Fuzzy set theory and its properties with examples

Fuzzy set is nothing but just a generalization of classical set theories. In this article you will get very preliminary concepts of fuzzy set theory.
You will be able to learn about alpha-cut, strong alpha-cut, support, core, normality fuzzy set, sub normality, cardinality and nucleus of a fuzzy set.
Also, you will find the very basic operation union, intersection and complement of fuzzy sets and explained with examples.
In general, a fuzzy set can be defined as combination of a crisp set and a membership function.

What is characteristic function of a crisp set?

Before proceeding to the basic concept of fuzzy set theory we need to know about the characteristic function of a crisp set.  The characteristic function of a crisp set can be defined as the assignment of values of either 1 or 0 to each element in the universal set, depending upon the condition that whether the element is included in that set or not.

The definition of a characteristic function can be written as follows:

Let \(A\) be a set on a universal set\(X\), the characteristic function \({\chi _A}:X \to \left\{ {0,1} \right\}\) is defined as

\({\chi _A}(x) = 0,x \notin A;\)

Or \({\mkern 1mu} {\chi _A}(x) = 1,x \in A\)

In case of the crisp set an element either belongs to the set or does not belongs to the set. Therefore the inclusion in the set can be assigned as the value 1 and the opposite, i.e. exclusion can be assigned as 0.

Example of a characteristic function:

Consider \(A = \left\{ {x,y,z} \right\}\) be a set on a universal set \(X\).
So, by the definition of characteristic function

\({\chi _A}(x) = 1\) since \(x \in A\)

Similarly, \({\chi _A}(y) = 1\) and \({\chi _A}(z) = 1\)

But, \({\chi _A}(t) = 0\) since \(t \notin A\)

What is fuzzy set?

In 1965 Lotif Ali Zadeh introduced the concept of fuzzy set. Fuzzy set theory provides a mathematical frame work in which uncertainty may arises due to incomplete information about  any problem, or information which is not reliable, or due to the linguistic nature in which the problem is defined.
Fuzzy set theory is becoming a major area of study for implementation in decision making under uncertain environment, artificial intelligence, automated systems etc.
We can generalize the characteristic function in such a way that the value can be assigned to an elements of the universal set which fall within the range 0 and 1.
The generalized characteristic function is termed as the membership function. Larger value of membership function corresponding to an element will denote higher degree of inclusion in that set. Such set together with the membership function is called a fuzzy set.

Definition of a fuzzy set

A fuzzy set is one which assigns grades of membership between 0 and 1 to objects within its universe of discourse. If \(X\) is a universal set then a fuzzy set A is defined by its membership function

\({\mu _A}:X \to (0,1)\)

An alternative definition of a fuzzy set is

Le \(X\) be a universal set of discourse and \(x\)be any element of \(X\) then a fuzzy set \(A\) is defined on \(X\)can be written as a collection of ordered pairs

\(A = \left\{ {(x,{\mu _A}(x)):x \in X} \right\}\)

Example of fuzzy set

 \(X = \left\{ {a,b,c} \right\}\) be a universal set, then a fuzzy set on \(X\) can be denoted as

\(A = \left\{ {(a,.7),(b,1),(c,.3)} \right\}\) or \(A = \left\{ {\frac{a}{{.7}},\frac{b}{1},\frac{c}{{.3}},} \right\}\)

This notation implies that the degree of inclusion of the elements  \(a,{\rm{ }}b,{\rm{ }}c\)  in the set is .7, 1, .3 respectively.


Like crisp sets, there are some terminologies associated with fuzzy sets which are given below:

\(\alpha \)-Cuts (alpha cut)of a fuzzy set

Consider a fuzzy set\(A\) in\(X\) and suppose \(\alpha  \in \left( {0,1} \right)\) be any real number.  Then the \(\alpha \)-cut of\(A\) , denoted by \({}^\alpha A\) and is defined as

\({}^\alpha A = \left\{ {x:{\mu _A}(x) \ge \alpha } \right\}\)

Example of \(\alpha \)-Cut 

\(A = \left\{ {\left( {a,.6} \right),\left( {b,.5} \right),\left( {c,1} \right),\left( {d,.2} \right)} \right\}\) be any fuzzy set defined on a universal set \(X = \left\{ {a,b,c,d} \right\}\).

Let \(\alpha  = .5\), then

\({}^\alpha A = \left\{ {x:{\mu _A}(x) \ge \alpha } \right\}\)

\( \Rightarrow {}^{.5}A = \left\{ {x:{\mu _A}(x) \ge .5} \right\}\)

\( \Rightarrow {}^{.5}A = \left\{ {a,b,c} \right\}\)

Strong \(\alpha \)-Cut(strong alpha cut)of a fuzzy set

Consider a fuzzy set\(A\) in\(X\) and a real number\(\alpha  \in \left( {0,1} \right)\). Then the strong \(\alpha \)-cut, denoted by \({}^{\alpha  + }A\) and is defined by

\({}^{\alpha  + }A = \left\{ {x:X;{\mu _A}(x) > \alpha } \right\}\)

Example of Strong \(\alpha \)-Cut

\(A = \left\{ {\left( {a,.6} \right),\left( {b,.5} \right),\left( {c,1} \right),\left( {d,.2} \right)} \right\}\) be any fuzzy set defined on a universal set \(X = \left\{ {a,b,c,d} \right\}\).

Let \(\alpha  = .5\), then

\({}^{\alpha  + }A = \left\{ {x:{\mu _A}(x) > \alpha } \right\}\)

\( \Rightarrow {}^{.5 + }A = \left\{ {x:{\mu _A}(x) > .5} \right\}\)

\( \Rightarrow {}^{.5 + }A = \left\{ {a,c} \right\}\)

Support of a Fuzzy

The support of a fuzzy set \(A\) is defined on \(X\) as

\(Support(A) = {}^{0 + }A = \left\{ {x:{\mu _A}(x) > 0} \right\}\)

Core of a Fuzzy Set

The core of a fuzzy set\(A\) is defined on \(X\) is a crisp set defined as

\(Core(A) = \left\{ {x:{\mu _A}(x) \ge 1} \right\} = {}^1A\)

Example of Core of a fuzzy set

\(A = \left\{ {\left( {a,.3} \right),\left( {b,.5} \right),\left( {c,1} \right),\left( {d,.2} \right)} \right\}\) be any fuzzy set defined on a universal set \(X = \left\{ {a,b,c,d} \right\}\).

