Friday, May 21, 2010

Binomial Coefficient

Let Us Understand What Is Binomial Coefficient.

the binomial coefficient  \tbinom nk is the coefficient of the x k term in the polynomial expansion of the binomial power (1 + x) n.

A binomial coefficient equals the number of combination of r items that can be selected from a set of n items. It also represents an entry in Pascal's triangle. These numbers are called binomial coefficients because they are coefficients in the binomial theorem.

Example:

Binomial Theorem

Let Us Learn What Is Binomial.


The sum of two monomial is called Binomial.

Binomial Theorem for Positive Integral Indices

Let us have a look at the following identities done earlier:

(a+ b)0 = 1 a + b ≠ 0
(a+ b)1 = a + b
(a+ b)2 = a2 + 2ab + b2
(a+ b)3 = a3 + 3a2b + 3ab2 + b3
(a+ b)4 = (a + b)3 (a + b) = a4 + 4a3b + 6a2b2 + 4ab3 + b4

In these expansions, we observe that

(i) The total number of terms in the expansion is one more than the index. For
example, in the expansion of (a + b)2 , number of terms is 3 whereas the index of
(a + b)2 is 2.

(ii) Powers of the first quantity ‘a’ go on decreasing by 1 whereas the powers of the
second quantity ‘b’ increase by 1, in the successive terms.

(iii) In each term of the expansion, the sum of the indices of a and b is the same and
is equal to the index of a + b.

Examples Using Binomial Theorem


Thursday, May 20, 2010

Factorial Notation

Let Us Learn About Factorial Notation.


Question:What is Factorial Notation?
Answer: Let n be a positive integer. The continued product of first n natural numbers is called factorial n and is denoted as n

The notation n! represents the product of first n natural
numbers
, i.e., the product 1 × 2 × 3 × . . . × (n – 1) × n is denoted as n!. We read this
symbol as ‘n factorial’. Thus, 1 × 2 × 3 × 4 . . . × (n – 1) × n = n !
1 = 1 !
1 × 2 = 2 !
1× 2 × 3 = 3 !
1 × 2 × 3 × 4 = 4 ! and so on.

We define 0 ! = 1
We can write 5 ! = 5 × 4 ! = 5 × 4 × 3 ! = 5 × 4 × 3 × 2 !
= 5 × 4 × 3 × 2 × 1!

Clearly, for a natural number n
n ! = n (n – 1) !
= n (n – 1) (n – 2) ! [provided (n ≥ 2)]
= n (n – 1) (n – 2) (n – 3) ! [provided (n ≥ 3)]
and so on.

Permutations when all the objects are distinct

Topic: Permutations are solved when all the objects are distinct

Question: How to Permutations are solved when all the objects are distinct?
Answer:
The number of permutations of n different objects taken r at a time,
where 0 < r ≤ n and the objects do not repeat is n ( n – 1) ( n – 2). . .( n – r + 1), which is denoted by nPr. Proof There will be as many permutations as there are ways of filling in r vacant

places . . . by ← r vacant places →

the n objects. The first place can be filled in n ways; following which, the second place
can be filled in (n – 1) ways, following which the third place can be filled in (n – 2)
ways,..., the rth place can be filled in (n – (r – 1)) ways. Therefore, the number of
ways of filling in r vacant places in succession is n(n – 1) (n – 2) . . . (n – (r – 1)) or
n ( n – 1) (n – 2) ... (n – r + 1)

This expression for nPr is cumbersome and we need a notation which will help to
reduce the size of this expression. The symbol n! (read as factorial n or n factorial )
comes to our rescue. In the following text we will learn what actually n! means.

Thursday, February 4, 2010

How to Solve Simultaneous Linear Equation Graphically

The graph of a linear equation ax+ by = c is a straight line equation.

Two distinct lines always intersect at exactly one point unless they are parallel (have the same slope).

The coordinates of the intersection point of the lines is the solution to the simultaneous linear equations describing the lines. So we would normally expect a pair of simultaneous equations to have just one solution.

Let's look at an example graphically:
2x + 3y = 7
4x + y = 9



From the graph we see that the point of intersection of the two lines is (2, 1)
Hence, the solution of the simultaneous equations is x = 2, y =1.

Tuesday, December 1, 2009

Example for complex conjugates

In grade 10 math , complex conjugates are a pair of complex numbers, both having the same real part, but with imaginary parts of opposite signs. For example, 3 + 4i and 3 - 4i are complex conjugates.
The conjugate of the complex number z
 z=a+ib, \,
where a and b are real numbers, is
\overline{z} = a - ib.\,
An alternate notation for the complex conjugate is z * . However, the \bar z notation avoids confusion with the notation for the conjugate transpose of a matrix, which can be thought of as a generalization of complex conjugation. If a complex number is represented as a 2×2 matrix, the notations are identical.We also can use scientific notation converter to find the number.

Let's an example problem from numeric and algebraic operations


Question:-

How do you use complex conjugates to find (3+7i)/(2-i) ?


Answer:-


In the Given problem the denominator is (2-i)

So ,it's conjugate is (2+i)

Multiplying both numerator and denominator with (2+i)

= (3+7i)(2+i)/(2-i)(2+i)

= (5+17i)/(4-i2)

We know that i2 = -1

= (5+17i)/4+1

= (5+17i)/5 Answer

Wednesday, September 16, 2009

Problem on Order of Operations

In mathematics and computer programming, an expression or string of symbols is intended to represent a numerical value; a properly-formed expression may be evaluated in an unambiguous way. But in practice, an expression with multiple terms and operators may be written without parentheses, in which case the intended value of the expression is determined by convention. When a term in the expression is both preceded and followed by an operator such as minus or times, a convention is needed to clarify which operator should be applied first; this rule is known as a precedence rule, or more informally order of operation. From the earliest use of mathematical notation[citation needed], multiplication took precedence over addition, whichever side of a number it appeared on. Thus 3 + 4 × 5 = 5 × 4 + 3 = 23. When numeric and algebraic operations were first introduced, in the 16th and 17th centuries, exponents took precedence over both addition and multiplication, and could be placed only as a superscript to the right of their base. Thus 3 + 5 2 = 28 and 3 × 5 2 = 75. To change the order of operations calculator , a vinculum (an over line or underline) was originally used. Today one uses parentheses (). Thus, if one wants to force addition to precede multiplication, one writes (3 + 4) × 5 = 35.

Let's see an example from algebra answers

Question:-

solve 24-(24+4+2)+2 x (4 x 2)

by using PEMDAS rule

Answer:-

In PEMDAS

P - Parentheses

E - Exponents

M - Multiplication

D - Division

A - Addition

S - Subtraction

let's solve the given equation step by step
in the same order ...

24-(30)+2 x (16)

24-32+30

combine the same terms

24+32-30

56 - 30

26 is the answer.