A: -1≤x≤1-2-n
B: -1＜x≤1-2-n
C: -1≤x＜1-2-n
D: -1＜X＜1-2-n
举一反三
- \( {1 \over {1 + x}} \)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: \( {1 \over {1 + x}} = 1 - x - { { {x^2}} \over 2}- \cdots - { { {x^n}} \over {n!}} + o\left( { { x^n}} \right) \)
- 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().
- \( {1 \over {1 + x}} \)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: \( {1 \over {1 + x}} = 1 - x + {x^2} - \cdots + {( - 1)^n}{x^n} + o\left( { { x^n}} \right) \)
- 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\(f(x) = {1 \over {1 + {x^2}}}\)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: \({1 \over {1 + {x^2}}} = \sum\limits_{n = 0}^\infty { { x^{2n}}} \matrix{ {} & {} \cr } ( - 1 < x < 1)\)
内容
- 0
负整数的补码可通过_____mod 2^(n+1)得到。 A: 2^n - x B: 2^(n+1) + x C: 2^n + x D: 2^(n+1) -1 + x
- 1
负整数的补码可用_____mod (2^(n+1))求得 A: 2^n – x B: 2^n + x C: 2^(n+1) + x D: 2^(n+1) -1 + x
- 2
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: \(\frac{2\pi^2}{3}+4\Sigma_{n=1}^{\infty} \frac{(-1)^n}{n^2} \cos nx ,\; x \in [-\pi,\pi]\)
- 3
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) $
- 4
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}