\(Core(A) = \left\{ {x:{\mu _A}(x) \ge 1} \right\} = {}^1A = \left\{ c \right\}\)

Height of a Fuzzy Set

The height of a fuzzy set \(A\) is the largest membership grade obtained by any element in that set.

\(h(A) = \mathop {\sup }\limits_{x \in X} \left\{ {{\mu _A}(x)} \right\}\)

Example of Height 

Consider \(A = \left\{ {\left( {a,.3} \right),\left( {b,.5} \right),\left( {c,0} \right),\left( {d,.8} \right)} \right\}\) be any fuzzy set defined on a universal set \(X = \left\{ {a,b,c,d} \right\}\).

\(h(A) = \mathop {\sup }\limits_{x \in X} \left\{ {{\mu _A}(x)} \right\} = \sup \left\{ {.3,.5,0,.8} \right\} = .8\)

Normal fuzzy set

A fuzzy set \(A\) defined on a set \(X\) is said to be a normal fuzzy set if and only if
\(h(A) = \mathop {\sup }\limits_{x \in X} \left\{ {{\mu _A}(x)} \right\}\)for at least one \(x \in X\)
\(A\)is called subnormal fuzzy set if \(h(A) < 1\).

Example of Normal fuzzy set

\(A = \left\{ {\left( {a,.3} \right),\left( {b,.5} \right),\left( {c,1} \right),\left( {d,.2} \right)} \right\}\) is a normal fuzzy set and \(B = \left\{ {\left( {a,.3} \right),\left( {b,.5} \right),\left( {c,0} \right),\left( {d,.8} \right)} \right\}\) is a subnormal fuzzy set.

Cardinality of a Fuzzy Set

A fuzzy set \(A\) defined on a set\(X\), where\(X\) is finite. The scale cardinality of \(A\) is defined as
\(\left| A \right| = \sum\limits_{x \in X} {{\mu _A}(x)} \)

Example of cardinality of a fuzzy set

Cardinality of the fuzzy set \(A = \left\{ {\left( {a,.3} \right),\left( {b,.5} \right),\left( {c,1} \right),\left( {d,.2} \right)} \right\}\) on \(X = \left\{ {a,b,c,d} \right\}\) is

\(\left| A \right| = .3 + .5 + 1 + .2 = 2.0\)

Nucleus of a fuzzy set

The nucleus of a fuzzy set \(A\)is defined by

\(neucleous(A) = \left\{ {x \in X:{\mu _A}(x) = 1} \right\}\) 

If there is only one point such that its membership degree is equal to 1, then this point is called as the peak value of the fuzzy set\(A\).

Example of nucleus of a fuzzy set

The nucleus of the fuzzy set \(A = \left\{ {\left( {a,.3} \right),\left( {b,.5} \right),\left( {c,1} \right),\left( {d,.2} \right)} \right\}\) on \(X = \left\{ {a,b,c,d} \right\}\) is \(\left\{ c \right\}\) .

Equality of two Fuzzy sets

Two fuzzy sets are said to be equal, if they have same number of elements and their membership function are also equal. i.e., if \(A\) and\(B\) are two fuzzy sets on \(X\) then\(A\) and\(B\)
are said to be equal and is denoted by if and only if for every

\({\mu _A}(x) = {\mu _B}(x)\forall x \in X\) 

Example of inequality of two fuzzy set

\(A = \left\{ {\left( {a,.3} \right),\left( {b,.5} \right),\left( {c,1} \right),\left( {d,.2} \right)} \right\}\) and \(B = \left\{ {\left( {a,.3} \right),\left( {b,.5} \right),\left( {c,0} \right),\left( {d,.2} \right)} \right\}\) are not equal fuzzy sets, since \({\mu _A}(c) \ne {\mu _B}(c)\) .

Union of Two Fuzzy Sets

Let \(A\) and\(B\) be two fuzzy sets defined on\(X\). Union of \(A\) and\(B\) is defined by the membership function

\({\mu _{A \cup B}}(x) = \max \left\{ {{\mu _A}(x),{\mu _B}(x)} \right\}\forall x \in X\) 

Example of union of two fuzzy sets

Consider two fuzzy sets  \(A = \left\{ {\left( {a,.2} \right),\left( {b,.6} \right),\left( {c,.7} \right),\left( {d,.2} \right)} \right\}\) and \(B = \left\{ {\left( {a,.3} \right),\left( {b,.4} \right),\left( {c,.9} \right),\left( {d,.1} \right)} \right\}\) on \(X = \left\{ {a,b,c,d} \right\}\) .

 \({\mu _{A \cup B}}(a) = \max \left\{ {{\mu _A}(a),{\mu _B}(a)} \right\} = \max \left\{ {.2,.3} \right\} = .3\)

\({\mu _{A \cup B}}(b) = \max \left\{ {{\mu _A}(b),{\mu _B}(b)} \right\} = \max \left\{ {.6,.4} \right\} = .6\)

\({\mu _{A \cup B}}(c) = \max \left\{ {{\mu _A}(c),{\mu _B}(c)} \right\} = \max \left\{ {.7,.9} \right\} = .9\)

\({\mu _{A \cup B}}(d) = \max \left\{ {{\mu _A}(d),{\mu _B}(d)} \right\} = \max \left\{ {.2,.1} \right\} = .2\)

\(\therefore A \cup B = \left\{ {\left( {a,.3} \right),\left( {b,.6} \right),\left( {c,.9} \right),\left( {d,.2} \right)} \right\}\)

Intersection of Two Fuzzy Sets

Let \(A\) and\(B\) be two fuzzy sets defined on\(X\). Intersection of \(A\) and\(B\) is defined by

\({\mu _{A \cap B}}(x) = \max \left\{ {{\mu _A}(x),{\mu _B}(x)} \right\}\forall x \in X\) 

Example of intersection of two fuzzy sets

Consider two fuzzy sets  \(A = \left\{ {\left( {a,.2} \right),\left( {b,.6} \right),\left( {c,.7} \right),\left( {d,.2} \right)} \right\}\) and \(B = \left\{ {\left( {a,.3} \right),\left( {b,.4} \right),\left( {c,.9} \right),\left( {d,.1} \right)} \right\}\) on \(X = \left\{ {a,b,c,d} \right\}\) .

