泰勒 1/x log_{10}(1+x)

解答

泰勒

\frac{1}{x} lo g_{10} (1 + x)

lo g_{10} (2) + \frac{- 2 ln ( 2 ) + 1}{2 ln ( 10 )} (x - 1) + \frac{8 ln ( 2 ) - 5}{8 ln ( 10 )} (x - 1)^{2} + \frac{- 96 ln ( 2 ) + 64}{96 ln ( 10 )} (x - 1)^{3} + \frac{384 ln ( 2 ) - 262}{384 ln ( 10 )} (x - 1)^{4} + \dots

求解步骤

应用泰勒公式

= lo g_{10} (2) + \frac{\frac{d}{d x} ( \frac{1}{x} lo g _{10} ( 1 + x ) ) ( 1 )}{1 !} (x - 1) + \frac{\frac{d ^{2}}{d x ^{2}} ( \frac{1}{x} lo g _{10} ( 1 + x ) ) ( 1 )}{2 !} (x - 1)^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( \frac{1}{x} lo g _{10} ( 1 + x ) ) ( 1 )}{3 !} (x - 1)^{3} + \dots

求导数

= lo g_{10} (2) + \frac{- lo g _{10} ( 2 ) + \frac{1}{2 l n ( 10 )}}{1 !} (x - 1) + \frac{\frac{8 l n ( 2 ) - 5}{4 l n ( 10 )}}{2 !} (x - 1)^{2} + \frac{\frac{- 96 l n ( 2 ) + 64}{16 l n ( 10 )}}{3 !} (x - 1)^{3} + \frac{\frac{384 l n ( 2 ) - 262}{16 l n ( 10 )}}{4 !} (x - 1)^{4} + \dots

整理后得

= lo g_{10} (2) + \frac{- 2 ln ( 2 ) + 1}{2 ln ( 10 )} (x - 1) + \frac{8 ln ( 2 ) - 5}{8 ln ( 10 )} (x - 1)^{2} + \frac{- 96 ln ( 2 ) + 64}{96 ln ( 10 )} (x - 1)^{3} + \frac{384 ln ( 2 ) - 262}{384 ln ( 10 )} (x - 1)^{4} + \dots

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ma c l a u r in e^{s i n (x)}

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