) - \arctan \left( {1 - \sqrt 2 x} \right)}}{{\sqrt 2 }}} \right|_0^1 = \frac{\pi }{{2\sqrt 2 }}.\end{align*}

7.求$1 - \frac{1}{7} + \frac{1}{9} - \frac{1}{{15}} + \cdots$的和.

\begin{align*}&\sum\limits_{n = 1}^\infty {\left( {\frac{1}{{8n - 7}} - \frac{1}{{8n - 1}}} \right)} = \sum\limits_{n = 1}^\infty {\int_0^1 {\left( {{x^{8n - 8}} - {x^{8n - 2}}} \right)} } \\=& \int_0^1 {\sum\limits_{n = 1}^\infty {\left( {{x^{8n - 8}} - {x^{8n - 2}}} \right)} dx} = \int_0^1 {\frac{{1 - {x^6}}}{{1 - {x^8}}}dx} \\= &\left. {\frac{{2\arctan x + \sqrt 2 \arctan \left( {1 + \sqrt 2 x} \right) - \arctan \left( {1 - \sqrt 2 x} \right)}}{4}} \right|_0^1 = \frac{{\sqrt 2 + 1}}{8}\pi .\end{align*}

8.求$1 - \frac{1}{4} + \frac{1}{7} - \frac{1}{{10}} + \cdots$的和.

\begin{align*}&\sum\limits_{n = 1}^\infty {\left( {\frac{1}{{6n - 5}} - \frac{1}{{6n - 2}}} \right)} = \sum\limits_{n = 1}^\infty {\int_0^1 {\left( {{x^{6n - 6}} - {x^{6n - 3}}} \right)} } \\= &\int_0^1 {\sum\limits_{n = 1}^\infty {\left( {{x^{6n - 6}} - {x^{6n - 3}}} \right)} dx} = \int_0^1 {\frac{{1 - {x^3}}}{{1 - {x^6}}}dx} = \int_0^1 {\frac{1}{{1 + {x^3}}}dx} \\=& \left. {\left( { - \frac{1}{6}\ln \left( {{x^2} - x + 1} \right) + \frac{1}{3}\ln \left( {x + 1} \right) + \frac{{\arctan \frac{{2x - 1}}{{\sqrt 3 }}}}{{\sqrt 3 }}} \right)} \right|_0^1 = \frac{{\sqrt 3 \pi + 3\ln 2}}{9}.\end{align*}

9.设${a_n} = 1 + \frac{1}{2} + \cdots + \frac{1}{n},n = 1,2, \cdots$,求$\sum\limits_{n = 1}^\infty {\frac{{{a_n}}}{{n\left( {n + 1} \right)}}}$的和.

\begin{align*}&\sum\limits_{n = 1}^\infty {\frac{{{a_n}}}{{n\left( {n + 1} \right)}}} = \sum\limits_{n = 1}^\infty {\frac{{1 + \frac{1}{2} + \cdots + \frac{1}{n}}}{{n\left( {n + 1} \right)}}} \\=&\sum\limits_{n = 1}^\infty {\left( {\frac{{1 + \frac{1}{2} + \cdots + \frac{1}{n}}}{n} - \frac{{1 + \frac{1}{2} + \cdots + \frac{1}{{n + 1}}}}{{n + 1}}} \right)} + \sum\limits_{n = 1}^\infty {\frac{1}{{{{\left( {n + 1} \right)}^2}}}} \\= & 1 - \mathop {\lim }\limits_{n \to \infty } \frac{{1 + \frac{1}{2} + \cdots + \frac{1}{{n + 1}}}}{{n + 1}} + \left( {\frac{{{\pi ^2}}}{6} - 1} \right) = \frac{{{\pi ^2}}}{6} - \mathop {\lim }\limits_{n \to \infty } \frac{{\frac{1}{{n + 2}}}}{1} = \frac{{{\pi ^2}}}{6}.\end{align*}

10.求$\sum\limits_{n = 0}^\infty {\left( {\frac{1}{{4n + 1}} + \frac{1}{{4n + 3}} - \frac{1}{{2n + 2}}} \right)}$的和.