 \({\mu _{A \cap B}}(a) = \min \left\{ {{\mu _A}(a),{\mu _B}(a)} \right\} = \min \left\{ {.2,.3} \right\} = .2\)

\({\mu _{A \cap B}}(b) = \min \left\{ {{\mu _A}(b),{\mu _B}(b)} \right\} = \min \left\{ {.6,.4} \right\} = .4\)

\({\mu _{A \cap B}}(c) = \min \left\{ {{\mu _A}(c),{\mu _B}(c)} \right\} = \min \left\{ {.7,.9} \right\} = .7\)

\({\mu _{A \cap B}}(d) = \min \left\{ {{\mu _A}(d),{\mu _B}(d)} \right\} = \min \left\{ {.2,.1} \right\} = .1\)

\(\therefore A \cap B = \left\{ {\left( {a,.2} \right),\left( {b,.4} \right),\left( {c,.7} \right),\left( {d,.1} \right)} \right\}\)

Complement of a Fuzzy Set

The complement of a fuzzy set \(A\) is defined by the membership function

\({\mu _{\overline A }}(x) = 1 - {\mu _A}(x)\forall x \in X\)

Example of complement of fuzzy sets

Consider a fuzzy set  \(A = \left\{ {\left( {a,.2} \right),\left( {b,.6} \right),\left( {c,.7} \right),\left( {d,.2} \right)} \right\}\) and on \(X = \left\{ {a,b,c,d} \right\}\) .

 \({\mu _{\overline A }}(a) = 1 - {\mu _A}(a) = 1 - .2 = .8\)

\({\mu _{\overline A }}(b) = 1 - {\mu _A}(b) = 1 - .6 = .4\)

\({\mu _{\overline A }}(c) = 1 - {\mu _A}(c) = 1 - .7 = .3\)

\({\mu _{\overline A }}(d) = 1 - {\mu _A}(d) = 1 - .2 = .8\)

\(\therefore {A^c} = \left\{ {\left( {a,.8} \right),\left( {b,.4} \right),\left( {c,.3} \right),\left( {d,.8} \right)} \right\}\) 

So, we hope in this article you have learned some basics of fuzzy set theory.
The basic terminologies and properties related to fuzzy sets like, alpha-cut, strong alpha-cut, support of a fuzzy set, core of a fuzzy set, normal fuzzy set, subnormal fuzzy set, cardinality, nucleus of a fuzzy set etc. are defined and illustrated with very simple examples for better understanding.
In addition union of two fuzzy sets, intersection of two fuzzy sets and complement of a fuzzy set has been explained with examples for each case.

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.

Proof of Liouville’s theorem in Complex Analysis

Liouville’s Theorem:

The Liouville’s theorem in Complex Analysis can be considered as an improvement of Picard's little theorem. This theorem states that every bounded and entire function is constant. The theorem is can be interpreted as: every entire function whose image omits two or more complex numbers must be constant. In this article the statement and proof of the theorem is explained.

Statement of the theorem:

If \(f\) is entire and bounded in the complex plane, then \(f(z)\)is constant throughout the plane.

Proof of the theorem:

Cauchy’s inequality states that a function  \(f\) is analytic within and on a circle\(\left| {z - {z_0}} \right| = R\), taken in the positive sense, then

\(\left| {{f^n}({z_0})} \right| \le \frac{{n!{M_R}}}{R}\)

Where, \(n = 0,1,2,...\) and \({M_R}\)be the maximum value of \(f(z)\)in the circle\(\left| {z - {z_0}} \right| = R\).

Here, given that, \(f\) is entire and bounded. Therefore, \(\left| {f(z)} \right| \le M\) for some \(M\)and taking \(n = 1\)in Cauchy’s inequality we get

\(\left| {f'({z_0})} \right| \le \frac{M}{R}\)

Where \({z_0}\)is any fixed point in the plane and \(R\)can be arbitrarily large. In this inequality, \(M\)is independent of the value of \(R\). Therefore, \(R\)can be taken arbitrarily large only when\(\left| {f'({z_0})} \right| = 0\). Also, the choice of \({z_0}\)is arbitrary, which means \(f'(z) = 0\)everywhere in the complex plane. Which implies that \(f(z)\)is constant throughout the complex plane.

This proves the theorem.

The proof of Liouville’s theorem is a simple one, but its applicability in Complex Analysis is very vast. Mostly, this theorem is used in questions where we need to check whether a function is constant or not.

Proof of Cauchy’s integral formula and evaluation of complex integrals

Cauchy’s integral formula expresses that a holomorphic function defined on a disk can be determined by its values on the boundary of the disk. Also, other integral formulas can be obtained for all derivatives of a holomorphic function from Cauchy's integral formula. Cauchy's integral formula implies in complex analysis that differentiation is equivalent to integration. In addition, it has many more applications in Complex Analysis. In this post, statement and proof of Cauchy's integral formula is given. Also, some questions related to Cauchy’s integral formula  and their solution is explained in a simple and step by step ways.

