Category: Maths Notes PPT

Proportionality is one of the important topics of Mathematics, and thus, a better understanding of this topic is required for the students, not only in order to get good marks but also for future mathematics endeavours. But there is one more thing that the students need to have in order to master the topic of Direct Proportion, and that one thing is an easy and understandable explanation of the same topic.

An Overview of the Direct Proportion

When the two numbers or the two quantities, say x and y are multiplicatively related by a fixed number, that is to say, a constant; then such relationship is regarded as the Proportionality relationship. It also means that if there is a proportional relationship between the two quantities, x and y then the ratio between the same is going to be constant.

For understanding it in an even simpler manner, let us assume you own a grocery shop. Now if the number of people who buy groceries from your store increases, then your profits are also going to increase, and if the numbers of buyers decrease, then your profits are going to decrease as well. From this, we can say that there is a direct relationship between both the quantities, that is to say, numbers of buyers and your profits are directly related.

Direct Proportion Meaning

Proportion is a concept of Mathematics that gives the relation between any two mathematical quantities. Two quantities are said to be proportional if they are multiplicatively connected by a constant. The proportional relationship between any two quantities can also be defined as the quantities whose product or ratio is constant. Two quantities are said to be directly proportional if their ratio is constant. 

If the product of any two quantities is a constant, then those two quantities are said to be inversely proportional. The two quantities which are directly proportional are related by a Direct Proportion symbol ‘∝’. The symbol for proportionality is removed by adding a Direct Proportion constant.

Direct proportion definition

Direct Proportion meaning is explained as follows. Two measurable quantities are said to be directly proportional if the increase in one quantity results in the increase of the other quantity and vice versa. Indirect variation, the ratio of two measurable quantities is constant. For example, if x and y are the two measurable quantities that are directly proportional to each other, then the direct proportion definition is Mathematically written as x ∝ y. If the direct proportion symbol is to be removed, a proportionality constant is added and the direct proportion symbol is replaced by an equal sign.

x = k y

[frac{x}{y}]  = k

In the above equation, ‘k’ is a proportionality constant.

If x1 and y1 are the initial values of any two quantities that are directly proportional to each other and x2 and y2 are the final values of those quantities. Then according to the direct proportionality relationship, 

[frac{x1}{y1}] = K and [frac{x2}{y2}] = k

So, we can infer that the ratio of initial values and the final values of any two quantities varying directly are equal and constant. 

[frac{x1}{y1}] = K and [frac{x2}{y2}] = k

Direct Proportion Example

• Marks scored is directly proportional to the performance in the test.

• Temperature is directly proportional to heat.

• Energy is directly proportional to work.

• Speed is directly proportional to distance.

• Earning is directly proportional to the amount of work done.

• The amount of food we consume is directly proportional to how hungry we are.

These are just a few real-world Direct Proportion examples.

Direct Variation Example Problems

1. In one of the real situations of direct proportion examples, a bus travels 150 km in 5 hours. What is the time taken by the bus to travel 700 km?

Solution:

Distance travelled and time taken are directly proportional to each other.

In the given question, the distance travelled in case 1 is x1 = 150 km

The distance travelled in case 2 is x2 = 700 km

The time taken in case 1 is y1 = 5 hours

Time taken in case 2 is y2 =?

The proportionality relationship can be stated as:

[frac{x1}{y1}] = [frac{x2}{y2}]

[frac{150}{5}] = [frac{700}{y2}]

y2 = [frac{700}{150}] × 5

y2 = 23.33

So, the time taken by the bus to travel 700 km is 23.33 hrs

2. Given that a and b are directly proportional to each other, complete the table given below.

Solution:

From the table x1 = 4, y1 = 6, x2 = 5, x3 = 12, x4 = 6

y2 = ? y3 = ? y4 = ?

