Question

How do you find the Maclaurin’s series of $ f(x) = \ln (1 + x) $ ?

Answer 1

Hint: To solve this equation, we must know the Maclaurin’s theorem which is,
 $
  f(x) = f(0) + \dfrac{{f'(0)}}{{1!}}x + \dfrac{{f''(0)}}{{2!}}{x^2} + ... \;
  $ ,
and by substituting the required value in the theorem we are able to find the series for the given function.

Complete step-by-step answer:
Let us consider the given question,
 $ f(x) = \ln (1 + x) $
Maclaurin’s theorem, it is the special form of law of mean,
 $
  f(x) = f(0) + \dfrac{{f'(0)}}{{1!}}x + \dfrac{{f''(0)}}{{2!}}{x^2} + ... \;
  $
We need to find the value of $ f(0),f'(0),f''(0),f'''(0) $ and substitute in the above series. For that we have to consider the given equation and put $ x = 0 $ to find the first term $ f(0) $ , by doing this we get,
 $ f(x) = \ln (1 + x),f(0) = \ln 1 = 0 $
And now differentiate the $ f(x) $ which will become $ f'(x) $ and put $ x = 0 $ . When we differentiate $ \ln (1 + x) $ with respect to $ x $ we get $ \dfrac{1}{{1 + x}} $ ,
 $ f'(x) = \dfrac{1}{{1 + x}},f'(0) = \dfrac{1}{{1 + 0}} = 1 $
Again differentiate $ f'(x) $ , which will be written as $ f''(x) $ and put $ x = 0 $ ,
 $ f''(x) = \dfrac{{ - 1}}{{{{(1 + x)}^2}}},f''(0) = - 1 $
Again differentiate $ f''(x) $ , which will be written as $ f'''(x) $ and put $ x = 0 $ ,
 $ f'''(x) = \dfrac{{1 \times 2}}{{{{(1 + x)}^3}}},f'''(0) = 2 \times 1 = 2! $
Again differentiate $ f'''(x) $ , which will be written as $ f''''(x) $ and put $ x = 0 $ ,
 $ f''''(x) = \dfrac{{ - 1 \times 2 \times 3}}{{{{(1 + x)}^4}}},f''''(0) = - 3 \times 2 \times 1 = - 3! $
Substituting the values, we get,
 $
  f(x) = \ln (1 + x) = 0 + \dfrac{1}{{1!}}x - \dfrac{1}{{2!}}{x^2} + \dfrac{{2!}}{{3!}}{x^3} - \dfrac{{3!}}{{4!}}{x^4} + ... \;
  $
 $ \ln (1 + x) = \dfrac{1}{{1!}}x - \dfrac{1}{{2!}}{x^2} + \dfrac{{{x^3}}}{3} - \dfrac{{{x^4}}}{4}+… $
 $ \ln (1 + x) = x - \dfrac{{{x^2}}}{2} + \dfrac{{{x^3}}}{3} - \dfrac{{{x^4}}}{4} + ... $
So, the correct answer is “ $ x - \dfrac{{{x^2}}}{2} + \dfrac{{{x^3}}}{3} - \dfrac{{{x^4}}}{4} + ... $ ”.

Note: Differentiate $ \dfrac{1}{{1 + x}} $ , to differentiate this we should know the basic differentiation formula which is $ {x^n} = n{x^{n - 1}} $ . Similarly, we take $ \dfrac{1}{{1 + x}} $ as $ {(1 + x)^{ - 1}} $ , and consider $ n = - 1 $ , by applying differentiation formula we get,
 $ \dfrac{d}{{dx}}\left( {\dfrac{1}{{1 + x}}} \right) = - 1{(1 + x)^{ - 1 - 1}} = \dfrac{{ - 1}}{{{{(1 + x)}^2}}} $ This is our answer. If we differentiate it again and again the power of the term $ (1 + x) $ is increased by a number $ 1 $ .