Statement of Cauchy’s integral formula:

Let \(f\)be analytic everywhere within and on a simple closed contour\(C\), taken in the positive sense. If \({z_0}\) is any point interior to\(C\), then

\(f({z_0}) = \frac{1}{{2\pi i}}\int_C {\frac{{f(z)}}{{z - {z_0}}}dz\,} \)

Proof of the formula:

Since, \(f\)is analytic everywhere, which implies\(f\)is continuous, therefore there corresponds to any positive number\(\varepsilon \), however small, a positive number \(\delta \)such that

 \(\left| {f(z) - f({z_0})} \right| < \varepsilon \)whenever \(\left| {z - {z_0}} \right| < \delta \) Eq. (1)

Let us choose a positive number \(\rho \)which is less than \(\delta \)and is so small that, the small positively oriented circle\(\left| {z - {z_0}} \right| = \rho \), denoted by \({C_0}\)is interior to\(C\), as shown in the following figure:

Then, Eq. (1) becomes

\(\left| {f(z) - f({z_0})} \right| < \varepsilon \)whenever \(\left| {z - {z_0}} \right| = \rho \) Eq. (2)

Now,\(\frac{{f(z)}}{{z - {z_0}}}\) is analytic in the closed region in between \(C\)and\({C_0}\), then

\(\int_C {\frac{{f(z)}}{{z - {z_0}}}dz = \int_{{C_0}} {\frac{{f(z)}}{{z - {z_0}}}dz} } \) Eq. (3)

(If \(C\)and \({C_0}\) denotes positively oriented simple closed contour, where \({C_0}\) is interior to \(C\)and \(f\)is analytic in the closed region consisting of these contours and all points between them, then \(\int_C {f(z)dz = \int_{{C_0}} {f(z)dz} } \) )

Now, \(\int_C {\frac{{f(z)}}{{z - {z_0}}}dz - \int_{{C_0}} {\frac{{f({z_0})}}{{z - {z_0}}}dz} } \)

\( = \int_{{C_0}} {\frac{{f(z)}}{{z - {z_0}}}dz - \int_{{C_0}} {\frac{{f({z_0})}}{{z - {z_0}}}dz} } \)  from Eq.(3)

\( = \int_{{C_0}} {\frac{{f(z) - f({z_0})}}{{z - {z_0}}}dz} \)

Now, \(\int_{{C_0}} {\frac{{f({z_0})}}{{z - {z_0}}}dz}  = f({z_0})\int_{{C_0}} {\frac{1}{{z - {z_0}}}dz}  = f({z_0})2\pi i\), since  \(\int_{{C_0}} {\frac{1}{{z - {z_0}}}dz}  = 2\pi i\)

Now,\(\int_{{C_0}} {\frac{{f(z) - f({z_0})}}{{z - {z_0}}}dz} \)

\( = \int_{{C_0}} {\frac{{f(z)}}{{z - {z_0}}}dz - } \int_{{C_0}} {\frac{{f({z_0})}}{{z - {z_0}}}dz} \)

\( = \int_{{C_0}} {\frac{{f(z)}}{{z - {z_0}}}dz - f({z_0})2\pi i} \)

From Eq. (2) it is clear that \({C_0}\) is of length \(2\pi \rho \)

\(\therefore \left| {\int_{{C_0}} {\frac{{f(z)}}{{z - {z_0}}}dz - f({z_0})2\pi i} } \right|\left\langle {\frac{\varepsilon }{\rho }} \right.2\pi \rho \) , (Since,\(\left| {\int_C {f(z)dz} } \right|\left\langle {ML} \right.\), where closed contour\(C\) is of  length \(L\), \(f\)is analytic within and on \(C\) and \(M\)is such that \(\left| {f\left( z \right)} \right| < M\) )

\( = \left| {\int_{{C_0}} {\frac{{f(z)}}{{z - {z_0}}}dz - f({z_0})2\pi i} } \right|\left\langle {2\pi \varepsilon } \right.\)

Since, left hand side of the inequality is a non-negative constant that is less than any arbitrary small positive number, therefore it must be equal to zero.

\(\therefore \int_{{C_0}} {\frac{{f(z)}}{{z - {z_0}}}dz - f({z_0})2\pi i}  = 0\)

\( \Rightarrow \int_{{C_0}} {\frac{{f(z)}}{{z - {z_0}}}dz = f({z_0})2\pi i} \)

\( \Rightarrow f({z_0}) = \frac{1}{{2\pi i}}\int_{{C_0}} {\frac{{f(z)}}{{z - {z_0}}}dz\,} \)

\( \Rightarrow f({z_0}) = \frac{1}{{2\pi i}}\int_C {\frac{{f(z)}}{{z - {z_0}}}dz\,} \)
This proves the theorem.

An important Note:

 Cauchy’s integral formula can be extended to  

\({f^n}({z_0}) = \frac{{n!}}{{2\pi i}}\int_C {\frac{{f(z)}}{{{{\left( {z - {z_0}} \right)}^{n + 1}}}}dz\,} \), \(n = 0,1,2,...\); 

\({f^n}({z_0})\)denotes the \({n^{th}}\)derivative of  \(f\)at \(z = {z_0}\)

Some problems and their solutions are given below to illustrate the utility of Cauchy’s integral formula:

Question 1: Evaluate \(\int_C {\frac{z}{{(9 - {z^2})(z + i)}}dz\,} \)where \(C\)is the positively oriented circle\(\left| z \right| = 2\).

Solution: Consider\(f(z) = \frac{z}{{(9 - {z^2})}}\), which is analytic and within and on the circle\(\left| z \right| = 2\). Now, the point \({z_0} =  - i\)lies within the circle\(\left| z \right| = 2\).
Therefore, from Cauchy’s integral formula \(f({z_0}) = \frac{1}{{2\pi i}}\int_C {\frac{{f(z)}}{{z - {z_0}}}dz\,} \)
We get that \(f( - i) = \frac{1}{{2\pi i}}\int_C {\frac{{\frac{z}{{9 - {z^2}}}}}{{z - ( - i)}}dz\,} \).

