What is taylor ln(1-3y) ?

The answer to taylor ln(1-3y) is -3y-9/2 y^2-9y^3-81/4 y^4-243/5 y^5+\ldots

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taylor ln(1-3y)

taylor

ln (1 - 3 y)

- 3 y - \frac{9}{2} y^{2} - 9 y^{3} - \frac{81}{4} y^{4} - \frac{243}{5} y^{5} + \dots

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

= 0 + \frac{- 3}{1 !} y + \frac{- 9}{2 !} y^{2} + \frac{- 54}{3 !} y^{3} + \frac{- 486}{4 !} y^{4} + \frac{- 5832}{5 !} y^{5} + \dots

Refine

= - 3 y - \frac{9}{2} y^{2} - 9 y^{3} - \frac{81}{4} y^{4} - \frac{243}{5} y^{5} + \dots

\int \frac{( x ^{2} )}{x ^{2} + 9} d x

\int \frac{5 - x}{x ^{2} + x - 2} d x

\frac{d}{d x} (e^{- 4 x} sin (4 x))

\int \frac{tan ^{2} ( θ )}{sec ( θ )} d θ

- y + \frac{d y}{d x} ln (x) = 0

What is taylor ln(1-3y) ?
The answer to taylor ln(1-3y) is -3y-9/2 y^2-9y^3-81/4 y^4-243/5 y^5+\ldots