LECTURE NOTES

TOPICS LECTURE NOTE 1INTRODUCTION:

Numerical Analysis isthe study of algorithms that use numerical approximation for the problems of mathematical analysis .The field of numerical analysis predates the invention of modern computers by many centuries.Linear application was already in use more than 2000 years ago.Many great mathematicians of the past were preocupied by numerical analysis,as is obvious from the names of important algorithms like newton’s method,Lagrange interpolation polynomial,Gauss elimination,or Euler’s method.Direct methods compute the solution to a problem in finite number of steps.These methods would give the precise answer if they were performed in infinite precision arithmetic.Example include Gaussian elimination.In contrast to direct methods,iterative methods are not expected to terminate in a finite number of steps.Starting from an initial guess iterative methods form successive approximations that converge tothe exact solution only in the limit.A convergence test,ofteninvolving the residual is specified in order to decide when a sufficiently accurate solution has been found.Even using infinite precision arithmetic these methods would not reach the solution within a finite number of steps .Examples include Newton’s method,bisection method,Jacobi iteration.In computational matrix algebra,iterative methods are generally needed for large problems.

Numerical methods are methods for solving problems numerically on a computer or calculator. In the decimal notation ,every real number is represented by

a finite or infinite sequence of decimal digits.For machine

computation the number must be replaced by a no.of finitely many

digits.

LECTURE NOTE 2

Most digital computers have two ways of representing

numbers,called fixed point and floating point.

In a fixed pont system all numbers are given with a fixed no of

decimal places.ex 56.358,0.008,2.000etc.

In a floating point system the no of significant digits is kept

fixed but the decimal point is floating.ex 0.62398x104 ,0.2417x10-12

Significant digit of a number c is any given digit of c,except possibly

for zeroes to the left of the first nonzero digit that serve only to fix

the position of the decimal point.

Error is the difference between the true value &

approximate value.There are 3 types of errors.

i)Absolute Error: If a is approximate value of a number x then error=x-

a is known as absolute error.

ii)Relative error =absolute error/true value

iii)Percent error=relative errorx100

Rounding off rule:Discard the (k+1)th and all subsequent decimals.

a)If all the number thus discarded is less than half a

unit(5) in the kth place leave the kth decimal unchanged.(rounding

down).

b)If it is greater than half a unit(5) in the kth place

add 1 to kth decimal(rounding up).

c)If it exactly half a unit(5) round off to the nearest

even decimal.

Absolute and Relative Errors:

Absolute Error: Suppose that and denote the true and approximate

values of a datum then the error incurred on approximating by is given by

and the absolute error i.e. magnitude of the error is given by

Relative Error: Relative Error or normalized error in representing a true

datum by an approximate value is defined by

and

Sometimes is defined by

LECTURE NOTE 3

TOPICS LECTURE NOTE 4

Numerical Errors:

Numerical errors arise during computations due to round-off errors and truncation errors.

Round-off Errors:

Round-off error occurs because computers use fixed number of bits and hence fixed number of binary digits to represent numbers. In a numerical computation round-off errors are introduced at every stage of computation. Hence though an individual round-off error due to a given number at a given numerical step may be small but the cumulative effect can be significant.

When the number of bits required for representing a number are less then the number is usually rounded to fit the available number of bits. This is done either by chopping or by symmetric rounding.

Chopping: Rounding a number by chopping amounts to dropping the extra digits. Here the given number is truncated. Suppose that we are using a computer with a fixed word length of four digits. Then the truncated representation of the number will be . The digits will be dropped. Now to evaluate the error due to chopping let us consider the

normalized representation of the given number i.e.

chopping error in representing .

So in general if a number is the true value of a given number and

is the normalized form of the rounded (chopped) number and is the normalized form of the chopping error then

Since , the chopping error

Symmetric Round-off Error :

In the symmetric round-off method the last retained significant digit is rounded up by 1 if the first discarded digit is greater or equal to 5.In other

words, if in is such that then the last digit in is raised by 1

before chopping . For example let

be two given numbers to be rounded to five

digit numbers. The normalized form x and y are and

. On rounding these numbers to five digits we get

and respectively. Now w.r.t here

In either case error .

Truncation Errors:

Often an approximation is used in place of an exact mathematical procedure. For instance consider the Taylor series expansion of say i.e.

Practically we cannot use all of the infinite number of terms in the series for computing the sine of angle x. We usually terminate the process after a certain number of terms. The error that results due to such a termination or truncation is called as 'truncation error'.

TOPICS LECTURE NOTE 5

ITERATION METHOD:

To solve the eq f(x)=0when there is no formula

for the exact soln ,we can use an approximation method ,in particular

an iteration method i.e.a method in which we start from an initial

approx x0,and compute step by step approximations x1,x2,….of an

unknown solution of f(x)=0.

In general ,iteration methods are easy to program because

the computational operations are the same in each step just the data change from step to step.There are various methods to solve by iteration method.

BISECTION METHOD: If a function f(x) is continuous in the closed

interval [a,b] and f(a)and f(b) are of opposite signs i.e f(a)f(b)¿0,then

there exists at least one root of f(x)=0between a&b.Let

f(a)f(b)¿0,h=(a+b)/2 the middle pt of [a,b].Now if f(h)=0then h is a

root of f(x)=0.If f(h)≠0 then either f(a)f(h)¿0 or f(b)f(h)¿0.

If f(a)f(h)¿0 then root lies between (a,h) and if f(b)f(h)¿0

Then root lies between (h,b).Now the interval [a,b] is reduced to (a,h)

or (h,b).In this way the process is repeated until the root is obtained

to the desired accuracy.

LECTURE NOTE 6

EXAMPLE: Find a root of x3-4x-9=0 by Bisection Method correct upto

3 decimal places.

Solution: Now f(1)=-12¿0 ,f(2)=-9¿0 ,f(3)=6¿0

Now f(2)f(3)¿0

Therefore a root lies between 2&3 . We take x0=(2+3)/2=2.5

f(2.5)A root lies between 2.5 &3 X1=2.75

In this way we will proceed to find the root to the desired accuracy.

TOPICS LECTURE NOTE 7

FIXED POINT ITERATION METHOD:

In this method we start with a

number x0 & a continuous function g(x) andcompute=g(x0),x2=g(x1),…..

In general xn+1=g(xn)………….(1)

The numbers x1,x2,x3,………..xn+1 are called iterates and the

function g is called an iteration function.

