Online Tutoring Standard Definite Integrals

The expression of definite integral with lower and upper limit is `int_a^b`f(x) dx . The definite integral is also known as Riemann integrals. In definite integral, `int_a^b`f(x) dx . where, a, b and x is the complex numbers and path of integration from a to b. The calculus fundamental theorem allows the definite integrals to be computed in terms of indefinite integrals. Where, F is the indefinite

integral and f(x) is continuous function. The standard form of indefinite integral is `int` f(x) dx. The standard form of definite integral with an interval a to b `int_a^b`f(x) dx = F(b) - F(a).

Standard integral formulas:

1.Standard integral formula for nth power ` int` xn dx  = `x^(n+1)/(n+1)` + c

2.Standard integral formula for exponential function `int` ex dx = ex + c

3.Standard integral formula for logarithmic function ` int ` log x dx = x (log x - 1) + c

4.Standard integral formula for trigonometric function ` int ` sec2x.dx = tan x + c

These formulas are helps to solve the definite integrals. Trained tutors are available in online to assist the learners for learning standard definite integrals at every time. The online tutors will solve every problem in step by step manner with the correct and neat explanation. Tutors are real person, who is working as a teaching staff in online with the help of Internet.

Online tutoring standard definite integrals problems:

Online tutoring standard definite integrals problem 1:

Find the value of definite integral sin x with the limits 0 to `pi`

Solution:

Given definite integral function is, `int_0^(pi) ` sin x dx

So, `int_0^(pi) ` sin x dx =  `[-cosx]_0^(pi)` .

Substitute the upper limit and lower limit value. So, we get

= - cos `(pi)` - (-cos 0)

= - (-1) - (-1)       (Because the trigonometric value of cos 0 = 1 and cos`pi` = -1)

= 1+ 1

= 2

Answer:    `int_0^(pi) ` sin x dx = 2.

Online tutoring standard definite integrals problem 2:

Find the value of definite integral : e-x with the limits 1 to 2

Solution:

The integral of e-x can be expressed as `int_1^2` e-x dx

Already we know the integral formula `int` eax dx = `1/a` eax + c

here,    a = -1

So, apply the above condition  ` int` e-x dx = `(1/(-1))` e(-1)x + c

`int_1^2`e-x dx   = `(1/(-1))` e(-1)x

= [-e-x ]`_1^2`

= [-e-2 ] - [-e-1]

= [-(0.135)] - [-(0.3678)]

= - 0.135 + 0.3678

we know  e-1 = 0.3678     and e-2 = 0.135

= 0.2328

Answer: 0.2328

Odd are Even Numbers and Math

Let us study about odd and even numbers in math. Odd numbers are defined as the numbers that are not divisible by 2 or any other even numbers but they are divisible by some other odd numbers.
Even numbers are defined as the numbers that are divisible by 2 and some other even numbers but not divisible by any odd numbers.
Examples for odd and even numbers are below.


Odd are even numbers and math:

Odd are even numbers and math example 1:

List out all the possible odd numbers from 1 to 50.



Solution:

All the possible odd numbers that are not divisible by 2 or any other even numbers between 1 and 50 are as follows:
1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47 and 49



Odd are even numbers and math example 2:

List out all the possible even numbers from 1 to 50.



Solution:

All the possible even numbers that are divisible by 2 or any other even numbers between 1 and 50 are as follows:
2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46 and 48



Odd are even numbers and math example 3:

Find the odd numbers from the following number series 34, 21, 76, 55, 81, 95, 111, 311, 512, 431, 224, 673 and 700



Solution:

The odd numbers present in the given number series 34, 21, 76, 55, 81, 95, 111, 311, 512, 431, 224, 673 and 700 are found to be as follows:
21, 55, 81, 95, 111, 311, 431 and 673 since these numbers cannot be divided by 2 or any other even numbers but can be divided by some other odd numbers.



Odd are even numbers and math example 4:

Find the even numbers from the following number series 34, 21, 76, 55, 81, 95, 111, 311, 512, 431, 224, 673 and 700



Solution:

The even numbers present in the given number series 34, 21, 76, 55, 81, 95, 111, 311, 512, 431, 224, 673 and 700 are found to be as follows:
34, 76, 512, 224 and 700 since these numbers cannot be divided by any odd numbers but can be divided by 2 or some other even numbers.


Odd are even numbers and math exercises:

List out all the possible odd numbers from 70 to 100. (Answer: 71, 73, 75, 77, 79, 81, 83, 85, 87, 89, 91, 93, 95, 97 and 99)
List out all the possible even numbers from 100 to 130. (Answer: 100, 102, 104, 106, 108, 110, 112, 114, 116, 118, 120, 122, 124, 126, 128 and 130)
Find the odd numbers from the following number series 423, 322, 333, 129 and 112. (Answer: 423, 333 and 129)
Find the even numbers from the following number series 423, 322, 333, 129 and 112. (Answer: 322 and 112)

Synthetic Methods

Synthetic division is a shorthand, or shortcut, method of polynomial division in the special case of dividing by a linear factor -- and it only works in this case. Synthetic division is generally used, however, not for dividing out factors but for finding zeroes (or roots) of polynomials.

Let's look at the quadratic equation: y = x2 + 5x + 6. From the rational root testwe  know that ± 1, 2, 3, and 6 are possible zeroes of the quadratic. (And, from the factoring, we know that the zeroes are, in fact, –3 and –2.) How would we use synthetic division to check the potential zeroes?  If we guess that x = 1 is a zero, then this means that x – 1 is a factor of the quadratic. And if it's a factor, then it will divide out evenly; that is, if we divide x2 + 5x + 6 by x – 1, we would get a zero remainder.

we divide x2 + 5x + 6 by x – 1 by synthetic method as follows:

1. write the coefficients ONLY inside an upside-down division symbol:

write coefficients in upside-down division symbol

Make sure you leave room inside, underneath the row of coefficients, to write another row of numbers later.


