Taylor Series Power

In mathematics, the Taylor series is a representation of a function as an infinite sum of terms calculated from the values of its derivatives at a single point. It is named after the English mathematician Brook Taylor. If the series is centered at zero, the series is also called a Maclaurin series, named after the Scottish mathematician Colin Maclaurin. It is common practice to use a finite number of terms of the series to approximate a function. The Taylor series may be regarded as the limit of the Taylor polynomials.

I like to share this taylor series expansions with you all through my article.

Definition of taylor series power:

Taylor series power:

Definition:   Taylor polynomial degree of function can be defined as the function of f. For an approximation of a function have a degree of polynomial. It can be differentiate for n times. The degree of Taylor polynomial for centered at a is

Pn(x) = f(a) + f′(a)(x - a) + (f″/ 2!)(x – a) ²+ …… + (f (n) (a)/ n!)(x - a)n

= ∑n k = 0 (f (k) (a) / k!) (x- a) k

Properties of taylor series power approximation:

In the interval (a + r, a - r), the series of converges for x and the result is equal to f(x) which function is analytic.
The power series representation can be used in the simplest form of Euler’s formula. It can be done in algebraic expressions
The result of expansions of Taylor series for cosine, sine and exponential functions are the one of fundamentals fields of harmonic analysis.
In singularity, the Taylor functions are sometimes cannot write as function.
The Taylor series is zero and also the function is not which is not analytic.


Example problem for Taylor series power:

Ex 1:   Solve: f(x) = (1 / (x + 1)), a = 1. Find the Taylor polynomial degree up to two.

Sol :    Given f(x) = (1 / x +1)          f (1) = 0.5

f’(x) = -1 / (x +1)2             f’ (1) = -0.25

f”(x) = 2 / (x +1)3             f’’(1) = 0.125

To find the Taylor polynomial:

General form of the Taylor polynomials is

Pn(x) = f (a) + f′ (a) (x - a) + (f″/ 2!)(x – a)2 + …… + (f (n) (a)/ n!)(x - a) n

Here we calculate up to third degree of Taylor polynomial

Pn(x)   = f (4) + f’ (4) (x – a) + f’’ (4) ((x – a)2/ 2!)

= 0.5 + (-0.25) (x - 1) + 0.125((x - 1)2 / 2!)

= 0.5 – 0.25x + 0.25 + (x2 – 2x + 1) (0.0625)

= 0.5 – 0.25x + 0.25 + 0.0625x2 – 0.125x + 0.0625

Pn(x) = 0.0625x2 – 0.375x +0.75.

Ex  2:  Solve: f(x) = (1 / (x - 3)) +1, a = 4. Find the Taylor polynomial degree up to three.

Sol :    Given f(x) = (1 / x - 3) +1          f (4) = 2

f’(x) = -1 / (x - 3)2             f’ (4) = -1

f”(x) = 2 / (x – 4)3             f’’(4) = 2

f’’’(x) = -6 / (x – 4)4          f’’’ (4) = -6

To find the Taylor polynomial:

General form of the Taylor polynomials is

Pn(x) = f (a) + f′ (a) (x - a) + (f″/ 2!)(x – a)2 + …… + (f (n) (a)/ n!)(x - a) n

Here we calculate up to third degree of Taylor polynomial

Pn(x)   = f (4) + f’ (4) (x – a) + f’’ (4) ((x – a)2/ 2!) + f’’’ (4) ((x – a)3/ 3!)

= 2 + - (1) (x - 4) + 2((x – 4)2 / 2!) + (-6) (x – 4)3/ 6

= 2 - x + 4 + (x2 – 8x + 16) - (x3 – 3(x2(4)) + 3(x (16)) + (43)

= 2 - x + 4 + x2 – 8x + 16 - x3 + 12x2 - 48x - 64

Pn(x) = - x3 + 13x2 - 57x – 46.

Taylor Series

                      Taylor series is a infinit number of terms.Taylor series is a  expansion of series and its function about a point.The Taylor series of a real or composite function ƒ(x) that is much differentiable in a neighborhood of a real or composite number a is the power series

             
    that can be written in the more compact sigma notation as 

              
     where n! indicates the factorial of n and ƒ ⁽ⁿ⁾(a) indicates the nth derivative of ƒ calculate at the point a.

I like to share this Taylor Series Expansion Example with you all through my article.

Explanation of taylor series:

Maclaurin Series:
If a=0, then it is said to be maclaurin series.


Derivation:
we define the power series as,


At x=0,


Differentiate the function,


 At x = 0,


Differentiating again will give,



At x=0, we will evaluate the equation as,



Generalizing the equation,we get


Substitute the  values of a_n in the power expansion,



Generalizing  f in a more general form, we have


Evaluating at x = a, we get


Sustitute above equation, we get taylor series.


        Taylor series can be used to evaluate the value of an whole function in each point, if the functional value and its derivatives are identified at single point. Uses of the Taylor series for whole functions are:
        1.The sum of partial series can be used as approximations of the entire function.
        2.The representation of series reduces many mathematical proofs.
       In Taylor series, algebraic functions are indicated using an algebraic equation, and transcendental functions are indicated using properties which holds them, namely differential equation. Example the exponential function is equal to its own derivative and its original value is 1.
       Taylor series are used to identify functions and operators in diverse areas of mathematics. Example: Analytical functions of matrices and operators can be defined as matrix exponential. In formal analysis, we directly work with the power series .

Example:

Example 1:
calculate the taylor series for

Using formula ,

Rewrite the equation,




substitutes


we get,

we get the new series as,


Example 2:

taylor series for f(x) = 1/ ( 1 + x )

the general case is 


hence the taylor series for f(x) is

Composite Functions

Students can learn about Composite Functions and How to Graph Composite Functions. Students can get help with Calculus problems involving Composite function from the online tutors.

In this article about Composite Functions, we will learn about Composite Functions and see about how to draw the graph for composite functions and how to get the composite functions. We shall discuss how to make the composite function and how to draw the resultant function on a graph sheet.

