数学分析拔尖随机推荐周民强教材习题第 5 章 幂级数、Taylor 级数5.4 多项式逼近连续函数级数不看本题本题讨论已掌握查看解析收入错题#18846试证明下列命题:(1) 设 f∈C(1)([0,1])f \in {C}^{\left( 1\right) }\left( \left[ {0,1}\right] \right)f∈C(1)([0,1]) ,则 Bn′(f,x){B}_{n}^{\prime }\left( {f,x}\right)Bn′(f,x) 在 [0,1]\left[ {0,1}\right][0,1] 上一致收敛于 f′(x)f'\left( x\right)f′(x) .(2) 设 f∈C((0,1))f \in C\left( \left( {0,1}\right) \right)f∈C((0,1)) (点 x=0x = 0x=0 或1为 fff 之瑕点),且瑕积分 ∫01f(x)dx{\int }_{0}^{1}f\left( x\right) \mathrm{d}x∫01f(x)dx 收敛. 若 ∫01f(x)xn dx= 0(n=0,1,2,⋯){\int }_{0}^{1}f\left( x\right) {x}^{n}\mathop{}\!\mathrm{d}x = \; 0\left( {n = 0,1,2,\cdots }\right)∫01f(x)xndx=0(n=0,1,2,⋯) ,则 f(x)≡0f\left( x\right) \equiv 0f(x)≡0 .(3) I=∫0+∞xn⋅sin(x1/4)e−x1/4 dx=0(n=0,1,2,⋯ )I = {\int }_{0}^{+\infty }{x}^{n} \cdot \sin \left( {x}^{1/4}\right) {\mathrm{e}}^{-{x}^{1/4}}\mathop{}\!\mathrm{d}x = 0\quad(n=0,1,2,\cdots)I=∫0+∞xn⋅sin(x1/4)e−x1/4dx=0(n=0,1,2,⋯) .(4) 设 f(x)=∑n=0∞anxnf\left( x\right) = \mathop{\sum }\limits_{n = 0}^{\infty }{a}_{n}{x}^{n}f(x)=n=0∑∞anxn 的收敛半径为 1,an>0(n∈N)1,{a}_{n} > 0\left( {n \in \mathbf{N}}\right)1,an>0(n∈N) 且 limx→1−(1−x)f(x)=1\mathop{\lim }\limits_{{x \rightarrow {1}^{ - }}}\left( {1 - x}\right) f\left( x\right) = 1x→1−lim(1−x)f(x)=1 ,则 limn→∞(a1+ a2+⋯+an)/n=1\mathop{\lim }\limits_{n \rightarrow \infty }\left( {{a}_{1} + }\right. \; \left. {{a}_{2} + \cdots + {a}_{n}}\right) /n = 1n→∞lim(a1+a2+⋯+an)/n=1 .