泰勒 f(x)=x^3-3x^2+4x+5,x=3

解答

泰勒

f (x) = x^{3} - 3 x^{2} + 4 x + 5, x = 3

17 + 13 (x - 3) + 6 (x - 3)^{2} + (x - 3)^{3}

求解步骤

应用泰勒公式

= 17 + \frac{\frac{d}{d x} ( x ^{3} - 3 x ^{2} + 4 x + 5 ) ( 3 )}{1 !} (x - 3) + \frac{\frac{d ^{2}}{d x ^{2}} ( x ^{3} - 3 x ^{2} + 4 x + 5 ) ( 3 )}{2 !} (x - 3)^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( x ^{3} - 3 x ^{2} + 4 x + 5 ) ( 3 )}{3 !} (x - 3)^{3} + \dots

求导数

= 17 + \frac{13}{1 !} (x - 3) + \frac{12}{2 !} (x - 3)^{2} + \frac{6}{3 !} (x - 3)^{3} + \frac{0}{4 !} (x - 3)^{4} + \frac{0}{5 !} (x - 3)^{5} + \frac{0}{6 !} (x - 3)^{6} + \frac{0}{7 !} (x - 3)^{7} + \frac{0}{8 !} (x - 3)^{8} + \frac{0}{9 !} (x - 3)^{9} + \frac{0}{10 !} (x - 3)^{10}

整理后得

= 17 + 13 (x - 3) + 6 (x - 3)^{2} + (x - 3)^{3}

ma c l a u r in (1 + x)^{- \frac{7}{3}}, 4

y^{^{''}} - 64 y = 0

a re a x^{2} - 18, 17 x

\frac{d y}{d x} = y x^{9} sin (x^{10}), y (0) = 6

\frac{d}{d s} (\frac{s}{s ^{2} + 16})