When two functions are f(X) and g(X), the composite function f(g) will be defined as f[g(x)].

Graphing composite functions

Below is given an example explained with graphical representation which will help you to understand the concept of graphing composite functions better:

Example:
Consider the following function f(x)=x+1 find `g(x)=x^2`
Solution:
To answer this f[g(x)]
Substitute the g(x) in f(x) function
`=f(x^2)`
Perform the f(x) function ie f(x)=x+1
here `x=x^2`
so `f(g(x))=x^2+1`

Since the function `y=f(x)=x^2+1`
The value of a y varies with respect to the x-value. Substitute different x-value to the function, the y value will be obtained. After getting y value make the values of x and y value in table format.
`f(x)=x^2+1`
`f(0)=0^2+1=1`
`f(1)=1^2+1=2`
`f(2)=2^2+1=5`
`f(3)=3^2+1=10`

x	0	1	2	3
y=f(x)	1	2	5	10

Assign the scale as per our co-ordinate points.
Scale:
In x-axis 1unit= 1 cm
In y-axis 1unit= 2 cm
Graph for the composite functions

Example Problems

Example: Consider the following function f(x)=2x+1 find `g(x)=x^2`
Solution: To answer this f[g(x)]
Substitute the g(x) in f(x) function
`=f(x^2)`
Perform the f(x) function ie f(x)=2x+1
here `x=x^2`
so `f(g(x))=2(x^2)+1`
Since the function `y=f(x)=2(x^2)+1`

The value of a y varies with respect to the x-value. Substitute different x-value to the function, the y value will be obtained. After getting y value make the values of x and y value in table format.
`f(x)=2(x^2)+1`
`f(0)=2(0^2)+1=1`
`f(1)=2(1^2)+1=3`
`f(2)=2(2^2)+1=9`
`f(3)=2(3^2)+1=17`

x	0	1	2	3
y=f(x)	1	3	9	17

Assign the scale as per our co-ordinate points.

My forthcoming post is on pu board karnataka and cbse syllabus 2012 will give you more understanding about Algebra.

Scale:
In x-axis 1unit= 1 cm
In y-axis 1unit= 3 cm
Graph for the composite functions

Reciprocal Function Tutor

Reciprocal function tutor is nothing but the inverse function tutor. Before knowing about the inverse function we have to know about the one – to – one function. Let us take the domain as X and range as Y of one to one function f. Thus, the reciprocal function of f has the y domain and X range and is represented as  f-1 (y) = x or f(x) = y.

one to one function

Let see about the reciprocal function with example:

Procedure to find the Reciprocal Function Tutor:

To find the formula of the reciprocal function tutor by following the given procedure:

Step 1: Change the given equation in form of function y = f(x).

Step 2: Solve the given equation for x in terms of y.

Step 3: Inter change the x and y variables and therefore y = f-1(x). Thus function of y is equal to reciprocal function of x.

Example Problems – Reciprocal Function Tutor:

Example 1:

Find the reciprocal of the function f(x) = `(3x - 6)/(3x+5)`.

Solution:

Step 1: Let write the given function as y = `(3x - 6)/(3x+5)`..

Step 2: Solve the function of x in term of y.

y = `(3x - 6)/(3x+5)`.

Multiply the denominator of the fraction 3x + 5 to y.

Now, y (3x + 5) = 3x – 6

3xy + 5y = 3x – 6

5y + 6 = 3x – 3xy

5y + 6 = x (3 - 3y)

Now, we get the value of x = `(5y + 6)/(3-3y)`.

Step 3: Now change the variable x in terms of y and vice versa.

Hence y =  `(5x + 6)/(3-3x)`

Thus the reciprocal of the function f(x) =  `(3x - 6)/(3x+5)`. is given by the reciprocal of function f-1 =  `(5x + 6)/(3-3x)`.

Answer: f-1 =  `(5x + 6)/(3-3x)`.

Example 2:

Find the reciprocal of the function f(x) = 7x – 6.

Solution:

Step 1: Let write the given function as y = 7x – 6.

Step 2: Solve the function of x in term of y.

y = 7x – 6

Add 6 on both sides, we get y + 6 = 7x - 6 +6

y + 6 = 7x

Divide 7 on both sides, we get  `(y +6)/(7)`. =  `(7x)/(7)`.

x = `(y +6)/(7)`

Algebra is widely used in day to day activities watch out for my forthcoming posts on neet medical 2013 and final syllabus for neet 2013. I am sure they will be helpful.

Step 3: Now change the variable x in terms of y and vice versa.

Hence y =`(x +6)/(7)`.

Thus the reciprocal of the function f(x) = 7x - 6 is given by the reciprocal of function f-1 = `(x +6)/(7)`.

Answer: f-1 = `(x +6)/(7)`.

Equivalence Statement

Logical Equivalence in simple word is the comparision of two logical statements. Meaning of the word logic is analysis. Analysis may be granted result or mathematical proof. Two statemnets satisfies the equivalance if the resultant truth table of both the given statements are exactly same. That is, the resultant truth values are the same. Equivalence statements satisfies the "if and only if" or biconditional operations. If A and B are two equivalant statements then one can be proved from another.
Symbol of Equivalence

We use '-= symbol to represent the term logical equivalence in discrete mathematics. Let us see about logical equivalence in this article.

Logical Equivalence Table

If the last column of two given components in the logical equivalence table are the same, then the two components are said to be logical equivalent.

We can  use'-= ' symbol to represent logical equivalences. Other name of logical equivalence is simple equivalence. These are the logical equivalence laws.

Solved Examples

Given below are some of the logical equivalence examples:

Example 1: Show ~ (Xvv Y) = (~X)^^ (~Y) simple equivalence.

Proof:

LHS:

~ (Xvv Y)
X Y Xvv Y
~ (Xvv Y)
T T T F
T F T F
F T T F
F F F T

RHS:

(~X)^^ (~Y)
X Y ~X ~Y (~X)^^ (~Y)
T T F F F
T F F T F
F T T F F
F F T T T

Last column of LHS table is equal to last column of RHS table. And so, it is a simple equivalence.

