下列定义的映射中, ___ 不是内积.
A: \(\langle x,y \rangle \triangleq xy ,x,y \in \mathbb{R}\)
B: \(\langle (x_1,\cdots,x_n),(y_1,\cdots,y_n) \rangle \triangleq \Sigma_{i=1}^{n}x_iy_i,(x_1,\cdots,x_n),(y_1,\cdots,y_n)\in \mathbb{R}^n\)
C: \(\langle f,g \rangle \triangleq \int_a^b f(x)g(x)\mathrm{d}x ,f,g \in C([a,b])\)(\([a,b]\)上连续实函数全体)
D: \(\langle (x_1,\cdots,x_n),(y_1,\cdots,y_n) \rangle \triangleq \Sigma_{i,j=1}^{n}a_{ij}x_iy_i,(x_1,\cdots,x_n),(y_1,\cdots,y_n)\in \mathbb{R}^n,A = (a_{ij})是实对称方阵\)
A: \(\langle x,y \rangle \triangleq xy ,x,y \in \mathbb{R}\)
B: \(\langle (x_1,\cdots,x_n),(y_1,\cdots,y_n) \rangle \triangleq \Sigma_{i=1}^{n}x_iy_i,(x_1,\cdots,x_n),(y_1,\cdots,y_n)\in \mathbb{R}^n\)
C: \(\langle f,g \rangle \triangleq \int_a^b f(x)g(x)\mathrm{d}x ,f,g \in C([a,b])\)(\([a,b]\)上连续实函数全体)
D: \(\langle (x_1,\cdots,x_n),(y_1,\cdots,y_n) \rangle \triangleq \Sigma_{i,j=1}^{n}a_{ij}x_iy_i,(x_1,\cdots,x_n),(y_1,\cdots,y_n)\in \mathbb{R}^n,A = (a_{ij})是实对称方阵\)
举一反三
- 求方程\(x = \cos x\)根的牛顿迭代公式是 。 A: \({x_{n + 1}} = {x_n} - { { {x_n} - \cos {x_n}} \over {1 + \sin {x_n}}},n = 0,1,2 \cdots \) B: \({x_{n + 1}} = {x_n} + { { {x_n} - \cos {x_n}} \over {1 + \sin {x_n}}},n = 0,1,2 \cdots \) C: \({x_{n + 1}} = {x_n} - { { {x_n} - \sin {x_n}} \over {1 + \sin {x_n}}},n = 0,1,2 \cdots \) D: \({x_{n + 1}} = {x_n} - { { {x_n} - \cos {x_n}} \over {1 + \cos{x_n}}},n = 0,1,2 \cdots \)
- \( {1 \over {1 + x}} \)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: \( {1 \over {1 + x}} = 1 - x + {x^2} - \cdots + {( - 1)^n}{x^n} + o\left( { { x^n}} \right) \)
- (1). 设总体 \( X \) 具有有限的数学期望 \( EX \) 和方差 \( DX \),\( X_1 ,X_2 ,\mbox{ }\cdots ,X_n \) 为总体 \( X \) 的样本,那么对样本均值 \( \bar {X}=\frac{1}{n}\sum\limits_{i=1}^n{X_i } \) 有()。
- 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); }