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

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

Explanation

Solution

Hint : To solve this equation, we must know the Maclaurin’s theorem which is,
f(x)=f(0)+f′(0)1!x+f′′(0)2!x2+...   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)f(x) = \ln (1 + x)
Maclaurin’s theorem, it is the special form of law of mean,
f(x)=f(0)+f′(0)1!x+f′′(0)2!x2+...   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)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=0x = 0 to find the first term f(0)f(0) , by doing this we get,
f(x)=ln⁡(1+x),f(0)=ln⁡1=0f(x) = \ln (1 + x),f(0) = \ln 1 = 0
And now differentiate the f(x)f(x) which will become f′(x)f'(x) and put x=0x = 0 . When we differentiate ln⁡(1+x)\ln (1 + x) with respect to xx we get 11+x\dfrac{1}{{1 + x}} ,
f′(x)=11+x,f′(0)=11+0=1f'(x) = \dfrac{1}{{1 + x}},f'(0) = \dfrac{1}{{1 + 0}} = 1
Again differentiate f′(x)f'(x) , which will be written as f′′(x)f''(x) and put x=0x = 0 ,
f′′(x)=−1(1+x)2,f′′(0)=−1f''(x) = \dfrac{{ - 1}}{{{{(1 + x)}^2}}},f''(0) = - 1
Again differentiate f′′(x)f''(x) , which will be written as f′′′(x)f'''(x) and put x=0x = 0 ,
f′′′(x)=1×2(1+x)3,f′′′(0)=2×1=2!f'''(x) = \dfrac{{1 \times 2}}{{{{(1 + x)}^3}}},f'''(0) = 2 \times 1 = 2!
Again differentiate f′′′(x)f'''(x) , which will be written as f′′′′(x)f''''(x) and put x=0x = 0 ,
f′′′′(x)=−1×2×3(1+x)4,f′′′′(0)=−3×2×1=−3!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+11!x−12!x2+2!3!x3−3!4!x4+...   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)=11!x−12!x2+x33−x44+…\ln (1 + x) = \dfrac{1}{{1!}}x - \dfrac{1}{{2!}}{x^2} + \dfrac{{{x^3}}}{3} - \dfrac{{{x^4}}}{4}+…
ln⁡(1+x)=x−x22+x33−x44+...\ln (1 + x) = x - \dfrac{{{x^2}}}{2} + \dfrac{{{x^3}}}{3} - \dfrac{{{x^4}}}{4} + ...
So, the correct answer is “ x−x22+x33−x44+...x - \dfrac{{{x^2}}}{2} + \dfrac{{{x^3}}}{3} - \dfrac{{{x^4}}}{4} + ... ”.

Note : Differentiate 11+x\dfrac{1}{{1 + x}} , to differentiate this we should know the basic differentiation formula which is xn=nxn−1{x^n} = n{x^{n - 1}} . Similarly, we take 11+x\dfrac{1}{{1 + x}} as (1+x)−1{(1 + x)^{ - 1}} , and consider n=−1n = - 1 , by applying differentiation formula we get,
ddx(11+x)=−1(1+x)−1−1=−1(1+x)2\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)(1 + x) is increased by a number 11 .