Taylor’s series and Maclaurian’s series representation with example

In this article some questions and their solutions on Taylor’s series as well as Maclaurian’s series are given. In complex analysis, Taylor’s theorem gives power series representation of an analytic function in an open disk. This representation of power series from theorem can be used to check differentiability of a function in an open subset of complex plane. However, without diving deep in this topic, the explanation of the theorem and simple examples of Taylor’s series as well as Maclaurian’s series representation of some elementary functions are elaborated.

Statement of Taylor’s theorem: 

Suppose a function \(f\)is analytic throughout an open disk\(\left| {z - {z_0}} \right| < {R_0}\). Then, at each point \(z\)in the disk, \(f(z)\) has the series representation as

\(f(z) = \sum\limits_{n = 0}^\infty  {{a_n}{{\left( {z - {z_0}} \right)}^n}} \) for \(\left| {z - {z_0}} \right| < {R_0}\)

Where \({a_n} = \frac{{{f^n}({z_0})}}{{n!}}\,;\,n = 0,1,2,...\)

 In short, it can be said that the power series \(\sum\limits_{n = 0}^\infty  {{a_n}{{\left( {z - {z_0}} \right)}^n}} \) converges to \(f(z)\) in the disk\(\left| {z - {z_0}} \right| < {R_0}\). This expression of \(f(z)\) is called the Taylor’s series of \(f(z)\) about the point\({z_0}\).

If a function \(f(z)\) is analytic at a point \({z_0}\), then \(f(z)\) must have has Taylor’s series representation about \({z_0}\). For this obvious reason, an entire function (analytic everywhere) can be expressed in Taylor’s series about each point in the plane.

When\({z_0} = 0\), the Taylor’s series becomes

\(f(z) = \sum\limits_{n = 0}^\infty  {{a_n}{z^n}} \)for \(\left| z \right| < {R_0}\)

Where \({a_n} = \frac{{{f^n}(0)}}{{n!}}\,;\,n = 0,1,2,...\)

This series is called the Maclaurina’s Series.

The following example shows, how we can express a function into its Taylors series and Maclaurian series expansion.

Example 1: The function \(f(z) = {e^z}\)is an entire function. Therefore, by Taylor’s theorem \(f(z)\) can be expressed in Taylor’s series about any point \(z = {z_0}\)on the complex plane.

We are considering \({z_0} = 0\)and wish to find the Maclaurina’s Series representation of \(f(z)\).
The Maclaurian’s series representation will be

\(f(z) = \sum\limits_{n = 0}^\infty  {{a_n}{z^n}} \)

Where \({a_n} = \frac{{{f^n}(0)}}{{n!}}\,;\,n = 0,1,2,...\)

For  \(\left| z \right| < \infty \), since \(f(z)\) is entire.

Now,

\(f(z) = {e^z}\)

\( \Rightarrow f'(z) = {e^z} \Rightarrow f'(0) = {e^0} = 1\)

\( \Rightarrow f''(z) = {e^z} \Rightarrow f''(0) = {e^0} = 1\)

\( \Rightarrow f'''(z) = {e^z} \Rightarrow f'''(0) = {e^0} = 1\)
…

\(i.e.\,{f^n}(z) = 1\,;\,n = 0,1,2,...\)

Therefore, Maclaurian’s series representation of \(f(z) = {e^z}\) for \(\left| z \right| < \infty \)will be
\(f(z) = \sum\limits_{n = 0}^\infty  {{a_n}{z^n}} \)

\( \Rightarrow f(z) = \sum\limits_{n = 0}^\infty  {\frac{{{f^n}(0)}}{{n!}}{z^n}} \)

\( \Rightarrow f(z) = \sum\limits_{n = 0}^\infty  {\frac{1}{{n!}}{z^n}} \)

\( \Rightarrow f(z) = 1 + \frac{z}{{1!}} + \frac{{{z^2}}}{{2!}} + \frac{{{z^3}}}{{3!}} +  \cdots \)

Example 2: The function \(f(z) = \cos z\)is an entire function. Therefore, by Taylor’s theorem \(f(z) = \cos z\)can be expressed in Taylor’s series about any point \(z = {z_0}\)on the complex plane.

