How do you find the third degree Taylor polynomial for f(x) = lnx, centered at a = 2?

Hint: In the above question we have to find the third-degree Taylor polynomial. Given a function f, a specific point x = a (called the centre), and a positive integer n, the Taylor polynomial of f at a, of degree n, is the polynomial T of degree n that best fits the curve y = f(x) near the point a, in the sense that T and all its first n derivatives have the same value at x = a as f does. So, let us see how we can solve this problem.

Complete step by step solution:
We have to solve this problem using Taylor polynomials. The general form of Taylor expansion centered at a of an analytical function f is :
 $\Rightarrow f(x) = \sum\limits_{n = 0}^\infty {\dfrac{{f{}^{(n)}(a)}}{{n!}}} (x - a){}^n$ . In this equation, f(x) is a derivative of f.
In the third degree of Taylor polynomial, the polynomial consists of the first four terms ranging from 0 to 3.
Therefore, the polynomial term is
 $= f(a) + f'(a)(x - a) + \dfrac{{f''(a)}}{2}{(x - a)^2} + \dfrac{{f''(a)}}{6}{(x - a)^3}.$
Given, f(x) = lnx
Therefore, $\Rightarrow f'(x) = \dfrac{1}{x},f''(x) = - \dfrac{1}{{{x^2}}},f'''(x) = \dfrac{2}{{{x^3}}}.$
So, the third-degree polynomial will be
 $=\ln (a) + \dfrac{1}{a}(x - a) - \dfrac{1}{{2{a^2}}}{(x - a)^2} + \dfrac{1}{{3{a^2}}}{(x - a)^3}$
So, now we have a = 2, then the polynomial will be

$=\ln (2) + \dfrac{1}{2}(x - 2) - \dfrac{1}{8}{(x - 2)^2} + \dfrac{1}{{24}}{(x - 2)^3}$

Note:
In the above solution we used the general form of the Taylor formula that is $\Rightarrow f(x) = \sum\limits_{n = 0}^\infty {\dfrac{{f{}^{(n)}(a)}}{{n!}}} (x - a){}^n$ . Then we find the third-degree Taylor polynomial in which we put a = 2 and f(x) = lnx. And therefore, we get $\ln (2) + \dfrac{1}{2}(x - 2) - \dfrac{1}{8}{(x - 2)^2} + \dfrac{1}{{24}}{(x - 2)^3}$ as the answer.