If the sequence x1,x2,x3,………..xn+1 converges to α,then

α =limn→ ∞

g¿¿n-1)=g(lim

n → ∞x

n-1)=g(α) since g is continuous.Therefore if the sequence { x1,x2,x3,………..xn+1 } is cgt then it

converges to a fixed point of iteration function.

The method of generating successive iterates with the

half of an arbitrary number x0 and a function g is known as fixed point

iteration.

Problem: Find a root of  x4-x-10 = 0                                Consider g1(x) = 10 / (x3-1) and the fixed point iterative scheme xi+1=10 / (xi

3 -1),      i = 0, 1, 2, . . .let the initial guess x0 be 2.0

i 0 1 2 3 4 5 6 7 8

xi 2 1.429 5.214 0.071 -10.004 -9.978E-3 -10 -9.99E-3 -10

So the iterative process with g1 gone into an infinite loop without converging.

Consider another function g2(x) = (x + 10)1/4   and the fixed point iterative scheme 

xi+1= (xi + 10)1/4,     i = 0, 1, 2, . . .

let the initial guess x0 be 1.0, 2.0 and 4.0

i 0 1 2 3 4 5 6

xi 1.0 1.82116 1.85424 1.85553 1.85558 1.85558

xi 2.0 1.861 1.8558 1.85559 1.85558 1.85558

xi 4.0 1.93434 1.85866 1.8557 1.85559 1.85558 1.85558

That is for g2 the iterative process is converging to 1.85558 with any initial guess.

Consider g3(x) =(x+10)1/2/x  and the fixed point iterative scheme 

xi+1=( xi+10)1/2 /xi, i = 0, 1, 2, . . .let the initial guess x0 be 1.8,

i 0 1 2 3 4 5 6 . 98

.

.

xi 1.8 1.90841.8082

51.9003

51.8152

91,8935

51.82129

.

.

.1.8555

TOPICS

LECTURE NOTE 8

NEWTON RAPHSON METHOD:

To find a simple (non repeated) real

root of the equation f(x)=0 we may use N-R Method.

Let x0 be an approx value of the desired root and h be the

small correction to it.Then x1=x0+h is the root of the given eq

f(x0+h)=0

By Taylor’s series expansion

f(x0+h)=f(x0)+h f’(x0)+h2 f’’(x0)/2 +……

Since h is very small neglecting higher orders of h we get

f(x0)+h f’(x0)=0

h=-f(x0)/ f’(x0)

Now x1= x0 - f(x0)/ f’(x0)

Taking successive approximations x2,x3,………..xn+1 we get

Xn+1=xn- f(xn)/ f’(xn

Examples

Square root of a number

Consider the problem of finding the square root of a number. Newton's method is one of many methods of computing square roots.

For example, if one wishes to find the square root of 612, this is equivalent to finding the solution to

The function to use in Newton's method is then,

with derivative,

With an initial guess of 10, the sequence given by Newton's method is

Where the correct digits are underlined. With only a few iterations one can

obtain a solution accurate to many decimal places.

Solution of cos(x) = x3

Consider the problem of finding the positive number x with cos(x) = x3. We can rephrase that as finding the zero of f(x) = cos(x) − x3. We have f'(x) = −sin(x) − 3x2. Since cos(x) ≤ 1 for all x and x3 > 1 for x > 1, we know that our solution lies between 0 and 1. We try a starting value of x0 = 0.5. (Note that a starting value of 0 will lead to an undefined result, showing the importance of using a starting point that is close to the solution.)

The correct digits are underlined in the above example. In particular, x6 is correct to the number of decimal places given. We see that the number of correct digits after the decimal point increases from 2 (for x3) to 5 and 10, illustrating the quadratic convergence.

TOPICS LECTURE NOTE 9

CONVERGENCE OF N-R METHOD:

Expanding f(x) about nthiterate xn in Taylor’s seriesf(x)=f(xn)+(x-xn) f’(xn)+(x- xn)2 f”(⋋)/2 ------------(1) where ⋋ lies between x & xn

[a,b] is the interval containing the root α and the iterates xn,n=0,1,2,..

For x= α eq (1) reduces to

0= f(xn)+( α -xn) f’(xn)+( α - xn)2 f”(⋋)/2

0= f(xn)+en f’(xn)+en2 f”(⋋)/2 (en= α -xn)

en +f(xn)/f’(xn)=- en2 f”(⋋)/2f’(xn) ……….(2)

But en+1 = α -xn+1

= α – [xn- f(xn)/ f’(xn)]

=( α -xn)+ f(xn)/ f’(xn)

=en+ f(xn)/ f’(xn)………………………..(3)From eq 2&3 en+1=- en

2 f”(⋋)/2f’(xn

we get N-R Method converges quqdratically

TOPICS LECTURE NOTE 10

REGULA FALSI METHOD:

In this method it requires to know the

Approximation location of the root in the given interval(a,b).Let at

the pts a=x0 ,b=x1 the values of the function f(x0) and f(x1) are of

opposite signs so that one root must lies between these 2 pts.Now

we replace the curve y=f(x) by the eq of chord joining [x0,f(x0)],[ x1,

f(x1)] as

y- f(x0)=[f (x1) - f(x0)]/(x1-x0)(x-x0)………(1)

The point where the above chord (1) cuts the x-axis is obtained by

putting y=0 in (1).

So from (1) - f(x0)=[f (x1) - f(x0)]/(x1-x0)(x-x0)

[f (x1) - f(x0)] (x-x0)=-f(x0)( x1-x0) X=x0 - f(x0)/ [f (x1) - f(x0)]…………(2) This value is taken as 2nd approximation X2=x0 - f(x0) (x1-x0)/ [f (x1) - f(x0)]

The iteration formula is

Xn+1=xn-1 - f(xn-1)/ [f (xn) - f(xn-1)]

The convergce process in the bisection method is very slow. It depends only on the choice of end points of the interval [a,b]. The function f(x) does not have any role in finding the point c (which is just  the mid-point of a and b). It is used only to decide the next smaller interval [a,c] or [c,b]. A better approximation to c can be obtained by taking the straight line L joining the points (a,f(a)) and (b,f(b)) intersecting the x-axis. To obtain the value of c we can equate the two expressions of the slope m of the line L. 

m =  f(b) - f(a)  =  0 - f(b)   (b-a)  (c-b)

      => (c-b) * (f(b)-f(a)) = -(b-a) * f(b)

   

c = b - f(b) * (b-a) f(b) - f(a)

Now the next smaller interval which brackets the root can be obtained by checking

f(a) * f(b) < 0 then b = c                 > 0 then a = c

                           = 0 then c is the root.

Selecting c by the above expression is called Regula-Falsi method or False position method.

