Monday, December 31

Solve Algebra Squared Fraction


Fractions:

In algebra, A certain part of the whole is called as fractions. The fractions can be denoted as a/b , Where a, b are integers. We can multiply two or more fractions. There are three types of fractions in math,

1) Proper fractions

2) Improper fractions

3) Mixed fractions

In this article we are going to see how to solve algebra squared fraction and some example solved problems on algebra squared fraction.

Solve Algebra Squared Fraction :

Steps to solve algebra squared fraction:

Step 1: We know that ( a/b)^n = a^n / b^n . So we can take square for the numerator and denominator.

Step 2: If it is possible we can simplify it.

Let us see the example problems.
Example Problems on Solve Algebra Squared Fraction :

Problem 1:

Solve (4/5)^2

Solution:

Given , (4/5)^2

We need to squared the given fraction .

(4/5)^2

We know that ( a/b)^n  = a^n / b^n .

We can apply the above rule,

(4/5)^2 = 4^2 / 5^2

= 16 / 25

Answer: (4/5)^2 = 16 / 25

Problem 2:

Solve (3x / 4xy) ^ 2

Solution:

Given, (3x / 4xy) ^ 2

We need to find the squared the given fraction,

We know that (a/b)^n = a^n / b^n .

(3x / 4xy) ^ 2 = ((3x) ^ 2) / ( (4xy)^2)

= (9x^2) / ( 16x^2y^2)

now we can simplify it,

Divide by x^2 on both numerator and denominator,

(9x^2) / ( 16x^2y^2) = (9x^2)/x^2 / ( 16x^2y^2) / x^2

= 9 / 16y^2

Answer: (3x / 4xy) ^ 2   = 9 / 16y^2

Problem 3:

Solve (( x^2 - 4 ) / ( x + 2) )^2

Solution :

Given , (( x^2 - 4 ) / ( x + 2) )^2

We need to find the squared the given fraction,

We know that ( a/b)^n = a^n / b^n .

(( x^2 - 4 ) / ( x + 2) )^2 = ( x^2 - 4 )^ 2 / ( x + 2)^2

Before that we can simplify the given fraction,

(( x^2 - 4 ) / ( x + 2) )^2 = (((x+2)(x-2))/ ( x + 2))^ 2

= ((x - 2)^2)

= x^2 - 2x + 4

Answer: (( x^2 - 4 ) / ( x + 2) )^2 = x^2 - 2x + 4

Thursday, December 27

Results and Discussion Section


Here we are going to see about the result and discussion, whether the solution to the given equation is correct or not. By simplifying the equation we find the value for the particular variable. when we substitute the value in the equation the equation balances on both sides.

For example: x – 4 = 0.

Here    x – 4 = 0

Add 4 on both sides. We get

x = 4

The solution is x = 4

According to the topic we have to check whether x = 4 is a solution to the given equation or not. By substituting the value the in equation we can see that both sides have same value.
Example Problem for Result and Discussion Section:

Given equation is discussion section  x + 7x + 5x + 3x – 5 + x = 29

Solution:

Given question is x + 7x + 5x + 3x – 5 + x = 29

Step 1:

Arrange the given question as per the x term and the numbers

x+ x + 3x + 5x + 7x – 5 = 29

Step 2:

Add the x term first

2x + 3x + 5x + 7x – 5 = 29

Step 3:

Add all the  ' x ' term in the equation

17x – 5 = 29

Step 4:

Add both sides by 5.Then we get

17x = 29 + 5

Step 5:

17 x = 34

Divide by 17 on both sides.

`(17x)/17` = `34/17`

x = 2

Discussion section about the result:

The result is 2. Here we are going to discuss about the result for given equation, whether the solution x =2 is correct or not.

Given question is x + 7x + 5x + 3x – 5 + x = 29

The result is x =2

Here we need to substitute the x value in the equation if that both side value is equal means the solution is correct. Otherwise the solution is not correct.

x + 7x + 5x + 3x – 5 + x = 29

Substitute the value x = 2 in the given equation

2 + 7( 2) + 5( 2) + 3(2) – 5 + 2 = 29

2 + 14 + 10 + 6 – 5 + 2 = 29

34 – 5 = 29

29 = 29

Both the sides are equal, so given solution is correct for the equation.
One more Example Problem for Result and Discussion Section:

Solve discussion section  16x - 12 = 20

Solution:

Add 12 on both sides. we get

16x - 12 + 12 = 20 + 12

16x  = 32

Divide by 16 on both sides.

`(16x )/ 16` = `32 / 16`

x = 2

whe we substitute x =2 in the equation 16x - 12 = 20. we get

16(2) - 12 = 20

32 - 12 = 20

20 = 20.

Thus the equation satisfies.

