泰勒 ln(x),5

解答

泰勒

ln (x), 5

ln (5) + \frac{1}{5} (x - 5) - \frac{1}{50} (x - 5)^{2} + \frac{1}{375} (x - 5)^{3} - \frac{1}{2500} (x - 5)^{4} + \dots

求解步骤

应用泰勒公式

= ln (5) + \frac{\frac{d}{d x} ( ln ( x ) ) ( 5 )}{1 !} (x - 5) + \frac{\frac{d ^{2}}{d x ^{2}} ( ln ( x ) ) ( 5 )}{2 !} (x - 5)^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( ln ( x ) ) ( 5 )}{3 !} (x - 5)^{3} + \dots

求导数

= ln (5) + \frac{\frac{1}{5}}{1 !} (x - 5) + \frac{- \frac{1}{25}}{2 !} (x - 5)^{2} + \frac{\frac{2}{125}}{3 !} (x - 5)^{3} + \frac{- \frac{6}{625}}{4 !} (x - 5)^{4} + \dots

整理后得

= ln (5) + \frac{1}{5} (x - 5) - \frac{1}{50} (x - 5)^{2} + \frac{1}{375} (x - 5)^{3} - \frac{1}{2500} (x - 5)^{4} + \dots

x \to - \infty lim (x^{n})

d er i v a t i v e F (x) = \frac{x ^{3} - 3 x ^{2} + 4 x - 5}{x ^{2}}

f (x) = e^{- x^{2}} + x

y^{^{'}} + 4 y = x^{2} e^{- 4 x}

l a pl a ce t r an s f or m 3 t + 1