Find the root of    tan(x)-x-1 = 0

   The graph of this equation is given in the figure. 

   Let a = 0.5 and b = 1.5   

Iteration No. a b c f(a) * f(c)

1 0.5 1.5 0.576 0.8836 (+ve)2 0.576 1.5 0.644 0.8274 (+ve)3 0.644 1.5 0.705 0.762 (+ve)4 0.705 1.5 0.76 0.692 (+ve)5 0.76 1.5 0.808 0.616 (+ve)6 0.808 1.5 0.851 0.541 (+ve)

. . . . .33 1.128 1.5 1.129 1.859E-4 (+ve)34 1.129 1.5 1.129 2.947E-6 (+ve)

   So one of the roots of tan(x)-x-1 = 0 is approximately 1.129.

TOPICS

LECTURE NOTE 11

MULLER METHOD:

This method takes a similar approach like secant method but projects a parabola through 3 points.It consists of deriving the coefficients of theparabola that goes through 3 points.These coefficients can then be substituted into the quadratic formula to obtain the point where the parabola intercepts the x-axis i.e the root estimate.The approach is facilitated by writing the parabolic eqn in a convenient form F2(x)=a(x-x2)2+b(x-x2)+c F(x0)=a(x0-x2)2+b(x0-x2)+c F(x1)= a(x1-x2)2+b(x1-x2)+c F(x2)= a(x2-x2)2+b(x2-x2)+c So c=f(x2)To solve the remaining coefficients i.e a,b algebraic manipulation can be used. Let us define a number of differences H0=x1-x0 h= x2-x1

δ0=[f(x1)-f(x0)]/ (x1-x0) δ1=[f(x2)-f(x1)]/ (x2-x1)Now a=¿1-δ0)/(h1+h0), b=ah1+δ1 , c= f(x2)f(x2) To find the root we use the formula

X3=x2+ (-2c)/b±√b2−4 ac

TOPICS LECTURE NOTE 12

LU DECOMPOSITION METHOD:

In this method, the coefficient matrix A of the system of eqns is decomposed or factorized into the product of a lower triangular matrix L and an upper triangular matrix U. We write the matrix A as A=LU Using the matrix multiplication rule to multiply the matrices L& U and comparing the elements of the resulting matrix with those of A we obtain

li1ui1+li2ui2+ …+linuin=aij, where lij=0 for j¿ i & uij=0 for i¿j The system of eqns involve n parameter family of solutions.To produce a unique soln it is convenient to choose either uii=1 or lii=1When we choose (i) lii=1the method is called the Doolittle’s method (ii) uii=1the method is called Crout’s method

Let A be a square matrix. An LU factorization refers to the factorization of A, with proper row and/or column orderings or permutations, into two factors, a lower triangular matrix L and an upper triangular matrix U,

In the lower triangular matrix all elements above the diagonal are zero, in the upper triangular matrix, all the elements below the diagonal are zero. For example, for a 3-by-3 matrix A, its LU decomposition looks like this:

Without a proper ordering or permutations in the matrix, the factorization may fail to materialize. For example, it is easy to verify (by expanding the matrix multiplication) that . If , then at least one of and has to be zero, which implies either L or U is singular. This is impossible if A is nonsingular. This is a procedural problem. It can be removed by simply reordering the rows of A so that the first element of the permuted matrix is nonzero. The same problem in subsequent factorization steps can be removed the same way, see the basic procedure below.

It turns out that a proper permutation in rows (or columns) is sufficient for the LU factorization. The LU factorization with Partial Pivoting refers often to the LU factorization with row permutations only,

where L and U are again lower and upper triangular matrices, and P is a permutation matrix which, when left-multiplied to A, reorders the rows of A. It turns out that all square matrices can be factorized in this form,[3] and the factorization is numerically stable in practice. This makes LUP decomposition a

useful technique in practice.

An LU factorization with full pivoting involves both row and column permutations,

where L, U and P are defined as before, and Q is a permutation matrix that reorders the columns of A.

An LDU decomposition is a decomposition of the form

where D is a diagonal matrix and L and U are unit triangular matrices, meaning that all the entries on the diagonals of L and U are one.

Above we required that A be a square matrix, but these decompositions can all be generalized to rectangular matrices as well. In that case, L and D are square matrices both of which have the same number of rows as A, and U has exactly the same dimensions as A. Upper triangular should be interpreted as having only zero entries below the main diagonal, which starts at the upper left corner.

Example

We factorize the following 2-by-2 matrix:

One way to find the LU decomposition of this simple matrix would be to simply solve the linear equations by inspection. Expanding the matrix multiplication gives

This system of equations is underdetermined. In this case any two non-zero elements of L and U matrices are parameters of the solution and can be set

arbitrarily to any non-zero value. Therefore to find the unique LU decomposition, it is necessary to put some restriction on L and U matrices. For example, we can conveniently require the lower triangular matrix L to be a unit triangular matrix (i.e. set all the entries of its main diagonal to ones). Then the system of equations has the following solution:

Substituting these values into the LU decomposition above yields

TOPICS LECTURE NOTE 13

INTERPOLATION:

In the mathematical field of numerical analysis, interpolation is a method of constructing new data points within the range of a discrete set of known data points.

In engineering and science, one often has a number of data points, obtained by sampling or experimentation, which represent the values of a function for a limited number of values of the independent variable. It is often required to interpolate (i.e. estimate) the value of that function for an intermediate value of the independent variable. This may be achieved by curve fitting or regression analysis.

A different problem which is closely related to interpolation is the approximation of a complicated function by a simple function. Suppose the formula for some given function is known, but too complex to evaluate efficiently. A few known data points from the original function can be used to create an interpolation based on a simpler function. Of course, when a simple function is used to estimate data points from the original, interpolation errors are usually present; however, depending on the problem domain and the interpolation method used, the gain in simplicity may be of greater value than the resultant loss in accuracy

LECTURE NOTE 14

LAGRANGE INTERPOLATION:

To find a polynomial Pn(x) satisfying the condition Pn(xi)=f(xi) where Pn is interpolating polynomial Let P(x)=a1x+a0 (where a0, a1are constants )Satisfying the interpolating conditions F(x0)=P(x0)= a1x0+a0

F(x1)=P(x1)= a1x1+a0

Elliminating a0, a1 from the above eqns we getP(x)=(x-x1)f(x0)/( x0-x1)+ (x-x0)f(x1)/( x1-x0)

=L0(x)f0+L1(x)f1

Where L0(x), L1(x) are called Lagrange fundamental polynomials satisfying the conditions Li(xj)={1for i=j,0 for i≠ jTherefore the linear Lagrange polynomial is P1(x)= L0(x)f0+L1(x)f1