2.Put the test zero, x = 1, at the left:

write test zero at left

3. Take the first number inside, representing the leading coefficient, and carry it down, unchanged, to below the division symbol:

carry down leading coefficient


4.Multiply this carry-down value by the test zero, and carry the result up into the next column:

multiply by test zero, and carry result up into next column


5.Add down the column:

add down the column

6. Multiply the previous carry-down value by the test zero, and carry the new result up into the last column:

multiply result by test zero, and carry result into next column


7.Add down the column:

This last carry-down value is the remainder.

that means (x2 + 5x + 6) ÷ (x – 1) =12

Hence x-1 is not a factor of the polynomial x2 + 5x + 6

Hermite Polynomials

The Hermite polynomials which are a classical orthogonal polynomial sequence arises in mathematics in probability, such as the Edgeworth series; in combinatorials, as an example of an Appell sequence, obeying the umbral calculus; and in physics, where they give rise to the eigenstates of the quantum harmonic oscillator. They are named in honor of Charles Hermite.

Definition

Hn(x) = (-1)nex2/2 [`d/dx`(e-x2/2)]n

(the "probabilists' Hermite polynomials"), or sometimes by

Hn(x) =(-1)nex2 [`d/dx`(e-x2)]n

(the "physicists' Hermite polynomials"). The above two definitions are not exactly equivalent; either is a rescaling of the other, to wit

Hnphys(x) = 2n/2Hnprob(`sqrt(2)`x )

Usually follow the first convention is followed. That convention is often preferred by probabilists because

`1/sqrt(2Pi)` e-x2/2

is the probability density function for the normal distribution in which expected value 0 and standard deviation 1.

Properties of Hermite polynomials

Hn is a polynomial of degree n. According to probabilists' definition it has leading coefficient 1, while according to physicists' definition it has leading coefficient 2n.

Orthogonality

Hn(x) is an nth-degree polynomial for n = 0, 1, 2, 3, .... These polynomials show orthogonality with respect to the weight function     (measure)

W(x) = e-x2/2   (probabilist)

or

W(x) = e-x2   (physicist)

Completeness

The Hermite polynomials (probabilist or physicist) form the basis of orthogonality of the Hilbert space of functions satisfying
`int_-oo^oo|f(x)|^2w(x)dx` < `oo`


Hermite's differential equation

The probabilists' version of Hermite polynomials are solutions of the differential equation

(e-x2/2 u')' + `lambda`e-x2/2 u = 0

Recursion relation

Hermite polynomials' sequences also satisfies the recursion

Hn+1(x) = xHn(x) - H'n(x) (probabilist)

Hn+1(x) = 2xHn(x) - H'n(x) (physicist)


Applications of Hermite polynomials

Hermite functions

One can define the Hermite functions from the physicists' version of polynomials:

`psi`n(x)  =   `1/sqrt(n!2^nsqrt(Pi))`  e-x2/2 Hn(x)


Recursion relation

Following recursion relations of Hermite polynomials, the Hermite functions satisfy

`psi`'n(x) = `sqrt(n/2)``psi`n-1(x) - `sqrt((n+1)/2)``psi`n+1(x)

Cramér's inequality

The Hermite functions satisfy the following inequality due to Harald Cramér

|`psi`n(x)|   `<=` K

for x real, where the constant K is less than 1.086435.

Hermite functions act as eigenfunctions of the Fourier transform

Area using Perimeter

Area:
Area is a measure of  the outermost surface of an object. Area is a bounded space with its surrounding. Area is the  measure of  plane region or  amount at surface which any object occupies is called the area for that object. Area for any object is a measure ofbounded surface to its bounded space

Perimeter:
The length of the boundary of any closed figure is called its perimeter. The distance around a two-dimensional shape,for a circle perimeter is a circumference.closed surface for inner space occupied it said to be the perimeter for an object.

Finding Area using Perimeter For Square:

Area of Square =a2
Perimeter of Square =4a
Example 1:
For square board the perimeter is 84cm find the area for square board.

Solution:
we know perimeter is 84cm
for square perimeter,4a=84
a=84/4
a=21
area of square =a2
=212
area of square =441cm

Example 2:
The perimeter of a square playground is 1200m. Find its area using perimeter.

Solution :
Perimeter of the square ground, p = 4a
4a = p
a =p/4

Hence, a = 1200
4 m [since p = 1200 m ]
∴ a = 300 m
Area of the square ground A = a2
= 300 m × 300 m
Area = 90000 sq.m.

Finding Area using Perimeter For Rectangle:

Area of Rectangle =Length `xx` Breadth
Perimeter of rectangle =2(Length+Breadth)

Example 1:
If perimeter of rectangle is 18 m and its breadth 4m. find its area using perimeter.

Solution:
2 length + 2 breadth = perimeter
2l + 2`xx` 4 = 18
2l + 8 = 18
2l = 18– 8
2 l = 10
l =5
l=5,b=4
Now the area of the rectangle A = l `xx` b
= 5 m × 4 m
= 20 sq.m.

Example 2:
The perimeter of the floor of a rectangular room is 24 m and the  length is 7 m. Find its area.