Taylor’s series representation about \(z = {z_0}\)is 

\(f(z) = \sum\limits_{n = 0}^\infty  {{a_n}{{\left( {z - {z_0}} \right)}^n}} \) for \(\left| {z - {z_0}} \right| < \infty \)

Where \({a_n} = \frac{{{f^n}({z_0})}}{{n!}}\,;\,n = 0,1,2,...\)

In this example, we will find Taylor’s series representation about the point \(z = \frac{\pi }{2}\)
Now,

\(f(z) = \cos z\)

\( \Rightarrow {f^n}(z) = \cos \left( {\frac{{n\pi }}{2} + z} \right)\)

\( \Rightarrow {f^n}\left( {\frac{\pi }{2}} \right) = \cos \left( {\frac{{n\pi }}{2} + \frac{\pi }{2}} \right)\)

\( \Rightarrow {f^n}\left( {\frac{\pi }{2}} \right) = \cos \left( {\frac{{(n + 1)\pi }}{2}} \right)\)

\( \Rightarrow {f^0}\left( {\frac{\pi }{2}} \right) = 0,\,{f^1}\left( {\frac{\pi }{2}} \right) =  - 1,\,{f^2}\left( {\frac{\pi }{2}} \right) = 0,\,{f^3}\left( {\frac{\pi }{2}} \right) = 1,{f^4}\left( {\frac{\pi }{2}} \right) = 0 \cdots \)

Therefore, Taylor’s series representation of \(f(z) = \cos z\) about \(z = \frac{\pi }{2}\) will be
\(f(z) = \sum\limits_{n = 0}^\infty  {{a_n}{{\left( {z - \frac{\pi }{2}} \right)}^n}} \)

\( \Rightarrow f(z) = \sum\limits_{n = 0}^\infty  {\frac{{{f^n}\left( {\frac{\pi }{2}} \right)}}{{n!}}{{\left( {z - \frac{\pi }{2}} \right)}^n}} \)

\( \Rightarrow f(z) = 0.{\left( {z - \frac{\pi }{2}} \right)^1} + \left( { - 1} \right)\frac{{{{\left( {z - \frac{\pi }{2}} \right)}^1}}}{{1!}} + \left( 0 \right)\frac{{{{\left( {z - \frac{\pi }{2}} \right)}^2}}}{{2!}} + \left( 1 \right)\frac{{{{\left( {z - \frac{\pi }{2}} \right)}^3}}}{{3!}} + \left( 0 \right)\frac{{{{\left( {z - \frac{\pi }{2}} \right)}^4}}}{{4!}} +  \cdots \)

\( \Rightarrow f(z) = \left( { - 1} \right)\frac{{{{\left( {z - \frac{\pi }{2}} \right)}^1}}}{{1!}} + \left( 1 \right)\frac{{{{\left( {z - \frac{\pi }{2}} \right)}^3}}}{{3!}} + \left( { - 1} \right)\frac{{{{\left( {z - \frac{\pi }{2}} \right)}^5}}}{{5!}} +  \cdots \)

\( \Rightarrow f(z) = \sum\limits_{n = 0}^\infty  {{{\left( { - 1} \right)}^{n + 1}}\frac{{{{\left( {z - \frac{\pi }{2}} \right)}^{2n + 1}}}}{{\left( {2n + 1} \right)!}}} \)

Putting \({z_0} = 0\)we can easily obtain the Maclaurian’s series for \(f(z) = \cos z\) as shown in the earlier example.

Example 3: The function \(f(z) = \frac{1}{{1 - z}}\)is not analytic at\(z = 1\). Therefore, by Taylor’s theorem \(f(z)\) can be expressed in Maclaurian’s series in the open disk \(\left| z \right| < 1\)  only.

The Maclaurian’s series representation is

\(f(z) = \sum\limits_{n = 0}^\infty  {{a_n}{z^n}} \) ; \(\left| z \right| < 1\)

Where \({a_n} = \frac{{{f^n}(0)}}{{n!}}\,;\,n = 0,1,2,...\)

Now,

\(f(z) = \frac{1}{{1 - z}} = {\left( {1 - z} \right)^{ - 1}}\)

\( \Rightarrow f'(z) = {\left( { - 1} \right)^2}{\left( {1 - z} \right)^{ - 2}} \Rightarrow f'(0) = {\left( { - 1} \right)^2}{\left( {1 - 0} \right)^{ - 2}} \Rightarrow f'(0) = 1\)

\( \Rightarrow f''(z) = {\left( { - 1} \right)^4}.1.2.{\left( {1 - z} \right)^{ - 3}} \Rightarrow f''(0) = {\left( { - 1} \right)^4}.1.2.{\left( {1 - 0} \right)^{ - 3}} \Rightarrow f''(0) = 1.2 = 2!\)

\( \Rightarrow f'''(z) = {\left( { - 1} \right)^6}.1.2.3.{\left( {1 - z} \right)^{ - 4}} \Rightarrow f'''(0) = {\left( { - 1} \right)^6}.1.2.3.{\left( {1 - 0} \right)^{ - 4}} \Rightarrow f'''(0) = 1.2.3 = 3!\)

…

\(i.e.\,{f^n}(0) = n!\,;\,\,\,\,n = 0,1,2,...\)

