Question

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

Answer 1

In order to calculate the third-degree Taylor polynomial of the $f(x)$which is centred at $x = a = 1$,use the formula of Taylor polynomial ${T_n}(x) = \sum\limits_{i = 0}^n {\dfrac{{{f^{(i)}}(a)}}{{i!}}{{(x - a)}^i}}$,up to the value of $n = 3$or to the term of having${(x -a)^3}$.

Find the derivative of the function up to the order 3 using the property of $\dfrac{d}{{dx}}(\ln x) = \dfrac{1}{x}$and chaining rule.

Putting all the values back into the formula will give your required result.

Formula Used:
${T_n}(x) = \sum\limits_{i = 0}^n {\dfrac{{{f^{(i)}}(a)}}{{i!}}{{(x - a)}^i}}$
$\dfrac{d}{{dx}}(\ln x) = \dfrac{1}{x}$

Complete step by step solution:
We are given a function $f(x) = \ln ({x^2})$

Let’s first rewrite the function using the property of logarithm that $\ln ({a^m}) = m\ln (a)$
$f(x) = 2\ln (x)$

Let’s have a look into the Taylor formula which we are going to use,
${T_n}(x) = \sum\limits_{i = 0}^n {\dfrac{{{f^{(i)}}(a)}}{{i!}}{{(x - a)}^i}}$

According to our question, Taylor Polynomial for $f$is centred at$x = a = 1$

In order to calculate the third degree Taylor polynomial of $f$,we have to extend summation up to the term having${(x - a)^3}$i.e. up to the value of $n = 3$

If we closely look the formula, we conclude that the each term in the formula requires:

1.A derivative of order (i). The $i$above the f looks like an exponent but its is not .It is actually representing the order of the derivative.

2.Putting the value of $a$in the derivatives.

3.Then Dividing every term by the factorial number $i!$

4.Finally Multiplying by $(x - a)$raised to the power $i$.

Let’s Expand our formula up to $n = 3$,we get
${T_3}(x) = f(a) + \dfrac{{{f^{(1)}}(a)}}{{1!}}{(x - a)^1} + \dfrac{{{f^{(2)}}(a)}}{{2!}}{(x - a)^2} + \dfrac{{{f^{(3)}}(a)}}{{3!}}{(x - a)^3} \\\ {T_3}(x) = f(1) + \dfrac{{{f^{(1)}}(1)}}{{1!}}{(x - 1)^1} + \dfrac{{{f^{(2)}}(1)}}{{2!}}{(x - 1)^2} + \dfrac{{{f^{(3)}}(1)}}{{3!}}{(x - 1)^3} \\\$----------(1)
Finding the values of ${f^{(1)}}(1)$, ${f^{(2)}}(1)$, ${f^{(3)}}(1)$, $f(1)$
$f(x) = 2\ln (x) \Rightarrow f(1) = 2\ln (1) = 0$
Let’s find out the derivatives with respect to x up to order 3, using the property of derivative that $\dfrac{d}{{dx}}(\ln x) = \dfrac{1}{x}$.
1 st derivative:
$f'(x) = {f^1}(x) = \dfrac{2}{x}\,\,\,\,\,\,\, \Rightarrow {f^1}(1) = 2$

2 nd derivative:
$f''(x) = {f^2}(x) = - \dfrac{2}{{{x^2}}}\,\,\,\,\,\,\, \Rightarrow {f^2}(1) = - 2$
3 rd derivative:
$f'''(x) = {f^3}(x) = \dfrac{4}{{{x^3}}}\,\,\,\,\,\,\, \Rightarrow {f^3}(1) = 4$
Putting values of ${f^{(1)}}(1)$, ${f^{(2)}}(1)$, ${f^{(3)}}(1)$, $f(1)$ all together in our original formula i.e. equation (1),we get

${T_3}(x) = f(1) + \dfrac{{{f^{(1)}}(1)}}{{1!}}{(x - 1)^1} + \dfrac{{{f^{(2)}}(1)}}{{2!}}{(x - 1)^2} + \dfrac{{{f^{(3)}}(1)}}{{3!}}{(x - 1)^3} \\\ = 0 + \dfrac{2}{1}{(x - 1)^1} + \dfrac{{ - 2}}{2}{(x - 1)^2} + \dfrac{4}{6}{(x - 1)^3} \\\ = 2(x - 1) - {(x - 1)^2} + \dfrac{2}{3}{(x - 1)^3} \\\$

Therefore, the Third degree Taylor polynomial of $f(x)$is equal to $2(x - 1) - {(x - 1)^2} + \dfrac{2}{3}{(x - 1)^3}$

Additional Information:
Factorial: The continued product of first n natural numbers is called the “n factorial “and denoted by $n!$.

Note: 1.Don’t forget to cross-check your answer at least once.
2.Meaning of Zero factorial is senseless to define it as the product of integers from 1 to zero. So, we define it as $0! = 1$.
3.Factorial of any number can be calculated as $n! = n(n - 1)(n - 2)(n - 3).......(2)(1)$