Example 2: Show   (X^^ Y) = ~ ((~X)vv (~Y)) simple equivalence.

Proof:

LHS:

(X^^ Y)
X Y (X^^ Y)
T T T
T F F
F T F
F F F

RHS:

~ ((~X)vv (~Y))
X Y ~X ~Y ((~X)vv (~Y)) ~ ((~X)vv (~Y))
T T F F F T
T F F T T F
F T T F T F
F F T T T F

Last column of LHS table is equal to last column of RHS table. And so, it is a simple equivalence.

Algebra is widely used in day to day activities watch out for my forthcoming posts on Multiply a Fraction and Determine Equation from Graph. I am sure they will be helpful.

Example 3: Show (Xvv Y)vv (~(Xvv Y)) =(Xvv (~Y))vv Y simple equivalence.

Proof:

LHS:

(Xvv Y)vv (~(Xvv Y))
X Y Xvv Y
(~(Xvv Y) (Xvv Y)vv (~(Xvv Y))
T T T F T
T F T F T
F T T F T
F F F T T

RHS:

(Xvv (~Y))vv Y
X Y ~Y (Xvv (~Y)) (Xvv (~Y))vv Y
T T F T T
T F T T T
F T F F T
F F T T T

Last column of LHS table is equal to last column of RHS table. And so, it is a simple equivalence.

Multiplying Dividing Decimals

Decimal number is nothing but number with base 10. The multiplying decimals is nothing but to multiply the given number considering it as a whole numbers. At the end of calculation put the decimal point into position. For dividing decimals number by which dividing by, must be a whole number.division by 10 or by powers of 10 is easy in a decimal system. Let us see sample problems for multiplying dividing decimals.

Please express your views of this topic Multiplying Integers Rules by commenting on blog.

multiplying dividing decimals:

Multiplying decimals:

The multiplying decimals is nothing but to multiply the given number considering it as a whole numbers. At the end of calculation put the decimal point into position.


How to work out the position of the decimal point:


Count up how many figures there are after the decimal point in the question. There must be the SAME number of figures after the decimal point in the answer.


Eg: 6.7 x 2

Write this as            67
x 2
-------------
134
--------------
There is one figure after the decimal point in the question, so there must be only one after the point in the answer.
6.7 x 2 = 13.4

Eg: 30.6 x 0.7

Write this as          306
x 7
-----------------
2142
----------------
There are two figures after the decimal point in the question, from both the numbers, so there must be two after the point in the answer.
30.6 x 0.7 = 21.42

Eg: 2.9 x 0.0351
Write this as            351
x 29
--------------
3159
7020 *
-------------
10179
------------

There are five figures after the decimal point in the question so    2.9 x 0.0351 = 0.10179

multiplying dividing decimals:

Dividing decimals:

Division by powers of 10

Like multiplication, division by 10 or by powers of 10 is easy in a decimal system.


A quick way to divide a decimal number by 10 is to move the decimal point ONE place to the LEFT.

←
Eg 24.185 / 10 = 2.4185

dont forget when we are dividing you are making a number smaller.


←
Eg 0.6 / 10 = 0.06

To divide by 100 move the decimal point TWO places to the left.


←
Eg 32.157 / 100 = 0.32157

←
Eg 0.6 / 100 = 0.06

Dividing decimals:

Eg: `4.48 / 0.4`

Procedure:

The number by which dividing by, must be a whole number. To make this as a whole number we multiply it by 10 or 100 or 1000 or a higher power of 10,  if necessary.

Eg: 0.4 x 10 = 4

Whatever you do to the number you are dividing by, do exactly the same to the number which you are dividing.

Eg: 4.48 x 10 = 44.8

Then do the division, putting the decimal point of the answer directly above the other decimal point when you write out the division.

Eg:                11.2
4 ) 44.8

You don’t have to do anything else. The answer is 11.2

Example: `0.36 / 0.6`

First multiply 0.6 by 10 to make it into a whole number: 6

Then do exactly the same to 0.36: 0.36 x 10 = 3.6

Now divide 3.6 by 6:           0.6
6 ) 3.6
The answer is: 0.6

Recurrence Relations Tutor

A recurrence relation is an equation with the intention of recursively describes a series: each term of the series is defined as a function of the above terms. The difference equations refer to a particular type of recurrence relation. Note but the "difference equation" is commonly used to refer to any recurrence relation. Tutoring is method of teaching. Tutoring is to teaching a single student or a group of students.

An example of a recurrence relation is the logistic map:

x n + 1 = r x n (1-xn)

Recurrence relations tutor – Examples:

Recurrence relations tutor  - Example 1:

Find the limiting ratio `lim_(n->oo) (x_{n+1})/(x_n)` , for the recurrence relation `x_n = x_{n-1}+x_{n-2}.`

Solution:

We find what the limit must be, assuming that it exists.

L = `\lim_{n\rightarrow\infty} \frac{x_{n+1}}{x_n} = \lim_{n\rightarrow\infty} \frac{x_n+x_{n-1}}{x_n} = 1 + \lim_{n\rightarrow\infty} \frac{x_{n-1}}{x_n} = 1+L^{-1}`

L = 1 + `L^{-1}`

`L^2=L+1`

`L^2-L-1=0`

L = `\frac{1 \pm \sqrt{ 5 }}{2}` , via the quadratic formula.
Recurrence relations tutor - Example 2:

Let
`x_n = { x_(n-1) + x_(n-2)`     if n > 1
`x_(1) epsi N`                       if n = 1
`x_(0) epsi N`                       if n = 0
0                                if n < 0

Show that `x_n = x_{1}F_{n-1} + x_{0}F_{n-2} \,\! "where" F_n\!` is the n-th Fibonacci number   (F0 = F1 = 1)

Solution:

Using the principle of induction we have:

BASIS: n=`2 \Rightarrow x_2 = x_1 + x_0 = x_{1}F_{1} + x_{0}F_{0}\!`

INDUCTIVE STEP: We have `x_{n-2} = x_{1}F_{n-3} + x_{0}F_{n-4}\,\! and x_{n-1} = x_{1}F_{n-2} + x_{0}F_{n-3}\!`

By definition we have:


`x_{n} := x_{n-1} + x_{n-2} = x_{1}F_{n-2} + x_{0}F_{n-3} + x_{1}F_{n-3} + x_{0}F_{n-4}`

`= x_{1}\(F_{n-2} + F_{n-3}\) + x_{0}\(F_{n-3}+F_{n-4}\) = x_{1}F_{n-1} + x_{0}F_{n-2} \mbox{ } \!`

Recurrence relations tutor – More Problems:

Recurrence relations tutor - Example 1:

Let  .