Wednesday, December 26

Raw Score Formula


Raw score is defined as the unique, unchanged transformation or calculation. When we bring together the information, in which the numbers we get hold of are known as raw score or raw data. Raw score can able to find the precise location of each data in a given distribution.

percentage of sampling error = (value of statistic-value of parameter)*100.

In this article we are going to see about raw score formula.
Raw Score Formula:

The raw score formula for variance can be given as follows,

Raw score formula

Where `sigma` and X indicates mean.
Raw Score Formula Related to Z Score Formula:

Z-score formula can be given as follows

Z = (x) - (X) / (S.D)

where x indicates a data value

X indicates mean (average of the given values)
S.D indicates standard deviation.

Using the formula of Z score we can find the raw score formula

But to find the raw score formula we have to solve this formula for x

hence the raw score formula for x can be given as follows


Z* (S.D) + X= x

Where x indicates a data value or raw score

X indicates mean (average of the given values)
S.D indicates standard deviation.

Example problem for raw score formula:

Let us take an example where the mean value X has to be 280 and the standard deviation is let to be 10.The data values are given as follows x1 = 300 and x2 = 285.Find the raw score value using raw score formula.

Solution:

Z = [(300) - (280)] / (10)


Z1= 2

Z= [(285) - (280)]/ (10)



Z2 = .5
Finding Raw Score using raw score formula.

Z = 2.2 hence


2.2 = (x-280)/10

(2.2) * 10 + 280 = x

x = 302

Z = 1.1

1.1 = (x- 280) /10.

(1.1) * 10 + 280 = x

x = 291

Thursday, December 20

Solve a Parabola with Fraction


Parabola is a half circle and a curve formed by an intersection of right circular cone. A set of all points are same distance from a fixed line is called as directrix and also fixed point called as focus not in the directrix. The midpoint between the directrix and focus of the parabola is called as vertex and the line passing through the vertex and focus is called as axis. Let us see about how to get the value of vertex, latus rectum, focus and axis of symmetry in parabola equation. Let us see about solve a parabola with fraction in this article

The general form the parabolic curve is y = ax^(2) + bx + c or  y^(2) = 4ax . Substitute the above formula to find the vertices, latus rectum, focus and axis of symmetry.

Example 1 for Solve a Parabola with Fraction – Vertex:

Find the vertex of a parabola where y = x^ (2) + 7/2 x + 3

Solution:

Given parabola equation is y = x^ (2) + 7/2 x + 3

To find the vertices of a given parabola, we have to plug y = 0 in the above equation, we get,

0 = x^ (2) + 7/2 x + 3

Now we have to factor the above equation, we get,

So x^ (2) + 3/2x + 2x + 3 = 0

x(x + 3/2) + 2 (x + 3/2) = 0

(x + 3/2) (x + 2) = 0

From this x + 3/2 = 0 and x + 2 = 0

Then x = - 3/2 and x = - 2

So, the vertices of given parabola equation is (-3/2, 0) and (-2, 0) .

Example 2 for Solve a Parabola with Fraction – Focus:

What is the focus of the following parabola equation where y^ (2) = 10x

Solution:

Given parabola equation is y^ (2) = 10x is of the form y^ (2) = 4ax

To find the focus of a parabola use the formula to find the value of focus value.

We know that the formula for focus, p = 1 / (4a)

Now compare the given equation y^(2) = 10x with the general equation y^ (2) = 4ax . So, that 4a = 10

From this p = 1 / (4a) = 1 / 10

So, the focus of a parabola equation is (0, 1/10) .

Other Example Problems to Solve a Parabola Fraction

Example 3 for Solve a Parabola with Fraction – Axis of Symmetry:

What is the axis of symmetry of the parabola where y = 4x^ (2) + 8x + 14 ?
Solution:

Given parabolic curve equation is y = 4x^ (2) + 8x + 14

From the above equation, a = 4 and b = 8

So the axis of the symmetry of the given parabola is -b/ (2a) = -8/ (2 xx 4) = -8/8 = - 1

Therefore, the axis of symmetry for a given parabolic curve equation is -1 .

Example 4 for Solve a Parabola with Fraction – Latus Rectum:

Find the latus rectum of the given parabola equation y^ (2) = 6x

Solution:

The given parabola equation is y^ (2) = 6x

To find the latus rectum, we have to find the value of p.

The parabola equation is of the form y^ (2) = 4a

Here 4a = 6

So, p = 1/ (4a) = 1/6

The formula for latus rectum is 4p .

From this, the latus rectum of the parabola is = 4p = 4 (1/6) = 4/6 = 2/3

Therefore, the latus rectum for the parabola equation is 2/3 .