Therefore, Maclaurian’s series representation of \(f(z) = \frac{1}{{1 - z}}\) for \(\left| z \right| < 1\)will be

\(f(z) = \sum\limits_{n = 0}^\infty  {{a_n}{z^n}} \)

\( \Rightarrow f(z) = \sum\limits_{n = 0}^\infty  {\frac{{n!}}{{n!}}{z^n}} \)

\( \Rightarrow f(z) = \sum\limits_{n = 0}^\infty  {{z^n}} \)

\( \Rightarrow f(z) = 1 + z + {z^2} + {z^3} +  \cdots \)

In a similar way, we can obtain the Maclaurian’s series representation for \(f(z) = \frac{1}{{1 + z}}\) in the open disk \(\left| z \right| < 1\)which appears as

\(f(z) = \sum\limits_{n = 0}^\infty  {{{( - 1)}^n}{z^n}}  = 1 - z + {z^2} - {z^3} +  \cdots \) 

Example 4: The function \(f(z) = \frac{1}{{1 - z}}\)is not analytic at\(z = 1\). Thus, in the open disk\(\left| {z - i} \right| < \sqrt 2 \), \(f(z)\) is analytic everywhere in the open disk and can be expressed in Taylor’s series about the point \(z = i\).

Taylor’s series representation about \(z = {z_0}\)is 

\(f(z) = \sum\limits_{n = 0}^\infty  {{a_n}{{\left( {z - {z_0}} \right)}^n}} \) for \(\left| {z - {z_0}} \right| < \infty \)

Where \({a_n} = \frac{{{f^n}({z_0})}}{{n!}}\,;\,n = 0,1,2,...\)

In this example, we will find Taylor’s series representation about the point \(z = i\) in the open disk  \(\left| {z - i} \right| < \sqrt 2 \).

Now,

\(f(z) = \frac{1}{{1 - z}} = {\left( {1 - z} \right)^{ - 1}}\)

\( \Rightarrow f'(z) = {\left( { - 1} \right)^2}{\left( {1 - z} \right)^{ - 2}} \Rightarrow f'(i) = {\left( { - 1} \right)^2}{\left( {1 - i} \right)^{ - 2}} \Rightarrow f'(i) = 1.{\left( {1 - i} \right)^{ - 2}}\)

\( \Rightarrow f''(z) = {\left( { - 1} \right)^4}.1.2.{\left( {1 - z} \right)^{ - 3}} \Rightarrow f''(i) = {\left( { - 1} \right)^4}.1.2.{\left( {1 - i} \right)^{ - 3}} \Rightarrow f''(0) = 2!{\left( {1 - i} \right)^{ - 3}}\)

\( \Rightarrow f'''(z) = {\left( { - 1} \right)^6}.1.2.3.{\left( {1 - z} \right)^{ - 4}} \Rightarrow f'''(0) = {\left( { - 1} \right)^6}.1.2.3.{\left( {1 - i} \right)^{ - 4}} \Rightarrow f''' = 3!{\left( {1 - i} \right)^{ - 4}}\)

\(i.e.\,{f^n}(i) = n!{\left( {1 - i} \right)^{ - (n + 1)}}\,;\,\,\,\,n = 0,1,2,...\)

Therefore, Taylor’s series representation of \(f(z) = \frac{1}{{1 - z}}\) in the open disk \(\left| {z - i} \right| < \sqrt 2 \)will be

\(f(z) = \sum\limits_{n = 0}^\infty  {{a_n}{{\left( {z - i} \right)}^n}} \)

\( \Rightarrow f(z) = \sum\limits_{n = 0}^\infty  {\frac{{n!{{\left( {1 - i} \right)}^{ - (n + 1)}}}}{{n!}}{{\left( {z - i} \right)}^n}} \)

\( \Rightarrow f(z) = \sum\limits_{n = 0}^\infty  {\frac{{{{\left( {z - i} \right)}^n}}}{{{{\left( {1 - i} \right)}^{(n + 1)}}}}} \)

In a similar way, we can obtain the Maclaurian’s series representation for \(f(z) = \frac{1}{{1 + z}}\) in the open disk \(\left| z \right| < 1\)which appears as

\(f(z) = \sum\limits_{n = 0}^\infty  {{{( - 1)}^n}{z^n}}  = 1 - z + {z^2} - {z^3} +  \cdots \) 

In this article the statement Taylor’s theorem is explained in a simple way. Also, the Taylor’s Series and Maclaurian’s series representation has been defined. Some elaborative examples are carried out in simple steps for better understanding. Finally, we are hoping that readers will be able to understand how an analytic function throughout an open disk can be represented in Taylor’s Series expansion and Maclaurian’s series expansions.