Question

How do you find the Taylor series for \[f(x) = \ln (1 + {x^2})\]?

Answer 1

Hint:In order to calculate the Taylor series of the $f(x)$which is centred at $x = a = 0$,we first derive the series for $\ln (x + 1)$ by use the formula of Maclaurin series \[f(x) = \sum\limits_{n = 0}^\infty {\dfrac{{{f^{(n)}}(0)}}{{n!}}{{(x)}^n}} \],up to the value of the nth term.

Find the derivative of the function up to the order n 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. At the end replacing all the $x$with the ${x^2}$we will get our required series.

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 \[f(x) = \ln ({x^2} + 1)\]

First, We will start working on the Taylor series for \[f(x) = \ln (x + 1)\]

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}} $

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

1.A derivative of order (i). The $i$above the f looks like an exponent but it 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 $i$.

According to our question, we have not given the centred point around which we can expand our series, so we will expand our Taylor series around $x = a = 0$.So, this is a Maclaurin Series.

Our formula becomes,
\[
f(x) = \sum\limits_{n = 0}^\infty {\dfrac{{{f^{(n)}}(0)}}{{n!}}{{(x - 0)}^i}} \\
f(x) = \sum\limits_{n = 0}^\infty {\dfrac{{{f^{(n)}}(0)}}{{n!}}{{(x)}^n}} \\
\]
This means, we have to work for the nth derivative as there is no range to the degree is given in question

Let’s find out some of the derivative up to order n

$f(x) = \ln (x + 1) \Rightarrow f(1) = \ln (1) = 0$

Let’s find out the derivatives with respect to x , using the property of derivative that

$\dfrac{d}{{dx}}(\ln x) = \dfrac{1}{x}$.

1 st derivative:
$f'(x) = {f^1}(x) = \dfrac{1}{{1 + x}}\,\,\,\,\,\,\, \Rightarrow {f^1}(1) = 1$

2 nd derivative:
$f''(x) = {f^2}(x) = - \dfrac{1}{{{{(x + 1)}^2}}}\,\,\,\,\,\,\, \Rightarrow {f^2}(1) = - 1$

3 rd derivative:
$f'''(x) = {f^3}(x) = \dfrac{2}{{{{(x + 1)}^3}}}\,\,\,\,\,\,\, \Rightarrow {f^3}(1) = 2$

4rd derivative
$f''''(x) = {f^4}(x) = - \dfrac{6}{{{{(x + 1)}^4}}}\,\,\,\,\,\,\, \Rightarrow {f^4}(1) = - 3$

Nth derivative
${f^n}(x) = {( - 1)^{n + 1}}\dfrac{{(n - 1)!}}{{{{(x + 1)}^n}}}\,\,\,\,\,\,\, \Rightarrow {f^n}(1) = {( -
1)^{n + 1}}(n - 1)!$

Putting vales all these values together in our original formula i.e. equation (1.),we get

\[\ln (1 + x) = \sum\limits_{n = 0}^\infty {\dfrac{{{f^{(n)}}(0)}}{{n!}}{{(x)}^n}} \]
\[\ln (1 + x) = \sum\limits_{n = 0}^\infty {\dfrac{{{{( - 1)}^{n + 1}}(n - 1)!}}{{n!}}{{(x)}^n}} \]

Simplifying this more, we get

\[\ln (1 + x) = \sum\limits_{n = 0}^\infty {\dfrac{{{{( - 1)}^{n + 1}}}}{n}{{(x)}^n}} \]

Now we have successfully defined the series for $\ln (1 + x)$ ,we can now replace all the
$x$with ${x^2}$to obtain the series for $\ln (1 + {x^2})$.
\[
\ln (1 + {x^2}) = \sum\limits_{n = 0}^\infty {\dfrac{{{{( - 1)}^{n + 1}}}}{n}{{({x^2})}^n}} \\
= \sum\limits_{n = 0}^\infty {\dfrac{{{{( - 1)}^{n + 1}}}}{n}{x^2}^n} \\
= {x^2} - \dfrac{{{x^4}}}{2} + \dfrac{{{x^6}}}{3} - \dfrac{{{x^8}}}{4}............... \\
\]
Therefore, the Taylor series for the $\ln (1 + {x^2})$is\[{x^2} - \dfrac{{{x^4}}}{2} + \dfrac{{{x^6}}}{3} - \dfrac{{{x^8}}}{4}...............\].

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 $n! = n(n - 1)(n - 2)(n - 3).......(2)(1)$