Friday, December 14

Short Definition of Geometry


Geometry (Ancient Greek: γεωμετρία; geo- "earth", -metria "measurement") "Earth-measuring" is a part of mathematics concerned with questions of size, shape, relative position of figures, and the properties of space. Initially a body of practical knowledge concerning lengths, areas, and volumes, in the 3rd century BC geometry was put into an axiomatic form by Euclid, whose treatment—Euclidean geometry—set a standard for many centuries to follow.( Source- Wikipedia)

Short Basic Geometry Definitions:
Short Definition of Geometry for Square:

In mathematical geometry, a square consists of 4 equal angles every side which includes 90° and 4 equal sides. Square area is calculated by-product of two opposite sides.
Short Definition of Geometry for Rectangle:

In mathematical geometry, a rectangle is equal to the square, it consists of 4 equal angles with 90° and 4 equal sides. Rectangle area is calculated by using the product of length and width.
Short Definition of Geometry for Triangle:

In mathematical geometry, triangles consist of 3 straight lines that are equal length and it is called as equilateral triangle. Triangle area is calculated by half of the product of bottom and height.
Example for Short Definition of Geometry:
To solve the Rectangle problems:

Area of rectangle= Length x width.

Perimeter of rectangle= 2 (Length + Width)

Example 1:

Find the area of rectangle that sides measure 6 centimeters and 4 centimeters respectively.

Solution:

Given that

Length – 6 centimeter.

Breadth – 4 centimeter.

The formula used to find out the area of rectangle is = length * breadth.

Area = length * breadth.

Substitute the given value in the formula

Area   = 6 * 4

The result of 6 and 4 = 24

Hence the area of the rectangle given is 24 square centimeter.

Example 2:

Find out the volume of the cube that sides measure 6 centimeter.

Solution:

Given that

Measure of cube length = 6.

The formula used to find out the volume of the cube is = a3.

While we insert the value in the formula we get the result is

Volume = 63.

= 216

Hence the volume of the given cube is 216 cubic centimeters.

To solve Square problems:

Perimeter of square= 2 * side

Area of square= side * side

Problem 3:

To find out the area and perimeter of square while side length is 4cm?

Solution:

Area of square = (side*side)

=4 * 4

=>16 cm2

Perimeter of square = 4 * side

= 4 * 4

= 16 cm

Thursday, December 13

The Multiplication Rules


In algebra fundamental arithmetic procedure (addition, subtraction, multiplication, and division) in general used in day-to-day life. The multiplication defined as product of numbers. Multiplication is usually expressed in a*b or ab form. The multiplying numbers are having two terms i) Multiplicand and ii) Multiplier. In this article, we are going to discuss about the multiplication rules

The Multiplication Rules – Rules and Example Problems:

The multiplication rules - Rules:

Rule 1: The multiplication of two integers by the connected signs resolve be positive sign

a) Positive `xx` positive = positive

b) Negative `xx` negative = positive

Rule 2:  The multiplication of two integers by the different signs will be negative

a) Positive `xx` negative = negative

b) Negative `xx` positive = negative.

The multiplication rules – Example problems:

Example 1:

Evaluate 6802 `xx` 322

Solution:

6802 `xx` 322

------------------

13604

13604

20406

--------------

2190244

--------------

Therefore, the final answer is 2190244.

Example 2:

Find the solution for given problem

(- 9667) `xx` (- 26)

Solution:

(- 9667) `xx` (- 26)

---------------------------

58002

19334

-------------------

251342

-----------------

Therefore, the final answer is 251342.

Example 3:

Find the solution for given problem

8655 `xx` (- 124)

Solution:

8654 `xx` (- 124)

-----------------

34616

17308

8654

----------------

- 1073096

-----------------

So, the final answer is (- 1073096)
The Multiplication Rules – Practice Problems:

Practice problem 1:

Find the solution for given problem

(- 7897) `xx` (- 53)

Answer: So, the final answer is 418541

Practice problem 2:

Find the solution for given problem

(- 7790) `xx` (- 48)

Answer: So, the final answer is 373920

Practice problem 3:

There are 280 students in every class at the performing arts school. If there are 62 complete classrooms, how many students attend this school?

Answer: 17360 students attend this school

Practice problem 4:

Jesse wants his giant chocolate chip cookies to have 28 chips each. Today he is make 355 cookies. How many chips does he need?

Answer: 9940 chips does he need.

Friday, November 23

Variance Covariance Formula


Variance: Variance is used in the statistical analysis to find the the extent to which a single variable is varying from its mean value given a set of values.

Variance of a random variable X is often denoted as VAR ( X). It is also denoted by the symbol `sigma` 2X

Covariance: Co-variance is used in statistical analysis to find the extent to which 2 variable are varying together given a set of values for both these variables.

Covariance of two random variable X and Y is denoted as COV ( X,Y). Unlike variance which has the mathematical symbol sigma ( `sigma` ), Covariance doesn't have any kind of symbol to depict it.
Formulae for Variance and Covariance.

