Question

f(x)=x^2sinx 的在零处的n阶导数,用泰勒公式,复习全书上面的答案全是零,为什么?

Answer 1

f(x) = (x^2)sinx,

要用到求高阶导数的 Leibniz 公式

(uv)^(n) =∑C(n,k)[u^(k)][v^(n-k)] （翻翻书）

注意到

(x^2)' = 2x,(x^2)" = 2,(x^2)^(k) = 0 (k≥3),

(sinx)^(k) = sin(x+kπ/2) (k=0,1,2,…,n),

则

[(x^2)sinx]^(n)

= C(n,0)(x^2)[(sinx)^(n)]+C(n,1)(2x)[(sinx)^(n-1)]+C(n,2)*2*[(sinx)^(n-2)]+0

= (x^2)sin(x+nπ/2)+2nxsin[x+(n-1)π/2]+n(n+1)sin[x+(n-2)π/2].