$ {1 \over {1 + x}} $的麦克劳林公式为（）。 A: $ {1 \over {1 + x}} = 1 + x + { { {x^2}} \over 2} + \cdots + { { {x^n}} \over {n!}} + o\left( { { x^n}} \right) $ B: $ {1 \over {1 + x}} = 1 + x + {x^2} + \cdots + {x^n} + o\left( { { x^n}} \right) $ C: $ {1 \over {1 + x}} = 1 - x + {x^2} - \cdots + {( - 1)^n}{x^n} + o\left( { { x^n}} \right) $ D: $ {1 \over {1 + x}} = 1 - x - { { {x^2}} \over 2}- \cdots - { { {x^n}} \over {n!}} + o\left( { { x^n}} \right) $

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2022-06-12 问题

$ {1 \over {1 + x}} $的麦克劳林公式为（）。 A: $ {1 \over {1 + x}} = 1 + x + { { {x^2}} \over 2} + \cdots + { { {x^n}} \over {n!}} + o\left( { { x^n}} \right) $ B: $ {1 \over {1 + x}} = 1 + x + {x^2} + \cdots + {x^n} + o\left( { { x^n}} \right) $ C: $ {1 \over {1 + x}} = 1 - x + {x^2} - \cdots + {( - 1)^n}{x^n} + o\left( { { x^n}} \right) $ D: $ {1 \over {1 + x}} = 1 - x - { { {x^2}} \over 2}- \cdots - { { {x^n}} \over {n!}} + o\left( { { x^n}} \right) $

2022-06-12 问题

$ {1 \over {1 + x}} $的麦克劳林公式为（）. A: $ {1 \over {1 + x}} = 1 + x + { { {x^2}} \over 2} + \cdots + { { {x^n}} \over {n!}} + o\left( { { x^n}} \right) $ B: $ {1 \over {1 + x}} = 1 + x + {x^2} + \cdots + {x^n} + o\left( { { x^n}} \right) $ C: $ {1 \over {1 + x}} = 1 - x + {x^2} - \cdots + {( - 1)^n}{x^n} + o\left( { { x^n}} \right) $

$ {1 \over {1 + x}} $的麦克劳林公式为（）. A: $ {1 \over {1 + x}} = 1 + x + { { {x^2}} \over 2} + \cdots + { { {x^n}} \over {n!}} + o\left( { { x^n}} \right) $ B: $ {1 \over {1 + x}} = 1 + x + {x^2} + \cdots + {x^n} + o\left( { { x^n}} \right) $ C: $ {1 \over {1 + x}} = 1 - x + {x^2} - \cdots + {( - 1)^n}{x^n} + o\left( { { x^n}} \right) $

2022-06-06 问题

将$f(x) = {1 \over {2 - x}}$展开成$x $的幂级数为（）。 A: ${1 \over {2 - x}} = \sum\limits_{n = 0}^\infty { { { { x^n}} \over { { 2^{n }}}}} $，$( - 2,2)$ B: ${1 \over {2 - x}} = \sum\limits_{n = 0}^\infty { { { { x^n}} \over { { 2^{n }}}}} $，$\left( { - 2,2} \right]$ C: ${1 \over {2 - x}} = \sum\limits_{n = 0}^\infty { { { { x^n}} \over { { 2^{n + 1}}}}} $，$( - 2,2)$ D: ${1 \over {2 - x}} = \sum\limits_{n = 0}^\infty { { { { x^n}} \over { { 2^{n + 1}}}}} $，$\left( { - 2,2} \right]$

将$f(x) = {1 \over {2 - x}}$展开成$x $的幂级数为（）。 A: ${1 \over {2 - x}} = \sum\limits_{n = 0}^\infty { { { { x^n}} \over { { 2^{n }}}}} $，$( - 2,2)$ B: ${1 \over {2 - x}} = \sum\limits_{n = 0}^\infty { { { { x^n}} \over { { 2^{n }}}}} $，$\left( { - 2,2} \right]$ C: ${1 \over {2 - x}} = \sum\limits_{n = 0}^\infty { { { { x^n}} \over { { 2^{n + 1}}}}} $，$( - 2,2)$ D: ${1 \over {2 - x}} = \sum\limits_{n = 0}^\infty { { { { x^n}} \over { { 2^{n + 1}}}}} $，$\left( { - 2,2} \right]$