Solution :
2 length + 2 breadth = perimeter
2 × 7 + 2b = 24
14 + 2b = 24
2b = 24 – 14
2 b = 10
b =`10/2`
= 5
Now the area using perimeter of  rectangle A = l` xx` b
= 7 m × 5 m
= 35 sq.m.

Dividing Fractions Solver

As division of whole numbers shows how often one integer is contained in another integer, to division of fraction shows how often one fraction is contained in another. If the numerator of any fraction be made a denominator, and the denominator a numerator, the fraction of made is called the reciprocal of the former. Thus 4 / 3 is the reciprocal of 3 / 4.

In division * as the divisor is to the dividend, of is a unit of the quotient (both in whole numbers and fraction).

Dividing fractions solver – Steps and methods:

To dividing algebraic fraction, follow these steps:

Write the given fractions.
Change the division sign to a multiplication sign and invert the second fraction.
Write the given fraction and cancel any common factors.
Multiply the numerators.
Multiply the denominators.

Dividing fractions solver – Methods:

Dividing common fraction
Dividing a fraction and a whole number
Dividing mixed number


Dividing fractions solver – Three methods and example problem:

Dividing common fractions:

To divide common fraction, follow these steps:

Invert the divisor.
Then multiply the numerator by the numerator and the denominator by the denominator.
Reduce to lowest terms when possible.

Example:

Dividing the fractions using fraction solver

(5 / 6) / (3 / 4)

Solution:

(5 / 6) / (3 / 4) (divisor)

= (5 /6) * (4 / 3)

= (5 * 4) / (6 * 3)

= 20 / 18

= 10 / 9

Dividing a fractions and a whole number:

To divide a fraction and a whole number, follow these steps:

Change the whole number to a fraction by the placing the whole number over one.
Invert the divisor.
Then multiply the numerator by the numerator and the denominator by the denominator.
Reduce to lowest terms when possible.

Example:

Dividing the fractions using fraction solver

16 / (5 / 8)

Solution:

16 / (5 / 8)

= (16 / 1) / (5 / 8)

= (16 / 1) * (8 / 5)

= (16 * 8) / (1 * 5)

= 128 / 5

= 25 3/5

Dividing mixed number fractions:

To divide mixed numbers, follow these steps:

Change the mixed number to an improper fraction
Invert the divisor.
Then multiply the numerator by the numerator and the denominator by the denominator.
Reduce to lowest terms when possible.

Example:

Dividing the fractions using fraction solver

(2 7/8) / (4 / 5)

Solution:

(2 7/8) / (4 / 5)

= (2 7/8) = ((2 * 8 + 7) / 8) = 23 / 5

= (23 / 8) / (4 / 5)

= (23 / 8) * (5 / 4)

= (23 * 5) / (8 * 4)

= (115 / 32)

= 3 19/ 32.

Hard Subtraction Problems

Subtraction is one of the arithmetic operations; it is the inverse of addition, meaning that if we start with any number and add any number and then subtract the same number we added, we return to the number we started with. Subtraction is denoted by a minus sign in infix notation. The traditional names for the parts of the formula

c − b = a

(Source: Wikipedia)

Hard subtraction example problems:

Hard subtraction problem 1:

Dylan made five hundred twenty-three cupcakes. Two hundred ten of them have already been put into packages. How many cupcakes are left to be packaged?

Solution:

Dylan made five hundred twenty-three cupcakes.

Two hundred ten of them have already been put into packages.

Cupcakes are left to be packaged = 523 – 210

= 313 cupcakes.

Answer: 313 cupcakes are left to be packaged.

Hard subtraction problem 2:

There are four hundred fifty employees working in an office building. Hundred an sixteen of them are about to leave to go out to lunch. How many employees will be left in the building?

Solution:

There are four hundred fifty employees working in an office building.

Hundred an Sixteen of them are about to leave to go out to lunch.

= 450 – 116

= 334

Answer: 334 employees will be left in the building.

Hard subtraction problem 3:

The pet store had three hundred twenty-eight bags of bird feed. Two hundred six will be eaten today. How many bags will be left?

Solution:

The pet store had three hundred twenty-eight bags of bird feed.

Two hundred six will be eaten today.

= 328 – 206

= 122 bags.

Answer: 122 bags will be left.

Hard subtraction practice problem:

Practice problem 1:

There were twenty-three magazines on a bookshelf this morning. Students came and borrowed ten of them. How many magazines are left?

Answer: 13 magazines are left

Practice problem 2:

There are fifty-two colored pens in a desk. There are twenty students and each will need one. How many colored pens will be left?

Answer: 32 colored pens will be left.

Practice problem 3:

There are forty-two birds sitting in two trees. One tree has twenty-two birds in it. How many birds are in the other tree?

Answer: 20 birds are in the other tree.

Parallelograms

Parallelogram is defined as the " four sided closed planar figure whose opposite sides are parallel to each other  " .

           Here ABCD is a parallelogram

There are some conditions to be satisfied by a quadrilateral so that it could be identified as parallelograms and they are as follows :
1 . The two pair of sides it consist of should be parallel to each other.
2 . The two pairs of opposite sides should be of same length .
3 . The angles of the two pair of opposite sides must be same .
4 . The diagonals drawn inside the parallelogram should bisect each other .
5 . One pair of opposite sides should be parallel and equal in length .

Types of Parallelograms

There are three types of parallelograms and they are as follows :

•  Square :

                    
                  
It is considere to be a type of parallelogram because it satisfies the condition required for a figure to be considered as the parallelogram such as
1 . The opposite pair of sides are parallel.
2 . The opposite sides are of same length .
3 . The opposite pair of angles are same .
4 . The diagonals inside the square bisect each other .
5 . One pair of opposite sides are parallel and they are equal in length .