\begin{align*}&\sum\limits_{n = 0}^\infty {\left( {\frac{1}{{4n + 1}} + \frac{1}{{4n + 3}} - \frac{1}{{2n + 2}}} \right)} = \sum\limits_{n = 0}^\infty {\int_0^1 {\left( {{x^{4n}} + {x^{4n + 2}} - {x^{2n + 1}}} \right)} } \\= &\int_0^1 {\sum\limits_{n = 0}^\infty {\left( {{x^{4n}} + {x^{4n + 2}} - {x^{2n + 1}}} \right)} dx} = \int_0^1 {\left( {\frac{{1 + {x^2}}}{{1 - {x^4}}} - \frac{x}{{1 - {x^2}}}} \right)dx} \\=& \int_0^1 {\frac{1}{{1 + x}}dx} = \ln 2.\end{align*}

11.求$1 - \frac{1}{4} + \frac{1}{6} - \frac{1}{9} + \frac{1}{{11}} - \frac{1}{{14}} + \cdots$的和.

\begin{align*}&\sum\limits_{n = 1}^\infty {\left( {\frac{1}{{5n - 4}} - \frac{1}{{5n - 1}}} \right)} = \sum\limits_{n = 1}^\infty {\int_0^1 {\left( {{x^{5n - 5}} - {x^{5n - 2}}} \right)dx} } = \int_0^1 {\sum\limits_{n = 1}^\infty {\left( {{x^{5n - 5}} - {x^{5n - 2}}} \right)} dx} \\=& \int_0^1 {\frac{{1 - {x^3}}}{{1 - {x^5}}}dx} = \int_0^1 {\left( {\frac{{\left( {5 - \sqrt 5 } \right)/10}}{{{x^2} + \frac{{\sqrt 5 + 1}}{2}x + 1}} + \frac{{\left( {5 + \sqrt 5 } \right)/10}}{{{x^2} + \frac{{ - \sqrt 5 + 1}}{2}x + 1}}} \right)dx} \\=& \left. {\left[ {\frac{{5 - \sqrt 5 }}{{10}}\sqrt {\frac{{10 + 2\sqrt 5 }}{5}} \arctan \frac{{4x + \sqrt 5 + 1}}{{\sqrt {10 - 2\sqrt 5 } }} + \frac{{5 + \sqrt 5 }}{5}\sqrt {\frac{2}{{5 + \sqrt 5 }}} \arctan \frac{{4x - \sqrt 5 + 1}}{{\sqrt {10 + 2\sqrt 5 } }}} \right]} \right|_0^1 \\=& \frac{{\sqrt {25 + 10\sqrt 5 } }}{{25}}\pi .\end{align*}

12.求$\frac{{{x^3}}}{{3!}} + \frac{{{x^9}}}{{9!}} + \frac{{{x^{15}}}}{{15!}} + \cdots$的和函数.

\begin{align*}&\frac{{\sinh x + \sinh \left( { - \frac{1}{2} + \frac{{\sqrt 3 i}}{2}} \right)x + \sinh \left( { - \frac{1}{2} - \frac{{\sqrt 3 i}}{2}} \right)x}}{3}\\= & - \frac{2}{3}\sinh \frac{x}{2}\cos \frac{{\sqrt 3 x}}{2} + \frac{{\sinh x}}{3} = \frac{{{x^3}}}{{3!}} + \frac{{{x^9}}}{{9!}} + \frac{{{x^{15}}}}{{15!}} + \cdots .\end{align*}

\begin{align*}&{\frac{{\sin x +\sin \left( { - \frac{1}{2} + \frac{{\sqrt 3 i}}{2}} \right)x + \sin \left( { - \frac{1}{2} - \frac{{\sqrt 3 i}}{2}} \right)x}}{{ - 3}}}\\= &\frac{2}{3}\sin \frac{x}{2}\cosh \frac{{\sqrt 3 x}}{2} - \frac{{\sin x}}{3} = \frac{{{x^3}}}{{3!}} - \frac{{{x^9}}}{{9!}} + \frac{{{x^{15}}}}{{15!}} - \frac{{{x^{21}}}}{{21!}} + \cdots .\end{align*}

13.求$\sum\limits_{n = 1}^\infty {\frac{{{{\left[ {\left( {n - 1} \right)!} \right]}^2}}}{{\left( {2n} \right)!}}{{\left( {2x} \right)}^{2n}}}$的和函数.