Quadratic interpolationis given by P2(x)= L0(x)f0+L1(x)f1+ L2(x)f2

In general Pn(x)= L0(x)f0+L1(x)f1+ L2(x)f2+… Ln-1(x)fn-1

Compute f(0.3) for the data

x 0 1 3 4 7f 1 3 49 129 813

using Lagrange's interpolation formula (Analytic value is 1.831)

 

  (x - x1) (x - x2)(x- x3)(x - x4)   (x - x0)(x - x1) (x - x2)(x - x3)  f(x)  = 

f0+ . . . +   f4

  (x0 - x1) (x0 - x2)(x0 - x3)(x0 - x4)

  (x4 - x0)(x4 - x1)(x4 - x2)(x4 - x3) 

 

 

  (0.3 - 1)(0.3 - 3)(0.3 - 4)(0.3 - 7)   (0.3 - 0)(0.3 - 3)(0.3 - 4)(0.3 -

7)  

       =   1+   3

+   (-1) (-3)(-4)(-7)   1 x (-2)(-3)(-6)   

   (0.3 - 0)(0.3 - 1)(0.3 - 4)(0.3 -

7)   (0.3 - 0)(0.3 - 1)(0.3 - 3)(0.3 - 7)  

 49 + 

129 + 

3 x 2 x (-1)(-4)   4 x 3 x 1 (-3)  

   (0.3 - 0)(0.3 - 1)(0.3 - 3)(0.3 -

4)  

8137 x 6 x 4 x 3  

         = 1.831

LECTURE NOTE 15

NEWTON DIVIDED DIFFERENCE:

Let us assume that the function f(x) is linear then we have

f(xi) - f(xj)

(xi - xj)where xi and xj are any two tabular points, is independent of xi and xj. This ratio is called the first divided difference of f(x) relative to xi and xj and is denoted by f [xi, xj]. That is

f [xi, xj] = 

f(xi) - f(xj)   =  f [xj, xi](xi - xj)

Since the ratio is independent of  xi and xj we can write   f [x0, x] = f [x0, x1]f(x) - f(x0)  

  = 

f [x0, x1]

(x - x0)  f(x) = f(x0) + (x - x0) f [x0, x1]

= 

1

|

f(x0)

 x0

- x

|

f1 - f0 f0x1 - f1x0

 =   x + 

x - x0f(x1)

 x1

- x

x1 - x0 x1 - x0

So if f(x) is approximated with a linear polynomial then the function value at any point x can be calculated by using f(x) P1(x) = f(x0) + (x - x1) f [x0, x1]where f [x0, x1] is the first divided difference of  f  relative to x0 and x1.Similarly if f(x) is a second degree polynomial then the secant slope defined above is not constant but a linear function of x. Hence we have 

f [x1, x2] - f [x0, x1]

x2 - x0

is independent of x0, x1 and x2. This ratio is defined as second divided difference of f relative to x0, x1 and x2. The secind divided difference are denoted as 

f [x1, x2] - f [x0, x1]

f [x0, x1, x2]

  = 

  x2 - x0

Now again since f [x0, x1,x2] is independent of x0, x1 and x2 we havef [x1, x0, x] = f [x0, x1, x2]

f [x0, x] - f [x1, x0]

 

  = 

f [x0, x1, x2]

x - x1  f [x0, x] = f [x0, x1] + (x - x1) f [x0, x1, x2]

f [x] - f [x0]    = 

f [x0, x1] + (x - x1) f [x0, x1, x2]

x - x0  f(x) = f [x0] + (x - x0) f [x0, x1] + (x - x0) (x - x1) f [x0, x1, x2]

LECTURE NOTE 16

NEWTON FORWARD INTERPOLATION:

Consider the equation of the linear interpolation optained in the earlier section : f1 - f0 f0x1 - f1x0

f(x)P1(x) = ax-1b = 

x + 

x1 - x0 x1 - x0

 

=              1   [(x1 - x)f0 + (x-x0)f1](x1 - x0)

  x1 - x x - x0 x - x0

f0

+ 

(f1- f0)   + 

f0

x1 - x0 x1 - x0 x1 - x0

x - x0

= f0 + 

(f1- f0) 

x1 - x0

= f0 + r f0    [ r = (x - x0) / (x1 - x0)  f0 = f1 - f0 ]

since x1 - x0 is the step lenght h, r can be written as (x - x0)/h and will be between (0, 1).

Consider the function value (xi, fi)  i = 0,1,2,--5 then the forward difference table is

  xi   fi fi 2fi 3fi 4fi 5fi

x0   f0              f0 = f1- f0        

x1   f1   2f0 = f1- f0

     

    f1 = f2 - f1   3f0 = 2f1-    

2f0

x2  f2   2f1 = f2 - f1

  4f0 = 3f1- 3f0

 

    f2 = f3 - f2   3f1 = 2f2 - 2f1

  5f0 = 4f1- 4f0

x3  f3   2f2 = f3 - f2

  4f1 = 3f2 - 3f1

 

    f3 = f4 - f3   3f2 = 2f3 - 2f2

   

x4  f4   2f3 = f4 - f3

     

    f4 = f5 - f4        x5  f5          

If  f(x) is known at the following data points xi 0 1 2 3 4fi 1 7 23 55 109

then find f(0.5) and f(1.5) using Newton's forward difference formula.Solution :Forward difference table

xi   fi fi 2fi 3fi 4fi

0   1            6      1   7   10        16   6  2  23   16   0

    32   6  3  55   22        54      4  109    

LECTURE NOTE 17

NEWTON BACKWARD DIFFERENCE TABLE:

As a particular case, lets again consider the linear approximation to f(x)f1 - f0 f0x1 - f1x0

f(x)P1(x) =

x + 

x1 - x0 x1 - x0

  x1 - x x - x0

 f0

+ f1 

x1 - x0 x1 - x0

  x - x1

(

x - x0

)= f1 -  f0+

 - 1 f1 

x0 - x1 x - x0

  x - x1

= f1

- (f1- f0) 

x1 - x0

= f1 + sf1

where s = (x - x1) / (x1 - x0) and f1 is the backward difference of  f  at x1.