`x_n := x_{n-1}*x_{n-2}`  if n>1
`x_1`                        if n=1
`x_0`                        if n=0
0                           if n<0 p="">
Show that `x_n = x_{1}^{F_{n-1}}*x_{2}^{F_{n-2}} \!` where `F_n\!` is the n-th Fibonacci number   (F0 = F1 = 1 and F(n < 0) = 0)

Solution:

As before: induction is da way!

Between, if you have problem on these topics T Score Distribution Table, please browse expert math related websites for more help on fraction of a set worksheet.

BASIS: For n = 2! we have `x_2 = x_1^{F_1}*x_0^{F_{0}} = x_1*x_0 \!`

INDUCTIVE STEP: `x_{n-2} = x_1^{F_{n-3}}*x_0^{F_{n-4}} \,\! and x_{n-1} = x_1^{F_{n-2}}*x_0^{F_{n-3}} \!` and

so `x_n = x_{n-1}*x_{n-2} = (x_1^{F_{n-2}}*x_0^{F_{n-3}}) * (x_1^{F_{n-3}}*x_0^{F_{n-4}}) `

= `x_1^{F_{n-2}+F_{n-3}}*x_0^{F_{n-3}+F_{n-4}}`

= `x_{1}^{F_{n-1}}*x_{2}^{F_{n-2}} !`

All Kinds of Fractions

A fraction means a part of a whole group or its region.A fraction written as  number  with the bottom part (denominator) showing how many parts the whole is divided into, and the top part (numerator). Fraction in all kinds can be written as many types which depends upon the numerator and denominator by its value.
Fraction in all kinds are

Proper fractions
Improper fractions
Mixed fractions

I like to share this Converting Fractions with you all through my article.

Classification for all kinds of Fractions:

Proper Fractions:
proper fraction is explained as  the denominator shows the number of parts into which whole divided and the numerator expressed  the number of parts which  we taken out. In other words numerator less than denominator.

Example:
`3/10,2/5`
Improper Fractions:
Improper fractions is the fraction whose  numerator is greater  than the denominator are called improper fractions.

Example:
`7/2, 9/7`

Mixed Fractions:
A  mixed fraction is typical fraction which  has  combination of a whole and its part.

Example:
2` 3/4,` 7` 2/9` ,



Addition and Subtraction fractions in all kinds:

Procedure for addition:

For addition,fraction numbers with same denominator, denominator remain same number and we add only the numerator.
For addition with different denominator fraction  we have to take lcm for all denominator and convert different denominator  into like denominator by taking LCM and add.


Example 1:
Add `1/5` and  `3/5`

Solution :
In this proper fraction we have same denominator
`1/5 +3/5` =` (1+3)/5`

=`4/5`
Example 2:

Add `2/3` and `1/5`
Solution:
In this fraction we have different denominator  so,we take LCM
The LCM of 3 and 5 is 15.

Therefore,`2/3+1/5` =`(2xx5)/(3xx5)+(1xx3)/(5xx3)`

=`10/15+3/15`

=`13/15`

Example 3:
Add `1 4/5` and `3 5/6`

Solution:
`1 4/5 +3 5/6` =  `(5+4)/5+(18+5)/6`


Now `9/5+23/6` =`(9xx6)/(5xx6)+(23xx5)/(6xx5) `    since LCM of 5,6 =30

=`36/30+115/30`

=`151/30`

I am planning to write more post on fraction of a set and Surface Integral of a Sphere. Keep checking my blog.


Rules for subtraction:

For subtraction,fraction numbers with same denominator, denominator remain same number and we subtract only the numerator.



For subtraction with different denominator fraction  we have to take lcm for all denominator and convert different denominator  into like denominator by taking LCM and subtract.


Example 1:

subtract  `1/3` from `2/3`
Solution:
here same denominator
`2/3-1/3`    =`(2-1)/3`

Example 2:
Find fraction for `4 2/5-2 1/3`

Solution :
Taking LCM for 5 and 3 is 15 because of different denominator
`4 2/5 -2 1/3` =  `22/5-7/5`
= `(22xx3)/(5xx3) -(7xx5)/(3xx5)`

=`66/15 -35/15`

=`(66-35)/15`

=`31/15`

Word Problems Review

In mathematics education, the term word problem is often used to refer to any mathematical exercise where significant background information on the problem is presented as text rather than in mathematical notation. As word problems often involve a narrative of some sort, they are occasionally also referred to as story problems and may vary in the amount of language used.(Source - Wikipedia )
In this article of word problems review, we are going to review some of the basic word problems.

I like to share this Statistics Example Problems with you all through my article.


Review on word problems through examples:

Example 1:

An Auditorium has 1020 seats. 875 of them are occupied. What percentage of the seats are occupied?

Solution:

Number of seats in Auditorium  = 1020

Number of seats occupied        =  870

Percentage of the seats occupied  = `<< 870/1020>>`   x 100

=  `<< 87/102>>` x 100

=  85.29

Example 2:

In a Theatre, there are 400 persons consists of men, women, and children.There are four times as many men as children, and thrice as many women as children. How many of children are there?

Solution:

Let 'x' be the number of children.