${\left( {\sqrt {1 - {x^2}} S'\left( x \right)} \right)^\prime } = \frac{4}{{\sqrt {1 - {x^2}} }},$故

$S\left( x \right) = \frac{{4\arcsin x}}{{\sqrt {1 - {x^2}} }} + \frac{1}{{\sqrt {1 - {x^2}} }},\quad \left| x \right| < 1.$

14.求$\sum\limits_{n = 1}^\infty {\frac{{{x^{n + 1}}}}{{\left( {1 - {x^n}} \right)\left( {1 - {x^{n + 1}}} \right)}}}$的和函数.

\begin{align*}&\left( {1 - \frac{1}{x}} \right)\sum\limits_{n = 1}^\infty {\frac{{{x^{n + 1}}}}{{\left( {1 - {x^n}} \right)\left( {1 - {x^{n + 1}}} \right)}}} \\=& \sum\limits_{n = 1}^\infty {\frac{{{x^{n + 1}}}}{{\left( {1 - {x^n}} \right)\left( {1 - {x^{n + 1}}} \right)}}} - \sum\limits_{n = 1}^\infty {\frac{{{x^n}}}{{\left( {1 - {x^n}} \right)\left( {1 - {x^{n + 1}}} \right)}}} \\= &\sum\limits_{n = 1}^\infty {\frac{{{x^{n + 1}} - {x^n}}}{{\left( {1 - {x^n}} \right)\left( {1 - {x^{n + 1}}} \right)}}} = \sum\limits_{n = 1}^\infty {\left( {\frac{1}{{1 - {x^{n + 1}}}} - \frac{1}{{1 - {x^n}}}} \right)} \\=& \mathop {\lim }\limits_{n \to \infty } \frac{1}{{1 - {x^{n + 1}}}} - \frac{1}{{1 - x}} = \begin{cases}\frac{1}{{x - 1}},&\left| x \right| > 1\\\frac{x}{{x - 1}},&\left| x \right| < 1\end{cases} .\end{align*}

$\sum\limits_{n = 1}^\infty {\frac{{{x^{n + 1}}}}{{\left( {1 - {x^n}} \right)\left( {1 - {x^{n + 1}}} \right)}}} = \begin{cases}\frac{x}{{{{\left( {x - 1} \right)}^2}}}, &\left| x \right| > 1\\\frac{{{x^2}}}{{{{\left( {x - 1} \right)}^2}}}, &\left| x \right| < 1\end{cases} .$

15.设$\sum\limits_{n = 1}^\infty {\frac{1}{{{a_n}}}}$为发散的正项级数, $x>0$,求$\sum\limits_{n = 1}^\infty {\frac{{{a_1}{a_2} \cdots {a_n}}}{{\left( {{a_2} + x} \right) \cdots \left( {{a_{n + 1}} + x} \right)}}}$的和函数.

\begin{align*}&\sum\limits_{n = 1}^\infty {\frac{{{a_1}{a_2} \cdots {a_n}}}{{\left( {{a_2} + x} \right) \cdots \left( {{a_{n + 1}} + x} \right)}}} \\=& \frac{{{a_1}}}{{{a_2} + x}} + \frac{1}{x}\sum\limits_{n = 2}^\infty {\left[ {\frac{{{a_1}{a_2} \cdots {a_n}}}{{\left( {{a_2} + x} \right) \cdots \left( {{a_n} + x} \right)}} - \frac{{{a_1}{a_2} \cdots {a_{n + 1}}}}{{\left( {{a_2} + x} \right) \cdots \left( {{a_{n + 1}} + x} \right)}}} \right]} \\=& \frac{{{a_1}}}{{{a_2} + x}} + \frac{1}{x}\left[ {\frac{{{a_1}{a_2}}}{{{a_2} + x}} - \mathop {\lim }\limits_{n \to \infty } \frac{{{a_1}{a_2} \cdots {a_{n + 1}}}}{{\left( {{a_2} + x} \right) \cdots \left( {{a_{n + 1}} + x} \right)}}} \right].\end{align*}