Find f(0.15) using Newton backward difference table from the data

x f(x)  f 2f   3f  4f0.1 0.09983            0.09884      

0.2 0.19867    -0.00199        0.09685   -0.00156  

0.3 0.29552   -0.00355   0.00121    0.0939    -0.00035  

0.4 0.38942    -0.0039        0.09      

0.5 0.47943        s  =  ( x - xn ) / h  =  (0.15 - 0.5) / 0.1  =  -3.5

f(0.15)  = fn + sfn +  s(s + 1)   2fn + s(s + 1)(s + 2) 3fn +    s(s + 1)(s + 2)(s + 3) 4f2! 3!

              = .97943 + (-3.5)*.09 +

  (-3.5)(-3.5 + 1)   (-.0039)+ 

(-3.5)(-3.5 + 1)(-3.5 + 2) (-.00035)

+ 2! 3! 

(-3.5)(-3.5 + 1)(-3.5 + 2)(-3.5 + 3) (.00121) 4!=  0.97943 - 0.315 - 0.01706 + 0.000765625 + 0.00033086= 0.14847

LECTURE NOTE 18

GAUSS SIEDAL METHOD:

Example 1: Solving a system of equations by the Gauss-Seidel method

Use the Gauss-Seidel method to solve the system 4x1 + x2 - x3 = 3

2x1 + 7 x2 + x3 = 19

x1 - 3 x2 +12 x3 = 31

<=>

x1 = -1/4 x2 + 1/4 x3 + 3/4

x2 = -2/7 x1 - 1/7 x3 + 19/7

x3 = -1/12 x1 + 1/4 x2 + 31/12

Solution:

We have

=

0 -1/4 1/4

-2/7 0 -1/7

-1/12 1/4 0

x1

x2

x3

+

3/4

= G + 19/7

31/12

The iteration formulas are

<=>

x1(k+1) = -1/4 x2

(k) + 1/4 x3(k) + 3/4

x2(k+1) = -2/7 x1

(k+1) - 1/7 x3(k) + 19/7

x3(k+1) = -1/12 x1

(k+1) + 1/4 x2(k+1) + 31/12

The difference between the Gauss-Seidel method and the Jacobi method is that here we use the coordinates x1

(k),...,xi-1(k) of x(k) already known to compute its ith

coordinate xi(k).

If we start from x1(0) = x2

(0) = x3(0) = 0 and apply the iteration formulas, we obtain

k x1(k) x2

(k) x3(k)

0 0 0 0

1 0,75 2,50 3,15

2 0,91 2,00 3,01

3 1,00 2,00 3,00

4 1,00 2,00 3,00

The exact solution is: x1 = 1, x2 = 2, x3 = 3.

For instance, when k=2, we have x2(2)= 2,00:

x2(2) = -2/7 x1

(2) - 1/7 x3(1) + 19/7 = -2/7 · 0,72 - 1/7 · 3,15 + 19/7 = 2,00

LECTURE NOTE 19

TAYLOR’S SERIES:

Consider the one dimensional initial value problem

y' = f(x, y),   y(x0) = y0

where  f is a function of two variables x and y and (x0 , y0) is a known point on the solution curve.

If the existence of all higher order partial derivatives is assumed for y at x = x0, then by Taylor series the value of y at any neibhouring point x+h can be written as 

y(x0+h) = y(x0) + h y'(x0) + h2 /2 y''(x0) + h3/3! y'''(x0) +  .  .  .  .  .  .

where ' represents the derivative with respect to x.  Since at x0, y0 is known,  y' at x0 can be found by computing f(x0,y0).  Similarly higher derivatives of y at x0 also can be computed by making use of the relation  y'  = f(x,y) 

      y''  = fx + fyy'       y''' = fxx + 2fxyy' + fyy y'2 + fyy''  

and so on.   Then 

y(x0+h) = y(x0) + h f + h2  ( fx + fyy' ) / 2! + h3 ( fxx + 2fxyy' + fyy y'2 + fyy'' ) / 3! + o(h4)

Q Using Taylor series method, find y(0.1) for y' = x - y2 ,  y(0) = 1 correct upto four decimal places.

Given y' = f(x,y) = x - y2

y''   = 1 - 2yy',     y''' = -2yy'' - 2y'2,     yiv  = -2yy''' - 6y'y'',    yv  =  -2yyiv -8y'y''' -6y''2

Since at x = 0, y = 1;

y' = -1,     y'' = 3,     y''' = -8,     yiv  = 34    and    yv  = -186

The forth order Taylor's formula is

y(x) = y(x0) + (x-x0) y'(x0, y0) + (x-x0)2 y''(x0, y0)/2! + (x-x0)3 y'''(x0, y0)/3! +                                                                          (x-x0)4 yiv(x0, y0)/4! + h5yv(x0, y0)/5! +. . .

       = 1 - x + 3 x2/2! - 8 x3/3! + 34 x4/4! - 186 x5/5!                         (since x0 = 0)

       = 1 - x + 3 x2/2 - 4 x3/3 + 17 x4/12 - 31 x5/20  

Now y(0.1) = 1 - (0.1) + 3 (0.1)2/2 - 4 (0.1)3/3 + 17 (0.1)4/12 - 31 (0.1)5/20           = 0.9 + 3 (0.1)2/2 - 4 (0.1)3/3 + 17 (0.1)4/12 - 31 (0.1)5/20           = 0.915  - 4 (0.1)3/3 + 17 (0.1)4/12 - 31 (0.1)5/20           = 0.9137 + 17 (0.1)4/12 - 31 (0.1)5/20           = 0.9138 - 31 (0.1)5/20           = 0.9138

Since the value of the last term does not add upto first four decimal places, the Taylor series formula of order four is sufficient to find y(0.1) accurate upto four decimal places. 

LECTURE NOTE 20

TRAPEZOIDAL RULE:

In numerical analysis, the trapezoidal rule (also known as the trapezoid rule or trapezium rule) is a technique for approximating the definite integral

The trapezoidal rule works by approximating the region under the graph of the

function as a trapezoid and calculating its area. It follows that

For a domain discretized into N equally spaced panels, or N+1 grid points a = x1 < x2 < ... < xN+1 = b, where the grid spacing is h=(b-a)/N, the approximation to the integral becomes

Error analysis

The error of the composite trapezoidal rule is the difference between the value of the integral and the numerical result:

There exists a number ξ between a and b, such that

LECTURE NOTE 21

SIMPSON’S 1/3RD RULE:

In numerical analysis, Simpson's rule is a method for numerical integration, the numerical approximation of definite integrals. Specifically, it is the following approximation:

If the interval of integration is in some sense "small", then Simpson's rule will provide an adequate approximation to the exact integral. By small, what we really mean is that the function being integrated is relatively smooth over the interval . For such a function, a smooth quadratic interpolant like the one used in Simpson's rule will give good results.

