第3章 3.8

(1)

$f = x^4 - 2x^3 + 4 \qquad (x = 2)$

$f' = 4x^3 - 6x^2 \qquad x = 2 \text{ のとき} \qquad f' = 8$

$f'' = 12x^2 - 12x \qquad x = 2 \text{ のとき} \qquad f'' = 24$

$f''' = 24x - 12 \qquad x = 2 \text{ のとき} \qquad f''' = 36$

$f^{(4)} = 24$

これから

$f = 4 + 8(x-2) + \frac{24}{2!}(x-2)^2 + \frac{36}{3!}(x-2)^3 + \frac{24}{4!}(x-2)^4$

$= 4 + 8(x-2) + 12(x-2)^2 + 6(x-2)^3 + (x-2)^4$

(2)

$f = x^5 \qquad (x = -1)$

$f' = 5x^4 \qquad x = -1 \text{ のとき} \qquad f' = 5$

$f'' = 20x^3 \qquad x = -1 \text{ のとき} \qquad f'' = -20$

$f''' = 60x^2 \qquad x = -1 \text{ のとき} \qquad f''' = 60$

$f^{(4)} = 120x \qquad x = -1 \text{ のとき} \qquad f^{(4)} = -120$

$f^{(5)} = 120$

これから

$f = -1 + 5(x+1) - \frac{20}{2!}(x+1)^2 + \frac{60}{3!}(x+1)^3 - \frac{120}{4!}(x+1)^4 + \frac{120}{5!}(x+1)^5$

$= -1 + 5(x+1) - 10(x+1)^2 + 10(x+1)^3 - 5(x+1)^4 + (x+1)^5$