设函数f 0 (x)在(-∞,+∞)内连续,f n (x)=∫ 0 x f n-1 (t)dt(n=1,2,…). (1)证明:f n (x)= 0 x (t)(x-t) n-1 dt(n=1,2,…); (2)证明:
【正确答案】正确答案:(1)n=1时,f 1 (x)=∫ 0 x f 0 (t)dt,等式成立; 设n=k时,f k (x)= 0 x f 0 (t)(x-t) k-1 dt, 则n=k+1时,f k+1 (x)=∫ 0 x f k (t)dt=∫ 0 x dt∫ 0 t f 0 (u)(t-u) k-1 du = 0 x du∫ u x f 0 (u)(t-u) k-1 dt= 0 x f 0 (u)(x-u) k du 由归纳法得f n (x)= 0 x f 0 (t)(x-t) n-1 dt(n=1,2,…). (2)对任意的x∈(-∞,+∞),f 0 (t)在[0,x]或[x,0]上连续,于是存在M>0(M与x 有关),使得|f 0 (t)|≤M(t∈[0,x]或t∈[x,0]),于是 |f n (x)|≤ |∫ 0 x (x-t) n-1 dt|= |x| n 因为 收敛,根据比较审敛法知
【答案解析】