Then the number of men = 4x and the number of women = 3x

So, the equation is      x + 4x + 3x = 400

8x = 400

x = 400 / 8

x = 50

Example 3:

A train 300 m long is running at a speed of 70 km/hr. What time will it take to cross a 50 m long bridge?

Solution:

In order to cross the bridge, the train will have to cover

distance = ( 300 + 50 )

= 350 m

Speed     = 70 x `<< 5/18>>`

= `<< 175/9>>`

= 19.44 m/s

Required time    = Distance / Speed

= `<< 350/19.44>>`

= 18 sec

Example 4:

A customer deposits an amount of $3000 in his account. The bank offers a interest of 2.5%. Calculate the simple interest for the customer earn in 7 years?

Solution:

Interest rate (r) = 2.5% = 0.025

Amount   (P) = $3000

time (t) = 7 years

Simple interest formula

I = P r t

I = 3000 × 0.025 × 7

= 5000 × 0.15

= $ 525

Practice problems for Review:

1) Adam deposits an amount of $6500 in his account with an interest of 4.5%. Calculate the simple interest that he earn in 5 years?

2) An Auditorium has 950 seats. 855 of them are occupied. What percentage of the seats are occupied?

Answer key:        1) $ 1462.5         2) 90 percent

Mixed Numbers

Mixed numbers the proper numbers is said to be numerator is less than denominator. Improper numbers says numerator is greater than denominator. A number consist of natural number and a fraction is called mixed number.Round mixed numbers as show in the below 2 `1/4` , 3 `4/5` , and 5 `7/8` . These are the mixed number. 2 `1/4` , In this mixed number the numerator is 1 and denominator is 4. 2 is out side of the given mixed number.First we have to find how to round the mixed number, here multiply 4 into the number 2 and add to the numerator, we get `(4xx2+1)/4` = `9/4`.

Problem in round mixed numbers

Problems 1 Round  mixed numbers and Adding mixed number:

(i)                  2 `1/5 ` + 3 `5/6`

Solution:

In the question `1/5` is the improper number because numerator is greater than denominator. Here 2 is out of the number. 2 `1/5` first we have to change in the proper number that is `(5*2+1)/5`

=` 11/5` is the proper number

The second number 3 `5/6`

In this number `(6*3+5)/6 ` we get

=`23/6`

Here we round  mixed numbers.

`11/5` + `23/6`

In this equation we have to take the Least common factor for 5 and 6 is 30

`11/5` +`23/6`

Here LCM is 30

`(11*6+23*5)/30 ` cross multiplication

`(66+115)/30` Adding  the number in the numerator we get

`=181/30.`

This is the answer for the adding mixed number for 2 `1/5 ` + 3 `5/6`

(ii)                2 `2/5` + 3 `1/6`

Solution: In the question `2/5` is the improper number because numerator is greater than denominator. Here 2 is out of the number.

2 `2/5` first we have to change in the proper number that is `(5*2+2)/5`

= `12/5` is the proper number

The second function is  3 `1/6`

In this function `(6*3+1)/6` we get

`=19/6`

Here we round  mixed numbers.

`12/5` + `19/6`

In this equation we have to take the Least common factor for 5 and 6 is 30

`12/5` +`19/6`

Here LCM is 30

`(12*6+19*5)/30` cross multiplication

`(72+95)/30` Adding the numbers in the numerator we get

`=167/30.`

This is the answer for the adding mixed numbers for 2` 2/5` + 3 `1/6.`

Round mixed number and adding mixed number with same denominator:

Problem for round mixed numbers and Add with same denominator

(i)  2 `1/8 ` + 3 `2/8`

Solution:

In the question `1/8` is the improper numbers because numerator is greater than denominator.

Here 2 is out of the number.2 `1/8` first we have to change into the proper numbers

`(2xx8+1)/8` ` = 1 7/8` is the proper numbers

The next fraction 3 `2/8`

The same we have to do this also we get

`(3xx8 +1)/8 = 25/8` is the proper number

Here we round mixed numbers and adding both functions

` = 17/8` + `25/8`

In this step adding mixed numbers is the easier way to add because the denominator is same so we can add directly

`(17+25)/8 =42/8`

So  `48/7` is the answer for the adding mixed fraction of 2 `1/8 ` + 3 `2/8`

Example Problem for round mixed numbers and Add with same denominator

2 `3/8` + 3 `5/8`

Solution:

In the question `3/8` is the improper numbers because numerator is greater than denominator.

Here 2 is out of the number. 2 `3/8` first we have to change into the proper numbers

`(2xx8+3)/8` = `19/8` is the proper numbers

The next fraction is 3 `5/8`

The same we have to do this also we get

`(3xx8 +5)/8 =29/8`

Here we round  mixed numbers

`=19/8` + `29/8`

In this step adding mixed numbers is the easier way to add because the denominator is same so we can add directly.

`(19+29)/8=``48/8`  is the answer for the Adding mixed numbers of 2 `3/8` + 3 `5/8`

Positive and Negative Chart

The integers are the obligatory in mathematics. There are the first topics in mathematics. The numbers are the basics of each person’s daily life.The integers are can be devided by tow parts.

The integers are may be seperated as positive integers  and negative integers.  The numbers should not have any end, it is normally endless. Normally, the integers are many types like whole numbers, counting numbers, positive integers,negative integers.

Defintion:

Positive intergers are defined as, each and every natural numbers are all called as positve numbers. The positive numbers should be like, 1, 2, 3, 4, 5,…….. . In the number chart the positive integers are represented as the right side of zero.

Negative integers are defined as, the opposite of every positive numbers are all called as negative numbers.The negative numbers should be like , -1, -2, -3, -4,-5,…… . In the number chart the negative integers are represented as the left side of zero. The negative integers are using (-) symbol.

positive and negative chart:

The chart should explains the operations. They are,

Adding Intergers chart:

Positive integer + Positive integer= Positive integer.
Positive integer + Negative integer= Higher integer symbol.
Negative integer +Positive integer= Higher integersymbol.
Negative integer+Negative integer= Negative integer.