Case 1: To find y2

[frac{x1}{y1}] = [frac{x2}{y2}]

[frac{4}{6}] = [frac{5}{y2}]

y2 = [frac{5}{4}] × 6

y2 = 7.5

 

Case 3: To find y4

[frac{x1}{y1}] = [frac{x4}{y4}]

[frac{4}{6}] = [frac{6}{y4}]

y4 = [frac{6}{4}] × 6

Y4 = 9

So, the completed table is as below:

 

3. Sumanth has Rs. 400/- with him. If he can purchase 5 kgs of ghee for 2180, how much ghee can he purchase with the amount he has?

Solution:

Total amount for 5 kg ghee is x1 = Rs. 2180/-

Ghee purchased with Rs. 2180/- is y1 = 5 kg

Amount with Sumanth is x2 = Rs. 400/-

Ghee purchased with Rs. 400/- is y2 = ? 

The money and amount of ghee purchased are directly proportional to each other.

The direct proportionality relationship can be written as: 

[frac{x1}{y1}] = [frac{x2}{y2}]

[frac{2180}{5}] = [frac{400}{y2}]

y2 = [frac{400}{2180}] × 5

y2 = 0.917

Sumanth can purchase 0.917 kgs of ghee with Rs. 400.

Fun Quiz:

1.Time and work are directly proportional to each other. Is this statement true?

1. Yes,

2. No

2.Which of
the following are directly proportional measurements?

1. Current flow and resistance

2. Volume and temperature

3. Mass and weight

3. From the given figure, identify the graphs that indicate direct proportion definition.

Conclusion

This is all about the explanation of the concept of direct proportion and how it is used to solve problems. Focus on how this concept has been used to derive the formula and used in the solved examples.

You must have read about Quantum Theory or Matrix Theory or Probability Theory in your math class or in some sci-fi movie. But have you ever heard of Divergence Theory? Well, here we are today to learn more about this theory. Divergence Theorem is a theorem that talks about the flux of a vector field through a closed area to the volume enclosed in the divergence of the field. It is a part of vector calculus where the divergence theorem is also called  Gauss’s divergence theorem or Ostrogradsky’s theorem.

State and Prove the Gauss’s Divergence Theorem

The divergence theorem is the one in which the surface integral is related to the volume integral. More precisely, the Divergence theorem relates the flux through the closed surface of a vector field to the divergence in the enclosed volume of the field. It states that the outward flux through a closed surface is equal to the integral volume within the surface of the divergence over the area. 

The net flow of an area will be received by subtracting the sum of all sources by the sum of every sink. The result describes the flow by a surface of a vector and the behavior of the vector field within it. 

To state Gauss’s divergence theorem in an easier way, let’s break it into parts. The surface integral of a vector field over a sealed area is known as flux through the surface. Therefore, the flux through the surface is equivalent to the volume integral of the divergence over the area inside the surface. Thus, the total sum of all sources of the field in an area gives the net flux out of the area.

This is an essential result for mathematics in engineering and physics. It is one of the most important theorems and is used to solve tough integral problems in calculus. It is used particularly in the field of electrostatic and fluid dynamics.

Divergence Theorem is generally applied in 3 dimensions, but it can be used in any number of dimensions. When you use it in 2 dimensions, it becomes equivalent to Green’s theorem which states that the line integral around any simple closed curve is equal to a double integral over the plane region. When you use it in 1 dimension, it becomes equivalent to integration by parts. 

The Divergence Theorem Proof 

Let us consider a surface denoted by S which encloses a volume denoted by V. 

Suppose vector A is the vector field in the given region. Suppose this volume is made up of a large number of parallelepipeds (1- 6 parallelepipeds) which represent elementary volumes.

Consider the volume of the jth parallelepiped is [ Delta V_{j} ]which is bounded by a surface [ S_{j} ] of area [ d overrightarrow{S_{j}}] . The surface integral of [ overrightarrow{A} ] over the surface [ S_{j} ]  will be-

[ oint ointlimits_{S} overrightarrow{A} . d overrightarrow{S_{j}}]

Now consider if the whole volume is divided into elementary volumes I, II, and III as shown below. 

