The post How to integrate the function? Explained with examples appeared first on Math Academy.

]]>Students face difficulties while integrating the functions manually due to larger and more complex calculations. In this post, we are going to explain the methods to integrate the functions along with examples and solutions.

In calculus, a term that is used to evaluate a new function or the numerical value of the function with respect to the integrating variable is known as integral. The process of integrating the function is known as integration.

It is generally used to evaluate the area under the curve. The rules and formulas of the integration are helpful in integrating the functions with the help of the corresponding variable of the function.

The new function that is evaluated by integrating the function whose original function is derivative is known as the indefinite integral. The constant of integration must be written along with the new function in this type of integral.

The numerical value of the function that is evaluated by placing the upper and lower limit values to the integrated function is known as the indefinite integral. The fundamental theorem of calculus is applied to enter the upper and lower limit values of the function.

Follow the below steps to integrate the function.

- First of all, take the given function and integrate the variable.
- Apply the notation of integration to the function.
- Apply the sum and difference rules of integration and write the notation of integral to each function separately.
- Take out the constant coefficients outside the integral notation.
- Integrate the functions by using power and trigonometry rules of integral.
- Apply the limit values to the integrated function in the case of the definite integral.

The formula for evaluating the new function (indefinite integral) is:

The formula for evaluating the numerical value of the function (definite integral) is:

An integral calculator can be used to integrate the function according to the above formulas to calculate the new function or the numerical value of the function.

Here are some examples of integral calculus to learn how to integrate the functions. The rules are very essential in calculating the numerical values and new functions.

Integrate the given function with respect to .

**Solution**

**Step 1:** First of all, take the function f(y) and the integrating variable that is corresponding to the function. After that apply the notation of integral to the given function.

Integrating variable =

**Step 2:** Now apply the notation of integral to each function separately to each function with the help of the sum and difference rules of integral calculus.

**Step 3:** Now take out the constant coefficients outside the integral notation.

**Step 4:** Now integrate the above expression with the help of power and trigonometry laws of integration.

Integrate the given function with respect to

Solution

**Step 1:** First of all, take the function f(z) and the integrating variable that is corresponding to the function. after that apply the notation of integral to the given function.

integrating variable

**Step 2:** Now apply the notation of integral to each function separately to each function with the help of the sum and difference rules of integral calculus.

**Step 3:** Now take out the constant coefficients outside the integral notation.

**Step 4:** Now integrate the above expression with the help of the power law of integration.

**Step 5:** Now use the fundamental theorem of calculus to apply the upper and lower limit values to the above-integrated value.

Now you can integrate any function by following the above formulas and solved examples of the integral calculus. We have discussed almost every basics of the integral calculus and take care of the intent of the article.

The post How to integrate the function? Explained with examples appeared first on Math Academy.

]]>The post Laplace Transformation appeared first on Math Academy.

]]>Integral transformation data back to the Work of Leonard Euler (1763 & 1769), who considered them essentially in the form of the inverse Laplace transform in solving second-order, linear ordinary differential equations. Even Laplace, in his great work, “Theorie Analytique Des Probabilities”(1812), credits Euler with introducing integral transforms. It is Spitzer (1878) who attached the name of Laplace to the expression

Employed by Euler. In this form, it is substituted into the differential equation where is the unknown function of the variable .

In the late 19th century, the Laplace transform was extended to his complex form by Poincare and Pincherle, Redis covered by Petzval, and extended to two variables by Picard, with the further investigation conducted by Abel and many others.

The first application of the modern Laplace transformations equations arose of Batemer (1910), who was working on radioactive decay

By setting

and obtaining the transformed equation. Bernstein (1920) used the expression

called it the Laplace transformation, in his work on theta functions. The modern approach was given particular impetus by Doetsch in the 1920s and 30s; he applied the Laplace transform to differential, integral, and integro-differential equations. This body of work culminated in his foundational 1937 text, “Theorie and Anwendungender Laplace Transformation.”

No account of the Laplace transformation would be complete without mention of the work of Oliver Heaviside, who produced a vast body of what is termed the “Operational Calculus”. The material is scattered throughout his three volumes, Electromagnetic Theory (1894, 1899, 1912), and bears many similarities to the Laplace transform method. Although Heaviside’s calculus was not the theory of Laplace Transformation, he did find favour with electrical engineers as a useful technique for solving their problems. Considerable research went into trying to make the Heaviside calculus rigorous and connecting it with the Laplace transform. One such effort was that of Bromwich, who, among others, discovered the Inverse Laplace Transformation!

Where lying to the right of all singularities of the function .

In this article, we shall consider theoretic aspects of the Laplace transform, reserving for the application of the subject to the solution of linear differential equations. This transformation is defined by the equation;

A function is called sectionally continuous or piece-wise continuous in any interval if it is continuous and has finite left and right-hand limits in every subinterval as shown in the graph of the function

A function is said to be of exponential order as

if

i.e., if given a positive integer, there exists a real number

Or

Briefly, we say that is of exponential order.

Sometimes we write,

For example. is of exponential order as being any positive integer.

So we are going to do L’Hospital Rule where we differentiate numerator and denominator individually,

hence,

(by L. Hospital’s rule)

=

Since.

Hence we can say that is an exponential order as.

Another example. is not of exponential order as

Taking,

Therefore,

is not of exponential order

An improper integral of the form is called Integral Transform of if it is convergent. Sometimes it is denoted by or Thus

The function appearing in the integrand is called Kernel of the transform. Here’s a parameter and is independent of may be real or complex number.

If we take

Then becomes

This transform is known as Laplace transform.

Some examples of well-known transformations

Hankel Transform of is

Mellin Transform of is

Fourier transform of is

Suppose is a real valued function defined over the interval

The Laplace Transform of, denoted by, is defined as

We also write

Here L is called Laplace transformation operator. The parameter s is a real or complex number. In general, the parameter is taken to be a real positive number. Sometimes we use symbol for parameter .

The Laplace transform is said to exist if the integral (1) is convergent for some value of .

The operation of multiplying by and integrating from is called Laplace Transformation.

Now we are going to see the existence of Laplace transformation on a function

Existence Theorem of Laplace Transform.

