Addition and Subtraction Facts

A fact is a group of a declaration of  each of which stand for the same meaning. All the declaration has the same meanings. Addition and subtraction facts contain 2 or 3 numbers. The types of facts operations are addition and subtraction facts, multiplication and division facts. In here we shall discuss about addition and subtraction facts. 

Addends:

The facts being added in an addition operation like 4 and 5 in 4 + 5 = 9 or 5 + 4 = 9.

Reverse operation:

Reverse of a process is call as inverse operation of addition and subtraction.

I like to share this Addition and Subtraction Word Problems with you all through my article. 

Sample problem for addition and subtraction facts:

First we take an addition operation like 3 + 5 = 8.
 In the next operation the addends are reversed like 5 + 3 = 8.
Now we can write the inverse operations of both.
8 - 5 = 3
8 - 3 = 5
Ex1: What are the four members of the fact family of 7 + 5 = 12?
Solution: 7 + 5 = 12
Reversing the addends:
       7 + 5 = 12;
Inverse operations:
        12 - 5 = 7;
        12 - 7 = 5;
Ex 2: Write the four facts of the family with numbers 9, 5 and 14.
Solution:9 + 5 = 14;
5 + 9 = 14;
14 - 9 = 5;
14 - 5 = 9;
When 2 of the numbers are same, then we have only 2 facts. Because, the 1st two and last two fact values are same

Some more examples

Ex 3 :Write the four facts of the family with numbers 11, 5 and 16.
Solution:In the problem we find the facts value for addition and subtraction.
11 + 5 = 16
5 + 11 = 16
16 - 11 = 5
16 - 5 = 11
When 2 of the numbers are same, then we have only 2 facts. Because, the 1st two and last two fact values are same
Ex 4: Write the four facts of the family with numbers 2, 7 and 9.
Solution:7 + 2 = 9
2 + 7 = 9
9 - 2 = 7
9 - 7 = 2
When 2 of the numbers are same, then we have only 2 facts. Because, the 1st two and last two fact values are same

Solving Online Vertical Asymptotes

Solving online vertical asymptotes mean we are going to solve the vertical asymptotes of the curve through online. Normally asymptotes mean the distance from the line to the curve tends to zero. Vertical asymptotes mean the distance from the curve to the vertical line which tends to the zero. If a rational functions denominator tends to zero mean we will get the vertical asymptotes equation. We will see some example for solving vertical asymptotes.

Examples for solving online Vertical Asymptote:

Example 1 for solving online vertical asymptotes:
          Find the vertical asymptotes of the given rational function `(x^2 + x + 1) / (x^2 + 3x + 2)`
Solution:
        The given rational function is function `(x^2 + x + 1) / (x^2 + 3x + 2)`
        If we want to find the vertical asymptotes of the given function we have to equal the given rational functions denominator to zero. So we get
                                  x² + 3x + 2 = 0
                                  x² + 2 x + x + 2 = 0
                                  x (x + 2) + (x + 2) = 0
                                  (x + 2) (x + 1) = 0
                                   x + 2 = 0 and x x + 1 = 0
                                   x = -2 and x = -1
            So the vertical line equations of the given rational functions are x = -2 and x = -1. The solutions for the rational functions are except -2 and -1.
And the vertical asymptotes of the rational function are -2 and -1.

Between, if you have problem on these topics sample papers for class 9 cbse, please browse expert math related websites for more help on sample papers for class 9 cbse 2011.

More examples for solving online vertical asymptotes:

Example 2 for solving online vertical asymptotes:
                     Find the vertical asymptotes of the given rational function `(x^2 + 9x + 5) / (x^2 + 2)`
Solution:
        The given rational function is function `(x^2 + 9x + 5) / (x^2 + 2)`
        If we want to find the vertical asymptotes of the given function we have to equal the given rational functions denominator to zero. So we get
                       x² + 2 = 0
                       x²  = -2
                 So there is no solution for x. It means the denominator has no zeros and there is no vertical asymptote for this function.
       If the x value is imaginary there are no vertical asymptotes for this function.

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.