()

The elementary volume I outward is the elementary volume II inward and the elementary volume II outward is the elementary III inward and so on. 

Thus, the sum of the elementary volumes integrals will cancel each other out and the surface integral arising from the surface S will be left. 

[ sum oint oint limits _{S_{j}}  overrightarrow {A}. d overrightarrow {S_{j}} = oint oint limits _{S} overrightarrow {A}. d overrightarrow {S}  …. (1) ]

Multiplying and dividing the left-hand side of the equation (1) by [ Delta V_{i} ], we get

[ oint limits _{S_{j}} overrightarrow {A} . d overrightarrow {S} = sum frac{1}{Delta V_{i}} (oint oint limits _{S_{i}} overrightarrow {A} . d overrightarrow {S_{i}} ) Delta V_{i} ]

Now, suppose the volume of surface S is divided into infinite elementary volumes such that [ Delta V_{i} rightarrow 0].

[ oint oint limits _{S} overrightarrow {A} . d overrightarrow{S} = limlimits_{Delta V_{i} rightarrow {0}}sum frac{1}{Delta V_{i}} (oint oint limits _{S_{i}} overrightarrow {A} . d overrightarrow {S} ) Delta V_{i} ….. (2) ]

Now,

[ lim limits_ {Delta V_{i} rightarrow {0}} (frac{1}{Delta V_{i}} (oint oint limits _{S_{i}} overrightarrow {A} . d overrightarrow {S})) = (overrightarrow {V}.overrightarrow {A} ) ]

Therefore, eq (2) becomes

[ oint oint overrightarrow {A} . d overrightarrow {S} = sum (overrightarrow{nabla} . overrightarrow {A} ) Delta V_{i} …. (3) ]

We know that [ Delta V_{i}rightarrow{0} ], thus[sum Delta V_{i} ] will become the integral over volume V.

This is the Gauss divergence theorem.

Gauss’s Divergence Theorem History

Lagrange was the first one to discover the Divergence Theorem in 1762. Later on in 1813, it was rediscovered independently by Carl Friedrich Gauss. He also gave the first proof of the general theorem in 1826. Many other mathematicians like Green, Simeon-Denis Poisson, and Frédéric Sarrus also discovered this theorem.

Conclusion

This article entails detailed information on Divergence Theory and its derivation. You can go through it for a comprehensive understanding. Also, download the PDFs to read at your ease. 

Vectors can be multiplied in two different ways, namely, scalar product or dot product in which the result is a scalar, and vector product or cross product in which the result is a vector. The dot product of two vectors means the scalar product of the two given vectors. It is a scalar number that is obtained by performing a specific operation on the different vector components. The dot product is applicable only for the pairs of vectors that have the same number of dimensions. The symbol that is used for the dot product is a heavy dot. This dot product is widely used in Mathematics and Physics. In this article, we would be discussing the dot product of vectors, dot product definition, dot product formula, and dot product example in detail.

Dot Product Definition

The dot product of two different vectors that are non-zero  is denoted by a.b and is given by:

a.b = ab cos θ

wherein θ is the angle formed between a and b, and,

0 ≤ θ ≤ π

()

If a = 0 or b = 0, θ will not be defined, and in this case,

a.b= 0

Dot Product Formula

You can define the dot product of two vectors in two different methods: geometrically and algebraically.

Dot Product Geometry Definition

The geometric meaning of dot product says that the dot product between two given vectors a and b is denoted by:

a.b = |a||b| cos θ

Here, |a| and |b| are called the magnitudes of vectors a and b and θ is the angle between the vectors a and b.

If the two vectors are orthogonal, that is,  the angle between them is 90, then a.b = 0 since cos 90 = 0.

If the two vectors are parallel to each other, then a.b =|a||b| since cos 0 = 1.