Lets consider 2 sets of data namely X and Y such that the values in X are x1 , x2 , x3 , x4 , ......xn and similarly the values in Y are y1 , y2 , y3 , y4 , .......yn

Now lets denote the mean of the X set of variables to be    Xm

Mean Xm = X1 + X2 + X3 +........ + Xn / n

Similarly lets denote the mean of the Y set of variables to be Ym

Mean Ym = Y1 + Y2 + Y3 +.......+ Yn / n

Formulae for Variance of X = ( x1 - xm)2 + (x2 - xm)2 + ( x3 - xm)2 + .......+ (xn - xm)2 / n

VAR ( X ) = (1/n) `sum` (xi - xm)2

Formulae for Variance of Y = ( y1 - ym)2 + (y2 - ym)2 + ( y3 - ym)2 + .......+ (yn - ym)2 / n

VAR ( Y ) = (1/n) `sum` (yi - ym)2

Formulae for Covariance of (X,Y) = ( x1 - xm) ( y1 - ym) + (x2 - xm)(y2 - ym) + ( x3 - xm)( y3 - ym) + .......+ (xn - xm)(yn - ym) / n

COV ( X , Y ) = (1/n) `sum` (xi - xm)(yi - ym)
Example Problem 1 on Variance and co Variance

Lets assume the below set of marks received by Bill and Bob in Maths, Physics, Chemistry, English and Biology for the purpose of solving Variance and Co Variance

table

Based on the Formulae given in the previous paragraph, Lets calculate the Mean Marks for Bill and Bob

Mean Marks for Bill = 16+12+14+18+20 /5 = 16.00

Mean Marks for Bob = 14+18+15+18+20 /5 = 17.00

Now lets apply the formulae of variance and find out the Variance of Bill and Bob

Variance of Bill = (16-16)2 + (12-16)2 + (14-16)2 + (18-16)2 + (20 - 16)2 / 5 = 40 / 5 = 8

Variance of Bob = (14-17)2 + ( 18-17)2 + (15-17)2 + (18-17)2 + (20-17)2 / 5 = 24 / 5 = 4.8

Co variance of ( Bill , Bob ) = (16-16)*(14-17) + (12-16)*(18-17) + (14-16)*(15-17) + (18-16)(18-17) + (20-16)*(20-17) / 5

Co Variance of ( Bill , Bob ) = 21/5 = 4.2

I am planning to write more post on What are Line Segments and Perpendicular and Parallel Lines. Keep checking my blog.

Example Problem 2 on Variance and co Variance

Lets assume a class of 4 student who have been asked to rate their liking toward 2 musical instruments Guitar and Piano on a scale of 10




Based on the above data, lets calculate the Mean of the people liking Guitar and Mean of people liking Piano.

Mean(Guitar) = ( 5+3+7+9)/4 = 24/4 = 6

Mean(Piano) = (5+9+9+5)/4 = 28/4 = 7

Var(Guitar) = ((5-6)2 + (3-6)2 + (7-6)2 + (9-7)2 )/ 4 = (1+9+1+4)/4 = 15/4 = 3.75

Var(Piano) = ((5-7)2 + (9-7)2 + (9-7)2 + (5-7)2 ) / 4 = (4+4+4+4)/4 = 16/4 = 4

Cov(Guitar,Piano) = [(5-6)(5-7) + (3-6)(9-7) + (7-6)(9-7) + (9-7)(5-7) ] / 4 = (2+(-6)+2+(-4))/4 = -6/4 = -1.5

Friday, November 9

Binomial Probability Function


Binomial probability Problems represented by B(n,p,x). It gives the probability of exactly x successes in ‘n’ Bernoullian trials, p being the probability of success in a trial. The constants n and p are called the parameters of the distribution. A Binomial distribution can be used under the following condition.

(i) any trial, result in a success or a failure

(ii) There are a finite number of trials which are independent.

(iii) The probability of success is the same in each trial.

In a Binomial distribution function mean is always greater than the variance. The binomial probability function example problems and practice problems are given below.
Example Problems - Binomial Probability Function:

Ex 1:  By using the binomial distribution. If the sum of mean and variance is 4.8 for  5 trials find the distribution

Solution:  np + npq = 4.8 , np(1 + q) = 4.8

5 p [1 + (1 − p) = 4.8

p2 − 2p + 0.96 = 0 , p = 1.2 , 0.8

p = 0.8 ; q = 0.2 [p cannot be greater than 1]

The Binomial distribution is P[X = x] = 5Cx (0.8)x (0.2)5−x,  x = 0 to 5

Practice Problem - Binomial Probability Function:

Pro 1: Find the value for p by using the Binomial distribution if n = 5and P(X = 3) = 2P(X = 2).

Ans: p = `(2)/(3)`

Pro 2: Find the probability values by using the binomial distribution.  A pair of dice is thrown 10 times. If getting a doublet is considered a success (i) 4 success (ii) No success.

