1/(1+x) 嗎 ?
泰勒 = Σ f微分n次(a) / n! * (x-a)^n
f(x) = 1/(1+x) → f(0) = 1
f'(x) = -1/(1+x)^2 → f'(0) = -1
f''(x) = (-1)*(-2)/(1+x)^3 → f''(0) = 2!
f'''(x) = (-1)*(-2)*(-3)/(1+x)^4 → f'''(0) = -3!
f''''(x) = (-1)*(-2)*(-3)*(-4)/(1+x)^5 → f''''(0) = 4!
在 0 展開的話
1/(1+x) = (1/0!)x^0 - (1/1!)x^1 + (2!/2!)x^2 - (3!/3!)x^3 + (4!/4!)x^4 …
=1 - x + x^2 - x^3 + x^4 |