• Rectangle :

                         
                          Rectangle is considered to be a type of parallelogram as it satisfies all the condition required to be considered as the parallelogram .

Types of parallelograms : Rhombus

• Rhombus :

                       
                             Rhombus is also considered as a type of parallelogram as it satisfies all the conditions of the parallelogram .

Secant Tangent Sine

Trigonometric functions are commonly defined as ratios of two sides of a right triangle containing the angle, and can equivalently be defined as the lengths of various line segments from a unit circle. The trigonometric functions are functions of an angle. It is also called as circular function. Trigonometric functions are important in the study of triangles and modeling periodic phenomena, among many other applications. The familiar trigonometric functions are the sine, cosine, and tangent. The other trigonometric function like cosecant, secant and cot are related to the familiar function.
                                                                                                                                                                             Source Wikipedia.

Some Trigonometric related functions:

1. `sin^2 theta + cos^2 theta = 1`
2. `tan theta = sin theta / cos theta`
3.` cot theta = 1/tan theta = cos theta/ sin theta `
4. `1+ tan^2 theta = sec^2 theta`
5.`sec theta ` = `1/cos theta`
6.` cosec theta = 1/ sin theta`
7. `sin 2theta = 2 sin theta cos theta`
8.`sin^2 theta ` = `(1 - cos 2theta)/2`

More details about trigonometric functions:

           In Right-angle triangle
               
Sine (Sin):
       In right-angle triangle, Ratio of the opposite side length and the hypotenuse of an angle is called as sine. It is reciprocal of cosecant.
             ` sin theta` = (opposite side) / (hypotenuse side)
Tangent (Tan):
         In right-angle triangle, Ratio of the opposite side length and the adjacent side length of an angle is called as tangent. In trigonometry relation is the ratio of sin and cosine. The tangent is reciprocal of cot.
                ` tan theta` = (opposite side) / (adjacent side)
Secant (Sec) = `1/cos` :
             In right-angle triangle, Ratio of the hypotenuse and the adjacent side length of an angle is called as secant. The secant is reciprocal of cosine.
                 `Sec theta` = (hypotenuse)/(adjacent) <br>
Cosine (Cos):
      In right-angle triangle, Ratio of the adjacent side length and the hypotenuse of an angle is called as cosine. It is a reciprocal of secant.
             .`cos theta` = (adjacent side)/(hypotenuse) 
Cosecant (cosec) = `1/sin` :
          In right-angle triangle, Ratio of  the hypotenuse and the opposite side length of an angle is called as cosecant. The cosecant is reciprocal of sine.
                ` cosec theta` = (hypotenuse)/(opposite side)
Cot` (1/ tan)` :
       In right-angle triangle, Ratio of the adjacent side length and the opposite side of an angle is called as cot. In trigonometry relation is the ratio of cosine and sin. The cot is reciprocal of tangent.
              `cot theta ` = (adjacent side)/(opposite side)

Trigonometric problems:

Trigonometric problem 1:
      If x = a cos t + b sin t and y = a sin t − b cos t,  show that   x² + y² = a² + b²
   Solution:
              x² + y² = (a cos t+ b sin t)² + (a sin t − b cos t)²                       
                          = a² cos²t + b² sin²t + 2ab cos t sin t+ a² sin²t + b² cos²t − 2ab sint cost                        
                          = a² (cos²t + sin²t) + b²(sin²t + cos²t)+ 2ab cos t sin t − 2ab sint cost
                          = a² + b²
                     Hence it has been proved.


Trigonometric problem 2:
      Prove that trigonometric relation: `(sec theta + 1/ (cot theta))^2 ` = `(1 + sin theta)/(1 - sin theta)`    
      Solution:
         Take left side, `sec theta + 1/cot theta `
         Squared the term, `sec theta + 1/cot theta` = `(sec theta + 1/cot theta)^2`   
       Now,
          `(sec theta + 1/cot theta)^2 = ( (1/cos theta) + (sin theta / cos theta) )^2`    ; we know, `sec theta = (1/cos theta) ; 1/cot theta = (sin theta /cos theta)`
                                     = `(1 + sin theta )^2/ (cos^2theta)`
                                     = `(1 + sin theta)^2 / (1 - sin^2 theta)`                                ; sin²θ + cos²θ = 1;  so,cos²θ = 1- sin²θ
                                     = `((1 + sin theta)(1 + sin theta)) / ((1 + sin theta) (1 - sin theta))`
      `(sec theta + tan theta)^2`     = `(1 + sin theta)/(1 - sin theta)`  
                                                  = Right side.
                  Hence, the given trigonometric relation is proved.

Value Chain Functions

Value chain Rule:

The derivative of the composition of the functions  which is computed as a product of their derivatives. To be precise:
Let f(x) and g(x) be two functions and let (g. f) (x) = g (f(x)) be their composition.

Value Chain rule 1:

If h(x) = g(f(x)), then h'(x) = g'(f(x))f'(x)

Note: In computing the composition g(f(x)), we apply f to x first, and then we apply g to f(x). According to the chain rule, when doing the derivative, we proceed in the opposite order: g is done first, and then f. One more thing, the f(x) part in g'f(x)) states that, while doing the derivative of g, we do not change f(x)(the f(x) term which is usually called the “inside”).

Sometimes we think of the composition in the following way:

y= g(u) and u =f(x) (i.e., y depends io u, and u depends on x). In that case, y depends on x and its derivative is,

Value Chain Rule 2:

If y= g(u) and u=f(x), then dy/dx = dy/du*du/dx

Examples of value chain function:

Value Chain Rule - Example1:

Compute the derivatives of the following functions.

