Learn Operations on Functions

Generally a function is represented in terms of  f(x).

Functions behave exactly as one would expect with regard to the four basic operations of algebra (addition, subtraction, multiplication, and division). When functions are combined by these operations, though, the domain of the new combined function is only the elements that were shared by the domains of the original functions.

Operations on functions :

As we add, subtract, multiply the numbers, we can add , subtract, multiply, divide the functions. There is one operation more called the composition of functions.

Let us learn through examples, that would be easier to understand.

Consider the two functions,

f(x) = x +2

g (x) = x2 -5


Operations on functions : Addition


f(x) = x +2

g (x) = x^2 -5

(f+g) (x) = x +2 +x2 -5

= x2+x -5 +2

= x2 + x -3

(f+g) (2) = 22 +2 -3 = 6 -3 =3


Operations on functions : Subtraction


f(x) = x +2

g (x) = x^2 -5

(f-g) (x) = x+2 – (x2 -5)

= x2+x +2 -5

=x2 +x -3

(f-g)(2) = 2 +2 - 22 -5 = -5


Operations on functions : Multiplication


f(x) = x +2

g (x) = x^2 -5

(f)(g)(x) = (x+2) (x2-5)

= x3 -5x +2x2 -10

= x3 +2x2 -5x -10

(f)(g)(2) = (2+2) (22-5) = -4


Operations on functions : Division


f(x) = x +2

g (x) = x2 -5

g(x) /f(x) =

=  [x2 -5] / [x+2]

g (2) / f (2 ) = (22 - 5)/ (2+2)

=(4 -5)/ (2+2) = -1 / 4


Operations on functions : Composition of Function


f(x) = x +2

g (x) = x^2 -5

(f o g)(x) read as f of g of x, we can say  that first, we find g(x) and then we find f(x) for the result that we got from g(x)

The concept is simple. First, the value of g at x is taken, and then the value of f at that value is taken

g(x) = x2 -5

(f o g)(x)= f (x2-5) = x2 -5 +2 = x2 -3

(f o g)(2) =

= g(2) =22 – 5 = -1

= f(-1) = -1+2 =1

Trigonometric Functions Unit Circle

Unit circle:

In mathematics, a unit circle is a circle with a radius of one. Frequently, except in trigonometry, the unit circle is the circle of radius one centered at the origin (0, 0) in the Cartesian coordinate system in the Euclidean plane. The unit circle is denoted as s1; the generalization to higher dimensions is the unit sphere.

Trigonometric functions on the unit circle:

All of the trigonometric functions of the angle θ can be modified by geometrically in terms of a unit circle centered at O.

The trigonometric functions of cosine and sine can be defined on the unit circle as follows. If (x, y) is a point of the unit circle, and if the ray from the origin (0, 0) to (x, y) makes an angle t from the positive x-axis, (where counterclockwise turning is positive), then

cos(t) = x

sin(t) = y

The equation x2 + y2 = 1 gives the relation

cos2(t) + sin2(t) = 1

The unit circle also operated that sine and cosine are periodic functions, with the identities

cos t = cos(2πk + t)

sin t = sin(2πk + t)

For any integer k

Area & Circumference:

1. Length of circumference:

The circumference’s length is related to the radius (r) by

c = 2*π*r

For unit circle circumference is 2*(because r = 1).

2. Area enclosed:

Area of the circle = π × area of the shaded square

Main article: Area of a disk

The area of the circle is π multiplied with the radius squared:

A = π *r2

For unit circle area is


Application:


Circle group:

The complex numbers can be pointed with points in the Euclidean plane, namely the number a + bi is identified with the point (a, b). Under this identification, the unit circle is a group under multiplication, called the circle group. It has main applications in mathematics and science.

Application of the Unit Circle:

• It is used to understand the relationship between the unit circle and the trigonometric (sine and cosine) functions.

• Because of this reason it is used to solve a real-world application.

Multiplication Solver

Multiplication (symbol "×") solver is the mathematical operation of scaling one number by another. It is one of the four basic operations in a elementary arithmetic (the others being addition, subtraction and division) (Source.Wikipedia). Multiplication solver  there are two jobs such that one of them can be completed in m ways, and when it has been completed in any one of these m ways, second job can be completed in n ways; then the two jobs in succession can be completed in m × n ways.

Examples for multiplication solver:


Rules for multiplication solver:

•    The multiplication solver of two positive integers is a positive integer.

•    The multiplication solver of a positive integer and the negative integer is a negative integer.

•    The multiplication solver of two negative integers is a positive integer.

