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 fis analytic throughout an open disk|zz0|<R0. Then, at each point zin the disk, f(z) has the series representation as

f(z)=n=0an(zz0)n for |zz0|<R0

Where an=fn(z0)n!;n=0,1,2,...


 In short, it can be said that the power series n=0an(zz0)n converges to f(z) in the disk|zz0|<R0. This expression of f(z) is called the Taylor’s series of f(z) about the pointz0.

If a function f(z) is analytic at a point z0, then f(z) must have has Taylor’s series representation about z0. For this obvious reason, an entire function (analytic everywhere) can be expressed in Taylor’s series about each point in the plane.

Whenz0=0, the Taylor’s series becomes

f(z)=n=0anznfor |z|<R0

Where an=fn(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)=ezis an entire function. Therefore, by Taylor’s theorem f(z) can be expressed in Taylor’s series about any point z=z0on the complex plane.

We are considering z0=0and wish to find the Maclaurina’s Series representation of f(z).
The Maclaurian’s series representation will be

f(z)=n=0anzn

Where an=fn(0)n!;n=0,1,2,...

For  |z|<, since f(z) is entire.

Now,

f(z)=ez

f(z)=ezf(0)=e0=1

f(z)=ezf(0)=e0=1

f(z)=ezf(0)=e0=1


i.e.fn(z)=1;n=0,1,2,...

Therefore, Maclaurian’s series representation of f(z)=ez for |z|<will be
f(z)=n=0anzn

f(z)=n=0fn(0)n!zn

f(z)=n=01n!zn

f(z)=1+z1!+z22!+z33!+

Example 2: The function f(z)=coszis an entire function. Therefore, by Taylor’s theorem f(z)=coszcan be expressed in Taylor’s series about any point z=z0on the complex plane.

Taylor’s series representation about z=z0is 

f(z)=n=0an(zz0)n for |zz0|<

Where an=fn(z0)n!;n=0,1,2,...

In this example, we will find Taylor’s series representation about the point z=π2
Now,

f(z)=cosz

fn(z)=cos(nπ2+z)

fn(π2)=cos(nπ2+π2)

fn(π2)=cos((n+1)π2)

f0(π2)=0,f1(π2)=1,f2(π2)=0,f3(π2)=1,f4(π2)=0

Therefore, Taylor’s series representation of f(z)=cosz about z=π2 will be
f(z)=n=0an(zπ2)n

f(z)=n=0fn(π2)n!(zπ2)n

f(z)=0.(zπ2)1+(1)(zπ2)11!+(0)(zπ2)22!+(1)(zπ2)33!+(0)(zπ2)44!+

f(z)=(1)(zπ2)11!+(1)(zπ2)33!+(1)(zπ2)55!+

f(z)=n=0(1)n+1(zπ2)2n+1(2n+1)!

Putting z0=0we can easily obtain the Maclaurian’s series for f(z)=cosz as shown in the earlier example.

Example 3: The function f(z)=11zis not analytic atz=1. Therefore, by Taylor’s theorem f(z) can be expressed in Maclaurian’s series in the open disk |z|<1  only.

The Maclaurian’s series representation is

f(z)=n=0anzn ; |z|<1

Where an=fn(0)n!;n=0,1,2,...

Now,

f(z)=11z=(1z)1

f(z)=(1)2(1z)2f(0)=(1)2(10)2f(0)=1

f(z)=(1)4.1.2.(1z)3f(0)=(1)4.1.2.(10)3f(0)=1.2=2!

f(z)=(1)6.1.2.3.(1z)4f(0)=(1)6.1.2.3.(10)4f(0)=1.2.3=3!



i.e.fn(0)=n!;n=0,1,2,...

Therefore, Maclaurian’s series representation of f(z)=11z for |z|<1will be

f(z)=n=0anzn

f(z)=n=0n!n!zn

f(z)=n=0zn

f(z)=1+z+z2+z3+

In a similar way, we can obtain the Maclaurian’s series representation for f(z)=11+z in the open disk |z|<1which appears as

f(z)=n=0(1)nzn=1z+z2z3+ 

Example 4: The function f(z)=11zis not analytic atz=1. Thus, in the open disk|zi|<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=z0is 

f(z)=n=0an(zz0)n for |zz0|<

Where an=fn(z0)n!;n=0,1,2,...

In this example, we will find Taylor’s series representation about the point z=i in the open disk  |zi|<2.

Now,

f(z)=11z=(1z)1

f(z)=(1)2(1z)2f(i)=(1)2(1i)2f(i)=1.(1i)2

f(z)=(1)4.1.2.(1z)3f(i)=(1)4.1.2.(1i)3f(0)=2!(1i)3

f(z)=(1)6.1.2.3.(1z)4f(0)=(1)6.1.2.3.(1i)4f=3!(1i)4

i.e.fn(i)=n!(1i)(n+1);n=0,1,2,...

Therefore, Taylor’s series representation of f(z)=11z in the open disk |zi|<2will be

f(z)=n=0an(zi)n

f(z)=n=0n!(1i)(n+1)n!(zi)n

f(z)=n=0(zi)n(1i)(n+1)

In a similar way, we can obtain the Maclaurian’s series representation for f(z)=11+z in the open disk |z|<1which appears as

f(z)=n=0(1)nzn=1z+z2z3+ 

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.

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