(a)  Y= (x2+1)14

(b) Y = `sqrt (sin x +1)`

(c)  Y = `1/(e^x +2)`

(d) Y = sin(x2 + 1)

(e)  Y = cos(sec x)

( f )   Y = (sin x)2 + sin (x2).

Solution:

(a)  We start by computing the derivative of the power of 14:

Y' = 14(x2+1)13(x2+1)`

= 14(x2+1)132x

= 28(x2+1)13

(b) Writing y = `(sin x+1)^(-1/2)` , we get

Y` =`1/2` `(sinx + 1)^(-1/2)` (sinx+1)`

=  `1/2`  (sinx + 1)-1/2 cos x

=  `1/2` cos x(sin x +1) -1/2.

(c)  Y = (ex+2)-1.thus,

Y` = (-1)(ex+2)-2(ex+2)`

= -ex(ex+2)-2

= -ex/(ex+2)2.

(d) We start by computing the derivative if sin:

Y` = cos (x2+1)(x2+1)`

= 2x cos(x2+1)

(e)  Y` =-sin(sec x) (sec x)`

= - sin (sec x ) sec x tan x.

( f )   We have to be careful about the order:

Y` = 2(sin x)1(sin x)` + cos(x2)(x2)`

= 2sin x cos x +2x cos(x2)

= sin 2x +2x cos(x2)

In simplifying, we used the formula 2 sin x cos c  = sin 2x.

Value chain Rule - Example 2:

Find `dy/dx` for the following functions,

(a)  Y = 4u2 – 3u +2, u = ex+2e2x

(b) Y  = ln u, u=sin x + cos x

Solution:

(a)  By the chain Rule ,

`dy/dx = (dy/(du)) . ((du)/dx)`

= (8u -3)(ex + 4e2x)

= 8((ex + 2e2x) – 3) (ex + 4e2x).

(b) As in (a),

`dy/dx = dy/(du) . (du)/dx`   = `1/u(cos x - sin x) =(cos x -sin x)/(sin x + cos x)`

Algebra Logarithms Solve

The algebra logarithms solve represents the problems in logarithms that uses the algebra. The logarithm for the given number to a given base is generally the power or exponent to which the base must be raised in order to produce that number.Sometimes the logarithmic table  are used in the problems for evaluating the values in the forms of logarithmic algebra.


Examples to explain algebra logarithms solve

solve the natural logarithm for ln ((ab)2/ca)

solution:

ln ((ab)2/ca)

=ln ((ab)2) -  ln (ca)    by property Quotient formula

=ln a2 + ln b2  - (ln c+ ln a)  by property Product formula

=2 ln a + 2 ln  b- ln c- lna

= ln a +  2 ln b - ln c

Consider the same problem without using natural logarithms.

solve log ((ab)2/ca)

=log ((ab)2) -  log (ca)    by property Quotient formula

=log a2 + log b2  - (log c+ log a)  by property Product formula

=2 log a + 2 log  b- log c- log a

= log a +  2 log b - log c

solve the following log x expansions for logarithmic algebra  log1213 + 13 log2222



Given:

log12 13 + 13 log22 22

= log1213 + 13 (1)          ( log aa = 1)

= log1213 + 13

This is the required log x expansion. Hence we solved the problem.

Problems to explain algebra logarithms solve

Solve log51003

Solution:

log51003=3 log5100

= 3 log5(10*10)

= 3 (log510 + log510)

= 3(2 log510)

Log51003= 6 log510.

Solve the following log x expansions for logarithmic algebra log17y -  log17 x +  log1715

Solution:

log17 y - log17 x + log17 15              (given)

= log17`(y / x)` + log17 15                     ( log a - log b = log `a/b` )

= log17( `y/x`  . 15)                                (log a + log b = log a.b)


This is the required log x expansion. Hence we solved the above problem.

Prove that  log3 4 x log4 5 x log5 6 x log6 7 x log7 8 x log8 9 = 2

Solution:
Left Hand Side = ( log3 4 x log4 5 ) x ( log5 6 x log6 7 )  x ( log7 8 x log8 9  )
=log3 5  x  log5 7  x  log7 9
= log3 5 x  (  log5 7  x  log7 9)
= log3 5 x log5 9
= log3 9 = log3 32
= 2log3 3
= 2 x 1
= 2 = Right Hand Side
This problem explains the change of base rule.

Linear Algebra Vector

Linear algebra vector is one of the most important topics in linear algebra or Algebraic theory. In this linear algebra having some of the sub topics, like vector, matrix and so on.  This linear algebra having some more important application in matrices and vector algebra. Now we are going to see  about vector algebra and vector spaces in linear algebra theory.

Vector space in Algebra:

A vector space is a concept of a group of vectors or set of vectors. Which are working on two operations, like1.Vector addition and 2. Scalar multiplication. This is general concept of vector. And it is having more subtopics: Vector spaces, sub spaces, fundamental sub spaces, inner product spaces, span, basis and dimension, change of basis, linear independence, least squares, orthogonal matrices, QR- Decomposition, and orthonormal basis.

Example for vector space: the complex number 4 + 5i can be considered a vector space, it is way of explained by

`[[4],[5]]`

the vector space is a “space”, such as abstract objects.  Which is we called as vectors. Now we are study about the vectors, and see the vectors like S2, S3… Sn and so on. These all are like a vector spaces.
Vector addition is denoted by = x, y `in`  V, (x + y).


