将函数f(x)=e-x2/2展开为x的幂级数，并计算f(9)(0)．
利用间接展开法，已知 ex=1+x+x2/2!+…+xn/n!+…，(-∞，+∞) 将-x2/2代换ex中的x而得之． ex=∑n=0∞xn/n!，(-∞，+∞) 故e-x2/2=∑n=0∞(-x2/2)n/n!= ∑n=0∞(-1)n[x2n/(2n•n！) 即 f(x)=e-x2/2=∑n=0∞(-1)n[x2n/(2n•n!) =1-x2/2+x4/(22•2！)+…+(-1)nx2n/(2n•n!)+…, 又f′(x)=(∑n=0∞(-1)n[x2n/(2n•n!))′ =-x+x3/2!-x5/(22•2！)+…+(-1)nx2n-1/[2n-1(n-1)]+…, =∑n=0∞[(-1)n2n/(2n•n!)]x2n-1 f′′ (x)=∑n=0∞[(-1)n2n(2n-1)/(2n•n!)]x2n-2 f′′′ ∑n=2∞[(-1)n2n(2n-1)(2n-2)/(2n•n!)]x2n-3…, f (9)(x)=∑n=0∞[(-1)n2n(2n-1)…(2n-8)/(2n•n!)]x2n-9 故 f (9)(0)=0．