Statement: – If is a function of class , then Laplace transform of exists.

Or,

Suppose is piece-wise continuous in every finite interval and is of exponential order an as. Then exists , i.e., Laplace transform exists.

Proof. Let be piece-wise continuous in every finite interval and of exponential order as

To show that

Let Then

Continuity of in the finite interval implies that

exists. It remains to show that

exists is of exponential order a implies

is finite, i.e., given a number , there exists a real number

i.e.,

can be made as small as we please choosing sufficiently large. Hence exists

Fact: The condition given in this part is sufficient for the existence of, but are not the necessary conditions.

Result 1.

Proof.

Result 2.

Proof.

Also if is positive integer.

Result 3.

For

Result 4.

Proof.

Result 5. Find

Proof.

Linear Property.

Suppose and are Laplace forms of and respectively. Then

Where and are any constants.

First Shifting Theorem (First Translation)

Proof. Let

Then,

Fact: This theorem can also be restated as:

If is the Laplace transform of and a is any real, or complex number, then is the Laplace transform of). That is to say,

Some examples related with the previous properties and we are going to know how to use these properties to solve a problem

Example.

Solution.

Solution. We know that,

Example. Let’s we are going to prove,

Solution. We know that

Using the result, , we get

Theorem 4. Second shifting Theorem (Second Translatation).

Statement:- If

then

Or

And

Example . We are going to find Laplace transform of where

Solution. then

Also

By second shifting theorem,

Theorem 5. Change of Scale Property.

If

Laplace Transform of Derivatives:

An Alternate form of Theorem 6.

To express the Laplace transform of the function in terms of the Laplace transform of the function

Generalising the results (1) and (2), we get the required result.

Theorem 7. Laplace Transform of Integral.

and

… (1)

We prove that

From (1) it is clear that

Periodic Functions :

Problem . Using Laplace transform,

solve where

Solution. Taking Laplace transform of given question,

A few worked examples should convince the reader that the Laplace transform furnishes a useful technique for solving liner differential equations. Lastly thanks to all of you and hope you all enjoy this article. We will meet next time with a new article.

Thank You.

References/Bibliography ⟶

This is merely a brief selection. An extensive list is given in reference –

Laplace Transforms and their applications to Differential equations. – N. W. MCLACHLAN

The Laplace Transform: Theory and Applications – Joel L. Schiff

(iii) Ordinary & Partial Differential

(iv) Advanced Calculus – Davis V. Widder

The post Laplace Transformation appeared first on Math Academy.

]]>The post Compact Sets and Continuous Functions on Compact Sets appeared first on Math Academy.

]]>In advance analysis, the notion of ‘Compact set’ is of paramount importance. In , Heine-Borel theorem provides a very simple characterization of compact sets. The definition and techniques used in connection with compactness of sets in are extremely important. In fact, the real line sets the platform to initiate the idea of compactness for the first time and the notion of compactness plays its important role in topological spaces.

The definition of compactness of sets in uses the notation of open cover of sets in . For this propose we need some definitions and illustrative examples to clear the meaning of cover of a set in .

**Definition (Cover):** Let be subset of and be a collection of sub sets of . is said to cover or, in other words, is said to be a covering of if

i.e. if for some .

If , for each , is an open set and then is said to be an open cover of .

**For example **

- (i) The family is an open cover of , since and the sets in family are open sets.
- (ii) The family is a collection of open sets in , but is not a cover of , since and do not belong to .
- (iii) The family is a collection of open sets in and for some . Hence is an open cover of .

**Note:** If be a collection of open intervals in such that then is also an open cover of .

**Definition (Sub-Cover):** Let and be a collection of sets in which covers . If be a sub-collection of such that itself is a cover of then is said to be a sub cover of . If is a finite sub collection of such that is a cover of then is said to be a finite sub cover of .

**For example** if then is an open cover of .

If then and ; this implies is an open sub cover of .

**Definition (Countable set):** set in is said to be a countable set if either it is finite or if it is infinite, it is enumerable i.e. there exists a bijective mapping from to .

**For example **

- (i) every finite set is countable,
- (ii) are all countable sets,
- (iii) the sets are not countable sets.

**Definition (Countable Sub cover):** Let and be a collection of sets in such that covers . If be a countable sub-collection of such that covers then is said to be a countable sub cover of .

**For example**, if then is an open cover of and is a countable sub-collection of , since if , the set has one-to-one correspondence with . also covers . Hence is a countable sub cover of . Note that there are infinitely many countable sub covers of , since is a countable set and is the set of open intervals with rational end points and hence itself is a countable family of open sets so that every sub cover of is countable.

We now give some examples of open cover of a set in which has no finite sub cover.

**Example 1.** Let and where . Show that is an open cover of but it has no finite sub cover.

**Solution:** Let . Then and by the Archimedean property of , there exists a natural number such that for some . Hence . This shows that is a collection of open sets in which covers i.e is an open cover of .

If possible, let where are natural numbers such that i.e. covers .

Let then and for

Since but , so we have a contradiction. Hence is not a cover of .

Thus there exists no finite subcollection of that will cover .

**Example 2.** Let and . Show that is an open cover of , but it has no finite sub cover.

**Solution:** Let . Then . By the Archimedean property of there exists a natural number such that for some , which shows that is an open cover of ( is also a countable cover of ).

If possible, let where are natural numbers such that i.e. is a sub cover of .

Let . Then and for all .

Thus but , a contradiction.

Thus it is proved that no finite subcollection of can cover .

**Example 3.** Let and Let . Show that is an open cover of but no finite subcollection of can cover .

**Solution:** Let . Then . For . Let . Then . By the Archimedean property of , there exists a natural number such that . Since , for for some natural number .

Hence is an open cover of .

If possible, let be a finite subcollection of , where are natural numbers , such that .

Let and then for natural numbers .

Thus . Since . Hence but , which is a contradiction. Hence no finite subcollection of covers .

**Example 4.** Let and . Let where . Show that is an cover of but it has no finite sub cover.

**Solution:** Let . Then .

Hence which implies is an open cover of .

If possible, let where for and .

Let

And .

Then

. Since but they do not belong to , we have a contradiction. Hence has no finite sub-collection that can cover , i.e. has no finite sub cover.