Ans: (a) (35/216)(5/6)6

(b) (5/6)10

Monday, November 5

Proportion Equation


The data is obtaining by the comparison of two ratios is called proportion data. Proportion data is represented as a:b = c:d. This proportion data can be written in the form of fraction as `a/b` = `c/d` . Where the pairs of data (a,b) and (c,d) are in proportion. When the proportions are equal, the cross product of the proportion will be also equal. That is, `a/b` = `c/d` can be written as ad=bc.

Examples for Proportion Equation:

Example 1 for proportion equation:

Martin read 63 pages of the book in 33 minutes. How many pages will he be able to read in 43 minutes?

Solution:

Martin takes 33 minutes to read 63 pages.

Martin will take 43 minutes to read x pages.

This can be written as,

`63/33` = `x/43`

Now we have to do the cross multiplication.

63 `xx ` 43 = x `xx` 33

2709 = 33x

This can be written as,

33x = 2709

Now we have to divide both sides by 33.

`(33x)/33` = `2709/33`

x = 82.09 now we have to round it to the unit place.

x = 82

Therefore, Martin will read 90 pages in 45 minutes.

Example 2 for proportion equation:

Paul bought 12 apples for dollar 48.  How many apples will he be able to buy in $ 93?

Solution:

Paul spends $48 for 12 apples.

Paul will spend $93 for x apples.

This can be written as,

`12/48` = `x/93`

Now we have to do the cross multiplication.

12 `xx` 93 = x `xx` 48

1116 = 48x

This can be written as,

48x = 1116

Now we have to divide both sides by 48.

`(48x)/48` = `1116/48`

x = 23

Therefore, Paul can buy 23 apples for $ 93.

Practice Problems for Proportion Equation:

Problem 1 for proportion equation:

Martin read 40 pages of the book in 28 minutes. How many pages will he be able to read in 52 minutes?

Solution: Martin will read 74 pages in 52 minutes.

Problem 2 for proportion equation:

Paul bought 8 apples for dollar 22. How many apples will he be able to buy in $ 66?

Solution: Paul can buy 24 apples for $ 66

Friday, October 19

Polynomials Calculator


An Algebraic expression is of the form axn is called a monomial. The variable a is called the coefficient of xn and n, the degree of monomial. For example, 7x3 is monomial in x of degree 3 and 7 is the coefficient of x3. The combination of two monomials is called a binomial and the combination of three monomials is called a trinomial. For example, 2x3 + 3x is a binomial and 2x5 – 3x2 + 3 is trinomial. The sum of n number of monomials, where n is finite and x is called a polynomial in x.
Illustration to Polynomials:

Polynomial Calculator Example 1:

The polynomial calculator of the equation x2 + ax + b gives the remainder 18, when divided by x – 2 and leaves the polynomial calculator of remainder –2 when that is been divided by (x + 3).

Find the values of a and b.

Solution to the polynomial calculator:

P(x) = x2 + ax + b.

In tyh is polynomial calculator,

When x – 2 divides P(x) then the remainder is P (2).

∴P (2) = 4 + 2a + b.

But remainder = 18 ⇒ 4 + 2a + b = 18;

2a + b = 14 (1)

When (x + 3) divides P(x)

, the remainder is P (–3).     ∴ P (–3) = (–3)2 + a (–3) + b

= 9 – 3a + b.

But remainder = –2;      ∴ 9 – 3a + b = –2;

⇒ –3a + b = –11 (2)

(1) ⇒   2a + b = 14

(2) ⇒ –3a + b = –11 (subtracting)

5a        = 25

(Or) a     = 5

Substituting a = 5 in equation (1) we get

10 + b = 14; b = 4, ∴ a = 5, b = 4
Subtraction of Polynomials Calculator:

Example for Polynomials calculator:

Subtract        2x3 – 3x2 – 1 from x3 + 5x2 – 4x – 6.

Solution:

Using associative and distributive properties, we have

( x3 + 5x2 – 4x – 6) – (2x3 – 3x2 – 1) = x3 + 5x2 – 4x – 6 – 2x3 + 3x2 + 1

= x3 – 2x3 + 5x2 + 3x2 – 4x – 6 + 1

= (x3 – 2x3) + (5x2 + 3x2) + (–4x) + (–6+1)

= –x3 + 8x2 – 4x – 5.

The subtraction can also be performed in the following way:

Line (1): x3 + 5x2 – 4x – 6.

Line (2): 2x3 – 3x2 – 1.

Changing the signs of the polynomial in Line (2), we get

Line (3): –2x3 + 3x2 + 1.

Adding the polynomials in Line (1) and Line (3), we get

–x3 + 8x2 – 4x – 5.

