What is the maclaurin series of ln(1-3x)?

Jul 29, 2015

Maclaurin series of $\ln \left(1 - 3 x\right)$ for the first four terms is

$0 - 3 x - \frac{9 {x}^{2}}{2} - 9 {x}^{3} + O \left(4\right) = - 3 x - \frac{9 {x}^{2}}{2} - 9 {x}^{3} + O \left(4\right)$

Explanation:

A Maclaurin series is just a Taylor series expansion of a function about 0:

f(x) = f(0) + f'(0)x + (f''(0)x^2)/(2!) + (f^3(0)x^3)/(3!) + ... + (f^n(0)x^n)/(n!)

In our case,

$f \left(x\right) = \ln \left(1 - 3 x\right)$ and $f \left(0\right) = 0$

Differentiate using the chain rule to get

$f ' \left(x\right) = - \frac{3}{1 - 3 x}$ and $f ' \left(0\right) = - 3$

Differentiate the first derivative using the quotient rule to get

$f ' ' \left(x\right) = - \frac{9}{1 - 3 x} ^ 2$ and $f ' ' \left(0\right) = - 9$

Differentiate the second derivative using a combination of the quotient rule and the chain rule for the derivative of the denominator to get

${f}^{3} \left(x\right) = \frac{- 54}{1 - 3 x} ^ 3$ and ${f}^{3} \left(0\right) = - 54$

Plug these values into the definition of the Maclaurin series and you'll get the answer for the first four terms:

$0 - 3 x - \frac{9 {x}^{2}}{2} - 9 {x}^{3} + O \left(4\right) = - 3 x - \frac{9 {x}^{2}}{2} - 9 {x}^{3} + O \left(4\right)$