What is taylor ln(1-4y) ?

The answer to taylor ln(1-4y) is -4y-8y^2-64/3 y^3-64y^4-1024/5 y^5+\ldots

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taylor ln(1-4y)

taylor

ln (1 - 4 y)

- 4 y - 8 y^{2} - \frac{64}{3} y^{3} - 64 y^{4} - \frac{1024}{5} y^{5} + \dots

Solution steps

Apply the Taylor Formula

= 0 + \frac{\frac{d}{d y} ( ln ( 1 - 4 y ) ) ( 0 )}{1 !} y + \frac{\frac{d ^{2}}{d y ^{2}} ( ln ( 1 - 4 y ) ) ( 0 )}{2 !} y^{2} + \frac{\frac{d ^{3}}{d y ^{3}} ( ln ( 1 - 4 y ) ) ( 0 )}{3 !} y^{3} + \dots

Evaluate Derivatives

= 0 + \frac{- 4}{1 !} y + \frac{- 16}{2 !} y^{2} + \frac{- 128}{3 !} y^{3} + \frac{- 1536}{4 !} y^{4} + \frac{- 24576}{5 !} y^{5} + \dots

Refine

= - 4 y - 8 y^{2} - \frac{64}{3} y^{3} - 64 y^{4} - \frac{1024}{5} y^{5} + \dots

\frac{\partial}{\partial x} (cos (8 x) sin (4 y))

\frac{1}{( 2 x ^{3} + 3 x + 5 ) ^{\frac{3}{4}}}

\int_{1}^{e} (ln (x))^{2} d x

\frac{d}{d x} (- sec (x))

\int \frac{x ^{2}}{x ^{3} + 1} d x

What is taylor ln(1-4y) ?
The answer to taylor ln(1-4y) is -4y-8y^2-64/3 y^3-64y^4-1024/5 y^5+\ldots