Example 1:

Multiply 283 by 104

Solution:

283 × 104 means 104 times 283

or 100 times 283 + four time 283

or 28300 + [4* 283]

or 29432

∴ 283 × 104 = 29432 [multiplication rules: The product of two positive integers is a positive integer.]

Example 2:

Multiply 750 × 98

Solution:

750 × 98 means 98 times 750

or 100 times 750 – two time 750

or 75000 – 1500

or 73500

∴ 750 × 98 = 73500      [multiplication rules: The product of a positive integer and the negative integer is a negative integer.]

Example 3:

Find the value of 10 × 5 + 12 × 5 + 14 × 5

Solution:

10 × 5 + 12 × 5 + 14 × 5 means

10 times 5 + 12 times 5 + 14 times 5

or 36 times 5

or 36 × 5

or 180

∴ The value = 180


Understanding Example 4:

Multiply 843 by 55 by ordinary method

Solution:

8 4 3 ×

5 5
..............
4215
...................
4 6 3 6 5              [multiplication rules: The product of two positive integers is a positive integer.]
.......................

Example 5:

Find the product of (a) (+ 6) × (– 6) (b) (– 13) × (+ 13) (c) (– 15) × (– 4)

Solution:

(a) (+ 6) × (– 6) = – 36 (positive × negative = negative)

(b) (– 13) × (+ 13) = – 169 (negative × positive = negatives)

(c) (– 15) × (– 5) = + 75 (negative × negative = positive)

Example 5:

Find the product of (a) (– 4) × (+ 5) × (+ 2) (b) (– 2) × (– 5) × (+ 10) (c) (– 4) × (– 2) × (– 6)

Solution:

(a) (– 4) × (+ 5) × (+ 2) (b) (– 2) × (– 5) × (+ 10) (c) (– 4) × (– 2) × (– 6)

= (– 20) × (+ 2) = (+ 10) × (+ 10) = (+ 8) × (– 6) [The product of a positive integer and the negative integer is a negative integer. multiplication rules]

= – 40 = + 100 = – 48

Example 6 :

Multiply  2 and 3

Solution:

By repeated addition we get the following:

2 × 4 = 2 + 2 + 2+2 = 8.

Practice problem for multiplication solver:


1.Multiply  2 and 5

Answer:10

2.Multiply  100 and 108

Answer:10800

learning area theorems

There are so many theorems on areas and deductions from that main theorem.  Any planar closed figure has an area.  If the figure is a noted one like triangle, rectangle, circle, polygon etc. we have specific formula so that we can find out areas by applying formula.

For Geometric closed figures like quadrilateral, irregular polygons we use intersecting lines to divide into triangles and add all areas of the triangle, so that all the triangles taken for this purpose are different segments of the polygon.

In this article, we study about how to calculate area for regular polygons.

Area of a polygon:

A polygon is a closed curve consisting of line segments in a plan which are called sides of the polygon.  There are no restrictions to the number of sides of a polygon in a plane.  Starting from 3, it can have any no of sides up to infinity.

Regular polygons are polygons which have equal sides and equal angles.  Regular polygons are special types of polygons with certain extra properties.

For an irregular polygon, no standard formula but can be considered as the sum of all triangles formed by joining the centroid of the polygon to each vertex.  If there are n sides, n triangles will be formed.  The full area of the irregular polygon will be the sum of area of all triangles.

Corollary to the above for a regular polygon:

For a regular polygon of n sides we know there exists a centre and from centre lines are drawn to each vertex, the polygon will be a combination of n congruent triangles.  In other words, for a regular polygon the lines drawn from centre divides the polyton into n number of congruent triangles.  If we calculate one triangle area, then total area is calculated by multiplying by number of sides.

So Area of the polygon = n(area of each triangle) = n(1/2*side*height from centre of polygon to any side)

= 1/2 (n*side)*apothem    (Since apothem is the altitude of the side from the centre)

= 1/2 * Perimeter*apothem  (since n multiplied by no of sides give perimeter)


Learning problems solved for regular polygons - area theorem


Learning area theorems:

Problem1:  Find area of a regular hexagon with side 6.

Solution:

Area of hexagon of 6 equal sides = 1/2 (perimeter)(apotnem)

We know that a hexagon when divided into 6 triangles, each triangle is equilateral as central angle is 60.

Area of an equilateral triangle =√ 3 (a^2)/4 = 1.732*6*6/4 = 15.588

Hence area of hexagon = 6*15.588 = 93.528


Applications of regular area theorems in squares and triangles:

This theorem applies for equilateral triangle and square also.