Dot Product Algebra Definition

The dot product algebra says that the dot product of the given two products – a = (a1, a2, a3) and b= (b1, b2, b3) is given by:

a.b= (a1b1 + a2b2 + a3b3)

Properties of Dot Product of Two Vectors 

Given below are the properties of vectors:

1. Commutative Property

a .b = b.a

a.b =|a| b|cos θ

a.b =|b||a|cos θ

2. Distributive Property

a.(b + c) = a.b + a.c

3. Bilinear Property

a.(rb + c) = r.(a.b) + (a.c)

4. Scalar Multiplication Property

(xa) . (yb) = xy (a.b)

5. Non-Associative Property

Since the dot product between a scalar and a vector is not allowed.

6. Orthogonal Property

Two vectors are orthogonal only when a.b = 0

Dot Product of Vector-Valued Functions

The dot product of vector-valued functions, that are r(t) and u(t), each gives you a vector at each particular time t, and hence, the function r(t)⋅u(t) is said to be a scalar function.

Solved Examples

Example 1:

Find the dot product of a= (1, 2, 3) and b= (4, −5, 6). What kind of angle the vectors would form?

Solution:

Using the formula of the dot products,

a.b = (a1b1 + a2b2 + a3b3)

You can calculate the dot product to be

= 1(4) + 2(−5) + 3(6)

= 4 − 10 + 18

= 12

Since a.b is a positive number, you can infer that the vectors would form an acute angle.

Example 2:

Two vectors A and B are given by:

A = 2i − 3j + 7k and B= −4i + 2j −4k

Find the dot product of the given two vectors.

Solution:

A.B = (2i − 3j +7k) . (−4i + 2j − 4k)

= 2 (−4) + (−3)2 + 7 (−4)

= −8 − 6 − 28

= −42

Key Points to Remember

• When two vectors are cross-products, the output is a vector that is orthogonal to the two provided vectors.

• The right-hand thumb rule determines the direction of the cross product of two vectors, and the magnitude is determined by the area of the parallelogram generated by the original two vectors.

• A zero vector is the cross-product of two linear vectors or parallel vectors.

Conclusion

Vector is a quantity that has both magnitude as well as direction. Few mathematical operations can be applied to vectors such as addition and multiplication. The multiplication of vectors can be done in two ways, i.e. dot product and cross product. The dot product of two vectors is the sum of the products of their corresponding components. It is the product of their magnitudes multiplied by the cosine of the angle between them. A vector’s dot product with itself is the square of its magnitude.

The equation of a line is an algebraic method to represent a set of points that together form a line in a coordinate system. The various points that together form a line in the coordinate axis can be represented as a set of variables (x, y) in order to form an algebraic equation, also referred to as the equation of a line. By using the equation of a line, it is possible to find whether a given point lies on the line.

The equation of any line is a linear equation having a degree of one. Let us read through the entire article to understand more about the different forms of an equation of a line and how we can determine the equation of a line.

A line segment can be defined as a connection between two points. Any two points, in two-dimensional geometry, can be connected using a line segment or simply, a straight line. The equation of a line can be found in the following three ways.

1. Slope Intercept Method 

2. Point Slope Method 

3. Standard Method

When two points that lie on a particular line are given, usually, the point-slope method is followed.

The equation of a line is [y – y_{1} = m(x – x_{1})] where [y_{1}] is the coordinate of the Y-axis, m is the slope, and [x_{1}] is the coordinate on the X-axis.

Finding the Slope of the Line Passing through Two Given Points

The slope or gradient of a line is the changing height of the line from the X-axis. For every unit of X, a change in Y on the line is known as the slope of a line. 

()

To calculate the slope, the formula used is [m = frac{y_{2} – y_{1}}{x_{2} – x_{1}}].

Here, the points are (2,5) and (6,7).

