THE KENYA METHODIST UNIVERSITY

P.O. Box 267MERU

MATHS 320: NUMERICAL ANALYSIS 1

Course Instructor: Winifred MutukuDepartment of Pure and Applied Sciences.

MATHS 320: NUMERICAL ANALYSIS 1

INTRODUCTION

This course will introduce the students to the general concepts and skills of mathematics

essential for application in problem solving and computational disciplines.

COURSE OBJECTIVES

During the course, the undergraduate student will be able to:

- Understand the algebra of the number systems, introduction to errors, sources of

errors and propagation of errors.

- Use Taylor’s series and Lagrange’s Interpolation to reconstruct the function f(x)

and determine the root of a function f(x) = 0 by bisection method, Secant,

Newton-Raphson and False-position methods.

- Interpolation based on finite difference operators; forward, backward and central

and form their respective difference tables.

- Understand the concept of Lagrange and Newton interpolation, find an

approximate value by Lagrange Interpolation; find unique polynomial by

Lagrange interpolation and Newton divided difference.

- State and use the least squares and Chebychev’s approximations

- Know the various numerical methods used to find derivative of a function f(x)

like methods based in interpolation, finite differences and undetermined

coefficient. Extrapolation to the limit.

1

COURSE CONTENT

- Number System.

- Introduction to errors

- Lagrange interpolation

- Finite difference

- Interpolation based on finite differences.

- Least square approximations

- Chebychev’s approximations

- Numerical differentiation.

- Extrapolation to the limit.

2

TEACHING AND LEARNING METHOD’S

Lecture method consisting of stating, illustrating, proving various numerical

methods. Application of these methods are given by use of examples.

Questions and answer method.

Tutorials in which problem solving techniques and general discussion are

conducted regularly.

Tests and assignments are continuously given to test students progress and

understanding and discussed together after grading.

STUDENTS ASSESSMENT

Assignment

Course Examination

Individual student’s marks of these assignments will contribute to the final examination

mark and grade for the course.

3

COURSE OUTLINE FOR THE TRIMESTER

WEEK 1 and 2

Number Systems

WEEK 3

Approximations and errors

WEEK 4

Lagrange interpolation

WEEK 5 and 6

Finite difference

Interpolation based on finite differences.

WEEK 7

Least square approximations

WEEK 8

Chebychev’s approximations

WEEK 9 and 10

Numerical differentiation.

WEEK 11 and 12

4

Extrapolation to the limit.

WEEK 13 and 14

Assignments

Course Examinations

LEARNING RESOURCES

Hilderbrand, F.B: Introduction to Numerical Analysis; 2nd Edition, McGraw-Hill, Inc., 1991.

Hultquist, P.F.: Numerical Methods for Engineers and Scientists; Benjamin/Cummings Publishing Company, Inc., 1988

Germund, D.; Numerical Methods; Prentice-Hall, Inc., 1974

5

Students Learning Objectives

WEEK 1 and 2

1) State the various number systems

2) Perform the basic operation of the different number systems

3) Convert between different number system

4) Represent data using floating point and fixed point methods.

5) Evaluate the complement of denary and binary number systems and use them to

perform subtraction of numbers.

Introduction

Numerical analysis or computing is an approach for solving complex mathematical

problems using only simple arithmetic operations. It involves formulation of

mathematical models of physical situations that can be solved with arithmetic operations.

It can be used in the following areas.

i. Finding roots of equations

ii. Solving systems of linear algebraic equations.

iii. Interpolation and regression analysis.

iv. Numerical integration.

v. Numerical differentiation.

vi. Solution of differential equations

vii. Boundary value problems

viii. Solution of matrix problems.

6

NUMBER SYSTEMS

There are different number systems depending on the base/radix. The number systems are

1. Decimal Number System

2. Binary Number System

3. Octal Number System

4. Hexadecimal Number System

5. Duodecimal Number System

Decimal Number System

This is also known as denary number system. It has 10 digits/symbols; 0, 1, 2, ……., 9.

The digits are 10, hence its base is 10. Every number in the system is interpreted as a sum

of multiples of integer powers of 10.

For example,

1. = 4 x 103 + 9 x 102 + 8 x 101+ 7 x 101

2.

The expanded notation for the integer part of a decimal number is

Where is the decimal digits 0, 1, 2, ……, 9 and m is the number of digits.

The fractional part of a decimal number is represented as

, where n is the number of digits in the fractional part.

7

BINARY NUMBER SYSTEM

The binary number system has a base of 2 with digits 0 and 1. These digits/symbols are

known as bits. It is represented as a sum of multiples of integer powers of 2.

For example,

i.

In general, the integer part of a binary number can be written as

, where

are the binary symbols/digits 0 and 1.

This can be generalized to any base b as;

, where b is an integer greater than 1 and digits are between 0

and b-1.

Note: the base is also called radix and the fractional point is called the radix point.

Exercise

Convert to denary number system.

HEXADECIMAL NUMBER SYSTEM

The base to the hexadecimal number system is 16 and the digits/symbols used are 0, 1, 2,

……..9 and A, B, C, ….., F. The letters A to F represent the digits 10, 11,…. To 15

respectively. The hexadecimal number system is a sum of multiples of integer powers of

16.

For example,

8

Exercise

1. Convert the following hexadecimal numbers to Denary number system

a.

b.

Converting Binary Number to Hexadecimal

i. Group the binary digits in sets of four

ii. Convert each group to its equivalent.

Example

Convert to hexadecimal number system

Solution

0111 1010 0001 0010 0001

0111 =

Similarly,

1010 = 10 = A

0001 = 1

0010 = 2

0001 = 1

Thus

9

OCTAL NUMBER SYSTEM

The octal number system has 8 digits and hence is of base 8. The digits /symbols are 0, 1,

2, ……, 7. Any number is a sum of multiples of integer powers of 8.

