Question

How do you find the third degree Taylor polynomial of $f\left( x \right)=\ln {{\left( x \right)}^{2}}$ at x = 1?

Answer 1

Hint : In the above question, we have to find the third-degree Taylor polynomial given a function f, a specific point x = c which is known as the centre and a positive integer ‘n’ such that $T\left( x \right)=\sum\limits_{n=0}^{\infty }{\dfrac{{{f}^{n}}\left( c \right){{\left( x-c \right)}^{n}}}{n!}}$ . Later we need to find the three derivatives of the given function as it is given in the question that we need to find the third degree polynomial. Thus we need to extend to ${{\left( x-c \right)}^{3}}$ term. Finding the substituting and then simplifying the polynomial we will get the required answer.

Complete step-by-step answer :
We have given that,
$f\left( x \right)=\ln {{\left( x \right)}^{2}}$
As we know that,
The Taylor polynomial for f centered at x = c is given by,
$T\left( x \right)=\sum\limits_{n=0}^{\infty }{\dfrac{{{f}^{n}}\left( c \right){{\left( x-c \right)}^{n}}}{n!}}$
As it is given in the question that we need to find the third degree polynomial.
Thus we need to extend to ${{\left( x-c \right)}^{3}}$ term.
Therefore,
We need to find three derivatives.
Now,
$f\left( x \right)=\ln {{\left( x \right)}^{2}}$
Using the definition of logarithm function i.e. $\log {{x}^{a}}=a\log x$
$f\left( x \right)=2\ln \left( x \right)$
Taking x = 1, we will get
$\Rightarrow f\left( 1 \right)=2\ln \left( 1 \right)=0$
Now,
$f'\left( x \right)=\dfrac{2}{x}$
Taking x = 1, we will get
$\Rightarrow f'\left( 1 \right)=\dfrac{2}{1}=2$
Now,
The second derivative will be,
$f''\left( x \right)=-\dfrac{2}{{{x}^{2}}}$
Taking x = 1, we will get
$\Rightarrow f''\left( 1 \right)=-\dfrac{2}{{{1}^{2}}}=-2$
Now,
The third derivative is,
$f'''\left( x \right)=\dfrac{4}{{{x}^{3}}}$
Taking x = 1, we will get
$\Rightarrow f'''\left( 1 \right)=\dfrac{4}{{{1}^{3}}}=4$
Therefore,
$\ln {{\left( x \right)}^{2}}\approx f\left( c \right)=\dfrac{{{f}^{1}}\left( 1 \right){{\left( x-c \right)}^{1}}}{1!}+\dfrac{{{f}^{2}}\left( 2 \right){{\left( x-c \right)}^{2}}}{2!}+\dfrac{{{f}^{3}}\left( 3 \right){{\left( x-c \right)}^{3}}}{3!}$
Now substituting the values from the above and solving,, we will get
$\ln {{\left( x \right)}^{2}}\approx 0=\dfrac{2\left( x-1 \right)}{1}+\dfrac{-2{{\left( x-1 \right)}^{2}}}{2}+\dfrac{4{{\left( x-1 \right)}^{3}}}{6}$
Simplifying the above, we will get
$\ln {{\left( x \right)}^{2}}\approx 2\left( x-1 \right)-1{{\left( x-1 \right)}^{2}}+\dfrac{2}{3}{{\left( x-1 \right)}^{3}}$
Hence, the required third degree Taylor polynomial of $f\left( x \right)=\ln {{\left( x \right)}^{2}}$ at x = 1 is $2\left( x-1 \right)-1{{\left( x-1 \right)}^{2}}+\dfrac{2}{3}{{\left( x-1 \right)}^{3}}$ .
So, the correct answer is “ $2\left( x-1 \right)-1{{\left( x-1 \right)}^{2}}+\dfrac{2}{3}{{\left( x-1 \right)}^{3}}$ ”.

Note : In order to solve these types of questions, students must need to know about the concept of Taylor polynomial. In the above solution we used the general form of the Taylor formula that is $T\left( x \right)=\sum\limits_{n=0}^{\infty }{\dfrac{{{f}^{n}}\left( c \right){{\left( x-c \right)}^{n}}}{n!}}$ .
Then we find the third-degree Taylor polynomial in which we substitute c = 1 and we have $f\left( x \right)=\ln {{\left( x \right)}^{2}}$ . Therefore we get the third degree Taylor polynomial of $f\left( x \right)=\ln {{\left( x \right)}^{2}}$ at x = 1 is $2\left( x-1 \right)-1{{\left( x-1 \right)}^{2}}+\dfrac{2}{3}{{\left( x-1 \right)}^{3}}$ .