So, comparing the point to the general notation of coordinates on a Cartesian plane, i.e., (x, y), we get [x_{1}, y_{1} = (2, 5) and x_{2}, y_{2} =(6, 7) ]

Substituting the values into the formula, 

[m = frac{7 – 5}{6 – 2}]

[m = frac{2}{3}]

Did You Know?

What happens if we interchange the values of [x_{1}, y_{1} and x_{2}, y_{2}]?

The value of m remains unchanged. The positioning of the coordinates does not affect the value of the slope. 

Taking the same example as above but interchanging the values of [x_{1}, y_{1} and x_{2}, y_{2}], we get [x_{1}, y_{1} = (6,7) and x_{2}, y_{2} = (2,5)]. 

[m = frac{5 – 7}{ 2 – 6}]

[m = frac{-2}{-3} = frac{2}{3}]

Hence, any one of the two coordinates can be used as [ x_{1}, y_{1} ] and the other as [ x_{2}, y_{2} ].

Finding the Equation of the Line Passing through Two Given Points

Steps to find the equation of a line passing through two given points is as follows:

1. Find the slope/gradient of the line.

2. Substitute the values of the slope and any one of the given points into the formula.

3. Simplify to obtain an equation resembling the standard equation of the line, i.e., Ax + By + C = 0, where A, B, and C are constants.

Taking the above example, where [x_{1}, y_{1} and x_{2}, y_{2}], we get [x_{1}, y_{1} = (2,5) and x_{2}, y_{2} = (6,7)] and the slope is calculated as [m = frac{2}{3}], substitute the value of m and any one point in the formula [y – y_{1} = m(x – x_{1})].

[y – y_{1} = m(x – x_{1})]

[y – 5 =  frac{2}{3} (x – 2)]

Cross-multiply and simplify:

[y – 5 =  frac{2}{3} (x – 2)]

[ Rightarrow 3 (y – 5) = 2 (x – 2)]

[ Rightarrow 3y – 15 = 2x – 4]

[ Rightarrow 3y – 2x = 15 – 4]

[ Rightarrow 3y – 2x = 11]

The same equation can be expressed in slope-intercept form by making the equations in terms of y as shown below.

[ Rightarrow 3y – 2x = 11]

[ Rightarrow 3y = 2x + 11]

[ Rightarrow y = frac{2}{3}x + frac{11}{3}]

Solved Examples

1. Find the equation of the line passing through the points (2,3) and (-1,0).

For calculating the slope, the formula used is [m = frac{y_{2} – y_{1}}{x_{2} – x_{1}}].

Here, the points are (2,3) and (-1,0)

So, comparing the point to the general notation of coordinates on a Cartesian plane, i.e., (x, y), we get (x1,y1) = (2,3) and (x2,y2) = (-1,0).

Substituting the values into the formula, 

[ Rightarrow m = frac{0 – 3}{-1 – 2}].

[ Rightarrow m = frac{-3}{-3}].

[ Rightarrow m = 1 ].

Substitute the value of m and any coordinate into the formula [y – y_{1} = m(x – x_{1})]. 

[y – y_{1} = m(x – x_{1})]

[y – 0 = 1(x – (-1)]

Simplify the equations: 

[y – 0 = 1(x – (-1)]

[ Rightarrow y = x + 1 ]

[ Rightarrow y – x = 1 ]

The same equation can be expressed in slope-intercept form by making the equations in terms of y. 

y = x + 1

The equation of the line passing through the points (2,3) and (-1,0) is y = x + 1 or y – x = 1.

2. Find the Equation of the Line Passing through the Point (1,3) and Having a Slope [frac{1}{3}].

Substitute the value of m and the coordinate into the formula [y – y_{1} = m(x – x_{1})].