$\mathop {\lim }\limits_{n \to \infty } \frac{{{a_1}{a_2} \cdots {a_{n + 1}}}}{{\left( {{a_2} + x} \right) \cdots \left( {{a_{n + 1}} + x} \right)}} = \mathop {\lim }\limits_{n \to \infty } \frac{{{a_1}}}{{\left( {1 + \frac{x}{{{a_2}}}} \right) \cdots \left( {1 + \frac{x}{{{a_{n + 1}}}}} \right)}} = 0.$

$\sum\limits_{n = 1}^\infty {\frac{{{a_1}{a_2} \cdots {a_n}}}{{\left( {{a_2} + x} \right) \cdots \left( {{a_{n + 1}} + x} \right)}}} = \frac{{{a_1}}}{{{a_2} + x}} + \frac{{{a_1}{a_2}}}{{x\left( {{a_2} + x} \right)}} = \frac{{{a_1}}}{x}.$

16.设$x>1$,求$\frac{x}{{x + 1}} + \frac{{{x^2}}}{{\left( {x + 1} \right)\left( {{x^2} + 1} \right)}} + \frac{{{x^4}}}{{\left( {x + 1} \right)\left( {{x^2} + 1} \right)\left( {{x^4} + 1} \right)}} + \cdots$的和函数.

## 解题过程中碰到的几个特殊数列

1.Catalan数

$a_n=a_0a_{n-1}+a_1a_{n-2}+\cdots+a_{n-1}a_0,\quad n\geq1,$

\begin{align*}F\left( x \right) = &xf\left( x \right) = {a_0}x + {a_1}{x^2} +  \cdots + {a_n}{x^{n + 1}} +  \cdots \\{F^2}\left( x \right) = &a_0^2{x^2} + \left( {{a_0}{a_1} + {a_1}{a_0}} \right){x^3} +  \cdots \\&+ \left( {{a_0}{a_{n - 1}} + {a_1}{a_{n - 2}} +  \cdots  + {a_{n - 1}}{a_0}} \right){x^{n + 1}} +  \cdots \\= &{a_1}{x^2} + {a_2}{x^3} +  \cdots  + {a_n}{x^{n + 1}} +  \cdots \\= & F\left( x \right) - {a_0}x,\end{align*}

$\sqrt {1 - 4x} = 1 + C_{1/2}^1\left( { - 4x} \right) + C_{1/2}^2{\left( { - 4x} \right)^2} + \cdots + C_{1/2}^{n + 1}{\left( { - 4x} \right)^{n + 1}} + \cdots ,$

\begin{align*}{a_n} &=  - \frac{1}{2}C_{1/2}^{n + 1}{\left( { - 4} \right)^{n + 1}}\\&=  - \frac{1}{2}\frac{{\frac{1}{2} \cdot \left( { - \frac{1}{2}} \right) \cdot \left( { - \frac{3}{2}} \right) \cdots \left( { - \frac{{2n - 1}}{2}} \right)}}{{n!}}{\left( { - 1} \right)^{n + 1}}{4^{n + 1}} = \frac{{C_{2n}^n}}{{n + 1}}.\end{align*}

2.Bell数

\begin{align*}{b_k} &= \sum\limits_{n = 0}^\infty  {\frac{{{n^k}}}{{n!}}}  = \sum\limits_{n = 1}^\infty  {\frac{{{n^{k - 1}}}}{{\left( {n - 1} \right)!}}}  = \sum\limits_{n = 0}^\infty  {\frac{{{{\left( {n + 1} \right)}^{k - 1}}}}{{n!}}} \\&= {b_{k - 1}} + C_{k - 1}^1{b_{k - 2}} +  \cdots  + C_{k - 1}^{k - 2}{b_1} + {b_0},\end{align*}

$B\left( x \right) = \sum\limits_{n = 0}^\infty {\frac{{B\left( n \right)}}{{n!}}{x^n}} = {e^{{e^x} - 1}}.$

$$\mathrm e^{\mathrm e^x} = \sum_{k=0}^\infty\frac{\mathrm e^{kx}}{k!} = \sum_{k=0}^\infty\frac 1{k!}\biggl(\sum_{n=0}^\infty\frac{(kx)^n}{n!}\biggr) = \sum_{n=0}^\infty\frac{x^n}{n!}\biggl(\sum_{k=0}^\infty\frac{k^n}{k!}\biggr),$$

