泰勒 arctan(x),1

解答

泰勒

arctan (x), 1

\frac{π}{4} + \frac{1}{2} (x - 1) - \frac{1}{4} (x - 1)^{2} + \frac{1}{12} (x - 1)^{3} - \frac{1}{40} (x - 1)^{5} + \dots

求解步骤

应用泰勒公式

= \frac{π}{4} + \frac{\frac{d}{d x} ( arctan ( x ) ) ( 1 )}{1 !} (x - 1) + \frac{\frac{d ^{2}}{d x ^{2}} ( arctan ( x ) ) ( 1 )}{2 !} (x - 1)^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( arctan ( x ) ) ( 1 )}{3 !} (x - 1)^{3} + \dots

求导数

= \frac{π}{4} + \frac{\frac{1}{2}}{1 !} (x - 1) + \frac{- \frac{1}{2}}{2 !} (x - 1)^{2} + \frac{\frac{1}{2}}{3 !} (x - 1)^{3} + \frac{0}{4 !} (x - 1)^{4} + \frac{- 3}{5 !} (x - 1)^{5} + \dots

整理后得

= \frac{π}{4} + \frac{1}{2} (x - 1) - \frac{1}{4} (x - 1)^{2} + \frac{1}{12} (x - 1)^{3} - \frac{1}{40} (x - 1)^{5} + \dots

l a pl a ce t r an s f or m 4 t e^{- 3 t}

n = 1 \sum \infty n^{n} x^{n}

y^{^{'}} + 8 (tan (8 x)) y = - 5 cos (8 x)

x \to \infty lim (\frac{x 7 + 1}{x 7})

t an g e n t y = x^{2} + 7