Thursday, October 18

Integral Part Definition


In geometrical and other applications of integral calculus it becomes necessary to find the difference in the valves of an integral of a function f(x) for two assigned values of the independent variable x, a and b. this difference is called the definite integral of f(x) over the range (a,b) and is denoted by $\int_{a}^{b}f(x)dx=F(b)-F(a)$, where F(x) is an integral of f(x), F(b) is the variable of F(x) at x=b, F(a) is the value of F(x) at x=a.

It is often written thus  : $\int_{a}^{b}f(x)dx=[F(x)]_{a}^{b}=F(b)-F(a)$.

Note 1. The integral $\int_{a}^{b}f(x)dx $ is read as the integral of f(x) from a to b.The number a is called the lower limit and the number b, the upper limit of integration.

Note 2. It should be seen that the value of a definite integral is perfectly unique and is independent of the particular form of integral which may employ to calculate it. Considering F(x)+c instead of F(x), we get

$\int_{a}^{b}f(x)dx=[F(x)+c]_{a}^{b}=[F(b)+c]-[F(a)+c]=F(b)-F(a)$

so that the arbitrary constant disappears in the process and we get the same values as on considering F(x). This is why the name is given as definite integral.

Note 3. It is assumed that a and b are finite.
Definition of Integration by Parts

When the given function cannot be integrated directly by using standard formulae, we try other methods. The process of integration is largely of tentative nature and no systematic procedure can be given as in differentiation. However, the following are two important methods opf integration.

1. Integration by Substitution           2. Integration by Parts

Here, we will discuss about Integration by parts

An Integration is the inverse process of differentiation.By differentiaon we find the derivative of the given function, whereas by integration we find the function whose derivative is known.

If the derivative of F(x) is f(x) then we say that the antiderivative or integral of f(x) is F(x),such that

int f( x ) dx= F( x )

Thusd/dx F(x) = f(x) => intf(x) dx = F(x)

Integration by Parts:

Theorem: if u and v are two differentiable function of x then

int ( u v ) dx = [ u * int v dx ] - int { du/dx * int v dx } dx .

We can express this result as given below:

Integral of product of two function

= ( 1st function ) * ( integral of 2nd function ) - int { ( derivative of 1st function ) * ( integral of 2nd function ) } dx .

We should choose u and v in such a way that the second function v on the right hand side is easy to integrate. Sometimes, this rule has to be used repeatedly. This rule is also useful in integrating logarithmic and inverse t - functions of the type log x, log(ax2 + bs + c), sin-1 x, tan-1 x etc.The following guidelines will help you to see which of the two functions in the product should be taken as the first function.

Notes:

If the integrand is of the form f(x) * xn, we consider xn as the first function and f(x)as the second function.
If the integrand contains a logarithmic or an inverse trigonometric function, we take it as the first function. If the second function is not given in any case we take that as one.

Example Problems on Integration by Parts:

Pro 1: Evaluate, int xsin 2x dx

Sol : Given,

int   xsin 2x dx

=     int sin 2x dx - int { d/dx ( x ) * int sin 2x dx } dx

=     x * ( -(cos 2x)/2 ) - int 1 * ( -(cos 2x)/2 ) dx

=  (-x cos 2x)/2 + 1/2 int cos 2x dx

"-(x cos  + 1/2 * "(sin

= -(x cos 2x)/2 + 1/4 sin 2x + C , which is the Answer.

Pro 2: int logx/x^2 dx

Sol :   Integrating by parts, taking logx as the fi rst function and 1/x^2  as the second function, we get

int "log   dx

=  int (log x) * 1/x^2 dx

=   ( log x ) * int 1/x^2 dx - int { d/dx ( log x ) * int 1/x^2 dx } dx

=    ( log x )(-1/x )   - int 1/x  * ( -1/x ) dx

= - log x/x + int 1/x^2 dx

= - log x/x - 1/x + C, which  is the required Answer.

Practice problems on Integration by Parts:

Pro 1: Evaluate, int (  x cos x) dx    ( Answer:   x sinx +cosx+c )

Pro 2: Evaluate, int e2x sin x dx            ( Answer: 1/5 e2x ( 2 sin x - cos x ) + C )

Thursday, October 4

Natural Logarithmic Calculator


In mathematics, the natural logarithmic function is defined as the function contains three ports, namely the number, the base and logarithm itself,  now the natural logarithmic calculator is the function of  the logarithm function calculator which means we are using to the base of ‘e’ where e is constant, base of  ‘e’ value is given by

Understanding Definition of Logarithm is always challenging for me but thanks to all math help websites to help me out.

e = 2.718281828. and the numbers, and the logarithmic itself.

The logarithmic function calculator is represented as y = log_b^(x).

