Question

How can you find Taylor expansion $f(x) = \ln (1 - x)$ of about $x = 0$?

Answer 1

Hint:
First, calculate the first few derivatives of f(x) .evaluate the function and its derivatives at x=a. fill in the right-hand side of the Taylor series expression using the Taylor formula given below. Write the result using a summation.

Formula used:
For calculating Taylor series of $f(x) = \ln (1 - x)$ at $x = 0$ we are using formula
$f(x) = \sum\limits_{n = 0}^\infty {\dfrac{{{f_n}(a)}}{{n!}}{{(x - a)}^n}} $
$f(x) = f(0) + f'(0)(x - a) + f''(0)\dfrac{{{{(x - a)}^2}}}{{2!}} + f'''(0)\dfrac{{{{(x - a)}^3}}}{{3!}}....$

Complete step by step answer:
Let us find the Taylor series for $f(x) = \ln (1 - x)$
First we will find $f(0)$,
$f(x) = \ln (1 - x) \Rightarrow f(0) = \ln (1) = 0$
By taking the derivatives,
$
  f'(x) = \dfrac{{ - 1}}{{1 - x}} \Rightarrow f'(0) = - 1 \\
  f''(x) = \dfrac{{ - 1}}{{{{(1 - x)}^2}}} \Rightarrow f''(0) = - 1 \\
  ...... \\
$
After putting the values of $f(0),f'(0),f(''0),....$ we get,
$f(x) = 0 + ( - 1)x + ( - 1)\dfrac{{{x^2}}}{2} + ...$
After simplifying we get,
$f(x) = - x - \dfrac{{{x^2}}}{2} - \dfrac{{{x^3}}}{3}...$
The Taylor series of $f(x) = \ln (1 - x)$ at $x = 0$ is
$f(x) = - x - \dfrac{{{x^2}}}{2} - \dfrac{{{x^3}}}{3}...$

Note:
For solving questions of such type, you have to remember formulas and also some important derivations. Then we have to take care if the small mistakes are not happening.