Monday, May 13, 2013

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 .

Saturday, May 11, 2013

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.

Thursday, May 9, 2013

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)`

Wednesday, May 8, 2013

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.

Tuesday, May 7, 2013

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)`

Monday, May 6, 2013

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

Sunday, May 5, 2013

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