$$a_n=\frac 1{n!}\sum_{k=0}^\infty\frac{k^n}{k!} > \frac{k^n}{n!\,k!},$$

$$\frac {k^n}{n!\,k!}=\frac{\displaystyle\Bigl(\frac n{\ln n}\Bigr)^n}{n!\,\displaystyle\Bigl(\frac n{\ln n}\Bigr)!} \sim\frac{\displaystyle\Bigl(\frac n{\ln n}\Bigr)^n}{\displaystyle\sqrt{2\pi n}\Bigl(\frac n{\mathrm e}\Bigr)^n\cdot \sqrt{\frac{2\pi n}{\ln n}}\Bigl(\frac n{\mathrm e\ln n}\Bigr)^{\frac n{\ln n}}} =\frac{(\mathrm e\ln n)^{\frac n{\ln n}}\sqrt{\ln n}}{2\pi n(\ln n)^n}.$$

## 一个求特征值的问题

$A = \left( {\begin{array}{*{20}{c}}{a_{11}^2}&{{a_{11}}{a_{12}} + 1}&{{a_{11}}{a_{13}} + 1}& \cdots &{{a_{11}}{a_{1n}} + 1}\\{{a_{11}}{a_{12}} + 1}&{a_{22}^2}&{{a_{22}}{a_{23}} + 1}& \cdots &{{a_{22}}{a_{2n}} + 1}\\{{a_{11}}{a_{13}} + 1}&{{a_{22}}{a_{23}} + 1}& \ddots & \ddots & \vdots \\\vdots & \vdots & \ddots &{a_{n - 1,n - 1}^2}&{{a_{n - 1,n - 1}}{a_{n - 1,n}} + 1}\\{{a_{11}}{a_{1n}} + 1}&{{a_{22}}{a_{2n}} + 1}& \cdots &{{a_{n - 1,n - 1}}{a_{n - 1,n}} + 1}&{a_{nn}^2}\end{array}} \right)$的特征值.

## Euler-Maclaurin求和公式估计梯形积分公式的误差

$\mathop {\lim }\limits_{n \to \infty } {n^4}\left( {\frac{1}{{24}} - n\left( {n\left( {\frac{\pi }{4} - {A_n}} \right) - \frac{1}{4}} \right)} \right).$

Euler-Maclaurin求和公式

\begin{align*}\frac{{b - a}}{n}\sum\limits_{i = 1}^n {\frac{1}{2}\left[ {f\left( {{x_{i - 1}}} \right) + f\left( {{x_i}} \right)} \right]} - \int_a^b {f\left( x \right)dx} = &\sum\limits_{k = 1}^m {\frac{{{B_{2k}}}}{{\left( {2k} \right)!}}{h^{2k}}\left[ {{f^{\left( {2k - 1} \right)}}\left( b \right) - {f^{\left( {2k - 1} \right)}}\left( a \right)} \right]} \\&+ \frac{{{B_{2m + 2}}}}{{\left( {2m + 2} \right)!}}{h^{2m + 2}}{f^{\left( {2m + 2} \right)}}\left( \xi \right)\left( {b - a} \right),\end{align*}

\begin{align*}&{A_n} + \frac{1}{{4n}} - \frac{\pi }{4} = \frac{1}{2}\left[ {\left( {{A_n} - \frac{1}{{2n}} + \frac{1}{n}} \right) + {A_n}} \right] - \frac{\pi }{4} = \frac{{{B_2}}}{{2!}} \cdot \frac{1}{{{n^2}}}\left[ {f'\left( 1 \right) - f'\left( 0 \right)} \right]\\+ &\frac{{{B_4}}}{{4!}} \cdot \frac{1}{{{n^4}}}\left[ {f'''\left( 1 \right) - f'''\left( 0 \right)} \right] + \frac{{{B_6}}}{{6!}} \cdot \frac{1}{{{n^6}}}\left[ {{f^{\left( 5 \right)}}\left( 1 \right) - {f^{\left( 5 \right)}}\left( 0 \right)} \right] + \frac{{{B_8}}}{{8!}} \cdot \frac{1}{{{n^8}}}{f^{\left( 8 \right)}}\left( \xi \right),\end{align*}