However, it is often the case that the function we are trying to integrate is not smooth over the interval. Typically, this means that either the function is highly oscillatory, or it lacks derivatives at certain points. In these cases, Simpson's rule may give very poor results. One common way of handling this problem is by breaking up the interval into a number of small subintervals. Simpson's rule is then applied to each subinterval, with the results being summed to produce an approximation for the integral over the entire interval. This sort of approach is termed the composite Simpson's rule.

Suppose that the interval is split up in subintervals, with an even number. Then, the composite Simpson's rule is given by

where for with ; in particular, and . The above formula can also be written as

The error committed by the composite Simpson's rule is bounded (in absolute value) by

where is the "step length", given by [

Error

The error in approximating an integral by Simpson's rule is

where is some number between and .

LECTURE NOTE 22

GAUSS QUADRATURE FORMULA:

In numerical analysis, a quadrature rule is an approximation of the definite integral of a function, usually stated as a weighted sum of function values at specified points within the domain of integration. (See numerical integration for more on quadrature rules.) An n-point Gaussian quadrature rule, named after Carl Friedrich Gauss, is a quadrature rule constructed to yield an exact result for polynomials of degree 2n − 1 or less by a suitable choice of the points xi and weights wi for i = 1, ..., n. The domain of integration for such a rule is conventionally taken as [−1, 1], so the rule is stated as

Gaussian quadrature as above will only produce accurate results if the function f(x) is well approximated by a polynomial function within the range [−1, 1]. The method is not, for example, suitable for functions with singularities. However, if the integrated function can be written as , where g(x) is approximately polynomial and ω(x) is known, then alternative weights and points that depend on the weighting function ω(x) may give better results, where

LECTURE NOTE 23

EULER METHOD:

In mathematics and computational science, the Euler method is a SN-order numerical procedure for solving ordinary differential equations (ODEs) with a given initial value. It is the most basic explicit method for numerical integration of ordinary differential equations and is the simplest Runge–Kutta method. The Euler method is named after Leonhard Euler,

Formulation of the method

Suppose that we want to approximate the solution of the initial value problem

Choose a value for the size of every step and set . Now, one step of the Euler method from to

The value of is an approximation of the solution to the ODE at time : . The Euler method is explicit, i.e. the solution is an explicit

function of for .

While the Euler method integrates a first-order ODE, any ODE of order N can be represented as a first-order ODE: to treat the equation

,

we introduce auxiliary variables and obtain the equivalent

equation

This is a first-order system in the variable and can be handled by Euler's method or, in fact, by any other scheme for first-order systems.[4]

Example

Given the initial value problem

we would like to use the Euler method to approximate .

Illustration of numerical integration for the equation Blue is the Euler method; green, the midpoint method; red, the exact solution, The step size is h = 1.0.

The Euler method is

so first we must compute . In this simple differential equation, the function is defined by . We have

By doing the above step, we have found the slope of the line that is tangent to the solution curve at the point . Recall that the slope is defined as the change in

divided by the change in , or .

The next step is to multiply the above value by the step size , which we take equal to one here:

Since the step size is the change in , when we multiply the step size and the slope of the tangent, we get a change in value. This value is then added to the initial value to obtain the next value to be used for computations.

The above steps should be repeated to find , and .

Due to the repetitive nature of this algorithm, it can be helpful to organize computations in a chart form, as seen below, to avoid making errors.

0 1 0 1 1 1 2

1 2 1 2 1 2 4

2 4 2 4 1 4 8

3 8 3 8 1 8 16

The conclusion of this computation is that . The exact solution of the differential equation is so

LECTURE NOTE 24

IMPROVED EULER METHOD:

In mathematics and computational science, Heun's method may refer to the improved[1] or modified Euler's method (that is, the explicit trapezoidal rule[2]), or a similar two-stage Runge–Kutta method. It is named after Karl Heun and is a numerical procedure for solving ordinary differential equations (ODEs) with a given initial value. Both variants can be seen as extensions of the Euler method into two-stage second-order Runge–Kutta methods.

The procedure for calculating the numerical solution to the initial value problem via the improved Euler's method is:

by way of Heun's method, is to first calculate the intermediate value and then the final approximation at the next integration point.

where is the step size and

LECTURE NOTE 25

RUNGE KUTTA METHOD:

One member of the family of Runge–Kutta methods is often referred to as "RK4", "classical Runge–Kutta method" or simply as "the Runge–Kutta method".

Let an initial value problem be specified as follows.

Here, y is an unknown function (scalar or vector) of time t which we would like to approximate; we are told that , the rate at which y changes, is a function of t and of y itself. At the initial time the corresponding y-value is . The function f and the data , are given.

Now pick a step-size h>0 and define

for n = 0, 1, 2, 3, . . . , using

[1]

LECTURE NOTE 26

Solve

with step size h = 0.025by using Runge kutta method.

The method proceeds as follows:

The numerical solutions correspond to the underlined values

LECTURE NOTE 27

Matrix inverse:

The general system of linear equations in n variables can be written in matrix form Ax = b, when a vector x is sought which satisfies this equation. We will now use the inverse matrix A-1 to find this vector.

1. The inverse matrix

In Step 11, we observed that if the determinant of A is non-zero, it has an inverse matrix A-1, when the solution of the linear system can be written as x = A-1b, i.e., the solution of the system of linear equations can be obtained by first finding the inverse of the coefficient matrix A, and then forming the product A-1

1. As the 2 x 2 example, consider

and seek the inverse matrix

such that

This is equivalent to solving the two systems

The method proceeds as follows:

1. Form the augmented matrix:

2. Apply elementary row operations to the augmented matrix such that A is transformed into an upper triangular matrix:

3. Solve the two systems:

using back-substitution. Note how the systems have been

constructed, using and columns of . From the first system, 3v1 = -2, 3v1u = -2/3, and 2� u1 + v<SUB.1< SUB> = 1, whence 2u1 = 1 + 2/3, u1 = 5/6. From the second system, 3v2 = 1, v2 = 1/3, and 2u2 + v2 = 0, whence 2u2 = -1/3, u2 = -1/6. Thus the required inverse matrix is:

4. Check: AA-1 should be equal to I. Multiplication yields:

so that A-1 is correct

LECTURE NOTE 28

Definition ExampleAn experiment is a situation involving chance or probability that leads to results called outcomes.

In the problem above, the experiment is spinning the spinner.

An outcome is the result of a single trial of an experiment.

The possible outcomes are landing on yellow, blue, green or red.

An event is one or more outcomes of an experiment.

One event of this experiment is landing on blue.