 [y – y_{1} = m(x – x_{1})]

[ Rightarrow y – 3 = m(x – x_{1})]

[ Rightarrow y – 3 = frac{1}{3}(x – 1)]

Cross multiply and simplify the equations:

[ Rightarrow y – 3 = frac{1}{3}(x – 1)]

[ Rightarrow 3(y – 3) = 1(x – 1)]

Simplify the equations further:

[ Rightarrow 3(y – 3) = 1(x – 1)]

[ Rightarrow 3y – 9 = x – 1]

[ Rightarrow 3y – x = 8]

The same equation can be expressed in slope-intercept form by making the equations in terms of y. 

[ Rightarrow 3y – x = 8]

[ Rightarrow 3y = x + 8]

[ Rightarrow y = frac{1}{3}x + frac{8}{3} ]

The equation of the line passing through the point (1,3) and having a slope of [frac{1}{3}] is [  Rightarrow 3y – x = 8 or frac{1}{3
}x + frac{8}{3}]. 

Conclusion

The equation of a line can be easily understood as a single representation for numerous points on the same line. The equation of a line has a general form, that is, ax + by + c = 0, and it must be noted that any point on this line satisfies this equation. There are two absolutely necessary requirements for forming the equation of a line, which are the slope of the line and any point on the line.

What is a Fraction?

A fraction represents a part of a whole or, more generally, any number of equal parts of a value. When we are working with fractions we assume about the relationship between the part of any number or a whole sum number. Something we do not even realize that we are working on a fraction because fractions are everywhere. A common or simple fraction consists of two parts, an integer numerator displayed above a line, and a non-zero integer denominator, displayed below that line. It tells us how many parts a whole is divided into. For example:

5/7 = means we have 5 parts out of the whole of 7.

5 is the numerator, it tells us how many parts we have.

7 is the denominator, it tells us how many parts the whole is divided into.

Properties Obeyed by the Fractions

Fraction obeys the commutative, associative, and distributive laws, and the rule against division by zero, like whole numbers.

Commutative Property

In mathematics, a binary operation is commutative if changing the sequence of the operands in an operation do not change the result of it. It is a fundamental property for many binary operations, and many mathematical proofs depend on this property.

For example:

Suppose that there are two fractions A and B and they are multiplied together. Then the order of A and B during multiplication cannot change the result. Same goes for addition too.

A × B = B × A

A + B = B + A

Associative Property

In mathematics, the associative property is a property of binary operations. In propositional logic, associativity is a valid rule of replacement for expressions with logical proofs. Within an operation containing two or more operands in a row of some associative operator, the way in which the operations are performed does not matter as long as the sequence of the operands in the operation remains the same. It means that rearranging the parentheses in such an expression will not change its result. Consider the following equations:

(4 + 5) + 6 = 4 + (5 + 6) = 15

(4 × 5) × 6 = 4 × (5 × 6) = 120

Distributive Property

The distributive property of binary operations is widely applicable in the distributive law. Distribution refers to two valid rules of replacement. The rules allow one to put conjunctions and disjunctions differently within logical proofs. By observing the example given below, you can understand the distributive property easily.

If there are three operands A, B, and C are in such an operation where A × (B + C) then it must be equal to A × B + A × C.

Division by Zero

In mathematics, division by zero is division where the denominator is zero. In ordinary mathematics, this expression has no meaning, as there is no number which, when multiplied by 0, gives another number, and so division by zero is undefined.

Equivalent Fraction

In mathematics, equivalent fraction can be defined as the fractions with different numerators and denominators that represent the same value or proportion of the whole. Here are some examples of equivalent fraction, like1/2, 2/4, 8/16. They all contain the different values in their numerators and denominators, but at last after evaluation they all give the same value 0.5 as the answer. Equivalent fractions are those types of fractions which give us the same value at last, even though they may look different.

Facts About the Equivalent fractions

Equivalent fractions represents the same amount of distance or points on a number line to one another.

All equivalent fractions are reduced to the same fractions in their simplest forms by dividing both numerator and denominator by their greatest common factor.