**Example 5.** The collection of open intervals is an uncountable cover of but where is the set of integers, is countable sub cover of .

**Solution:** Let . Since is both unbounded above and unbounded below, thee always exist two real numbers and such that such that in such that . Hence .

Thus . As open interval is an uncountable subset of , so is an uncountable cover of .

Thus . Then by the Archimendeon property of real numbers, there exists an integer such that . This implies for some . Hence . Thus is a countable sub cover of .

**Definition (Compact set): ** set () is said to be a compact set in if every open cover of has a finite sub cover. More explicitly, is said to be compact if for any open cover of , there is a finite sub collection of such that i.e. is a finite sub-cover of .

**NOTE:** To prove that a set is compact in , we must examine an arbitrary collection of open sets whose union contains , and show that is contained in the union of some finite number of sets in the given collection, i.e. we must have to show that any open cover of has a finite sub-cover. But to prove that a set is not compact, it is sufficient to choose one particular open cover has no finite sub-cover, i.e. union of any finite number of sets in fails to contain .

** Statement:**– close and bounded subset of is a compact set in , or in other words every open cover of a closed and bounded subset of has a finite sub cover.

** Proof**, Let be a closed and bounded subset of .

Let be an open cover of . We assume that has no finite sub-cover. Then is not a subject of the union of finite number of open sets in .

Since H is a bounded subject of , there exist real number such that .

Let . If then at least one of the two subsets and are subset of the union of finite number of open sets in , for otherwise both and are subsets of the union of finite number of open sets in contains , contradicting our assumption that has no finite sub-cover.

We call or according as and it is not a subset of the union of finite number of open sets in or and it is not a subset of the union of finite number of open sets in .

Let and . The at least one of the subsets and is non-empty and it is not a subset of the union of finite number of open sets in . If the first subset is non-empty and it is not a subset of the union of finite number of open sets in , we call , otherwise we call .

Let , and .

Continuing this process of bisection of intervals, we have a family of close and bounded intervals such that

, for all ,

For all is non-empty and it is not a subset of the union of finite number of open sets in .

such That as .

Then by Nested Interval Theorem, , a singleton set. We shall show that .

Since , for any positive , there exists a natural number such that i.e. and . Hence . Since and it is not a subset of the union of finite number of open sets in , contains infinite number of elements of is a limit point of . Since is closed, .

Now for some . is an open set, hence there exists a positive and hence . Since is an open cover of , for some , which goes against the construction of (.

Hence our assumption that is not a subset of the union of finite number of sets in is wrong and it is established that if is closed and bounded, any open cover of has a finite sub cover so that is a compact set in .

**Remark:** In the Heine-Borel theorem neither of the two conditions (i) is closed (ii) is bounded can be dropped. The theorem fails if one of the two conditions is withdrawn – this is evident if we go through the example 1.2.3 and the example 1.2.1. In example 1.2.3, is closed but bounded and in example 1.2.1, is closed but no bounded.

Thus both the conditions (i) and (ii) are necessary for a set in to be compact. Next we shall show that these two conditions are also sufficient for a set to be compact in .

** Statement:**– compact subset of is closed and bounded in .

** Proof.** Let be a compact in . First we shall prove that is a closed set in .

Let and . Then exist two positive numbers and such that .

Let .

Then is an open cover of is compact, has a finite sub cover i.e. there exist elements of and positive numbers such that . For each there exists positive numbers such that .

Let .

Then . Therefore is an interior point of .

Since is arbitrary point of , is open. Hence is closed.

Nest we shall prove that is bounded.

Let be a fixed positive number. Then is an open cover of . Since is compact, has a finite sub-cover. Then there exist points of such that is a finite cover of . If and then is bounded.

Hence it is proved that if is a compact set in , it is closed and bounded in . This completes the proof.

Combining the theorems 1 and 2 we have the following theorem which gives a complete characterization of compact sets in .

**Note:** – set in is compact if and only if is closed and bounded in .

**Definition (Heine-Borel Property):** set () is said to possess Heine-Borel property if every open cover of has a finite sub cover.

set is said to be compacted if it has the Heine-Borel property.

**Example 6.** Using the definition of compact set, prove that the set is not compact although it is a closed set in .

**Solution:** In example 1.2.1, it is shown that , where , is an open cover of and has no finite sub cover. Hence from definition is not compact.

is a closed set in , since is open.

**Note:** In the example 1.3.1, does not satisfy Heine-Borel property, since is not bounded in .

**Example 7.** Using definition of compact set show that a finite subset of is a compact set in .

**Solution:** Let be a finite subset of . Let be an open cover of . Then each is contained in some open set of for some . Let . Then . Thus also covers .

Hence is a finite sub cover of . Therefore, by definition, is a compact set in .

**Statement:**– If be a compact subset of , then every infinite subset of has a limit point belonging to .

Let be an infinite subset of the compact subset of of such that has no limit point belonging to .

Let . Then is not a limit point of . There exists a positive such that where , called deleted .

Let , which is a collection of open sets in . Since so is an open cover of .

Since is compact, there exists a finite sub collection of where and such that covers i.e.

i.e.

[Since and for ]

which shows that is a finite subset of a compact set in has a limit point in .

**Statement:** – If be such that every infinite subset of has a limit point in then is closed and bounded in .

**Proof.** First we shall prove that is bounded. If possible, let be unbounded above. Let be any point of . Since is unbounded above, there exists a point in such that . By similar argument there exists a point in such that and so on. Continuing this process indefinitely we ultimately have an infinite subject of , which being a discrete set, has no limit point in is a bounded above subset of . Similarly, If is unbounded below we can construct an infinite subset of which has no limit point. Hence is also bounded below so that is a bounded subset of .

Next we shall prove that is closed in .

Since S is an infinite and bounded subset of , by the Bolzano-Weierstrass theorem on set, has a limit point in .

Let be a limit point of . Then for any is infinite.

For is infinite. Let us take a point .

For is infinite. Let us take a point ,

such that , Continuing this process, we have an infinite subset

of such that

for . We shall show that has a unique limit point which is .

Let be any positive number. Then by the Archimedean property of , there exists a natural number such that and contains infinite subset of . Thus for every positive , is infinite which proves that is a limit point of .