Subracting Integerts chart:

Positive integer – Positive integer = Positive integer.
Positive integer– Negative integer = Positive integer.
Negative integer– positive integer = Higher integer symbol.
Negative integer –Negative integer=Higher integer symbol.

Multiplication integerts chart:

Positive integer * Positive integer = Positive integer.
Positive integer * negative integer = Negative integer.
Negative integer*Positive integer= Negative integer.
Negative integer*Negative integer= Positive integer.

Dividing integers:

Positive integer /  Positive integer= Positive integer.
Positive integer / Negative integer= Negative integer.
Negative integer / positive integer = Negative integer.
Negative integer /Negative integer = Positive integer.

Examples:

Problem 1:

Find the addtion of integers which is following given chart:

(i)           (-5) + ( 10)

(ii)          (-7) +(-8)

Solution:

(i)           (-5) + (10):

According to the addtion chart, we get higher value symbol as a result.

= (-5) + (10).

= (10).

(ii)          (-7) + (-8):

According to the addition chart, we get negative value as the result.

= (-7) + (-8).

= -15.

Problem 2:

Find the subraction of integers which is following given chart:

(i)           (-10) – (-25).

(ii)          (100) – ( 58).

Solution:

(i)           (-10) – (-25).

According  to the subraction chart, we get higher value symbol as the result.

= (-10) –(-25).

= 15.

(ii)          (100) –(58).

According to the subraction chart, we get positive value as the result.

= 100 -58.

= 42.

Problem 3:

Find the multiplication of integers which is the value is (-50) * (25).

Solution:

According to the multiplication chart, we get negative value as a result.

= (-50) * (25).

= - 1250.

Problem 4:

Find the division of integers which  the value is (-25) / (-5).

Solution:

According to the division chart, we get positive value as a result.

= (-25) /(-5).

= 5.

SOLID SHAPES VERTICES

Solid shapes are to have, length, depth, and height these are the three dimensions. These are Three dimensional shapes.  The examples for a Solid shape are sphere, cube, cuboids, cylinder, cone, Prism, Pyramid, Dodecahedron, Octahedron, and Tetrahedron. Here we are going to see in detail about Solid shapes. The real life example of a Solid shape is book, table, etc... Vertex means a corner or a point where lines meet.

I like to share this Solid Geometry with you all through my article. 

Sphere:

• A sphere is the set of all those points in the space which are equidistant from a fixed point.
• The sphere has no vertex.

 
Cube:

• The solid shape which has all the sides equal in length is known as cube.
• The point of intersection of three mutually perpendicular edges is known as vertex. A cube has 8 vertices

 

Cuboids:

• The Solid like a matchbox, an ordinary brick, a room, etc., having six rectangular faces are parallel and congruent, are known as cuboids. A cuboid is also known as rectangular parallelepiped.
• The point of intersection of three mutually perpendicular edges is known as vertex. A cuboid has 8 vertices.

 
Cylinder:

• Solid line measuring jars, circular pillars, circular pencils, Circular pipes, road rollers and gas cylinders are said to have a cylindrical shape.
• There is no vertices for the cylinder.

 

Cone:

• If an Right angled triangle is revolved about one of the sides containing the right angle, the solid thus generated is called a right circular cone. 
• A cone does not has a vertex

 
Pyramid

• A solid shaped object with a flat, often square, base and four flat triangular sides which slope inwards and meet to form a point at the top.
• There are five vertices in a pyramid

Algebra is widely used in day to day activities watch out for my forthcoming posts on use the substitution method to solve the system of equations and karnataka education board. I am sure they will be helpful.

Dodecahedron:

• A Dodecahedron is a solid shaped figure, which has twenty vertices, thirty edges and twelve equivalent pentagonal triangular faces.

Octahedron:

• An Octahedron is a solid shaped figure which has six vertices, twelve edges, and eight equivalent equilateral triangular faces.

Tetrahedron:

• A Tetrahedron is a solid shaped figure which has four vertices , six edges, and four equivalent equilateral triangular faces

Writing in Slope Intercept Form

The general equation of a line is y =mx+c  it is called slope intercept form of that line. Here m represent the slope of the line and c is the y intercept .here we are going to learn about how to write slope intercept form and its example problems.

For example:

y- 3x = 6

The slope intercept form is y =mx+b

-3x moves to right hand side so it will be 3x

Therefore slope intercept form y =3x+6

I like to share this slope of the line with you all through my article.

Example problems to write in slope intercept form:

Write the equation of a straight line in slope intercept form whose slope = 3 and passes through the point (-1,-6).

Solution:

We know that general equation of a line y=mx+b

Here

m=3

x=-1

y=-6

Substitute all the value in the general equation

-6 = 3(-1) +b

-6 =-3 +b

Add both sides +3b

-6 +3 =-3+3 +b

b=-3

Substitute the y intercept value in the general equation,

y=3x-3

There the equation of straight line is y=3x-3

Writing equations in slope intercept form:

Write the slope intercept form of the line that the line passes through the points (2, 3) (5, 6)

Solution:

We want to find the equation for a line that passes through the two points:

(2, 3) and (5, 6).

First of all, remember what the equation of a line is:

y = mx+b

Where:

m is the slope
slope intercept form

First, let's find what m is, the slope of the line...

For lines like these, the slope is always defined as "the change in y over the change in x" or, in equation form:

m = `(y2-y1) / (x2-x1)`

Here,

x1=2 and y1=3.

x2=5 and y2=6.

Now, just plug the numbers into the formula for m above, like this:

m=`(6-3)/(5-2)`

m=`3/3`

m=1

Therefore the slope of the given line is 1

Algebra is widely used in day to day activities watch out for my forthcoming posts on Solving Special Right Triangles and cbse syllabus for class 9. I am sure they will be helpful.

Slope intercept form = y=mx+b

x=2 and y =3

So

3=1*2+b

3=2+b

b=3-2

b=1

Therefore the slope intercept form y=x-1

Prime Factorization Table

In mathematics, prime factorization is one interesting topic in number theory. The process of the positive integers in which they are dividing the integer exactly and also without a remainder value is called as Prime factorization. These types of numbers can be divided by 1 or itself then it is called as prime numbers.