For triangle, centroid is the centre and we know the apothem for the triangle formed with two vertices and one side is 1/3 h where h is the height of the triangle.  So area of the triangle = 3*a*1/2*1/3h = a2h/2 = √3/2a/2*a = √3a^2/4 = Area of the equilateral triangle

For square, the intersection of diagonals is the centroid and the apothem for each triangle is a/2.

So area of square by applying this theorem = 4a*a/2*1/2 =a^2

Problem 2:   Find area of a regular pentagon with side 7.

Solution:  Here we know each angle for a regular pentagon = 540/5 =108.

Hence line joining each vertex to the centre forms an angle of 108/2 = 54 degree with the side.

So 5 triangles are formed with base as 7 and angles two equalling 54 degrees.

Now we can find apothem using this.  Apothem / 1/2 side = tan 54:   or apothem = 7tan 54/2 =7(1.3763)/2

= 4.8173

Area of pentagon = 5*7*4.8173/2 = 168.61/2 =84.303

Thus given side and no of sides, the area of a regular polygon we can find out by using area theorems for polygons.

CONCLUSION:

In this article, we learnt about the area theorems for polygons and worked out problems on that. An interesting thing in a regular polygon is if we know the side we can find out the area of the polygon.

Real Numbers and their Decimal Expansions

Notation and Terminology   :   

A real number has a decimal representation. It gives the approximate location of the number on the number line.

Examples:

The rational number 1/2 is real and has the decimal representation 0.5.  The rational number  has the representation . The number 1/3 is also real and has the infinite decimal representation 1.333… This means there is an infinite number of 3’s, or to put it another way, for every positive integer n, the nth decimal place of the decimal representation of 1/3 is 3.


The number  has a decimal representation beginning 3.14159…  So you can locate  approximately by going 3.14 units to the right from 0.  You can locate it more exactly by going 3.14159 units to the right, if you can measure that accurately.  The decimal representation of  is infinitely long so you can theoretically represent it with as much accuracy as you wish.  In practice, of course, it would take longer than the age of the universe to find the first 10 to the power of 10 to the power of 10 digits.

Bar notation:

It is customary to put a bar over a sequence of digits at the end of a decimal representation to indicate that the sequence is repeated forever.  For example,

42 (1/3) = 42, 3bar.

and 52.71656565… (65 repeating infinitely often) may be written 52.7165 . 65 for 65 the bar notation is assign. means we have to put bar.

A decimal representation that is only finitely long, for example 5.477, could also be written 5.4770. in this 5.4770 for zero we have to put bar on that means we have to write bar(-) on zero.


Terminology:
The decimal representation of a real number is also called its decimal expansion.  A representation can be given to other bases besides 10; more about that here.


Variations in usage:

Approximations:

If you give the first few decimal places of a real number, you are giving an approximation to it.  Mathematicians on the one hand and scientists and engineers on the other tend to treat expressions such as " 3.14159" in two different ways.

The mathematician  may think of it as a precisely given number, namely 314159 / 100000, so in particular it represents a rational number. This number is not pie, although it is close to it.

The scientist or engineer will probably treat it as the known part of the decimal representation of a real number. From their point of view, one knows 3.14159 to six significant figures.

Abstractmath.org  always takes the mathematician's point of view.  If I refer to 3.14159, I mean the rational number 314159 / 100000.  I may also refer to pie as “approximately 3.15159…”.

Decimal representation and infinite series:

The decimal representation of a real number is shorthand for a particular infinite series (MW, Wik).  Let the part before the decimal place be the integer n and the part after the decimal place be

α1,α2,α3........

where αi is the digit in the i th place.  (For example, for ∏,  n=3 , α1=1,α2=4,α3=1,and so forth.)  Then  the

the decimal notation n.α1,α2,α3.... .  represents the limit of the series

The decimal representations of two different real numbers must be different. However, two different decimal representations can, in certain circumstances, represent the same real number.   This happens when the decimal representation ends in an infinite sequence of 9’s or an infinite sequence of 0’s.

Example

<!--[if !ie]--><!--[endif]-->   

<!--[if !ie]-->

<!--[endif]-->

<!--[if !ie]--><!--[endif]-->

These equations are exact.  <!--[if !ie]--><!--[endif]--> is exactly the same number as 3.5.  (Indeed, <!--[if !ie]--><!--[endif]-->    , 3.5, 35/10 and 7/2 are all different representations of the same number.)

Two proofs that <!--[if !ie]--><!--[endif]-->    <!--[if !ie]--><!--[endif]-->

The fact that <!--[if !ie]--><!--[endif]-->   is notorious because many students simply don’t believe it is true.  I will give two proofs here.