2021-04-14 问题

已知()y()=()ln()x(),则()y()(()n())()=()。A.()(()−()1())()n()n()!()x()−()n()"()role="presentation">()(()−()1())()n()n()!()x()−()n();()B.()(()−()1())()n()(()n()−()1())()!()x()−()2()n()"()role="presentation">()(()−()1())()n()(()n()−()1())()!()x()−()2()n();()C.()(()−()1())()n()−()1()(()n()−()1())()!()x()n()"()role="presentation">()(()−()1())()n()−()1()(()n()−()1())()!()x()-n();()D.()(()−()1())()n()−()1()n()!()x()−()n()+()1()"()role="presentation">()(()−()1())()n()−()1()n()!()x()−()n()+()1().

已知()y()=()ln()x(),则()y()(()n())()=()。A.()(()−()1())()n()n()!()x()−()n()"()role="presentation">()(()−()1())()n()n()!()x()−()n();()B.()(()−()1())()n()(()n()−()1())()!()x()−()2()n()"()role="presentation">()(()−()1())()n()(()n()−()1())()!()x()−()2()n();()C.()(()−()1())()n()−()1()(()n()−()1())()!()x()n()"()role="presentation">()(()−()1())()n()−()1()(()n()−()1())()!()x()-n();()D.()(()−()1())()n()−()1()n()!()x()−()n()+()1()"()role="presentation">()(()−()1())()n()−()1()n()!()x()−()n()+()1().

2022-06-17 问题

函数$f(x) = x^2,\; x \in [-\pi,\pi]$的Fourier级数为 A: $\frac{\pi^2}{3}+4\Sigma_{n=1}^{\infty} \frac{(-1)^n}{n^2} \sin nx ,\; x \in [-\pi,\pi]$ B: $\frac{\pi^2}{3}+4\Sigma_{n=1}^{\infty} \frac{(-1)^n}{n^2} \cos nx ,\; x \in [-\pi,\pi]$ C: $\frac{2\pi^2}{3}+4\Sigma_{n=1}^{\infty} \frac{(-1)^n}{n^2} \sin nx ,\; x \in [-\pi,\pi]$ D: $\frac{2\pi^2}{3}+4\Sigma_{n=1}^{\infty} \frac{(-1)^n}{n^2} \cos nx ,\; x \in [-\pi,\pi]$

函数$f(x) = x^2,\; x \in [-\pi,\pi]$的Fourier级数为 A: $\frac{\pi^2}{3}+4\Sigma_{n=1}^{\infty} \frac{(-1)^n}{n^2} \sin nx ,\; x \in [-\pi,\pi]$ B: $\frac{\pi^2}{3}+4\Sigma_{n=1}^{\infty} \frac{(-1)^n}{n^2} \cos nx ,\; x \in [-\pi,\pi]$ C: $\frac{2\pi^2}{3}+4\Sigma_{n=1}^{\infty} \frac{(-1)^n}{n^2} \sin nx ,\; x \in [-\pi,\pi]$ D: $\frac{2\pi^2}{3}+4\Sigma_{n=1}^{\infty} \frac{(-1)^n}{n^2} \cos nx ,\; x \in [-\pi,\pi]$

2022-06-06 问题

将$f(x) = {1 \over {1 + {x^2}}}$展开成$x$的幂级数为( )。 A: ${1 \over {1 + {x^2}}} = \sum\limits_{n = 0}^\infty { { {( - 1)}^n}{x^{2n}}} \matrix{ {} & {} \cr } ( - \infty < x < + \infty )$ B: ${1 \over {1 + {x^2}}} = \sum\limits_{n = 0}^\infty { { {( - 1)}^n}{x^{2n}}} \matrix{ {} & {} \cr } ( - 1< x < 1)$ C: ${1 \over {1 + {x^2}}} = \sum\limits_{n = 0}^\infty { { {( - 1)}^n}{x^{2n}}} \matrix{ {} & {} \cr } ( - 1 < x < 1)$ D: ${1 \over {1 + {x^2}}} = \sum\limits_{n = 0}^\infty { { x^{2n}}} \matrix{ {} & {} \cr } ( - 1 < x < 1)$