\( \Rightarrow f( - i) = \frac{1}{{2\pi i}}\int_C {\frac{z}{{(9 - {z^2})(z + i)}}dz\,} \)

\( \Rightarrow \int_C {\frac{z}{{(9 - {z^2})(z + i)}}dz\,}  = f( - i)2\pi i\)

\( \Rightarrow \int_C {\frac{z}{{(9 - {z^2})(z + i)}}dz\,}  = \frac{{ - i}}{{9 - ( - {i^2})}}2\pi i\)

\( \Rightarrow \int_C {\frac{z}{{(9 - {z^2})(z + i)}}dz\,}  = \frac{{ - i}}{{10}}2\pi i\)

\( \Rightarrow \int_C {\frac{z}{{(9 - {z^2})(z + i)}}dz\,}  = \frac{{ - {i^2}}}{{10}}2\pi \)

\( \Rightarrow \int_C {\frac{z}{{(9 - {z^2})(z + i)}}dz\,}  = \frac{1}{{10}}2\pi i\)

\( \Rightarrow \int_C {\frac{z}{{(9 - {z^2})(z + i)}}dz\,}  = \frac{\pi }{5}\)

Therefore, value of the required integral is\(\frac{\pi }{5}\).

Question 2: Evaluate \(\int_C {\frac{{{e^{ - z}}}}{{z - \frac{{\pi i}}{2}}}dz\,} \)where \(C\)is the positively oriented boundary of the square \(x =  \pm 2\) and\(y =  \pm 2\).

Solution: Consider\(f(z) = {e^{ - z}}\), which is analytic on and within the boundary, given by the square \(x =  \pm 2\) and \(y =  \pm 2\) .

Now, the point \({z_0} = \frac{{\pi i}}{2}\)lies within the boundary\(C\).

Therefore, from Cauchy’s integral formula \(f({z_0}) = \frac{1}{{2\pi i}}\int_C {\frac{{f(z)}}{{z - {z_0}}}dz\,} \)

We get that \(f\left( {\frac{{\pi i}}{2}} \right) = \frac{1}{{2\pi i}}\int_C {\frac{{{e^{ - z}}}}{{z - \frac{{\pi i}}{2}}}dz\,} \)

\( \Rightarrow \int_C {\frac{{{e^{ - z}}}}{{z - \frac{{\pi i}}{2}}}dz\,}  = f\left( {\frac{{\pi i}}{2}} \right)2\pi i\)

\( \Rightarrow \int_C {\frac{{{e^{ - z}}}}{{z - \frac{{\pi i}}{2}}}dz\,}  = {e^{ - \frac{{\pi i}}{2}}}2\pi i\)

\( \Rightarrow \int_C {\frac{{{e^{ - z}}}}{{z - \frac{{\pi i}}{2}}}dz\,}  = \left( {\cos \frac{\pi }{2} - i\sin \frac{\pi }{2}} \right)2\pi i\)

\( \Rightarrow \int_C {\frac{{{e^{ - z}}}}{{z - \frac{{\pi i}}{2}}}dz\,}  = \left( {0 - i} \right)2\pi i\)

\( \Rightarrow \int_C {\frac{{{e^{ - z}}}}{{z - \frac{{\pi i}}{2}}}dz\,}  =  - 2\pi {i^2}\)

\( \Rightarrow \int_C {\frac{{{e^{ - z}}}}{{z - \frac{{\pi i}}{2}}}dz\,}  =  - 2\pi ( - 1)\)

\( \Rightarrow \int_C {\frac{{{e^{ - z}}}}{{z - \frac{{\pi i}}{2}}}dz\,}  = 2\pi \)

Therefore, value of the required integral is\(2\pi \).

Question 3: Evaluate \(\int_C {\frac{1}{{{z^2} + 4}}dz\,} \)where \(C\) is the positively oriented circle \(\left| {z - i} \right| = 2\).

Solution: Consider \(f(z) = \frac{1}{{z + 2i}}\)which is analytic within and on the circle \(\left| {z - i} \right| = 2\),

Now, the point \({z_0} = 2i\)lies within the circle\(\left| {z - i} \right| = 2\).

Therefore, from Cauchy’s integral formula \(f({z_0}) = \frac{1}{{2\pi i}}\int_C {\frac{{f(z)}}{{z - {z_0}}}dz\,} \) we get that \(f(2i) = \frac{1}{{2\pi i}}\int_C {\frac{{\frac{1}{{\left( {z + 2i} \right)}}}}{{z - 2i}}dz\,} \)

\( \Rightarrow f(2i) = \frac{1}{{2\pi i}}\int_C {\frac{1}{{{z^2} + 4}}dz\,} \)

\( \Rightarrow \int_C {\frac{z}{{{z^2} + 4}}dz\,}  = f(2i)2\pi i\)

\( \Rightarrow \int_C {\frac{z}{{{z^2} + 4}}dz\,}  = \frac{1}{{2i + 2i}}2\pi i\)

\( \Rightarrow \int_C {\frac{z}{{{z^2} + 4}}dz\,}  = \frac{1}{{4i}}2\pi i\)

\( \Rightarrow \int_C {\frac{z}{{{z^2} + 4}}dz\,}  = \frac{\pi }{2}\)

Therefore, value of the required integral is\(\frac{\pi }{2}\).

Question 4: If \(C\) is the positively oriented circle \(\left| z \right| = 1\), then evaluate\(\int_C {\frac{{{e^{2z}}}}{{{z^4}}}dz\,} \).

Solution: Consider \(f(z) = {e^{2z}}\)which is entire i.e. analytic everywhere.

Now, the point \({z_0} = 0\)lies within the circle\(\left| z \right| = 1\).

