Question

The fourth term in Taylor series of log x centered at a=1 is ?
$
  (a){\text{ }}\dfrac{{{{(x - 1)}^3}}}{3} \\
  (b){\text{ }}\dfrac{{{{(x - 1)}^3}}}{2} \\
  (c){\text{ }}\dfrac{{ - {{(x - 1)}^4}}}{4} \\
  (d){\text{ (x - 1)}} \\
 $

Answer 1

Hint – In this question use the direct formula for the Taylor series expansion of a function f(x) centered at any point z such that $f\left( x \right) = f\left( z \right) + f'\left( z \right)\left( {x - z} \right) + \dfrac{{f''\left( z \right)}}{{2!}}{\left( {x - z} \right)^2} + \dfrac{{f'''\left( z \right)}}{{3!}}{\left( {x - z} \right)^3} + .........$. Then search for the fourth term. This will give the answer.

Complete step-by-step solution -
As we know that the Taylor series expansion of f(x) which is centered at z is
$f\left( x \right) = f\left( z \right) + f'\left( z \right)\left( {x - z} \right) + \dfrac{{f''\left( z \right)}}{{2!}}{\left( {x - z} \right)^2} + \dfrac{{f'''\left( z \right)}}{{3!}}{\left( {x - z} \right)^3} + .........$
Now the given function is
$f\left( x \right) = \log x$ Centered at a = 1.
So the fourth term of the function according to Taylor series expansion is
Fourth term = $\dfrac{{f'''\left( a \right)}}{{3!}}{\left( {x - a} \right)^3}$
Now a is 1 so,
$ \Rightarrow \dfrac{{f'''\left( 1 \right)}}{{3!}}{\left( {x - 1} \right)^3}$......................... (1)
So first differentiate the function w.r.t. x three times we have,
Now as we know differentiation of log x is (1/x) so we have,
$ \Rightarrow \dfrac{d}{{dx}}f\left( x \right) = \dfrac{d}{{dx}}\log x = f'\left( x \right) = \dfrac{1}{x} = {x^{ - 1}}$
Now as we know differentiation of $\dfrac{d}{{dx}}{x^{ - n}} = \left( { - n} \right){x^{ - n - 1}}$
Therefore, $f''\left( x \right) = \left( { - 1} \right){x^{ - 1 - 1}} = - {x^{ - 2}}$
Now again re-differentiate we have,
Therefore, $f'''\left( x \right) = - \left( { - 2} \right){x^{ - 2 - 1}} = 2{x^{ - 3}} = \dfrac{2}{{{x^3}}}$
$ \Rightarrow f'''\left( 1 \right) = \dfrac{2}{1} = 2$
Now substitute this value in equation (1) we have,
$ \Rightarrow \dfrac{{f'''\left( 1 \right)}}{{3!}}{\left( {x - 1} \right)^3} = \dfrac{2}{{3!}}{\left( {x - 1} \right)^3} = \dfrac{2}{{3 \times 2}}{\left( {x - 1} \right)^3} = \dfrac{1}{3}{\left( {x - 1} \right)^3}$
So this is the required fourth term.
Hence option (A) is correct.

Note – Taylor series of a function is an infinite sum of terms that are expressed in terms of the function’s derivatives at a single point. For most common functions ,the function and the sum of its Taylor series are equal near this point. Taylor series is also called McLaurin series so we need to get confused if this term is used in question of this kind.