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

The answer to taylor ln(x+1),1 is ln(2)+1/2 (x-1)-1/8 (x-1)^2+1/24 (x-1)^3-1/64 (x-1)^4+\ldots

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taylor ln(x+1),1

taylor

ln (x + 1), 1

ln (2) + \frac{1}{2} (x - 1) - \frac{1}{8} (x - 1)^{2} + \frac{1}{24} (x - 1)^{3} - \frac{1}{64} (x - 1)^{4} + \dots

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

= ln (2) + \frac{\frac{1}{2}}{1 !} (x - 1) + \frac{- \frac{1}{4}}{2 !} (x - 1)^{2} + \frac{\frac{1}{4}}{3 !} (x - 1)^{3} + \frac{- \frac{3}{8}}{4 !} (x - 1)^{4} + \dots

Refine

= ln (2) + \frac{1}{2} (x - 1) - \frac{1}{8} (x - 1)^{2} + \frac{1}{24} (x - 1)^{3} - \frac{1}{64} (x - 1)^{4} + \dots

\frac{\partial}{\partial y} (x^{2} y cos (\frac{z}{t}))

y^{^{'}} - \frac{8}{x} y = (\frac{y}{x ^{6}})^{3}

- 5 x

y^{^{''}} + 6 y^{^{'}} + 13 y = 0

\int \frac{x ^{3}}{64} d x

What is taylor ln(x+1),1 ?
The answer to taylor ln(x+1),1 is ln(2)+1/2 (x-1)-1/8 (x-1)^2+1/24 (x-1)^3-1/64 (x-1)^4+\ldots