To prove uniqueness, let be a limit point of . Let . Then the neighborhoods and are disjoint (since either or ). By the Archimedean property of , there exists a natural number such that . Since each of belongs to , so contains all elements of expect some finite number of elements and hence can contain almost finite number of elements of . This implies is not a limit point of . Hence is the only limit point of . By the condition of the theorem . Hence is closed.

Thus it is proved that is closed and bounded in .

**Note:** – subset of a compact subset of is compact if and only if every infinite subset of has a limit point belonging to .

**Statement:** – A Subset of is compact if and only if every sequence in has a subsequence that converges to a point in .

**Proof.** Let be compact. Then is closed and bounded.

Let be a sequence of points in . Since is bounded, is bounded. By the **Bolzano-Weierstrass theorem** on sequence, there exists a subsequence of that converges to a point, say . Since is closed, if , and is open. Then there exists a neighborhood of which contains no point of . This implies contains no element of the sequence which contradicts that . Thus . Hence every sequence in has a subsequence converging to a point of .

Suppose is not closed. Then has a limit point, say which is not in . Since is a limit point of , there is a sequence in , where for all , such that . Then every subsequence of converges to . Since , there is no subsequence of that converges to a point of .

Suppose is not bounded. Then there exists a sequence in such that for all . Then every subsequence of unbounded sequence is unbounded and hence no subsequence of converges to a point in .

Hence, by contrapositive argument, it is proved that if every sequence in has a subsequence that converges to a point of the is closed and bounded and hence by Heine-Borel theorem is compact.

**Note:** Following **theorem 5**., an alternative definition of compact set can be given in the from:

**“A set in is called a compact set in if every sequence in has a subsequence that converges to a point of .”**

**Theorem 5** and the **Heine-Borel **theorem together prove the equivalence of the two definitions.

**Example 8.** If is a closed subset of a compact set in then using definition of compact set, prove that is compact.

**Solution:** is open, since is closed.

Let be an open cover of . Suppose is not an open cover of . Let . Then i.e., is an open cover of . Since , is also an open cover of . being compact, has a finite sub collection such that , where . must contain , otherwise which implies is an open cover of , which is contrary to our assumption.

Since we have .

Which shows that,

is a finite sub collection of and covers , which implies is a finite sub cover of . Therefore, by definition, is compact.

**Example 9. Every compact set in has greatest as well as least element.**

**Solution:** Let be any compact set in . If possible, let have no greatest element. Then for each element . let . Then is an open set. Let , a family of open sets in . Let . Since has no greatest element, there exists an element in such that . Thus . Hence is an open cover of . being compact, has a finite sub-cover, say .

Let Then for and . Let .

Then and . This leads to a contradiction, since . Hence our assumption is not tenable and has greatest element.

To prove the next part, let, if possible, has no least element. Then for each element , let is an open set. Let , a family of open sets in . Let . Since has no least element there exists an element in such that . Then and . Thus is an open cover of . Since is compact, has a finite sub collection that covers .

Let where and . If then .

Now but which contradicts our assumption. Hence has least element.

**Example 10.** If and are component sets in , show that is also compact. Give an example to show that union of an infinite number of compact sets in is not necessarily a compact set in .

**Solution:** Let be a family of open sets in such that i.e. is an open cover of .

Since and ,

is an open cover of both and . Since and are both compact sets in , then there exist two finite sub collections

and of such that

and where and .

Let . Then is a finite sub collection of such that

or

for some

Or for some

Hence

cover . Hence, from definition, is compact.

**Second Part.**

Let . Then for each , is closed and bounded set in and by Heine-Borel Theorem, is compact for every . Thus is an infinite collection of compact sets in .

But , which is not a compact set (see example 1.2.2). Hence union of infinite number of compact sets in is not necessarily compact.

**Example 11.** Let be a closed subset of and be a component subset of . Prove that is component.

**Solution:** Since is compact, by converse of Heine-Borel theorem is closed. being closed, is a closed subset of compact set . Then following exactly similar arguments given in example 1.3.3 (replacing \mathbb{R}\sigma = \{C\}\mathbb{R}S =\ \bigcap_{C\in\sigma} CC\sigmaSR\mathbb{R}-Sg’ = \{G : \alpha \in \Lambda\}\mathbb{R}SS \subset \ \bigcap_{\alpha\in\land} G_\alphagCC \in \sigmag’\{G : \alpha \in \Lambda\}\mathbb{R} – S\mathbb{R} \subset \left\lbrace\bigcup_{\alpha\in\land} G_\alpha\right\rbrace \cup \left(\mathbb{R} – S\right)C \subset \mathbb{R}C \subset \left\lbrace \bigcup_{\alpha\in\land} G_\alpha\right\rbrace \cup \left(\mathbb{R} – S\right)C \in \sigmag’CC \in \sigmaCC \in \sigmag”g’g”CC \in \sigmaf” = \{G_{\alpha_1},\ G_{\alpha_2}, \ldots,\ G_{\alpha_m},\ \mathbb{R}\ – S\}g”\mathbb{R} – SC \subset \bigcup_{i=1}^{m}G_{\alpha_i}gCC \in \sigmaC \subset \left\lbrace\bigcup_{i=1}^{m}G_{\alpha_i}\right\rbrace \cup (\mathbb{R} – S)C \in \sigmaS \subset CS \subset \bigcup_{i=1}^{m}G_{\alpha_i}g”’ = \{G_{\alpha_1},\ G_{\alpha_2}, \ldots,\ G_{\alpha_m}\}g”’gSgS(0, 1]\mathbb{R}I_n = \left(\frac{1}{n+1},\ \frac{n+1}{n}\right), n \in \mathbb{N}\sigma = \{I_n : n \in \mathbb{N}\}\sigma\mathbb{R}x \in (0, 1]x = 1x \in I_nn \in \mathbb{N}0< x < 1\mathbb{R}mm \le\ \frac{1}{x} < m + 1