.......................................

Factoring Polynomials using Algebra

The reverse process of multiplying polynomials is defined as factoring polynomials. Consider that when we factor a number, we are searching for prime factors that multiply together to give the number. When a polynomial is factorized, usually only the polynomials are broken down to have integer coefficients and constants. Simplest way for factoring is that a common factor for every term. So we can factor out the common factor in the polynomials.

For example,  8=4*2, or 16=4*4


Problems:


Factoring Polynomials using Algebra Tiles:

Example for Factoring Polynomials using Algebra Tiles 1:

Factorize the polynomial x3 – 5x2 – 12x + 36

Solution for Factoring Polynomials using Algebra Tiles 1:

Sum of the coefficients of terms: 1–5–12 + 36 = 18 ≠ 0. ∴ (x–1) is not a factor.

Sum of the coefficients of even degree terms = –5 + 36 = 31

Sum of the coefficients of odd degree terms = 1 – 12 = –11

Since they are not equal we guess that (x + 1) is also not a factor. Let us check whether x – 2

is a factor. By synthetic division method

                 2 | 1       -5     -12        +36

                    |

                    |          +2      -6          -36
                    ________________________

                      1        -3     -18     |    0      

                  _________________________



Since the remainder is 0, (x – 2) is a factor. To find other factors

                      x2 – 3x – 18 = x2 – 6x + 3x – 18

                                       = x (x–6) + 3 (x–6) = (x + 3) (x – 6)

                       Therefore, x3 – 5x2 – 12x + 36 = (x–2) (x–6) (x+3)


Sample problem

Example for Factoring Polynomials using Algebra Tiles 2:

Factorize 2x3 + x2 – 5x + 2

Solution for Factoring Polynomials using Algebra Tiles 2:

Since the sum of the coefficients of all the terms: 2 + 1 – 5 + 2 = 5 – 5 = 0

We guess that (x – 1) is a factor.

By synthetic division,

                              1 | 2         +2        -5        +2

                                |

                                |              2        +3        -2

                                 ________________________

                                     2         3         -2     |   0        

                                _________________________

                     Remainder is 0. Quotient is 2x2 + 3x – 2

To find other factors, factorize the quotient,

                   2x2 + 3x – 2 = 2x2 + 4x – x – 2

                                = 2x (x + 2) – 1 (x + 2) = (x + 2) (2x – 1)

                   ∴ 2x3 + x2 – 5x + 2 = (x – 1) (x + 2) (2x – 1)

Mean Median Average

Mean

The mean is the average of the numbers.

It is easy to evaluate: Just add up all the numbers, then divide by how many numbers there are.

Example:

what is the mean of 2, 7 and 9?

Solution: 2 + 7 + 9 = 18
= 18 ÷ 3

= 6
Mean is 6

Median


The middle number (in a sorted list of numbers). Half the numbers in the listing are less, and half the numbers are greater are called as the median.

To find the Median, place the numbers you are given in value arrange and find the middle number.

If there are two middle numbers then average those two numbers.

Average

Average - The middle or most general in a set of data. There are three types of standard in mathematics - the mean, the median and the mode.


Concept of mean median average


Average

The average is a calculated "central" value or rate of a set of numbers.

It is easy to calculate: add up all the numbers and divide by how many numbers there are and you will have the average.

Example:

the average of 4, 6 and 11

Solution:   (4+6+11)/3

= 21/3

= 7

Average is 7

Mean

The most general expression for the mean of a statistical distribution with a discrete random variable is the mathematical average of all the terms. To compute it, add up the values of all the terms and then divide by the number of terms. This expression is also called the arithmetic mean.

Median

The median of a distribution with a discrete random or chance variable depends on whether the number of terms in the distribution is even or odd. If the number of conditions is odd, then the median is the value of the term in the middle. This is the value such that the number of conditions having values greater than or equal to it is the same as the number of terms having values less than or equal to it. If the number of conditions is even, then the median is the average of the two terms in the middle, such that the number of terms having values greater than or equal to it is the same as the number of terms having values less than or equal to it.


Example of mean median average


A student has gotten the subsequent grades on his tests: 87, 95, 76, and 88. He wants an 85 or better overall. What is the least amount of grade he must get on the last test in order to achieve that average?

The unknown score is "x". Then the desired average is:

(87 + 95 + 76 + 88 + x) ÷ 5 = 85

Multiplying through by 5 and simplifying, I get:

87 + 95 + 76 + 88 + x = 425
346 + x = 425
x = 79

He needs to get at least a 79 on the last test.