What is taylor ln(1-4y),0 ?

The answer to taylor ln(1-4y),0 is -4y-8y^2-64/3 y^3-64y^4-1024/5 y^5+\ldots

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taylor ln(1-4y),0

taylor

ln (1 - 4 y), 0

- 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

x y^{^{'}} = y + x^{2} + y^{2}

\int - 5 x^{3} d x

x \to 0 lim (\frac{sin ( k t )}{t})

5 x^{3}

x \to - 1 lim (ln (x^{2} - 1))

What is taylor ln(1-4y),0 ?
The answer to taylor ln(1-4y),0 is -4y-8y^2-64/3 y^3-64y^4-1024/5 y^5+\ldots