\Rightarrow \frac{1}{m+1} < x \le\ \frac{1}{m} <\ \frac{m+1}{m} \Rightarrow x \in I_mm \in \mathbb{N}(0, 1] \subset \bigcup_{n\in\mathbb{N}} I_n\sigma(0, 1]\sigma’ = \{I_{r_1},\ I_{r_2}, \ldots,\ I_{r_m}\}\sigmar_1, r_2, \ldots, r_m(0, 1] \subset \bigcup_{k=1}^{m}I_{r_k}u = \max \{r_1 + 1, r_2 + 1,\ldots, r_m + 1\} \Rightarrow u \geq r_k + 1k = 1, 2, \ldots, mv = \min \{\frac{r_1}{r_1+ 1},\ \frac{r_2}{r_2+ 1}, \ldots,\ \frac{r_m}{r_m+\ 1}\} \Rightarrow v \leq \frac{r_k}{r_k+ 1}k = 1, 2, \ldots, m0 < \frac{1}{u} \leq \frac{1}{r_k+ 1} < \frac{r_k+ 1}{r_k} \leq \frac{1}{v},\ x = 1, 2,\ldots, m\Rightarrow I_{r_k} \subset \left(\frac{1}{u},\ \frac{1}{v}\right)k = 1, 2,\ldots, m\Rightarrow (0, 1] \subset \left(\frac{1}{u},\ \frac{1}{v}\right)0 < \frac{1}{u} < 1\frac{1}{u} \in (0, 1]\frac{1}{u} \left(\frac{1}{u},\ \frac{1}{v}\right)\sigma(0, 1](0, 1]a < x < b\Gamma = \{(x – \varepsilon, x + \varepsilon), \varepsilon > 0\}\Gamma[a, b]I_x = (x – \varepsilon, x + \varepsilon),\ \varepsilon > 0a < x < b\Gamma = \{I_x : a < x < b\}x(a, b)I_x\mathbb{R}x_1, x_2 \in (a, b)a < x_1 < a + \varepsilonb – \varepsilon < x_2 < bx_1 – \varepsilon < a,\ x_1 + \varepsilon > ab < x_2 + \varepsilon, x_2 – \varepsilon < ba \in I_{x_1}b \in I_{x_2}[a, b] \subset \bigcap_{a<x<b} I_x \Rightarrow \Gamma[a, b][a, b]\GammaS \subset\mathbb{R}SC = \{A : \alpha \in \Lambda\}A_\alpha\mathbb{R}\LambdaSS \subset \bigcup_{\alpha\in\land} A_\alphax \in S\alpha(x) \in \Lambdax \in A_{\alpha(x)}A_{\alpha(x)}xI(x)xx \in I(x) \subset A_{\alpha(x)}I(x)J(x) \subset I(x)x \in J(x)j(x)S \subset \bigcup_{x\in S} I\left(x\right)\{J(x) : x \in S\}S\{J(x) : x \in S\}C’ = \{j_1, j_2, j_3, \ldots\}SJ_n \in C’x_n \in Sx_n \in J_n \subset I(x_n) \subset A_{\alpha_n}J_n \in C’A_{\alpha_n} \in CC” = \{A_{\alpha_1},\ A_{\alpha_2}, \ldots,\ A_{\alpha_n}, \ldots\}C”CS\bm{\{0\} \cup \left\{ }\frac{\bm{1}}{\bm{n}}\bm{:n\in\ \mathbb{N}\right\rbrace}S =\left\lbrace 0,\ 1,\ \frac{1}{2},\ \frac{1}{3}, \ldots\right\rbraceTST = \left\lbrace 0,\ \frac{1}{m}\ :m\ \in\ N_1\right\rbraceT = \left\lbrace\frac{1}{m}\ :m\ \in\ N_1\right\rbraceN_1\mathbb{N}0T0 \in SSSSS\mathbb{R}\mathbb{R}A = \left\lbrace\frac{1}{m}+\frac{1}{n}\ :m\ \in\ \mathbb{N},\ n\ \in\mathbb{N}\right\rbraceB = \left\lbrace 0,\ \frac{1}{m}\ :m\ \in\ \mathbb{N}\right\rbraceC = \{0\}S = A \cup B \cup C\mathbb{R}S \subset [0, 2]A,\ B,\ CA’ =\left\lbrace 0,\ \frac{1}{m}\ :m\ \in\ \mathbb{N}\right\rbraceB’ = \{0\}C’ = \emptysetS’ = A’ \cup B’ \cup C’ = \{0\} \cup \left\lbrace\frac{1}{m}\ :m\ \in\ \mathbb{N}\right\rbrace = B \cup C \subset SSS\mathbb{R}S\mathbb{R}f : \mathbb{N} \to Bf(m) = \frac{1}{m},\ m \in\ \mathbb{N}BS’S’\mathbb{R}\{F_n\}_n\mathbb{R}\bigcap_{n\in\mathbb{N}} F_n\{F_n\}_nF_n \supset F_{n+1}n \in \mathbb{N}F_nn \in \mathbb{N}F_n \neq \emptysetn \in \mathbb{N}x_n \in F_n\{x_n\}_nx_n \in F_n \subset F_mm \leq nX\{x_n\}_nXX\alpha \in Xx_n = \alphan\mathbb{N}p \in \mathbb{N}k \geq px_k = \alpha\alpha = x_k \in F_k \subset F_pp \in \mathbb{N}\alpha \in \bigcap_{n\in\mathbb{N}} F_n \Rightarrow \bigcap_{n\in\mathbb{N}} F_n \neq \emptysetXF_nF_1 \supset XXX\betam\betaX\betaN(\beta)XN(\beta)x_nx_n \in F_mn \geq mN(\beta) \cap F_m\betaF_mF_m\beta \in F_mm \in \mathbb{N}\beta \in \bigcap_{n\in\mathbb{N}} F_n \Rightarrow \bigcap_{n\in\mathbb{N}}{F^\prime}_n \neq \emptyset\bigcap_{n\in\mathbb{N}}{F^\prime}_nE = \{r \in \mathbb{Q} : \sqrt{2} < r < \sqrt{3} \}E\mathbb{Q}E\mathbb{Q}\mathbb{R}x_1y_1E\subset \mathbb{Q}\sqrt{2} < x_1 < y_1 < \sqrt{3}x_2y_2E\sqrt{2} < x_2 < x_1 < y_1 < x_2 < \sqrt3E\{x_n\}_n\{y_n\}_n\{x_n\}_n\sqrt{2}\{y_n\}_n\sqrt3\sigma = \{I_n : n \in \mathbb{N}\}I_n = (x_n,\ y_n)\sqrt2 < x_n < y_n < \sqrt3n \in \mathbb{N}x_n \in \mathbb{Q}y_n \in \mathbb{Q}m \in \mathbb{N}x \in Ex \in I_mm \in \mathbb{N}E \subset\bigcup_{n\in\mathbb{N}} I_n\sigmaE\sigma’ = \{I_{r_1},\ I_{r_2}, \ldots,\ I_{r_m}\}\sigma\sigma’Er_1, r_2,\ldots, r_mE \subset \bigcup_{i=1}^{m}I_{r_i}p = \max \{r_1, r_2,\ldots, r_m\}p \in \mathbb{N}I_{r_i} \subset I_pI = 1, 2, \ldots, mE \subset I_p = (x_p,\ y_p)\sqrt2 < y_p < \sqrt3 \Rightarrow y_p \in Ey_p \notin I_p\sigmaE\mathbb{Q}E\mathbb{Q}x \in E \Rightarrow 1 < \sqrt2 < x < \sqrt3 < 212E\mathbb{Q}E\mathbb{Q}p < \sqrt2\mathbb{R}rp < r < \sqrt2\deltap\delta = r – p > 0(p – \delta, p + \delta)Ep + \delta = r < \sqrt2(p – \delta, p + \delta) \cap E = \emptyset \Rightarrow pEq > \sqrt3\mathbb{R}\sqrt3 <s < q\delta’ = q – s > 0\delta’q(p – \delta’, p + \delta’)Eq – \delta’ = s > \sqrt3(p – \delta’, p + \delta’) \cap E = \emptyset \Rightarrow qEEEEEE\mathbb{Q}\mathbb{R}\mathbb{Q}\bm{\mathbb{R}}E\mathbb{R}\sigma\mathbb{R}E\sigma\sigma’\sigma’\sigma’\sigma’ = \{I_n : n \in \mathbb{N}\}\sigma’EE \subset\bigcup_{n\in\mathbb{N}} I_nA_n = E \cap \left(\bigcup_{k=1}^{n}I_k\right)^c = E – \bigcup_{k=1}^{n}I_k (n = 1, 2, 3,\ldots.)A_n \supset A_{n+1}n \in \mathbb{N}\bigcup_{k=1}^{n}I_kE\mathbb{R}n \in \mathbb{N}A_n\mathbb{R}\{A_n\}A_n \neq \emptysetn \in \mathbb{N}\bigcap_{n\in\mathbb{N}} A_n \neq \emptyset \Rightarrowc \in \bigcap_{n\in\mathbb{N}} A_n \Rightarrow c \in Ec \notin I_nn \in \mathbb{N}c \notin \bigcup_{n\in\mathbb{N}} I_n\sigma’EA_m = \emptyset \Rightarrow E \subset \bigcup_{k=1}^{m}I_k\sigma” = \{I_1, I_2, \ldots, I_m\}\sigma’\sigmaEEEf : S \to \mathbb{R}S \subset \mathbb{R}Sc \in Sfc\varepsilon> 0(\varepsilon, 0) > 0f(c)\ – \varepsilon < f(x) < f(c) + \varepsilonc – \delta < x < c + \deltax \in Sf(x)\ \in\ N(f(c),\ \varepsilon)x \in N(c, \delta) \cap SfSSS \subset \mathbb{R}f : S \to \mathbb{R}SfSS\mathbb{R}f(S)\mathbb{R}f\left(S\right)\sigma = \{I : \alpha \in \Lambda\}f(S)\ \subset\ \bigcup_{\alpha\in\Lambda} I_\alpha\sigmaf(S)a \in Sf(a)\ \in\ f(S)I_{\alpha’}\sigmaf(S) \in I_{\alpha’}I_{\alpha’}f(a)\varepsilon_a\delta_af(x)\ \in\ N(f(a),\ \varepsilon_a)\ \subset\ I_\alpha’x \in N(a,\delta_a) \cap SC = \{N(a, \delta_a) : a \in S\}SSCC’ = \{N(a, \delta_a) : i = 1, 2, \ldots, m\}a_i \in S\delta_{a_i} > 0i = 1, 2, \ldots, mC’SS \subset \bigcup_{i=1}^{m}{N(a_i,\ \delta_{a_i})}f(S)f(x)x \in Sx \in N(a_p,\ \delta_{a_p})p = 1, 2, \ldots, mx \in N(a_p, \varepsilon_{a_p}) \cap Sp = 1, 2, \ldots, mf(x) \in\ N(f(a_p),\ \varepsilon_{a_p})\ \subset\ I_{{\alpha’}_p}p = 1, 2, \ldots, mf(x) \in\ f(S)\ \Rightarrow\ f(x) \in\ I_{{\alpha’}_p}p = 1, 2, \ldots, mf(S) \subset\ \bigcup_{P=1}^{m}{\ I_{{\alpha’}_p}}\sigma’ = \{{I}_{{\alpha’}_1},\ I_{{\alpha’}_2}, \ldots,\ I_{{\alpha’}_m}\}\sigmaf(S)\sigma’\sigmaf(S)f(S)\mathbb{R}\mathbb{R}\sup{f(S)}f(S)f(S)\mathbb{R}\mathbb{R}f : S \to \mathbb{R}S (\subset R)f(S)S\mathbb{R}(2, 3]\mathbb{R}S = \left\lbrace x \in (0, 1): x \neq \frac{1}{n}, n \in \mathbb{N}\right\rbrace\sigma = \left\lbrace \left(\frac{1}{n+1},\ \frac{1}{n}\right) : n \in \mathbb{N}\right\rbrace\sigmaS\sigmaSA = \left[\frac{1}{2},\ \frac{7}{2}\right]B =\left(1,\ \frac{9}{2}\right)A \cup BA\mathbb{R}B\mathbb{R}A \cup B\mathbb{Q}AB\mathbb{R}A \cap B\mathbb{R}\mathbb{R}$?

