A: [x^2 | x - [1..n]]
B: sum[x^2 | x - [1..n]]
C: sum [x | x^2 - [1..n]]
D: sum [x | x - [1..n^2]]
E: sum [y | y- [1..n], y = x^2]
F: sum [ x*y | (x,y) - zip [1..n] [1..n]]
G: sum [ x*x | (x,x) - zip [1..n] [1..n]]
举一反三
- 对于给定非负整数[img=11x14]17de86c7a4e3414.png[/img]，计算[img=132x24]17de86c7b1aebe2.png[/img]的方法是 A: [x^2 | x - [1..n]] B: sum[x^2 | x - [1..n]] C: sum [x | x^2 - [1..n]] D: sum [x | x - [1..n^2]] E: sum [y | y- [1..n], y = x^2] F: sum [ x*y | (x,y) - zip [1..n] [1..n]] G: sum [ x*x | (x,x) - zip [1..n] [1..n]]
- 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: \({1 \over {1 + {x^2}}} = \sum\limits_{n = 0}^\infty { { x^{2n}}} \matrix{ {} & {} \cr } ( - 1 < x < 1)\)
- 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: \({e^x} = \sum\limits_{n = 0}^\infty { { {( - 1)}^n} { { {x^n}} \over {n!}}} ( - 1 < x < 1)\)
- 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().
内容
- 0
下面两条if语句合并成一条if语句为( )。 if(a<=b) x=1; else y=2; if(a>b) printf("**y=%d\n",y); else printf("##x=%d\n",x); A: if(a<=b){ x=1; printf(" B: C: x=%d\n",x); } else{ y=2; printf("**y=%d\n",y); } D: if(a<=b) x=1; printf(" E: F: x=%d\n",x); else y=2; printf("**y=%d\n",y); G: if(a<=b){ x=1; printf("**y=%d\n",y); } else{ y=2; printf(" H: I: x=%d\n",x); } J: if(a>b){ x=1; printf(" K: L: x=%d\n",x); } else{ y=2; printf("**y=%d\n",y); }
- 1
设随机变量X和Y相互独立且X～N(0，1)，Y～N(1，1)，则（ ）． A: P{X + Y £ 0} = 1/2 B: P{X + Y £ 1} = 1/2 C: P{X - Y £ 0} = 1/2 D: P{X - Y £ 1} = 1/2
- 2
判断差分系统的因果性（1）y（n）=x（n+1）-x（n）（2）y（n）=x（n）-x（n-1）
- 3
\( {1 \over {1 + x}} \)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: \( {1 \over {1 + x}} = 1 - x - { { {x^2}} \over 2}- \cdots - { { {x^n}} \over {n!}} + o\left( { { x^n}} \right) \)
- 4
设 X ~ N(3, 12)，Y ~ N(2, 4)，且 X，Y 独立，则 X − Y ~ N(1, 8) .