We take the value of base of ‘e’ is approximately e = 2.71.
Natural Logarithmic Calculator:

Here, we introduce the natural logarithmic calculator. The function of natural logarithmic calculator is represented as  y = ln(x)

Explanation about natural logarithmic calculator:

Input:

calcutor             given value

The given value is conversions by logarithmic calculator

Conversions:

 x =  conversion value.

Solution:

 y = calculated value will be displayed

Here we can the change the equation by x and y.

y = ln(x) for y i n natural logarithm

y = ln(x) for x  in natural logarithm

Note: we can calculate or perform the positive numbers only not negative numbers by natural logarithmic calculator.
Some Problems about Natural Logarithmic Calculator:

Problem1:

To solve the natural logarithmic function of ln(2)

Solution:

Input:

x = 2  given value

Conversion:

x = 2  the given value conversion by logarithmic calculator.

Solution:

y = 0.69 approximately

Problem 2:

To solve natural logarithmic function of  span style="font-size: small; " mce_style="font-size: small; ln(4)

Solution:

Input:

 x = 4 given value

Conversion:

 x = 4  the given value conversion by logarithmic calculator.

Solution:

 y = 1.38 approximately

Problem 3:

To solve natural logarithmic function of ln(6)

Solution

Input:

 x = 6 given value

Conversion:

 x = 6  the given value conversion by logarithmic calculator.

Solution:

 y = 1.79 approximately

Problem 4:

To solve the natural logarithmic function of in ln(10)

Solution:

Input:

x = 10 given value

Conversion:

 x = 10 the given value conversion by logarithmic calculator.

Solution:

 y = 2.30 approximately

Problem 5:

To solve the natural logarithmic function of ln(15)

Solution:

Input:

x = 15 given value

Conversion:

 x = 15 the given value conversion by logarithmic calculator.

Solution:

 y = 2.70  approximately


Practices problems for natural logarithmic function:

Problem1;

To solve the natural logarithmic of ln (20)

Answer is y = 2.99

Problem2

To solve natural logarithmic function of ln (50)

Answer is y = 3.91

Problem3

To solve natural logarithmic function of ln(100)

Answer is y = 4.60

Problem 4

To solve the natural logarithmic function of in ln(99)

Answer is y = 4.59

Hence, here we obtain the natural logarithmic calculator.

Friday, September 7

Scientific Notation Rules


 Often we come across abnormally large numbers or abnormally small numbers like 400,000,000,000 or very, very small number like0.00000000002. Especially in the field of science such abnormal size of numbers is quite common.

In calculations involving such numbers, the work is cumbersome and possible errors may occur.

A special type of notation is used to simplify the dealings with these type of numbers and it is named as scientific notation as such numbers are very predominant in the field of science.

However a scientific notation must be used following a set of rules, universally accepted.

Let us take a closer look.
Scientific Notation Rules – Description

Since very large or very small numbers can be rounded to the appropriate place value, the power of 10 is used as base in scientific notation. The established general form is,

                                       a x 10n, where,  1= I a I < 10

If  ‘n’ is positive, it represents the number of zeroes from ‘a’ to the right of unit place and hence  it denotes large numbers.

For very, very small decimal numbers ‘n’ is negative and represents the number of zeroes before ‘a’ up to the decimal point.
Scientific Notation Rules – Examples

The velocity of light is about 310,000,000 kilometers per second.

We find there are seven 0s after the digits 1. That is the number is same as 31 times 10 to the power 7. But as per scientific notation rules, 31x 107 is not allowed and hence the given is modified as 3.1 x 108

Thus in scientific notation, 310,000,000 = 3.1 x 108

Similarly it is equally incorrect to mention (as a scientific notation) 310,000,000 as 0..31 x 109

Let us consider another example with a very  small number.

The wave lengths of a certain color is experimentally found to be 0.0000015 millimeter.

There are 6 decimal places to cross the digit 1from right. That the given number is (1/1000000)th of 1.5.

Hence in scientific notation it is written as 1.5 x 10-6

Friday, July 27

Importance of Derivatives of Trigonometric Functions


Trigonometric Functions 
In Mathematics, trigonometric functions or circular functions are ratios of the sides of a right angled triangle, having the given angle.  Trigonometry deals with functions of an angle present in a triangle. It relates the angles of the given triangle with the length of its three sides. It also helps to find the length and angle of a triangle. Trigonometric functions are widely used in real life in various fields such as carpentry, surveys, engineering, physics, navigation, astronomy etc.

Trigonometric functions are sin p, cos p, tan p, cot p, sec p, csc p.  Among them sine, cosine and tangent are the most familiar and basic trigonometric functions. Before going into derivatives of trigonometric functions, let us first see what these trigonometric functions refer to.

Let us consider a circle of radius 1. If a ray originates at the origin which makes an angle with x axis, what will the sine, cosine and tangent of this angle provide?
The sine of this angle gives the length to which the triangle has risen on the other side i.e. the y component

The cosine of this angle gives the x component length

The tangent of this angle gives the slope of the triangle, which is obtained by dividing Y component by the X component.

