泰勒 sqrt(8+x^2),1

解答

泰勒

8 + x^{2}, 1

3 + \frac{1}{3} (x - 1) + \frac{4}{27} (x - 1)^{2} - \frac{4}{243} (x - 1)^{3} - \frac{4}{2187} (x - 1)^{4} + \dots

求解步骤

应用泰勒公式

= 3 + \frac{\frac{d}{d x} ( 8 + x ^{2} ) ( 1 )}{1 !} (x - 1) + \frac{\frac{d ^{2}}{d x ^{2}} ( 8 + x ^{2} ) ( 1 )}{2 !} (x - 1)^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( 8 + x ^{2} ) ( 1 )}{3 !} (x - 1)^{3} + \dots

求导数

= 3 + \frac{\frac{1}{3}}{1 !} (x - 1) + \frac{\frac{8}{27}}{2 !} (x - 1)^{2} + \frac{- \frac{8}{81}}{3 !} (x - 1)^{3} + \frac{- \frac{32}{729}}{4 !} (x - 1)^{4} + \dots

整理后得

= 3 + \frac{1}{3} (x - 1) + \frac{4}{27} (x - 1)^{2} - \frac{4}{243} (x - 1)^{3} - \frac{4}{2187} (x - 1)^{4} + \dots

f (x) = 4 x^{5} ln (x)

ma c l a u r in ln (1 - x^{2})

\int t^{2} e^{\frac{t}{3}} d t

x \to \infty lim (\frac{( e ^{- x y - x + y} )}{x ^{2} y ^{2} + x y})

y^{^{''}} - 2 y^{^{'}} + 8 y = 320 t^{2}