Properties of vector algebra:

In this vector addition and scalar multiplication is having some important properties, there are given below,

Vector addition property:

Commutative addition property:` vecx` + `vecy` = `vecy + vecx.`
Associative addition property: `vecx + (vecy + vecz)` = `(vecx + vecy) + vecz` .
Addition identity property: Here it is a 0 vector, so 0 +` vecx ` = `vecx `   for all x.
Addition inverse property: for each x vector, there exist another y vector like`vecx + vecy` = 0.



Scalar multiplication property:

Scalar associative property: α (βx) = (vecαβ) x.
Scalar distributive property: (α + β) x = αx+βx.
Vector distributive property: α(x+y) = αx+αy.
Scalar identity property: 1x = x.

These all are the very important in linear algebra properties in vector.


Types of vector:

Types of Vectors

Zero Vector A vector whose initial and terminal points coincide, is called a zero Vector (or null vector), and denoted as `vec0` . Zero vector can not be assigned a definite direction as it has zero magnitude. Or, alternatively otherwise, it may be regarded as having any direction. The vectors `vec(A A) `, `vec(BB)` represent the zero vectors,

Unit Vector A vector whose magnitude is unity (i.e., 1 unit) is called a unit vector. The Unit vector in the direction of `veca ` given vector a is denoted by `hata` .

Co initial Vectors Two or more vectors having the same initial point are called co initial vectors.

Collinear Vectors Two or more vectors are said to be collinear if they are parallel to the same line, irrespective of their magnitudes and directions.

Equal Vectors Two vectors `veca` and `vecb` are said to be equal, if they have the same magnitude and direction regardless of the positions of their initial points, and written as `veca = vecb`

Negative of a Vector A vector whose magnitude is the same as that of a given vector (Say, `vec(AB)` ), but direction is opposite to that of it, is called negative of the given vector.

For example, vector `vec(BA)` is negative of the vector `vec(AB)` , and written as `vec(BA) = vec(-AB)`

Trinomial Squaring Twice

In elementary algebra, a trinomial is a polynomial consisting of three terms or monomials.(source : WIKIPEDIA)
The trinomial must be one of the following form .
Examples for trinomial:    
1.  x + y + z , where x , y, z are variables.
2. 2xy + 3x + 4y, Where x, y are variables.
3. x²+2x-7, where x is variable.
4. ax + by + c = 0 , Where a,b,c are constants and x,y are variables.
In the following section we are going to see some solved problems and practice problems on trinomial squaring twice .

Solved problems on trinomial squaring twice:

Trinomial squaring twice problem 1:
Solve by squaring twice (2x + 5y + 6)⁴
Solution:
Given , (2x + 5y + 6)⁴
We can write it as , (2x + 5y + 6)⁴  = (2x + 5y + 6)^{2^2}               
                                                             = (2x + 5y + 6)² × (2x + 5y + 6)²
Let us take (2x + 5y + 6)²
We can apply the following formula to square the trinomial,
(a+b+c)² = ( a² + b² + c² + 2ab + 2bc + 2ca )
(2x + 5y + 6)² = (4x² + 25 y² + 36 + 20xy + 60x + 24x)
(2x + 5y + 6)² × (2x + 5y + 6)² = (4x² + 25 y² + 36 + 20xy + 60x + 24x) × (4x² + 25 y² + 36 + 20xy + 60x + 24x)
Answer: 2x + 5y + 6)⁴ =   (4x² + 25 y² + 36 + 20xy + 60x + 24x) × (4x² + 25 y² + 36 + 20xy + 60x + 24x) 

Trinomial squaring twice problem 2:
Find the trinomial by squaring (3x +2).
Solution:
Given (3x +2)
We can get the trinomial by squaring the given binomial.
(3x+2)² = (3x + 2) ( 3x + 2)
              = 3x(3x + 2) + 2( 3x+2 )
              = 9x² + 6x + 6x + 4
              = 9x² + 12x + 4
Answer: (3x+2)² =  9x² + 12x + 4

Trinomial squaring twice problem 3 :
Factor the trinomial 2x²- 22 x + 48
Solution:
Given 2x²- 22 x + 48
Divide by 2 ,We get  x²- 11x + 24
                                                                     24   (product)
                                                                     /  \    
                                                                 - 8   - 3 
                                                                    \   / 
                                                                    - 11     (sum)
Factors:  (x-8) and (x-3).
Verification:
(x-8) (x-3)  = x(x-8) -3(x-8)
                = x²-8x -3x+24
(x-8) (x-3)  = x² -11x + 24

Practice problems on trinomial squaring twice:

Problems:
1. Find the trinomial by squaring (5x + 6 )
2.Find the trinomial by squaring (x-8)
Answers:
1.25x² + 60x + 36
2. x² -16x + 64

Study Correlation Coefficient

In this article we will discuss about study correlation coefficient. The correlation coefficient, an idea from statistics calculates of how well trends in the expected values follow trends in past real values. The correlation coefficient is a value between 0 and 1. If there is no association between the predicted values and the definite values the correlation coefficient is 0 or very low.

Formula for study correlation coefficient:

`"Correlation(r)"=[NsumXY-((sumX)(sumY))/((sqrt([(NsumX^2)-(sumX)^2][NsumY^2-(sumY^2)])) ]]`

Where
              N = Total number of values
              X = 1st achieve
              Y = 2nd achieve
            sum XY = Addition of the product of 1st and 2nd achieve
              sum X = Addition of 1st achieve
              sum y = Addition of 2nd achieve
              sum x² = Addition of square 1st achieve
              sum y² = Addition of square 2nd get achieve

Example problems for study correlation coefficient:

Study Correlation coefficient – Example 1:
Calculate the Correlation co-efficient of following table.

