用MATLAB计算两个有限长时域序列线性卷积的命令是（ ） A: yn=fft(x1n,N) B: yn=fft(n,m) C: yn=conv(x1n,x2n) D: yn=fft(x1n,x2n)

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: \( {1 \over {1 + x}} = 1 - x + {x^2} - \cdots + {( - 1)^n}{x^n} + o\left( { { x^n}} \right) \) D: \( {1 \over {1 + x}} = 1 - x - { { {x^2}} \over 2}- \cdots - { { {x^n}} \over {n!}} + o\left( { { x^n}} \right) \)

\( {1 \over {1 + x}} \)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: \( {1 \over {1 + x}} = 1 - x + {x^2} - \cdots + {( - 1)^n}{x^n} + o\left( { { x^n}} \right) \)

将函数\(f(x) = {e^x}\)展开成\(x\)的幂级数为（ ）。 A: \({e^x} = \sum\limits_{n = 0}^\infty { { { { x^n}} \over {n!}}} ( - \infty < x < + \infty )\) B: \({e^x} = \sum\limits_{n = 0}^\infty { { {( - 1)}^n} { { {x^n}} \over {n!}}} ( - \infty < x < + \infty )\) C: \({e^x} = \sum\limits_{n = 0}^\infty { { { { x^n}} \over {n!}}} ( - 1 < x < 1)\) D: \({e^x} = \sum\limits_{n = 0}^\infty { { {( - 1)}^n} { { {x^n}} \over {n!}}} ( - 1 < x < 1)\)

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: \({1 \over {1 + {x^2}}} = \sum\limits_{n = 0}^\infty { { x^{2n}}} \matrix{ {} & {} \cr } ( - 1 < x < 1)\)

M＝｛x丨－2＜x＜－1｝,N＝｛x丨x＞1｝,M∪N等于多少?

【单选题】设集合M={x | -1≤x≤3},N={x |1≤x≤6},则M∪N=( ) A. {x |-1≤x≤3} B. {x |-1≤x≤1} C. {x |-1≤x≤6} D. {x |-1≤x≤0}

函数$y = \ln x$，则${\left( {\ln x} \right)^{\left( n \right)}} = {\left( { - 1} \right)^{n - 1}}{{\left( {n - 1} \right)!} \over {{x^n}}}$。（ ）