Thus, trigonometric functions can also be defined as the lengths of all line segments from a given unit circle.

Derivative of Trigonometric Function
With this basic understanding on trigonometric functions, let us explore on the derivatives of trigonometric functions namely sin p, cos p, tan p, cot p, sec p, csc p.

The derivatives for each of these trigonometric functions can be represented as follows:

Derivative of Sinx
The symbolic expression of Sinx derivative is given by

d/dx (sin x) = cos x   or sin’ x = cos x


It is easy to find the derivatives of sin x. This is because, the derivative of sin x can be obtained by just calculating the cos x.

Derivative of Cosx
The Cosx derivative is given by

d/dx (cos x) = - sin x or cos’ x = - sin x


Derivative of Tan p
The derivative of tan p is given by “tan-1 p = sec^2   p”

Derivative of Cot p
The derivative of cot p is given by “cot-1 p = - coesc^2 p”

Derivative of Sec p
The derivative of sec p is given by “sec’ p = sec p tan p”

Derivative of Csc p
The derivative of csc p is given by “csc’ p = - csc p cot p”

Tuesday, June 26

Different types of Matrices


Matrix is a rectangular array of numbers or functions which are called the elements or the entries of the matrix. The elements or entries are arranged in rows and columns in either () or [] brackets. The elements of a matrix are denoted by aij where ‘i’ is the row number and ‘j’ is the column number. We can only add or subtract two matrices if they both have same number of rows and columns.

Order of matrices: An order of the matrices is written as m X n where m is the number of rows and n is the number of columns in it. That means a matrix of order 2 X 2 can only be added to a matrix of order 2 X 2 and a matrix of order 3 X 3 can only be subtracted from a matrix of order 3 x 3.

Types of matrices

1. Column matrix: A matrix is said to be a column matrix if has only one column. For example: - a matrix of order 4 X 1 is said to be a column matrix as it has four rows but only one column. In general, a matrix of order m X 1 is considered as column matrix where m can be any real number.
2. Row matrix: A matrix is said to be row matrix if it has only one row. For example: - a matrix of order 1 X 4 is said to be a row matrix as it has four columns but only one row. In general, a matrix of order 1 X n is considered as row matrix where n can be any real number. A example of row matrix is [2 , 6, 11, 7].
3. Zero matrix: A matrix is said to be zero matrix if all its elements or entries are zero. It is also called a null matrix. For example: - [0], [0 0] are zero matrices. We denote zero matrix by O.
4. Square matrix: A matrix is said to be square matrix if its number of rows are equal to number of columns. Thus m X n is a square matrix if m = n and is known as a square matrix of order ‘n’.
5. Unit matrix: A matrix is said to be unit matrix if all its elements or entries are one. For example: -
[1], [1 1], [1 1 1] are all unit matrices.

Monday, June 18

Polynomial Factoring


Factoring polynomial is similar to factoring numbers, but you will be dealing with expressions instead of numbers. In order to factor a polynomial, you have to determine the polynomial that divides the original polynomial evenly.

Polynomial Factoring
Polynomial Factoring
How to factor a polynomial?
Different approaches are used to factor the polynomial; most common approaches are simple factoring, and factoring in pairs. In this article, you will find examples for these two approaches of polynomial factoring.

Simple Factoring
In simple factoring, you identify a common factor for the expression and place it before the parenthesis. What is a common factor? To know that, look at the example below:
5x + 60 = 5 (x) + 5(6) = 5 (x+12)

Now the common factor of the above expression is 5. To cross verify, multiply the factor with each term in the expression and ensure if you are getting the original polynomial back. For example,

5(x+6) = 5 multiplied with x + 5 multiplied with 12 = 5x + 60 (the original polynomial)

Factoring in Pairs
“Factoring in Pairs” is yet another factoring polynomials solver. This approach is used when you cannot find any common factor for the terms in the expression. In this approach, the expression is split into pairs of terms and then each pair will be factored separately. Here is an example:

Find the factors of ab-5b-2a+10.

This polynomial algebraic expression includes 2 variables namely a and b. There is no common factor for these 4 terms. In this case, take the first two terms and find a common factor. Similarly, take the last 2 terms and find a common factor, as shown below:

ab-5b-2a+10 = b (a-5) -2 (a-5) = (b-2) (a-5)

In the above step, the first two terms ab-5b are taken into consideration. Take the common factor out, it is b. Now it becomes b (a-5). Similarly take the next two terms -2a+10. The common factor is 2 here. When you put it as -2, the expression becomes -2 (a-5). Now the expression can be further simplified as (b-2) (a-5).

If the consecutive terms do not have any common factors, then you can rearrange the terms based on the commonality and then do the factoring.