X	60	61	62	63	64
Y	3.1	3.6	3.8	4	4.1

Solution 1 for study correlation coefficient:
            Step 1:  Count the number of values.
                         N = 5
            Step 2:  Calculate XY, X², Y²
                         See the below table

X	Y	x*y	x*x = x^2	y*y = y^2
60	3.1	60*3.1=186	60*60=3600	3.1*.31=9.61
61	3.6	61*3.6=219.6	61*61=3721	3.6*3.6=12.96
62	3.8	62*3.8=235.6	62*62=3844	3.8*3.8=14.44
63	4	63*4   =252	63*63=3969	4*4 = 16
64	4.1	64*4.1=262.4	64*64=4096	4.1*4.1=16.81

       Step 3:

       Find sumX, sumy, sumxy, sumx², sumy².

        Sum x = 310
        sum y = 18.6
        sum xy = 1155.6
        sum x² = 19230
        sum y² = 69.82

      Step 4:

       Now, Substitute in the above formula given.

            `"Correlation(r)"=[NsumXY-((sumX)(sumY))/((sqrt([(NsumX^2)-(sumX)^2][NsumY^2-(sumY^2)])) ]]`

             = `[5(1155.6) - ((310)(18.6)) / ((sqrt([5(19230)-(310)^2][5(69.82)-(18.6)^2]))]]`
              = `5778 - 5766 / sqrt(([96150 - 96100] [349.1-345.96]))`
              = `12 / sqrt(50 x 3.14 )`
              = `12/ sqrt(157)`
            = `12/12.5299`
            = 0.9577
            Answer is 0.9577
Study Correlation coefficient – Example 2:
         Find the Correlation co-efficient of following table

X	Y
50	3
51	3
52	3
53	4
54	4

        


 Solution for study correlation coefficient:
        Step 1: Count the number of values.
            N = 5
        Step 2: Find XY, X², Y²
            See the below table

X value	Y value	x* y	x*x	y*y
70	3	70*3=210	70*70=4900	3*3=9
71	3	71*3=213	71*71=5041	3*3=9
72	3	72*3=216	72*72=5184	3*3=9
73	4	73*4=292	73*73=5329	4*4=16
74	4	74*4=296	74*74=5476	4*4=16

          Step 3: Find sum X, sum y, sum xy, sum x², sum y².
            sum x = 360  
            sum y = 17
            sum xy = 1227
            sum x² = 25930
            sum y² = 59
           Step 4: Now, Substitute in the above formula given.
            `"Correlation(r)"=[NsumXY-((sumX)(sumY))/((sqrt([(NsumX^2)-(sumX)^2][NsumY^2-(sumY^2)])) ]]`
              = `[5(1227) - ((360)(17)) / ((sqrt([5(25930)-(360)^2][5(59)-(17)^2]))]]`                
              = `(6135 - 6120)/((sqrt([129650 -129600] [295-289])))`
              = 1`5/sqrt(50 xx6)`
              = `15/sqrt(300) `
              = `15/17.32`
              = 0.8660
              Answer is: 0.8660

Least Common Multiple of 8

In mathematics, the least common multiple of two rational numbers a and b is the smallest positive rational number that is an integer multiple of both a and b. Since it is a multiple, it can be divided by a and b without a remainder. (Source: From Wikipedia).

Least common multiple of two numbers can be found by their multiples. Here we are going to learn how to find the least common multiple of two or more numbers.

Example problems to find the least common multiple of two numbers

Here we will see some example problems to find the least common multiple of two numbers.
Example 1
Find the least common multiple of 6 and 8
Solution
The least common multiple of 6 and 8 can be found by finding the multiples 6 and 8.
The list of multiples of 6 and 8 are given below
The multiples of 6 = 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 72, 78, 84.
The multiples of 8 = 8, 16, 24, 32, 40, 48, 56, 64, 72, 80.
Here 24 is the lowest common number. So, 24 is the lowest common multiple of 6 and 8.
Example 2
Find the least common multiple of 5 and 8
Solution
The least common multiple of 5 and 8 can be found by finding the multiples 5 and 8.
The list of multiples of 5 and 8 are given below
The multiples of 5 = 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60.
The multiples of 8 = 8, 16, 24, 32, 40, 48, 56, 64, 72, 80.

Here 40 is the lowest common number. So, 40 is the lowest common multiple of 5 and 8.
Example 3
Find the least common multiple of 7 and 8
Solution
The least common multiple of 7 and 8 can be found by finding the multiples 7 and 8.
The list of multiples of 7 and 8 are given below
The multiples of 7 = 7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98, 105.
The multiples of 8 = 8, 16, 24, 32, 40, 48, 56, 64, 72, 80.
Here 56 is the lowest common number. So, 56 is the lowest common multiple of 7 and 8.
Example 4
Find the least common multiple of 11 and 8
Solution
The least common multiple of 11 and 8 can be found by finding the multiples 11 and 8.
The list of multiples of 11 and 8 are given below
The multiples of 11 = 11, 22, 33, 44, 55, 66, 77, 88 , 99, 110
The multiples of 8 = 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88.
Here 88 is the lowest common number. So, 88 is the lowest common multiple of 11 and 8.

Practice problems to find the least common multiple of two numbers

Practice problem 1

Find the least common multiple of 10 and 8
Answer: The least common multiple of 10 and 8 is 40
Practice problem 2
Find the least common multiple of 12 and 8
Answer: The least common multiple of 12 and 8 is 24
Practice problem 3
Find the least common multiple of 15 and 8
Answer: The least common multiple of 15 and 8 is 120

Prime Number

• In mathematics, a prime number (or a prime) is a natural number that has exactly two distinct natural number divisors: 1 and itself. The first twenty-five prime numbers are: Prime Numbers:- 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.

