Question

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

Answer 1

Maclaurin series of `ln(1-3x)` for the first four terms is

`0 - 3x - (9x^2)/2 - 9x^3 + O(4)= -3x - (9x^2)/2 - 9x^3 + O(4)`

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(x) = ln(1-3x)` and `f(0) = 0`

Differentiate using the chain rule to get

`f'(x) = -3/(1-3x)` and `f'(0) = -3`

Differentiate the first derivative using the quotient rule to get

`f''(x) = -9/(1-3x)^2` and `f''(0) = -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(x)= (-54)/(1-3x)^3` and `f^3(0) = -54`

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

`0 - 3x - (9x^2)/2 - 9x^3 + O(4)= -3x - (9x^2)/2 - 9x^3 + O(4)`