What is taylor ln(1+x),3 ?

The answer to taylor ln(1+x),3 is 2ln(2)+1/4 (x-3)-1/32 (x-3)^2+1/192 (x-3)^3-1/1024 (x-3)^4+\ldots

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taylor ln(1+x),3

taylor

ln (1 + x), 3

2 ln (2) + \frac{1}{4} (x - 3) - \frac{1}{32} (x - 3)^{2} + \frac{1}{192} (x - 3)^{3} - \frac{1}{1024} (x - 3)^{4} + \dots

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

= 2 ln (2) + \frac{\frac{1}{4}}{1 !} (x - 3) + \frac{- \frac{1}{16}}{2 !} (x - 3)^{2} + \frac{\frac{1}{32}}{3 !} (x - 3)^{3} + \frac{- \frac{3}{128}}{4 !} (x - 3)^{4} + \dots

Refine

= 2 ln (2) + \frac{1}{4} (x - 3) - \frac{1}{32} (x - 3)^{2} + \frac{1}{192} (x - 3)^{3} - \frac{1}{1024} (x - 3)^{4} + \dots

\int \frac{3}{x ^{2}} + \frac{2}{x ^{3}} d x

x \frac{d y}{d x} = y (ln (y) - ln (x))

\frac{d}{d x} (\frac{ln ( x ) - 1}{ln ^{2} ( x )})

\frac{12}{1 + x ^{4}}, 6 x^{2}

2 x^{x}

What is taylor ln(1+x),3 ?
The answer to taylor ln(1+x),3 is 2ln(2)+1/4 (x-3)-1/32 (x-3)^2+1/192 (x-3)^3-1/1024 (x-3)^4+\ldots