• Condition: - If X is the prime number next the factors of the number x is 1 and X. (Source:Wikipedia)                                                                                                                                           

Prime number of 7?

• The prime number 7 has only two divisors 1 and 7.
• Here we are going to discuss about some prime number problems.

Example problems of prime number:

Example: - 1

          Find the number 17 is the prime number or not?

Solution:-

          Given number is 17 we have to find out it is prime number or not.

          The factors of the number 17 is the 1, 17 only.

          In the number 17 have only two factors so it is a prime number.

Example: - 2

          Find the number 331 is prime number or not?

Solution:-

         Given number is 331 we have to find out it is prime number or not.

          The factors of the number 331 is the 1, 331 only.

          In the number 331 have only two factors so it is a prime number.

Example:-3

        Find the number 1259 is prime number or not?

Solution:-

          Given number is 1259 we have to find out it is prime number or not.

          The factors of the number 1259 is the 1, 1259 only.

          In the number 1259 have only two factors so it is a prime number.

Example: - 4

        Find out the given number 8 is prime number or not?

Solution:-

        Given number is 8 we have to find out it is prime number or not.

        The factors of the number 8 are 1, 2, 4, and 8 so it is more than two.

        In the number 8 has two or more factor so it not a prime number.

Example: - 5

        Find the number 300 is prime number or not?

Solution:-

        Given number is 300 we have to find out it is prime number or not.

        The factors of the number 300 are 1, 3, 5, 15, 30 and 300 is more than two.

        In the number 300 have more than two factor hence it is not a prime number.

Ellipse Parabola Hyperbola

An ellipse is a plane curve that results from the intersection of a cone by a plane in a way that produces a closed curve.

A parabola is a conic section, the intersection of a right circular conical surface and a plane to a generating straight line of that surface. Given a point and a corresponding line on the plane, the locus of points in that plane that are equidistant from them is a parabola.

A hyperbola is a curve, specifically a smooth curve that lies in a plane, which can be defined either by its geometric properties or by the kinds of equations for which it is the solution set.(Source: Wikipedia)



Example ellipse problem :

Find the equation of ellipse with its center at the origin and having foci at (`+- 2sqrt(6)`, 0 )and an eccentricity

equal to the  ,`(2sqrt(6))/(7)`

Solution:

The focal points on the X axis is the ellipse is oriented, view A, and the standard form of the equation is ` (x^2)/(a) + (y^2)/(b)`

The center of the origin, the numerators of the fractions on the left is x2 and y2. The problem is find the values of a and b.

The distance from the center to be either of the foci is equal to c. so in this problem

C = `2sqrt(6)`

from the given coordinates of the foci.

The values of a, c, and e (eccentricity) are related by

c=a*e

or

` a = (c)/(e)`

Substitute the values from c and e,

a = `(2sqrt(6))/((2sqrt(6))/(7))`

and

a = 7

a2= 49

Then, using the formula

b = `sqrt(a^2-c^2)`

or

b2 = a2 - c2

and substituting for a2 and c2,

b2 = 49 - (`2sqrt(6)` )2

b2= 49 - (4)(6)

b2= 49 - 24

gives the final required value of

b2 = 25

Then the equation of the ellipse is

`(x^2)/(49) + (y^2)/(25) ` = 1

Example parabola problem :

Reduce the equation y2-6y = 8x – 1 to standard form.

Solution:

Rearrange the equation of the 2nd -degree term and any 1st -degree terms of the same unknown are on the left side. Then group the unknown term appearing only in the 1st degree and all constants on the right:

y2 - 6y = 8x -1


Then complete the square in y:

y2 - 6y + 9 = 8x – 1 + 9

(y - 3)2 = 8x + 8

To get the equation in the form

(y - k)2 = 4a(x-h)

Factor an 8 out of the right side. Thus,

(y - 3)2= 8(x + 1)

is the equation of the parabola with its vertex at (-1,3).

Example hyperbola problem :

Find the equation of the hyperbola with an eccentricity of 3/2, directories x = ± 4/3, and foci at ( ± 3,0).

Solution:

The foci of the X axis at the points are (3,0) and ( - 3,0), so the equation is of the form ` (x^2)/(a^2)+(y^2)/(b^2)` = 1

First, the foci are given as (± 3,0); and since the foci are also the points (± c,0), then c=3

The eccentricity is given and the value of a2 can be determined from the formula

c = a * e

a = c/(e)

a = `(3)/((3)/(2))`

a =` (6)/(3)`

a = 2

a2 = 4

The relationship of hyperbola a, b and c are b2 = c2 – a2

And

b2 = (3)2 – (2)2

b2 = 9 – 4

b2 = 5

When these values are substituted in the equation `(x^2)/(a^2)+(y^2)/(b^2)` = 1 the equation `(x^2)/(4)+(y^2)/(5)` = 1

results and is the equation of the hyperbola.

The directories are given as x = `(4)/(3)` and, since

d = `(a)/(e)` or

a=de

Substituting the values given for d and e results in

a = `(4)/(3)((3)/(2))`

therefore,

a=2

and

a2 = 4

The value of c can be determined by the given equation,

c = ae

and a has been found to equal 2 and e is given as(3)/(2); therefore,

c =` 2((3)/(2))`

With the values of a and c computed, the value of b is found as before and the equation can be written.