How to Find Equivalent Fractions:

By multiplying the numerator and denominator of a fraction by the same non-zero number, we can change the fraction into equivalent of the original fraction. This is true because for any non-zero number a, the fraction a/a = 1. n n = 1 Therefore, multiplying by a/a is equivalent to multiplying by one, and any number multiplied by one has the same value as the original number.

For Example:

To find the equivalent fraction of 2/5we multiply both the numerator and denominator by 2, then we get equivalent fraction 4/10.

How to Simplify the Equivalent Fractions?

Dividing an equivalent fractions numerator and denominator by the same non-zero number will also yields an equivalent fraction. This is called simplifying or reducing the fraction. A simple fraction in which the numerator and denominator both are prime number is said to be irreducible.

For Example:

To find the equivalent fraction of 6/18 we divide both the numerator and denominator by their greatest common factor, then we get equivalent fraction 1/3.

How to Check Two Fractions are Equivalent Fraction:

To check whether the fractions are equivalent or not, just simplify all the fractions. If they reduce into the same fraction, then the fractions are equivalent fractions.

For example:

We will check whether the fractions 5/10 and 12/36 are equivalent or not. We will simplify both the fractions –

[frac{5}{10}=frac{1*5}{2*5}=frac{1}{2}]

[frac{12}{36}=frac{4*3}{4*3*3}=frac{1}{3}]

The fractions ½ and ⅓ are not the same, hence the two fractions are not Equivalent fractions.

Addition and Subtraction

Equivalent fractions are an important tool when adding or subtracting fractions with different denominators. Let’s look at an example:

John bought one half of a cake. He didn’t know that while he was out, his wife, Linda, bought one fourth of a cake. When they got home, how much cake did they have altogether? 

We are adding fractions here:

When we are adding or subtracting fractions, they must have the same denominator. But in the above case, the denominator are two and four (2 and 4). As we read earlier by multiplying the numerator and denominator of a fraction, we get an equivalent fraction. So in this case, for making the denominator equal we have to perform this operation:

 [frac{1}{2}frac{2}{2}=frac{2}{4}]

Now we have got the same denominator and we can add them like:

Converting Between Decimals and Fractions

Decimal numbers are more useful to work with when performing calculations, but sometimes it lacks in precision that common fractions have, because sometimes an infinite repeating decimal is required to reach the exact precision. Thus, it is useful to convert repeating decimals into fractions. To change a common fraction to a decimal, do a long division of the decimal representations of the numerator by the denominator and round the answer to the desired accuracy. For example, to change ¼ to a decimal, divide 1 1.00 {displays
tyle 1.00} by 4 to obtain 0.25. To change 1/3 to decimal, divide 1 by 3 and stop when the desired accuracy is obtained, e.g. 4 digits after the decimal point. To change a decimal into a fraction, write 1 in the denominator 1 followed by as many zeroes as there are digits to the right of the decimal point, and write all the digits of the original decimal in the numerator, excluding the decimal point. Thus 12.346 = 12346/1000.

Summary:

• You can make equivalent fraction just by multiplying or dividing both top and bottom numerator and denominator) by the same value.

• You can only multiply or divide, never add or subtract, to get an equivalent fraction because the fraction we get from addition and subtraction will not be equivalent to the value we have.

• You can divide if and only if the top and bottom stay as whole numbers.

Even and Odd functions form an important aspect in several mathematical analyses. These are functions which when taking an additive inverse, satisfy a specific symmetry.  A basic understanding of these functions is crucial for someone who wants to apply Calculus to real-life situations. These functions have several important applications in Mathematics.  Odd and even functions can be depicted algebraically or in the form of graphs. Refer to the official website of for an elaborate and comprehensive explanation.

Functions are the basis of calculus. It is fundamental to the understanding of any real-life application of applied mathematics. A function indicates a correspondence between two variables. The dependence can be established in different forms such as tabular form, graphs and charts, or equational form.

 

To define, function from X to Y is a rule that maps each element in set X with a unique element in set Y. Set X is called the domain of the function whereas set Y is termed as co-domain.

