Question

The fourth term in Taylor series of \[\log x\] centered at \[a = 1\]is
A.\[\dfrac{{{{(x - 1)}^3}}}{3}\]
B. \[\dfrac{{{{(x - 1)}^2}}}{2}\]
C. \[\dfrac{{ - {{(x - 1)}^4}}}{4}\]
D. \[(x - 1)\]

Answer 1

Hint:Here we use the general expansion of a function using Taylor series. Open the summation of Taylor series up to fourth term and calculate the required differentiation of the function as needed. Find the value of differentiations of function at \[a = 1\] and substitute in the expansion.

Formula used: Taylor series expansion of a function \[f(x)\] at point a is given by \[f(x) = \sum\limits_{n = 0}^\infty {\dfrac{{{f^{(n)}}(a)}}{{n!}}{{(x - a)}^n}} \]where \[{f^{(n)}}\]is the nth differentiation of the function.
* Differentiation of\[{x^n}\] with respect to x is given by \[\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}\].

Complete step-by-step answer:
We know Taylor series expansion of a function \[f(x)\] at point a is given by \[f(x) = \sum\limits_{n = 0}^\infty {\dfrac{{{f^{(n)}}(a)}}{{n!}}{{(x - a)}^n}} \]
Since we have to find the fourth term in the expansion, we open the summation in RHS to four terms.
\[ \Rightarrow f(x) = \dfrac{{{f^{(0)}}(a)}}{{0!}}{(x - a)^0} + \dfrac{{{f^{(1)}}(a)}}{{1!}}{(x - a)^1} + \dfrac{{{f^2}(a)}}{{2!}}{(x - a)^2} + \dfrac{{{f^{(3)}}(a)}}{{3!}}{(x - a)^3}\]
Since we know any number having the power 0 is equal to 1
\[ \Rightarrow f(x) = \dfrac{{f(a)}}{{0!}} + \dfrac{{{f^{(1)}}(a)}}{{1!}}{(x - a)^1} + \dfrac{{{f^2}(a)}}{{2!}}{(x - a)^2} + \dfrac{{{f^{(3)}}(a)}}{{3!}}{(x - a)^3} …….… (1)\]
We are given the function\[f(x) = \log x\], then \[f(a) = \log a\]
We have the fourth term as\[\dfrac{{{f^{(3)}}(a)}}{{3!}}{(x - a)^3}\]. So we need to do three differentiations of the function
Differentiating the function \[f(a) = \log a\] with respect to a
\[ \Rightarrow \dfrac{d}{{da}}f(a) = \dfrac{d}{{da}}\log a\]
Since we know \[\dfrac{d}{{dx}}\log x = \dfrac{1}{x}\]
\[ \Rightarrow \dfrac{d}{{da}}f(a) = \dfrac{1}{a}\]
\[ \Rightarrow f'(a) = \dfrac{1}{a} ………..… (2)\]
Now differentiating the function \[f'(a) = \dfrac{1}{a}\]with respect to a
\[ \Rightarrow \dfrac{d}{{da}}f'(a) = \dfrac{d}{{da}}({a^{ - 1}})\]
Since we know \[\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}\]
\[ \Rightarrow \dfrac{d}{{da}}f'(a) = - {a^{ - 1 - 1}}\]
\[ \Rightarrow \dfrac{d}{{da}}f'(a) = - {a^{ - 2}}\]
\[ \Rightarrow \dfrac{d}{{da}}f'(a) = \dfrac{{ - 1}}{{{a^2}}}\]
\[ \Rightarrow f''(a) = \dfrac{{ - 1}}{{{a^2}}} ………..… (3)\]
Now differentiating the function \[f''(a) = \dfrac{{ - 1}}{{{a^2}}}\] with respect to a
\[ \Rightarrow \dfrac{d}{{da}}f''(a) = \dfrac{d}{{da}}( - {a^{ - 2}})\]
Since we know \[\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}\]
\[ \Rightarrow \dfrac{d}{{da}}f''(a) = - ( - 2{a^{ - 2 - 1}})\]
\[ \Rightarrow \dfrac{d}{{da}}f''(a) = 2{a^{ - 3}}\]
\[ \Rightarrow \dfrac{d}{{da}}f''(a) = \dfrac{2}{{{a^3}}}\]
\[ \Rightarrow f'''(a) = \dfrac{2}{{{a^3}}}\]
Now we substitute the value of a as 1.
\[ \Rightarrow f'''(1) = \dfrac{2}{{{1^3}}} = 2\]
The value in fourth term\[\dfrac{{{f^{(3)}}(a)}}{{3!}}{(x - a)^3}\]at \[a = 1\] is
\[ \Rightarrow \dfrac{{{f^{'''}}(1)}}{{3!}}{(x - 1)^3} = \dfrac{{2{{(x - 1)}^3}}}{{3!}}\]
Now we know that factorial of any number opens as \[n! = n(n - 1)(n - 2)........3.2.1\]
Therefore, the value of \[3! = 3 \times 2 \times 1\]
Substitute the value of \[3! = 3 \times 2 \times 1\]in the denominator.
\[ \Rightarrow \dfrac{{{f^{'''}}(1)}}{{3!}}{(x - 1)^3} = \dfrac{{2{{(x - 1)}^3}}}{{3 \times 2 \times 1}}\]
Cancel the same term i.e. 2 from both numerator and denominator.
\[ \Rightarrow \dfrac{{{f^{'''}}(1)}}{{3!}}{(x - 1)^3} = \dfrac{{{{(x - 1)}^3}}}{3}\]
So, the value of the fourth term in the Taylor series is \[\dfrac{{{{(x - 1)}^3}}}{3}\]

So, the correct answer is “Option A”.

Note:Students might start solving for the values of each term as they see the term Taylor expansion, we don’t need the terms we just need the differentiation of function for the next differentiation till fourth term. Also, many students by mistake put in value of x as 1 as they think we are finding the value of function \[f(x)\]so we have to substitute x as 1, but we are finding the expansion series which is centered at a so we substitute the value of a as 1