Therefore, from the formula\({f^n}({z_0}) = \frac{{n!}}{{2\pi i}}\int_{{C_0}} {\frac{{f(z)}}{{{{\left( {z - {z_0}} \right)}^{n + 1}}}}dz\,} \)and taking \(n = 3\)we get

\(f'''(0) = \frac{{3!}}{{2\pi i}}\int_C {\frac{{{e^{2z}}}}{{{{\left( {z - 0} \right)}^{3 + 1}}}}dz\,} \)

\( \Rightarrow f'''(0) = \frac{6}{{2\pi i}}\int_C {\frac{{{e^{2z}}}}{{{z^4}}}dz\,} \)

\( \Rightarrow \int_C {\frac{{{e^{2z}}}}{{{z^4}}}dz\,}  = \frac{{2\pi i}}{6}f'''(0)\)

\( \Rightarrow \int_C {\frac{{{e^{2z}}}}{{{z^4}}}dz\,}  = \frac{{\pi i}}{3}{\left( {8{e^{2z}}} \right)_{z = 0}}\)

\( \Rightarrow \int_C {\frac{{{e^{2z}}}}{{{z^4}}}dz\,}  = \frac{{8\pi i}}{3}\)

Therefore, value of the require integral is\(\frac{{8\pi i}}{3}\).

 Question 5: If \(C\) is a positively oriented circle is\(\left| z \right| = 2\) then evaluate\(\int_C {\frac{z}{{{{\left( {z - 1} \right)}^2}}}dz\,} \).

Solution: Consider\(f(z) = z\), which is analytic on and within  \(\left| z \right| = 2\) .

Now, the point \({z_0} = 1\)lies within the circle\(\left| z \right| = 2\).

Therefore, using the formula\({f^n}({z_0}) = \frac{{n!}}{{2\pi i}}\int_{{C_0}} {\frac{{f(z)}}{{{{\left( {z - {z_0}} \right)}^{n + 1}}}}dz\,} \)and taking \(n = 1\)we get

\(f'(1) = \frac{{1!}}{{2\pi i}}\int_C {\frac{z}{{{{\left( {z - 1} \right)}^{1 + 1}}}}dz\,} \)

\( \Rightarrow f'(1) = \frac{1}{{2\pi i}}\int_C {\frac{z}{{{{\left( {z - 1} \right)}^2}}}dz\,} \)

\( \Rightarrow \int_C {\frac{z}{{{{\left( {z - 1} \right)}^2}}}dz\,}  = f'(1)2\pi i\)

\( \Rightarrow \int_C {\frac{z}{{{{\left( {z - 1} \right)}^2}}}dz\,}  = 1.2\pi i\)

\( \Rightarrow \int_C {\frac{z}{{{{\left( {z - 1} \right)}^2}}}dz\,}  = 2\pi i\)

Therefore, value of the required integral  is\(2\pi i\).

So, we hope readers have grasped the overall concept of Cauchy’s integral formula and its proof. In addition, the questions and solutions on evaluation of integrals of analytic functions through closed contours will be beneficial to readers and will help in solving these types of problems.

Convergence of sequence and its examples

Convergence of a sequence

Convergence of sequence can be defined in the two ways: A sequence \(\left\{ {{S_n}} \right\}\) is said to converge to a real number \(l\) , if for each \(\varepsilon  > 0\) there exist a positive integer \(m\)(depends upon \(\varepsilon \)) such that

\(\left| {{S_n} - l} \right| < \varepsilon \) for all \(n \ge m\)

Same thing can be explained in the following form:

A sequence \(\left\{ {{S_n}} \right\}\) is said to converge to a real number \(l\)if

\(\mathop {\lim }\limits_{n \to \infty } {S_n} = l\)

Or one may write as \({S_n} \to l\) if \(n \to \infty \)


Let us consider a sequence \(\{ {S_n}\} \), where \({S_n} = \frac{{2n + 1}}{{n + 3}}\)
Here, \(\mathop {\lim }\limits_{n \to \infty } {S_n}\)

\( = \mathop {\lim }\limits_{n \to \infty } \frac{{2n + 1}}{{n + 3}}\)

 \( = \mathop {\lim }\limits_{n \to \infty } \frac{{\frac{{2n + 1}}{n}}}{{\frac{{n + 3}}{n}}}\)

\( = \mathop {\lim }\limits_{n \to \infty } \frac{{2 + \frac{1}{n}}}{{1 + \frac{3}{n}}}\)

\( = \frac{{2 + 0}}{{1 + 0}}\)

\( = 2\)

Therefore, \(\mathop {\lim }\limits_{n \to \infty } {S_n} = 2\) and we can say the sequence converges to the limit 2.

Let us consider another example.

Consider a sequence \(\{ {S_n}\} \), where \({S_n} = \frac{{{n^2}}}{{2n + 3}}\)

Here, \(\mathop {\lim }\limits_{n \to \infty } {S_n}\)

\( = \mathop {\lim }\limits_{n \to \infty } \frac{{{n^2}}}{{2n + 3}}\)

 \( = \mathop {\lim }\limits_{n \to \infty } \frac{n}{{2 + \frac{3}{n}}}\)

\( = \infty \)

 Here, \(\mathop {\lim }\limits_{n \to \infty } {S_n} = \infty \)

In this case the sequence does not converge to a limit. We say that the sequence diverges to \( + \infty \) .

Convergence of sequence is a vast topic and has tremendous usability in studying the real analysis. If you want to know the basic knowledge about sequence and its related terminologies like bounds of limit, limit point etc., you can follow the link:

Apart from convergence and divergence of sequences one more case arises in the nature of a sequence termed as oscillating sequence.

Oscillating sequence:

A bounded sequence which does not converge and has at least two limit points is said to be an oscillating sequence.

For example, consider the sequence \(\{ {S_n} = {( - 1)^n}\}  = \left\{ { - 1, + 1, - 1, + 1,...} \right\}\)

Here the sequence \(\{ {S_n}\} \) does not converge to a limit and has two limit points -1 and +1.

Therefore, \(\{ {S_n} = {( - 1)^n}\}  = \left\{ { - 1, + 1, - 1, + 1,...} \right\}\) is an oscillating sequence.