Variable in Functions

If f is a function, it is denoted as y = f(x); x is called the independent/exogenous variable and y is called the dependent/endogenous variable.

 

For a function to be defined, every element in the domain should be mapped to a unique element in the co-domain. However, not every element in the co-domain may be related to an element of the domain. The specific value of f(x) is known as the image of element x by function f. Therefore, the set of all elements in set Y that are images to at least one element in set X is called the range of the function. The range is a subset of co-domain as shown in the image below.

 

Odd functions

If f is a real-valued function on a real set, f is even if:

-f(x) =f (-x)

Or, f (-x) +f(x) =0

If any given function follows the above rule, it is said to be an odd function.

The graph of any even function is rotationally symmetric along with the origin.

 

Even functions

If f is a real-valued function on a real set, f is even if:

F(x)=f(-x)

Or, f(x)-f (-x) =0

If any given function follows the above rule, it is said to be an even function.

The graph of any even function is symmetric to the y-axis, i.e. it forms a mirror image.

 

Solved Examples

It is essential to go through some examples to understand the above-mentioned points properly. Some illustrations on how to know if a function is even:

Example 1

F(x) = x² + 1

Solution:

Replacing x with (-x) in f(x)

Therefore, F (-x) – (- x²) + 1

Or, F(-x) = x² + 1

Or, F (-x) = f (x)

Hence, f(x) is an even function.

 

Example 2

f(x)= cos x

Solution:

Trigonometry confirms us that cos(x) = cos(-x)

Therefore, F (-x) = f (-x)

The cosine function is even. 

 

Example 3

F(-x) = x³ – x

Now, substituting the value of x with (-x) in f(x)

Therefore, F (-x) = (-x³) – (-x)

Or,

F (-x) = (-x³) + x

Likewise,

–    F (x) = -(x³) – x

–    F (x) = -x³ + x

Thus, it appears that

f(-x) = f(x)

Hence, f(x) is an odd function.

 

 

Some Basic Properties of Even Odd Functions

As a rule of thumb, every real-values function can be decomposed using an even and odd function. Let fe (x) represent an even function while fo(x) denotes odd function. Thus,

Any even function fe(x) = f(x)+f(−x)f(x)+f(−x)/2 and

Every odd function fo(x) = f(x)−f(−x)f(x)−f(−x)/2 and

 f(x) = f_e(x) + f_o(x)

And, f(x) = f_e(x) + f_o(x)

There are instances of some functions satisfying the conditions of both even and odd functions. Such functions are defined everywhere in the real-value set.

The absolute value of an odd function is even.

 

Algebraic Properties Even Odd Functions

• The addition of two even functions produces an even function.

• The addition of two odd functions produces an odd function.

• The subtraction of two even functions is even.

• The subtraction of two odd functions is a function.

• The addition/ difference of even and odd is neither even nor odd, except for the cases where one function is zero. The product of two even functions is even.

• The multiplication of two odd functions will turn out to be an even function.

• The multiplication of two even functions will turn out to be an odd function.

• The division of two even functions is even.

• The division of two odd functions is an even function.

• The product/division of an even and odd function is an odd function.

Practical tips to master the concepts of odd and even functions:

Even and odd functions form part of usual calculus. Those who find the concepts difficult to master can follow these simple tips to excel in the subject:

• Understand the meaning of even and odd functions by going through the definitions.

• Carefully go through the properties of each function multiple times.

• Try to refer to multiple sources to study the same topic to get a better overall perspective

• Use ‘s solutions when solving questions from the topic to understand the ideal way to reach the solutions.

• Practice as many questions as possible from the topic. Calculus and all the other seemingly difficult concepts become easier with diligent practice. Draw graphs for each question.

• Understand the practical applications of odd and even functions. Understanding the objective behind learning a particular topic helps to better relate to the concepts and understand them in a more pro
  found way.