${n^4}\left( {\frac{1}{{24}} - n\left( {n\left( {\frac{\pi }{4} - {A_n}} \right) - \frac{1}{4}} \right)} \right) = \frac{1}{{2016}} + \frac{{{B_8}}}{{8!}} \cdot \frac{1}{{{n^2}}}{f^{\left( 8 \right)}}\left( \xi \right),$

## 2014年浙江大学数学分析考研试题解答

6.设空间体积为$V$的任意$\Omega,X_0\in \Omega ,0<\alpha<3$.证明

$\int_\Omega {{{\left| {X - {X_0}} \right|}^{\alpha - 3}}dX} \le C{V^{\alpha /3}}, \text{其中C只与\alpha有关}.$

\begin{align*}\int_D {{{\left| {X - {X_0}} \right|}^{\alpha - 3}}dX} &= \int_{{D_1}} {{{\left| {X - {X_0}} \right|}^{\alpha - 3}}dX} + \int_{{D_2}} {{{\left| {X - {X_0}} \right|}^{\alpha - 3}}dX} \\\int_\Omega {{{\left| {X - {X_0}} \right|}^{\alpha - 3}}dX} &= \int_{{D_1}} {{{\left| {X - {X_0}} \right|}^{\alpha - 3}}dX} + \int_{{\Omega _2}} {{{\left| {X - {X_0}} \right|}^{\alpha - 3}}dX} .\end{align*}

\begin{align*}\int_{{D_2}} {{{\left| {X - {X_0}} \right|}^{\alpha - 3}}dX} &= {\left| {\xi - {X_0}} \right|^{\alpha - 3}}{V_{{D_2}}},\xi \in {D_2}\\\int_{{\Omega _2}} {{{\left| {X - {X_0}} \right|}^{\alpha - 3}}dX} &= {\left| {\eta - {X_0}} \right|^{\alpha - 3}}{V_{{\Omega_2}}},\eta \in {\Omega _2}.\end{align*}

7.$f(x)$在$[0,1]$单增,证明:

$\mathop {\lim }\limits_{y \to + \infty } \int_0^1 {f\left( x \right)\frac{{\sin xy}}{x}dx} = \frac{\pi }{2}f\left( {{0_ + }} \right).$

$0 \le g\left( t \right) - g\left( {{0_ + }} \right) < M_1\varepsilon ,$

\begin{align*}\int_0^1 {\left[ {f\left( x \right) - f\left( {{0_ + }} \right)} \right]\frac{{\sin xy}}{x}dx} &= \left( {\int_0^\delta {} + \int_\delta ^1 {} } \right)\left[ {f\left( x \right) - f\left( {{0_ + }} \right)} \right]\frac{{\sin xy}}{x}dx\\&= {I_1} + {I_2}.\end{align*}

${I_1} = \left[ {f\left( \delta \right) - f\left( {{0_ + }} \right)} \right]\int_\eta ^\delta {\frac{{\sin xy}}{x}dx} = \left[ {f\left( \delta \right) - f\left( {{0_ + }} \right)} \right]\int_{y\eta }^{y\delta } {\frac{{\sin z}}{z}dz} ,$

$\left| {\int_0^z {\frac{{\sin z}}{z}dz} } \right| \le L\left( L \text{为常数}\right),$从而

$\left| {\int_{y\eta }^{y\delta } {\frac{{\sin z}}{z}dz} } \right| = \left| {\int_0^{y\delta } {} + \int_0^{y\eta } {} } \right| \le 2L.$

\begin{align*}&\mathop {\lim }\limits_{y \to + \infty } \int_0^1 {f\left( x \right)\frac{{\sin xy}}{x}dx} = \frac{\pi }{2}f\left( {{0_ + }} \right)\\=& \mathop {\lim }\limits_{y \to + \infty } \int_0^1 {\left[ {f\left( x \right) - f\left( {{0_ + }} \right)} \right]\frac{{\sin xy}}{x}dx} + f\left( {{0_ + }} \right)\mathop {\lim }\limits_{y \to + \infty } \int_0^1 {\frac{{\sin xy}}{x}dx} \\= &0 + f\left( {{0_ + }} \right)\int_0^{ + \infty } {\frac{{\sin z}}{z}dz} = \frac{\pi }{2}f\left( {{0_ + }} \right).\end{align*}