Similarly, the sequence \(\{ {S_n} = {( - 1)^n} + 1\}  = \left\{ {0,2,0,2,...} \right\}\) is an example of an oscillating sequence.  The sequence is not convergent and has two limit points 0 and 2.

Now we will discuss a very useful condition which can be used for determination of convergence of sequences very easily. The condition is given below:

The sequence \(\{ {r^n}\} \) converges if and only if \( - 1 < r \le 1\)

 Using the above result we can find some examples of sequences.
1. \(\left\{ {{1^n}} \right\}\) , \(\left\{ {{{\left( {\frac{1}{2}} \right)}^n}} \right\}\) , \(\{ {\left( { - .8} \right)^n}\} \) are few examples of convergent sequence.

2. \(\left\{ {{{1.2}^n}} \right\}\) , \(\left\{ {{{\left( { - \frac{3}{2}} \right)}^n}} \right\}\) , \(\{ {\left( { - 2} \right)^n}\} \) are few examples of non-convergent sequence.

Now we will prove the above statement. To prove the above statement we break down the proof into five cases. For the task, we need to prove the following five cases.

Case 1: Suppose \(r \in (1,\infty )\) or  \(r > 1\)
Then r can be written as

\(r = 1 + h\) where \(h > 0\) {e.g. 1.2=1+.2}

\( \Rightarrow {r^n} = {\left( {1 + h} \right)^n} = 1 + nh + \frac{{n(n - 1)}}{{2!}}{h^2} + ...\)

\( \Rightarrow {r^n} = 1 + nh + \frac{{n(n - 1)}}{{2!}}{h^2} + ... \ge 1 + nh\) for all \(n \in N\)

Now, let us consider \(G\) be any number however large,

we have \(1 + nh > G\)

if  \(nh > G - 1\)

or \(n > \frac{{G - 1}}{h}\)

Let \(m\) be any positive integer such that \(m > \frac{{G - 1}}{h}\) 

Therefore, for \(G > 0\) there exist a positive integer such that

\({r^n} = 1 + nh + \frac{{n(n - 1)}}{{2!}}{h^2} + ... \ge 1 + nh > G\)

\( \Rightarrow {r^n} > G\) for all \(n \ge m\)

Since, the term of the sequences are sufficiently large for \(n \ge m\) , therefore the sequence diverges to \(\infty \) for \(r > 1\).

Case 2: Suppose \(r = 1\)

\( \Rightarrow \left\{ {{r^n}} \right\} = \left\{ {1,1,1,...} \right\}\)

i.e. \(\mathop {\lim }\limits_{n \to \infty } {r^n} = 1\)

Therefore, \(\left\{ {{r^n}} \right\}\) converges for \(r = 1\) .

Case 3: Suppose \(r \in ( - 1,1)\)

Then \(r\) can be written as

\(\left| r \right| = \frac{1}{{1 + h}}\) where \(h > 0\) {e.g. .2=1/(1+4)}

\( \Rightarrow {\left| r \right|^n} = \frac{1}{{{{\left( {1 + h} \right)}^n}}} = \frac{1}{{1 + nh + \frac{{n(n - 1)}}{{2!}}{h^2} + ...}}\)

\( \Rightarrow {\left| r \right|^n} = \frac{1}{{1 + nh + \frac{{n(n - 1)}}{{2!}}{h^2} + ...}} \le \frac{1}{{1 + nh}}\) for all \(n \in N\)

\( \Rightarrow {\left| r \right|^n} \le \frac{1}{{1 + nh}}\) for all \(n \in N\)

Now, let us consider \(\varepsilon \) be any number however large,

we have \(\frac{1}{{1 + nh}} < \varepsilon \)

if  \(\frac{{1 - \varepsilon }}{{\varepsilon h}} < n\)  ; on simplification.

Let \(m\) be any positive integer such that \(m > \frac{{1 - \varepsilon }}{{\varepsilon h}}\)

Therefore, for \(\varepsilon  > 0\) there exist a positive integer  \(m\) such that

\( \Rightarrow {\left| r \right|^n} \le \frac{1}{{1 + nh}} < \varepsilon \)

\( \Rightarrow {\left| r \right|^n} < \varepsilon \) for all \(n \ge m\)

\( \Rightarrow {\left| {r - 0} \right|^n} < \varepsilon \) for all \(n \ge m\)

Therefore, the sequence \(\left\{ {{r^n}} \right\}\) converges to the point 0.

Case 4: Suppose \(r =  - 1\)

\( \Rightarrow \left\{ {{r^n}} \right\} = \left\{ { - 1, + 1, - 1,...} \right\}\)

i.e. \(\mathop {\lim }\limits_{n \to \infty } {r^n} =  - 1\,or\, + 1\)

Therefore, \(\left\{ {{r^n}} \right\}\) has two limit points and it is bounded. Hence it will be an oscillating sequence for \(r =  - 1\).

Case 5: Suppose \(r <  - 1\)

We consider \(r =  - z\) such that \(z > 0\)

\( \Rightarrow \left\{ {{r^n}} \right\} = \left\{ {{{\left( { - z} \right)}^n}} \right\} = \left\{ {{{\left( { - 1} \right)}^n}{{\left( z \right)}^n}} \right\} = \left\{ { - z,{z^2}, - {z^3},...} \right\}\)

Therefore, \(\left\{ {{r^n}} \right\}\) has both positive and negative terms and is also unbounded. Thus, the sequence will be an oscillating sequence.

So, finally we get
From Case 1 \(\left\{ {{r^n}} \right\}\)is divergent for  \(r > 1\)

From Case 2 \(\left\{ {{r^n}} \right\}\)is convergent for  \(r = 1\)

From Case 3 \(\left\{ {{r^n}} \right\}\)is convergent for \(r \in ( - 1,1)\)

From Case 4 \(\left\{ {{r^n}} \right\}\)is oscillating for \(r =  - 1\)

From Case 5 \(\left\{ {{r^n}} \right\}\)is oscillating for \(r <  - 1\)

Therefore, finally we can conclude that \(\left\{ {{r^n}} \right\}\)  converges if and only if \( - 1 < r \le 1\) .