Probability is the measure of how likely an event is.

The probability of landing on blue is one fourth.

In order to measure probabilities, mathematicians have devised the following formula for finding the probability of an event.

Probability Of An Event

P(A) =  The Number Of Ways Event A Can Occur The total number Of Possible Outcomes

The probability of event A is the number of ways event A can occur divided by the total number of possible outcomes. Let's take a look at a slight modification of the problem from the top of the page.

The probability of event A is the number of ways event A can occur divided by the total number of possible outcomes. Let's take a look at a slight modification of the problem from the top of the page.

Experiment 1:   A spinner has 4 equal sectors colored yellow, blue, green and red. After spinning the spinner, what is the probability of landing on each color?

Outcomes:   The possible outcomes of this experiment are yellow, blue, green, and red.

Probabilities:  P(yellow)  = 

# of ways to land on yellow  = 

1

total # of colors  4 

 

P(blue)  =  # of ways to land on blue

 =  1

total # of colors  4 

 

P(green)  =  # of ways to land on green  =  1

total # of colors  4 

 

P(red)  =  # of ways to land on red

 =  1

total # of colors  4 

Experiment 2:   A single 6-sided die is rolled. What is the probability of each outcome? What is the probability of rolling an even number? of rolling an odd number?

Outcomes:   The possible outcomes of this experiment are 1, 2, 3, 4, 5 and 6.

Probabilities:  

P(1)  =  # of ways to roll a 1

 = 

1

total # of sides 6

 

P(2)  =  # of ways to roll a 2

 = 

1

total # of sides 6

 

P(3)  =  # of ways to roll a 3  =  1

total # of sides 6

 

P(4)  =  # of ways to roll a 4

 = 

1

total # of sides 6

 

P(5)  =  # of ways to roll a 5

 = 

1

total # of sides 6

 

P(6)  =  # of ways to roll a 6

 = 

1

total # of sides 6

 

P(even)

 =  # ways to roll an even number

 = 

3  = 

1

total # of sides 6 2

 

P(odd)  =  # ways to roll an odd number

 = 

3  = 

1

total # of sides 6 2

Experiment 2 illustrates the difference between an outcome and an

event. A single outcome of this experiment is rolling a 1, or rolling a 2, or rolling a 3, etc. Rolling an even number (2, 4 or 6) is an event, and rolling an odd number (1, 3 or 5) is also an event.

In Experiment 1 the probability of each outcome is always the same. The probability of landing on each color of the spinner is always one fourth. In Experiment 2, the probability of rolling each number on the die is always one sixth. In both of these experiments, the outcomes are equally likely to occur. Let's look at an experiment in which the outcomes are not equally likely.

Experiment 3:   A glass jar contains 6 red, 5 green, 8 blue and 3 yellow marbles. If a single marble is chosen at random from the jar, what is the probability of choosing a red marble? a green marble? a blue marble? a yellow marble?

Outcomes:   The possible outcomes of this experiment are red, green, blue and yellow.

Probabilities:   P(red)  = 

# of ways to choose red  = 

  6  

 =    3  

total # of marbles 22 11

 

P(green)  = 

# of ways to choose green  = 

  5  

total # of marbles 22

 

P(blue)  = 

# of ways to choose blue  = 

  8    = 

  4  

total # of marbles 22 11

 

P(yellow)  = 

# of ways to choose yellow  = 

  3  

total # of marbles 22

The outcomes in this experiment are not equally likely to occur. You are more likely to choose a blue marble than any other color. You are least likely to choose a yellow marble.

Experiment 4:   Choose a number at random from 1 to 5. What is the probability of each outcome? What is the probability that the number chosen is even? What is the probability that the number chosen is odd?

Outcomes:   The possible outcomes of this experiment are 1, 2, 3, 4 and 5.

Probabilities:   P(1)  = 

# of ways to choose a 1  = 

1

total # of numbers 5

 

P(2)  =  # of ways to choose a 2  = 

1

total # of numbers 5

 

P(3)  =  # of ways to choose a 3  = 

1

total # of numbers 5

 

P(4)  =  # of ways to choose a 4  = 

1

total # of numbers 5

 

P(5)  =  # of ways to choose a 5  = 

1

total # of numbers 5

 

P(even)  =  # of ways to choose an even number  = 

2

total # of numbers 5

 

P(odd)  =  # of ways to choose an odd number  = 

3

total # of numbers 5

The outcomes 1, 2, 3, 4 and 5 are equally likely to occur as a result of this experiment. However, the events even and odd are not equally likely to occur, since there are 3 odd numbers and only 2 even numbers from 1 to 5.

Summary:   The probability of an event is the measure of the chance that the event will occur as a result of an experiment. The probability of an event A is the number of ways event A can occur divided by the total number of possible outcomes. The probability of an event A, symbolized by P(A), is a number between 0 and 1, inclusive, that measures the likelihood of an event in the following way:

If P(A) > P(B) then event A is more likely to occur than event B.

If P(A) = P(B) then events A and B are equally likely to occur

LECTURE NOTE 29RANDOM VARIABLES:

The outcome of an experiment need not be a number, for example, the outcome when a coin is tossed can be 'heads' or 'tails'. However, we often want to represent outcomes as numbers. A random variable is a function that associates a unique numerical value with every outcome of an experiment. The value of the random variable will vary from trial to trial as the experiment is repeated.

There are two types of random variable - discrete and continuous.

A random variable has either an associated probability distribution (discrete random variable) or probability density function (continuous random variable).

Examples

1. A coin is tossed ten times. The random variable X is the number of tails that are noted. X can only take the values 0, 1, ..., 10, so X is a discrete random variable.

2. A light bulb is burned until it burns out. The random variable Y is its lifetime in hours. Y can take any positive real value, so Y is a continuous random variable.

The expected value (or population mean) of a random variable indicates its average or central value. It is a useful summary value (a number) of the variable's distribution.

Stating the expected value gives a general impression of the behaviour of some random variable without giving full details of its probability distribution (if it is discrete) or its probability density function (if it is continuous).

Two random variables with the same expected value can have very different distributions. There are other useful descriptive measures which affect the shape of the distribution, for example variance.

The expected value of a random variable X is symbolised by E(X) or µ.