Some of the prime numbers are 2, 3, 5, 7, 9, 11, and 13…

A factor is a number that perfectly divides another number. For example, 2 is a factor of 4 and 3 is a factor of 6. Also, 4 is not a factor of 6 because 6 is not divisible by 4. Every number has a minimum of two factors - 1 and itself. A number which has only two factors (1 and itself) is known as a prime number. A number which has more than 2 factors is known as composite. 5 is an example of a prime number (factors : 1 and 5) and 10 is an example of a composite number (factors: 1, 2, 5 and 10).

Factorization, the method of reducing a number into a product of its prime factors, is a very handy tool. Factorization is essential in finding the greatest common divisor and the least common multiple. These are in turn used to add or subtract fractions. More advanced forms of factorization are made use of in algebraic addition and subtraction.

Prime factorization table - Definitions:

NUMBERS	FACTORS	NUMBER OF FACTORS
1	1	1
2	1, 2	2
3	1, 3	2
4	1, 2, 4	3
5	1, 5	2
6	1, 2, 3, 6	4
7	1, 7	2
8	1, 2, 4, 8	4
9	1, 3, 9	3
10	1, 2, 5, 10	4
11	1, 11	2
12	1, 2, 3, 4, 6, 12	6
13	1, 13	2
14	1, 2, 7, 14	4
15	1, 3, 5, 15	4
16	1, 2, 4, 8, 16	5
17	1, 17	2
18	1, 2, 3, 6, 9, 18	6
19	1, 19	2
20	1, 2, 4, 5, 10, 20	6
21	1, 3, 7, 21	4
22	1, 2, 11, 22	4
23	1, 23	2
24	1, 2, 3, 4, 6, 8, 12, 24	8
25	1, 5, 25	3
26	1, 2, 13, 26	4
27	1, 3, 9, 27	4
28	1, 2, 4, 7, 14, 28	6
29	1, 29	2
30	1, 2, 3, 5, 6, 10, 15, 30	8
31	1, 31	2
32	1, 2, 4, 8, 16, 32	6
33	1, 3, 11, 33	4
34	1, 2, 17, 34	4
35	1, 5, 7, 35	4
36	1, 2, 3, 4, 6, 9, 12, 18, 36	9
37	1, 37	2
38	1, 2, 19, 38	4
39	1, 3, 13, 39	4
40	1, 2, 4, 5, 8, 10, 20, 40	8
41	1, 41	2
42	1, 2, 3, 6, 7, 14, 21, 42	8
43	1, 43	2
44	1, 2, 4, 11, 22, 44	6
45	1, 3, 5, 9, 15, 45	6
46	1, 2, 23, 46	4
47	1, 47	2
48	1, 2, 3, 4, 6, 8, 12, 16, 24, 48	10
49	1, 7, 49	3
50	1, 2, 5, 10, 25, 50	6
51	1, 3, 17, 51	4
52	1, 2, 4, 13, 26, 52	6
53	1, 53	2
54	1, 2, 3, 6, 9, 18, 27, 54	8
55	1, 5, 11, 55	4
56	1, 2, 4, 7, 8, 14, 28, 56	8
57	1, 3, 19, 57	4
58	1, 2, 29, 58	4
59	1, 59	2
60	1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60	12
61	1, 61	2
62	1, 2, 31, 62	4
63	1, 3, 7, 9, 21, 63	6
64	1, 2, 4, 8, 16, 32, 64	7
65	1, 5, 13, 65	4
66	1, 2, 3, 6, 11, 22, 33, 66	8
67	1, 67	2
68	1, 2, 4, 17, 34, 68	6
69	1, 3, 23, 69	4
70	1, 2, 5, 7, 10, 14, 35, 70	8
71	1, 71	2
72	1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36,72	12
73	1, 73	2
74	1, 2, 37, 74	4
75	1, 3, 5, 15, 25, 75	6
76	1, 2, 4, 19, 38, 76	6
77	1, 7, 11, 77	4
78	1, 2, 3, 6, 13, 26, 39, 78	8
79	1, 79	2
80	1, 2, 4, 5, 8, 10, 16, 20, 40, 80	10
81	1, 3, 9, 27, 81	5
82	1, 2, 41, 82	4
83	1, 83	2
84	1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84	12
85	1, 5, 17, 85	4
86	1, 2, 43, 86	4
87	1, 3, 29, 87	4
88	1, 2, 4, 8, 11, 22, 44, 88	8
89	1, 89	2
90	1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90	12
91	1, 7, 13, 91	4
92	1, 2, 4, 23, 46, 92	6
93	1, 3, 31, 93	4
94	1, 2, 47, 94	4
95	1, 5, 19, 95	4
96	1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96	12
97	1, 97	2
98	1, 2, 7, 14, 49,98	6
99	1, 3, 9, 11, 33, 99	6
100	1, 2, 4, 5, 10, 20, 25, 50, 100	9
101	1, 101	2
102	1, 2, 3, 6, 17, 34, 51, 102	8
103	1, 103	2
104	1, 2, 4, 8, 13, 26, 52, 104	8
105	1, 3, 5, 7, 15, 21, 35, 105	8
106	1, 2, 53, 106	4
107	1, 107	2
108	1, 2, 3, 4, 6, 9, 12, 18, 27, 36, 54, 108	12
109	1, 109	2
110	1, 2, 5, 10, 11, 22, 55, 110	8
111	1, 3, 37, 111	4
112	1, 2, 4, 7, 8, 14, 16, 28, 56, 112	10
113	1, 113	2
114	1, 2, 3, 6, 19, 38, 57, 114	8
115	1, 5, 23, 115	4
116	1, 2, 4, 29, 58, 116	6
117	1, 3, 9, 13, 39, 117	6
118	1, 2, 59, 118	4
119	1, 7, 17, 119	4
120	1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120	16
121	1, 11, 121	3
122	1, 2, 61, 122	4
123	1, 3, 41, 123	4
124	1, 2, 4, 31, 62, 124	6
125	1, 5, 25, 125	4
126	1, 2, 3, 6, 7, 9, 14, 18, 21, 42, 63, 126	12
127	1, 127	2
128	1, 2, 4, 8, 16, 32, 64, 128	8
129	1, 3, 43, 129	4
130	1, 2, 5, 10, 13, 26, 65, 130	8
131	1, 131	2
132	1, 2, 3, 4, 6, 11, 12, 22, 33, 44, 66, 132	12
133	1, 7, 19, 133	4
134	1, 2, 67, 134	4
135	1, 3, 5, 9, 15, 27, 45, 135	8
136	1, 2, 4, 8, 17, 34, 68, 136	8
137	1, 137	2
138	1, 2, 3, 6, 23, 46, 69, 138	8
139	1, 139	2
140	1, 2, 4, 5, 7, 10, 14, 20, 28, 35, 70, 140	12
141	1, 3, 47, 141	4
142	1, 2, 71, 142	4
143	1,11, 13, 143	4
144	1, 2, 3, 5, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144	15
145	1, 5, 29, 145	4
146	1, 2, 73, 146	4
147	1, 3, 7, 21, 49, 147	6
148	1, 2, 4, 37, 74, 148	6
149	1, 149	2
150	1, 2, 3, 5, 6, 10, 15, 25, 30, 50, 75, 150	12
151	1, 151	2
152	1, 2, 4, 8, 19, 38, 76, 152	8
153	1, 3, 9, 17, 51, 153	6
154	1, 2, 7, 11, 14, 22, 77, 154	8
155	1, 5, 31, 155	4
156	1, 2, 3, 4, 6, 12, 13, 26, 39, 52, 78, 156	12
157	1, 157	2
158	1, 2, 79, 158	4
159	1, 3, 53, 159	4
160	1, 2, 4, 5, 8, 10, 16, 20, 32, 40, 80, 160	12
161	1, 7, 23, 161	4
162	1, 2, 3, 6, 9, 18, 27, 54, 81, 162	10
163	1, 163	2
164	1, 2, 4, 41, 82, 164	6
165	1, 3, 5, 11, 15, 33, 55, 165	8
166	1, 2, 83, 166	4
167	1, 167	2
168	1, 2, 3, 4, 6, 7, 8, 12, 14, 21, 24, 28, 42, 56, 84, 168	16
169	1, 13, 169	3
170	1, 2, 5, 10, 17, 34, 85, 170	8
171	1, 3, 9, 19, 57, 171	6
172	1, 2, 4, 43, 86, 172	6
173	1, 173	2
174	1, 2, 3, 6, 29, 58, 87, 174	8
175	1, 5, 7, 25, 35, 175	6
176	1, 2, 4, 8, 11, 16, 22, 44, 88, 176	10
177	1, 3, 59, 177	4
178	1, 2, 89, 178	4
179	1, 179	2
180	1, 2, 3, 4, 5, 6, 9, 10, 12, 15, 18, 20, 30, 36, 45, 60, 90, 180	18
181	1, 181	2
182	1, 2, 7, 13, 14, 26, 91, 182	8
183	1, 3, 61, 183	4
184	1, 2, 4, 8, 23, 46, 92, 184	8
185	1, 5, 37, 185	4
186	1, 2, 3, 6, 31, 62, 93, 186	8
187	1, 11, 17, 187	4
188	1, 2, 3, 47, 94, 188	6
189	1, 3, 6, 9, 21, 27, 63, 189	8
190	1, 2, 5, 10, 19, 38, 95, 190	8
191	1, 191	2
192	1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 64, 96, 192	14
193	1, 193	2
194	1, 2, 97, 194	4
195	1, 3, 5, 13, 15, 39, 65, 195	8
196	1, 2, 4, 7, 14, 28, 49, 98, 196	9
197	1, 197	2
198	1, 2, 3, 6, 9, 10, 18, 22, 33, 66, 99, 198	12
199	1, 199	2
200	1, 2, 4, 5, 8, 10, 20, 25, 40, 50, 100, 200	12