I hope, this article will help the readers to get a good idea about convergence of a sequence, divergence of a sequence and oscillating sequence. For each definition, examples are given for better understanding about this topic. Finally, an important result has been proved which is used more often for checking of convergence of sequences of the type \(\left\{ {{r^n}} \right\}\). 

Sequence of real numbers and related terminologies with examples

Sequence of real numbers:    

A function whose domain is the set of natural number and range is the set of the real number is defined as the sequence of real numbers. i.e. any function \(f:N \to R\) can be said to real sequence. In general language sequence can be understood as a collection of objects or events with some kind repetition or similarity. In Real Analysis, the concept of sequence is vaster, but the overall linguistic view is a little bit similar. In this article, the definition of sequence, bounds of sequence and the well-known Bolzano weierstrass property is discussed with simple examples.
Sequences are denoted by \(\left\{ {{S_n}} \right\}\) , for example \(f(x) = {x^2}\) defined from \(N\) to \(R\) can be considered  as a sequence. More conveniently, the sequence can be written as

\({S_n} = \{ {n^2}\}  = \{ {1^2},{2^2},{3^2},....\}  = \{ 1,4,9,....\} \) 

A sequence may have finite number of points, called finite sequence and may have or infinite number of points, called infinite sequence. The characteristic of a sequence can be determined by how bigger or how smaller its elements are. Therefore, the concept of boundedness is evolved in sequence which is elaborated below.

Bounds of a sequence:

A sequence \(\left\{ {{S_n}} \right\}\) is said to be bounded above if there exist a number \(k\)such that
\({S_n} \le k\)for all \(n \in N\)

For example: Let us consider the sequence \(\{ {S_n}\} \), where \({S_n} = \frac{1}{n}\)
Therefore, the terms of the sequence are \(\{ {S_n}\}  = \left\{ {1,\frac{1}{2},\frac{1}{3},..} \right\}\)
Here, \({S_1} = 1,{S_2} = \frac{1}{2},{S_3} = \frac{1}{3},..,{S_n} = \frac{1}{n},...\)
By simply looking at the terms we can see that  \(1,\frac{1}{2},\frac{1}{3},.. \le 1\)
i.e. \({S_1},{S_2},{S_3},..,{S_n},... \le 1\)
or we can say that \({S_n} \le 1\)for all \(n \in N\)
Therefore, the sequence \(\left\{ {{S_n} = \frac{1}{n}} \right\}\)is said to be bounded above.

A sequence \({S_n}\) is said to be bounded below if there exist a real number \(k\)such that
\(k \le {S_n}\)for all \(n \in N\)

For example: Let us consider the sequence \(\{ {S_n}\} \), where \({S_n} = {n^2}\)
Therefore, the terms of the sequence are \(\{ {S_n}\}  = \left\{ {1,4,9,..} \right\}\)
Here, \({S_1} = 1,{S_2} = 4,{S_3} = 9,..,{S_n} = {n^2},...\)
By simply looking at the terms we can see that  \(1 \le 1,4,9,..\)
i.e. \(1 \le {S_1},{S_2},{S_3},..,{S_n},...\)
or we can say that \(1 \le {S_n}\)for all \(n \in N\)
Therefore, the sequence \(\left\{ {{S_n} = {n^2}} \right\}\)is said to be bounded below.
More generally, a sequence is said to be bounded if the sequence is both bounded above and bounded below.

Limit point of a sequence:

A number \(\xi \)  is said to be a limit point of a sequence \(\left\{ {{S_n}} \right\}\) if every open interval consisting the point \(\xi \) contains an infinite number of elements of the sequence.
In a more general and simple way \(\xi \)  is said to be a limit point of a sequence \(\left\{ {{S_n}} \right\}\) if  \(\left( {\xi  - \varepsilon ,\xi  + \varepsilon } \right)\) contains an infinite number of points of the sequence, where \(\varepsilon \) is any positive number.

For example, let us consider the sequence \(\{ {S_n}\}  = \left\{ {1,\frac{1}{2},\frac{1}{3},..} \right\}\)
We consider \(\xi  = 0\) and \(\varepsilon \) be any positive number.
Then the open interval \(\left( {0 - \varepsilon ,0 + \varepsilon } \right) = \left( { - \varepsilon , + \varepsilon } \right)\)contains an infinite number of elements of the sequence \(\left\{ {{S_n}} \right\}\).
Therefore, \(\xi  = 0\) is a limit point of the sequence\(\left\{ {{S_n}} \right\}\).

Now, obviously one question may arise in the mind of the readers. “Does every sequence have limit points?” The answer of the question lies in the most fundamental theorem of the real sequence. The theorem is known as Bolzano weierstrass theorem. This theorem relates the existence of the limit point of a sequence with the boundedness of the sequence. The statement of the theorem is given below:

Bolzano weierstrass theorem:

Every bounded sequence has a limit point.
In a simple way, we can say that a sequence having its upper and lower bound must have a limit point.

For example, let us consider the sequence \(\{ {S_n}\} \), where \({S_n} = \frac{1}{n}\)
Here, \({S_1} = 1,{S_2} = \frac{1}{2},{S_3} = \frac{1}{3},..,{S_n} = \frac{1}{n},...\)
By simply looking at the terms we can see that  \(0 < 1,\frac{1}{2},\frac{1}{3},.. \le 1\)
i.e. the sequence is bounded above by 1 and bounded below by 0. Thus, the given sequence is bounded.
Therefore, by Bolzano weierstrass theorem one can say that the sequence has a limit point.
The limit point is 0, which is already shown in the earlier example.

In this article, the definition of real sequences and its related terminologies (limit point of a sequence, the bounds of a sequence and the well-known Bolzano weierstrass property or Bolzano weierstrass theorem) are discussed with examples. For each terms, one example is given for good understanding about the terms related to sequences.