If X is a discrete random variable with possible values x1, x2, x3, ..., xn, and p(xi) denotes P(X = xi), then the expected value of X is defined by:

where the elements are summed over all values of the random variable X. If X is a continuous random variable with probability density function f(x), then the expected value of X is defined by:

Example Discrete case : When a die is thrown, each of the possible faces 1, 2, 3, 4, 5, 6 (the xi's) has a probability of 1/6 (the p(xi)'s) of showing. The expected value of the face showing is therefore:

µ = E(X) = (1 x 1/6) + (2 x 1/6) + (3 x 1/6) + (4 x 1/6) + (5 x 1/6) + (6 x 1/6) = 3.5

Notice that, in this case, E(X) is 3.5, which is not a possible value of X

LECTURE NOTE 30

Frequency Distributions

Recall also that in our general notation, we have a data set with n points arranged in a frequency distribution with k classes. The class mark of the i'th class is denoted xi; the frequency of the i'th class is denoted fi and the relative frequency of th i'th class is denoted pi = fi / n.

Mean

The mean of a data set is simply the arithmetic average of the values in the set, obtained by summing the values and dividing by the number of values. Recall that when we summarize a data set in a frequency distribution, we are approximating the data set by "rounding" each value in a given class to the class mark. With this in mind, it is natural to define the mean of a frequency distribution by

The mean is a measure of the center of the distribution. As you can see from the

algebraic formula, the mean is a weighted average of the class marks, with the relative frequencies as the weight factors. We can compare the distribution to a mass distribution, by thinking of the class marks as point masses on a wire (the x-axis) and the relative frequencies as the masses of these points. In this analogy, the mean is literally the center of mass--the balance point of the wire.

Recall also that we can think of the relative frequency distribution as the probability distribution of a random variable X that gives the mark of the class containing a randomly chosen value from the data set. With this interpretation, the mean of the frequency distribution is the same as the mean (or expected value) of X.

Variance and Standard Deviation

The variance of a data set is the arithmetic average of the squared differences between the values and the mean. Again, when we summarize a data set in a frequency distribution, we are approximating the data set by "rounding" each value in a given class to the class mark. Thus, the variance of a frequency distribution is given by

The standard deviation is the square root of the variance:

LECTURE NOTE 31

In probability theory and statistics, the binomial distribution with parameters n and p is the discrete probability distribution of the number of successes in a sequence of n independent yes/no experiments, each of which yields success with probability p. A success/failure experiment is also called a Bernoulli experiment or Bernoulli trial; when n = 1, the binomial distribution is a Bernoulli distribution. The binomial distribution is the basis for the popular binomial test of statistical significance.

The binomial distribution is frequently used to model the number of successes in

a sample of size n drawn with replacement from a population of size N. If the sampling is carried out without replacement, the draws are not independent and so the resulting distribution is a hypergeometric distribution, not a binomial one. However, for N much larger than n, the binomial distribution is a good approximation, and widely used.

Probability mass function

In general, if the random variable X follows the binomial distribution with parameters n and p, we write X ~ B(n, p). The probability of getting exactly k successes in n trials is given by the probability mass function:

for k = 0, 1, 2, ..., n, where

is the binomial coefficient, hence the name of the distribution. The formula can be understood as follows: we want exactly k successes (pk) and n − k failures (1 − p)n − k. However, the k successes can occur anywhere among the n trials, and

there are different ways of distributing k successes in a sequence of n trials.

LECTURE NOTE 32

Definition

A discrete random variable X  is said to have a Poisson distribution with parameter λ > 0, if, for k = 0, 1, 2, …, the probability mass function of X  is given by:[6]

where

e is Euler's number (e = 2.71828...) k! is the factorial of k.

The positive real number λ is equal to the expected value of X and also to its variance

The Poisson distribution can be applied to systems with a large number of possible events, each of which is rare. How many such events will occur during a fixed time interval? Under the right circumstances, this is a random number with a Poisson distribution

LECTURE NOTE 33

The hypergeometric distribution applies to sampling without replacement from a finite population whose elements can be classified into two mutually exclusive categories like Pass/Fail, Female/Male or Employed/Unemployed. As random selections are made from the population, each subsequent draw decreases the population causing the probability of success to change with each draw.

The following conditions characterize the hypergeometric distribution:

The result of each draw can be classified into one of two categories. The probability of a success changes on each draw.

A random variable follows the hypergeometric distribution if its probability mass function (pmf) is given by:[1]

Where:

is the population size is the number of success states in the population

is the number of draws is the number of successes

is a binomial coefficient

LECTURE NOTE 34

The simplest case of a normal distribution is known as the standard normal distribution. This is a special case where μ=0 and σ=1, and it is described by this probability density function:

The factor in this expression ensures that the total area under the curve ϕ(x) is equal to one.[6] The 1/2 in the exponent ensures that the distribution has unit variance (and therefore also unit standard deviation). This function is symmetric around x=0, where it attains its maximum value ; and has inflection points at +1 and −1.

Authors may differ also on which normal distribution should be called the "standard" one. Gauss himself defined the standard normal as having variance σ2 = 1/2, that is

Stigler [7] goes even further, defining the standard normal with variance σ2 = 1/2π :

LECTURE NOTE 35

DISTRIBUTION OF SEVERAL RANDOM VARIABLES:

DISCRETE 2 DIMENSIONAL DISTRIBUTION:

CONTINUOUS 2 DIMENSIONAL DISTRIBUTION:

LECTURE NOTE 36

RANDOM SAMPLING:

In statistics, a simple random sample is a subset of individuals (a sample) chosen from a larger set (a population). Each individual is chosen randomly and entirely by chance, such that each individual has the same probability of being chosen at any stage during the sampling process, and each subset of k individuals has the same probability of being chosen for the sample as any other subset of k individuals.[1] This process and technique is known as simple random sampling, and should not be confused with systematic random sampling. A simple random sample is an unbiased surveying technique.

The sample mean is given by

Where n is sample size

The sample variance is given by

LECTURE NOTE 37Estimation of parameters:

MAXIMUM LIKELYHOD METHO DConsider a discrete(or continuous) random variable X whose probability functionf(x)depends on θ,let us consider n independent sample values

A solution of above equation

so

LECTURE NOTE 38

CONFIDENCE INTERVALS

LECTURE NOTE 39

TESTING OF HYPOTHESIS:

Steps for testing

LECTURE NOTE 40

ACCEPTANCE SAMPLING:

LECTURE NOTE 41

CHI-SQUARE METHOD:

STEPS FOR CHI-SQUARE TEST

LECTURE NOTE 42

REGRESSION ANALYSIS:

Steps to find confidence interval for regression coefficient

LECTURE NOTE 43

COORELATION ANALYSIS:

The coorelation coefficient is given by

Steps for test of coorelation coefficient

LECTURE NOTE 44

UNIVERSITY QUESTIONS

LECTURE NOTE 45

UNIVERSITY QUESTIONS