Steps to find the prime factorization with example:

Different steps to find the prime Factorization:

• Even number contains a factor of 2.
• Number which is ending in 5 has a factor of 5.
• Number which is ends with 0 and also it is above 0 contains the factors of 2 and 5.

My forthcoming post is on Sinusoidal Function Equation and 6th grade math problems online will give you more understanding about Algebra.

Example problem for prime factorization:
Example 1: Find the factorization for the given number 25.
Solution:
25 are divided by 5.
So, 5 are the factors of the given number 25.
Example 2: Find the prime factorization for the given number 28.
Solution:
28 is divided by 2 means, we get 14. Here 14 is not a prime factor. So we factor the 14.
                28 ÷ 2 = 14
14 is divided by 2 means, we get 7. Here 7 is a prime factor.
                14 ÷ 2 = 7
Each factor is a prime number.
So Prime factorization for the given number 28 is 2 × 2 × 7

Example 3: Find the prime factorization of 135
Solution:
             Given
             135
             135 is not a divided by 2. So we divide, 135 can be divided by 3

             135 ÷ 3 = 45

Then we factoring 45, and we find that 3 is the smallest prime number

             45 ÷ 3 = 15

Then we factoring 15, and we find that 3 is the smallest prime number

             15 ÷ 3 = 5

All the factors are prime numbers.

             135 = 3 × 3 × 3 × 5

3 × 3 × 3 × 5 is the prime factorization of 135.

Example 4: Find the prime factorization of 188
Solution:
             Given
             188
             188 can be divided by 2.

             188 ÷ 2 = 94

Then we factoring 94, and we find that 2 is the smallest prime number

             94 ÷ 2 = 47

             47 is a prime number.

All the factors are prime numbers.

             188 = 2 × 2 × 47

2 × 2 × 47 is the prime factorization of 188.