- An introduction to functional analysis, By Charles Swartz.
- Introduction to Topology, By Bert Mendelson.
- The Elements of Real Analysis, R.G
- Real Analysis, H.L. Royden

Thank you for reading this article. We will meet next time with a new article

Also, you may want to check our blog posts about Fourier Series, Riemann Integral, Lebesgue Outer Measure, and All The Logarithm Rules You Know and Don’t Know About

And don’t forget to join us on our Facebook page stay updated on any new articles and a lot more!!!!!

The post Compact Sets and Continuous Functions on Compact Sets appeared first on Math Academy.

]]>The post Fourier Series appeared first on Math Academy.

]]>The study of a special type of series whose terms are trigonometric functions of a variable was started in the 18^{th} century when **J.B.J. Fourier** (1768-1830) was successful in his attempt to prove that an arbitrary function given in the interval can be expressed under certain conditions in a trigonometric series of the form

and subsequently, such type of special series is given the name ‘Fourier Series’. It has great importance in the study of many branches of science including social sciences. The knowledge of Fourier series provides a powerful technique for solving problems of a periodic nature which occur in situations like theory of conduction of heat, electrical and mechanical vibrations, propagation of electromagnetic waves, acoustics, optics, etc. Fourier was the first person to initiate an elaborate study of trigonometric series and afterwards **Dirichlet, Parseval, Bessel** and many other mathematician made major contributions to the study of conditions of a function to be represented in a trigonometric series, convergence, integration, and differentiation of the Fourier series. In this article we are going to study Fourier series and their properties.

Before starting the article of Fourier series, it is essential to study the behaviour of periodic functions, piecewise monotone and piecewise continuous functions and some properties related to them.

**Definition:** Let , a function is said to be periodic if there exists a positive constant such that for all ; the least number possessing this property (if such a number exists) is called the period of the function. is called periodic of period or fundamental period . is periodic of period implies also for all .

The trigonometric functions , are periodic of period and , are periodic of period . is periodic of period Every constant function is trivially periodic with no definite Fundamental period.

The following observations are quite clear from the definitions of periodic functions.

- If is defined on an interval of length then can be extended to the whole of by making periodic of period .
- The graph of a periodic function over will be replicas of a portion over an interval of length if is period of the function.

Hence if a function is defined on it can be made to be a periodic function on of period . In fact, sum of two periodic function of same period is also a periodic function of that period.

If is periodic of period then the following results are quite easy to prove:

**Definition:** A function is said to be piecewise monotone if there exists a partition of into open sub-intervals in each of which is monotone.

Every monotone function is piecewise monotone but the converse is **not** true.

**For example**

is piecewise monotone on [0, 1], but is not monotone on [0, 1].

**Definition:** A function is said to be piecewise continuous if there exists a partition of into open sub-intervals in each of which is continuous.

Every continuous function is piecewise continuous but the converse is **not** true.

**For example**, the function defined on by

* *is piecewise continuous, since it is continuous on , , , but is not continuous on ; and are points of discontinuities of on .

Theorem 2.1.

Definition (Fourier Series): Let be an integrable function on or if is unbounded on , let the improper integral be absolutely convergent.

Then the trigonometric series

is called the Fourier series in corresponding to the function , where , called Fourier Coefficients, are given by

Definition: A sequence of Riemann integrable functions defined on is called orthogonal if

and it is said to be orthonormal if

From Theorem 2.1, it easily follows that are all **orthonormal** on . The function is **orthonormal** on

Not every function can be expanded into Fourier series. There are certain conditions which are to be satisfied by a function for its expansion into Fourier series. The most familiar one is **Dirichlet’s** conditions.

Let be a given function. is said to satisfy the Dirichlet’s conditions in if it satisfies one of the following two conditions:

- is bounded on and the interval can be broken into a finite number of open sunintervals of , in each of which is monotone i.e., in other words is bounded and
**piecewise monotone**on ; - has a finite number of points of infinite discontinuities on but when arbitrarily small neighbourhoods of these points of discontinuities are excluded from , is bounded and piecewise monotone in the remainder of the interval and also the improper integral is absolutely convergent.

**Remark:** Dirichlet’s conditions can also be put in a slightly different form:

is said to satisfy Dirichlet’s condition on if is a function of bounded variation on or when has a finite number of points of finite discontinuities on and when arbitrarily small neighbourhoods of these points are excluded from , is a function of bounded variation on the reminder of and is absolutely convergent on .

If the series converges uniformly to on then it is the Fourier series for in .

**Proof. **Since the series converges uniformly to on , is the sum function of the series and we can write

(1)

By integrating term-by-term on , we get using Theorem 2.1,

Multiplying (1) by on both sides and integrating term-by-term (which is valid, since is a bounded function and after multiplication the series is still uniformly convergent on , we have

Multiplying (1) by on both sides and integrating term-by-term, we have

Hence, from definition the series (1) represents the Fourier series for on .

**Note: **Fourier series exists for every Riemann integrable function. But that series may or may not converge to .

If **be bounded, integrable and monotone** on then

** Proof.** (i) Since is monotone on , applying

Since is convergent, by **Cauchy’s condition of convergence**

Since is

Hence

for .

** Proof: – **(ii) For the integra .

We write .

Since is bounded and monotone on , is also bounded and monotone on for and .

By the **second mean** value theorem,

Since is convergent, there exists such that for all .

Therefore .

Since , for any positive there exists a positive such that

whenever .

Hence whenever

by

,

Hence of , there exists such that whenever

This implies

\ben*}

\left\vert\int_{0}^{a}\varphi(x)\dfrac{\sin nx}{x}dx\right\vert & \leq\left\vert\int_{0}^{h}\varphi(x)\dfrac{\sin nx}{x}dx\right\vert+\left\vert\int_{h}^{a}\varphi(x)\dfrac{\sin nx}{x}dx\right\vert\\

& <\dfrac{\varepsilon}{2}+\dfrac{\varepsilon}{2}=\varepsilon\text{ when and .}

\end{align*}

Thus

**Corollary 1. **If is bounded integrable and monotone on where

,

then

** Proof. ** where

is bounded and integrable on . is monotone increasing on and hence on where .