将$f(x) = {1 \over {1 + {x^2}}}$展开成$x$的幂级数为( )。 A: ${1 \over {1 + {x^2}}} = \sum\limits_{n = 0}^\infty { { {( - 1)}^n}{x^{2n}}} \matrix{ {} & {} \cr } ( - \infty < x < + \infty )$ B: ${1 \over {1 + {x^2}}} = \sum\limits_{n = 0}^\infty { { {( - 1)}^n}{x^{2n}}} \matrix{ {} & {} \cr } ( - 1< x < 1)$ C: ${1 \over {1 + {x^2}}} = \sum\limits_{n = 0}^\infty { { {( - 1)}^n}{x^{2n}}} \matrix{ {} & {} \cr } ( - 1 < x < 1)$ D: ${1 \over {1 + {x^2}}} = \sum\limits_{n = 0}^\infty { { x^{2n}}} \matrix{ {} & {} \cr } ( - 1 < x < 1)$

2022-06-01 问题

M＝｛x丨－2＜x＜－1｝,N＝｛x丨x＞1｝,M∪N等于多少?

2022-06-01 问题

设置集合M={x丨-1≤x＜2}。N={x丨x≤1}，则集合M∩N=（） A: {x丨x > -1} B: {x丨x > 1} C: {x丨-1 ≤ x ≤ 1} D: {x丨1 ≤ x ≤ 2}

2021-04-14 问题

设X ~ N(2, 9)则Y = (X – 2 )/9 ~ N(0, 1).

2022-06-06 问题

下面级数求和错误的是 A: $\sum_{n=0}^\infty q^n = \frac{1}{1-q} (0\lt q\lt1) $ B: $\sum_{n=1}^\infty \frac{x^{2^{n-1}}}{1-x^{2^n}} = \frac{x}{1-x} (|x|\lt 1) $ C: $\sum_{n=1}^\infty \frac{1}{{n!}} = e $ D: $\sum_{n=1}^\infty \frac{x^{2^{n-1}}}{1-x^{2^n}} = \frac{1}{1-x} (x>1) $

下面级数求和错误的是 A: $\sum_{n=0}^\infty q^n = \frac{1}{1-q} (0\lt q\lt1) $ B: $\sum_{n=1}^\infty \frac{x^{2^{n-1}}}{1-x^{2^n}} = \frac{x}{1-x} (|x|\lt 1) $ C: $\sum_{n=1}^\infty \frac{1}{{n!}} = e $ D: $\sum_{n=1}^\infty \frac{x^{2^{n-1}}}{1-x^{2^n}} = \frac{1}{1-x} (x>1) $

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M＝｛x丨－2＜x＜－1｝,N＝｛x丨x＞1｝,M∪N等于多少?

设置集合M={x丨-1≤x＜2}。N={x丨x≤1}，则集合M∩N=（） A: {x丨x > -1} B: {x丨x > 1} C: {x丨-1 ≤ x ≤ 1} D: {x丨1 ≤ x ≤ 2}

设X ~ N(2, 9)则Y = (X – 2 )/9 ~ N(0, 1).

下面级数求和错误的是 A: $\sum_{n=0}^\infty q^n = \frac{1}{1-q} (0\lt q\lt1) $ B: $\sum_{n=1}^\infty \frac{x^{2^{n-1}}}{1-x^{2^n}} = \frac{x}{1-x} (|x|\lt 1) $ C: $\sum_{n=1}^\infty \frac{1}{{n!}} = e $ D: $\sum_{n=1}^\infty \frac{x^{2^{n-1}}}{1-x^{2^n}} = \frac{1}{1-x} (x&gt;1) $

下面级数求和错误的是 A: $\sum_{n=0}^\infty q^n = \frac{1}{1-q} (0\lt q\lt1) $ B: $\sum_{n=1}^\infty \frac{x^{2^{n-1}}}{1-x^{2^n}} = \frac{x}{1-x} (|x|\lt 1) $ C: $\sum_{n=1}^\infty \frac{1}{{n!}} = e $ D: $\sum_{n=1}^\infty \frac{x^{2^{n-1}}}{1-x^{2^n}} = \frac{1}{1-x} (x>1) $