Then by **Theorem 3.2**

**Corollary 2. **If be bounded and integrable and monotone on for then

** Proof. **From (ii) of

Since we have

**Remark:** the integrals in **Theorem 3.2** are known as Dirichlet’s Integrals.

If be bounded, integrable and monotone on and (not necessarily in the same sense), where , then

where stand for Fourier coefficients of .

** Proof. **From definition of we have

If **is bounded, integrable, periodic pf period** and it is piecewise monotone on then the Fourier series for in is equal to

**at** **in** , **when** **is continuous at** ,

**at** **in** , **when** **and** **exist,**

**at** **, when ** **and** **exist.**

**Proof.** The Fourier series for (since it satisfies Dirichlet’s condition in ) in is

where

and

Let,

Then

In the first integral, we will put and in the second integral, we are going to put .

Then

Since satisfies Dirichlet’s conditions in, if we consider as functions of then they satisfy Dirichlet’s conditions in and in respectively. If exist then and have the limits when tends to zero.

Hence, if , we have using Dirichlet’s integrals,

**Hence (ii) proved.**

If is continuous at where then .

Hence,

**which proves (i).**

At ,

If and exist, using Dirichlet’s integrals,

We have,

Similarly, it can be shown that

**Remark:** The **Theorem 3.4** establishes the point-wise convergence of the Fourier series for to the function and not the uniform convergence. The **Fourier series** expression of a function is unique.

If the Fourier series for in converges uniformly to on then

Where are the Fourier coefficients for in .

**Proof.** The Fourier series of in is

Where

Since the series (1) converges uniformly to on , then we can write (1) as

Multiplying both sides of (2) by and then integrating term-by-term (which is valid, since is bounded on or if unbounded, is absolutely convergent) on we have

If is bounded and integrable on or if is unbounded on but is convergent, then

Where are the Fourier coefficients for in .

**Proof.** The Fourier series expression for on is

Where are given by

for all .

Let

Since , we have

From (1)

Again from (1),

Hence using (2), (3) and (4) we have the inequality

**Note:** Taking in Bessel’s inequality we obtain

Since and hence is integrable on , the series

**Example 4.1.** We are going to solve the Fourier series representation of in where

and hence we will deduce that .

**Solution:** is continuous on and hence it is bounded and integrable on . Since for , is monotone increasing on . Thus satisfies Dirichlet’s conditions on so that can be represented by a Fourier series

Hence the Fourier series expansion for in is

Since is continuous on , so we can write

** Example 4.2.** We are going to find the Fourier series expansion of the function

Hence we will deduce that

**Solution:** Since is an even function, its derivative is odd and hence is symmetric about origin. in and . Hence is piecewise monotone in . Hence satisfies Dirichlet’s conditions in . Therefore can be represented by the Fourier series

For since is an odd function.

Hence, Fourier series corresponding to in is

Since is **continuous** on ,

**Example 4.3.** Obtain the Fourier series expansion of in , where

Then we are going to show that the sum of the series

**Solution:** is piecewise monotone on . is continuous on and hence is integrable on . Thus satisfies **Dirichlet’s conditions** in .

So, can be represented by the Fourier series

where

For

Thus, the Fourier series expansion of in is

i.e.,

Since is continuous on ,

Therefore,

**Observation.** If a function is not defined at then can be defined arbitrarily at those points. However, it is preferable to define at those points in such a way that makes continuous on .

**Example 4.4.** We will find Fourier series of the periodic function with period defined as

Then we will calculate the sum of the series at .

And we will prove that .

**Solution:** For the sake of continuity, we define

is piecewise monotone in . It is bounded and periodic with period . is continuous on except at . Hence is integrable on . Thus satisfies Dirichlet’s conditions on . Hence can be represented by Fourier series, which is

Where

Thus, Fourier series for in is

At , the Fourier series for converges to

Hence sum of the series at is .

Since is continuous at , we have

** Example 4.5.** If on , we will prove that the Fourier series of is given by

Then we will deduce that

and

**Solution:** is piecewise monotone on . is continuous on and hence is integrable on . Therefore satisfies Dirichlet’s conditions on . Hence can be represented by Fourier series in which is

where

Since is odd function, for

Hence the Fourier series for in is

Since is continuous on , we can write

Hence therefore,

Since

and

hence by **Weierstrass M-test**,

The 5series

**[ Remark: We can use comparison test for the test of convergence]**

This implies the infinite series

converges uniformly to in . Then by **Parseval’s Identity** (Theorem 3.5)

** Example 4.6.** Find the Fourier series of the periodic function with period defined by

The we will find the sum of the series at and then we are going to show that

**Solution:** We define at and at as , to make continuous at

is piecewise monotone on . It is continuous on except at and and is bounded on . Hence is a bounded periodic and integrable function on which validates the expansion of into Fourier series, which is

Where

Thus, the Fourier series for in is

i.e.

Since is periodic function of period , so . Since is not continuous at , the Fourier series for converges at .

Hence sum of the series at is equal to .

At , the Fourier series for converges to

Therefore

**Example 4.7.** Let be defined as follows:

We are going to obtain the Fourier’s coefficients and the Fourier series for the function and hence we will find that sum of the series can be represented as

**Solution:** is piecewise monotone in . It is periodic with period . It is bounded on . is continuous in except at . Hence it is integrable on . Thus satisfies Dirichlet’s conditions on which validates Fourier series expansion of in .

The Fourier coefficients are

For ,

when is odd, . When is even,

For ,

Thus, we have

and

Hence the Fourier series for in is

Since is continuous at , the sum of the series at is

Therefore,

When is an even function on

Then Fourier coefficients becomes zero, since is then odd and hence

Thus the Fourier series for becomes a **cosine series**.

If is odd i.e.,

then the Fourier coefficients

and

Thus, the Fourier series for will be a **sine series**.

**Example 5.1.** We are going to consider the even function on and we will see that it has a cosine series in Fourier’s form which is

Also the series converges to in and by using this series we will show that,

**Solution:**

is **piecewise monotone** on . is **bounded** on . is continuous on and hence is integrable on . Thus satisfies Dirichlet’s conditions on and it can be represented by Fourier series on , which is

Where

For

Hence

Since is odd function, .

Thus, Fourier series for is in form of the **cosine series**