Example

Question

Convert the following Octal numbers into their denary equivalent

a)

b)

Converting Binary Number to Octal

a) Group the binary digits into sets of 3

b) Convert each set to its denary equivalent.

Example

Convert to octal number system

Solution

001 011 010

001 =

Similarly,

011 = 3

010 = 2

Thus

10

DUODECIMAL NUMBER SYSTEM

The duodecimal number system has 12 symbols/digits.The digits are 0, 1, 2,….., 9 and X

and . The last 2 symbols represent 10 and 11 respectively. The duodecimal number

system is a sum of multiples of integer powers of 12.

Example

Question

Convert to its denary equivalent.

Exercise

Express the following numbers in denary number system.

a)

b)

c)

d)

CONVERSION OF NUMBERS

Non-Decimal System to Decimal Systems

The following algorithm is used to convert numbers in base2, base8, base 12 and base16

to a decimal number.

Integer Part

11

1. Multiply the left most digit by the base b.

2. Add the next digit to the right to the product.

3. Multiply the sum by the base and add the next digit.

4. Continue the process until the last (right most) digit is added.

The sum is the decimal equivalent of the given integer number.

Fractional Part.

1. Multiply the right most digit by b-1.

2. Add the next digit to the left to the product.

3. Multiply the sum by b-1 and add the next digit.

4. Continue the process until the last (left most) digit in the fractional part is added.

The product is the decimal equivalent of the given fractional number.

Example.

1. Convert to its decimal equivalent.

Solution

Integer Part

1 1 0 1

1x2=2

2+1=3x2

6+0= 6x2

12+1 = 13

Decimal part

1 1 0 1

12

1x2-1=05

0.5+0 = 0.5x2-1=0.25

0.25+1=1.25x2-1=0.625

0.625+1=1.625x2-1=0.8125

Thus

2. Convert to its denary equivalent.

Solution

1 2 A F

1 x 16 = 16

16 + 2 = 18 x 16 = 288

288 + 10 = 298 x 16 = 4768

4768 + 15 = 4783

Thus

Exercise

1. Convert the following numbers to denary number system.

a.

b.

c.

d.

13

Denary System to Non-Denary Systems.

The following algorithm is used to convert numbers in denary number system to base2,

base8, base 12 and base16.

Integer Part

1. Divide the integer part by the base b of the new system. The remainder will

constitute the right digit of the integer part of the new number.

2. Divide the quotient again by the base b. the remainder is the 2nd digit from the

right.

3. Continue the process until zero quotients is obtained. The last remainder is the left

most digit of the new number.

Example 1

Change to binary.

Thus

Solution

2 2452 12212 6102 30 12 1502 712 312 112 01

14

Example 2

Convert to octal.

Solution

8 524

8 65 4

8 81

8 10

8 01

Thus

Question

Convert to duodecimal.

Fractional part

1. Multiply the fractional part of the decimal number by the base b of the new

system. The integral part of the product constitutes the left most digit of the

fractional part of the new number.

2. Multiply the fractional part of the product by the base. The integral part of the

resultant product is the 2nd digit from the left.

3. Continue the process until a zero fractional part occurs. The integer part of the last

product will be the right most digit of the fractional part of the new number.

15

Example

1) Change octal.

Solution

0.526 x 8 = 4.208

0.208 x 8 = 1.664

0.664 x 8 = 5.312

0.312 x 8 = 2.496 and so on.

Thus the number

Change to binary.

Solution

Integer part

2 43

2 21 →1

2 10 →1

2 5 → 0

2 2 → 1

2 1 → 0

2 0 → 1

Decimal Part

0.375 x 2 = 0.75

0.75 x 2 = 1.50

16

0.50 x 2 = 1.00

Thus

Questions

1) Convert the following numbers to the stated number system.

a) to duodecimal

b) to octal

c) to hexadecimal

d) to binary

Octal and Hexadecimal Conversions

Using the binary system as an intermediate stage, we can convert octal numbers to

hexadecimal numbers and vice versa.

Octal to Hexadecimal

1) Evaluate the binary equivalent of each octal number.

2) Regroup them into quadruplets and convert each group to its hexadecimal equivalent.

Hexadecimal to Octal

1) Evaluate the binary equivalent of each hexadecimal number.

2) Regroup them into triplets and convert to its octal equivalent.

Example

1) Convert to hexadecimal

Solution

2 2

17

2 1→ 0

2 0 → 1

2 0 → 0

2 4

2 2 →0

2 1 →0

2 0 → 1

2 3

2 1 → 1

2 0 → 1

2 0 → 0

Thus

Regrouping into 4’s we have 0000 1010 0011

2) Convert to Octal

Solution

Converting each digit into binary gives

18

Regrouping into 3’s

00 111 001.101 110 00

Questions

1. Convert

a.

b.

c. to its octal, binary and hexadecimal forms

2. Express

a.

b. to its binary, octal and denary forms.

REPRESENTING NUMBERS

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Computers use binary digits to represent numbers and other information. The memory is

organized into strings of bits called words. Each string has the same length in a particular

computer, although different computers may use different word lengths. The largest

number a computer can store depends on its word length. For example the largest binary

number a 16bit word can hold is 16bits of 1. This binary number is equivalent to a

decimal value of 65535.

The largest decimal number that can be stored in a computer is given by the relation

, where n is the word length in bits. Decimal numbers are first

converted into the binary equivalent and then represented in either fixed point form or

floating point form.

Fixed Point Form

For integers, the decimal or binary point is always fixed to the right of the least

significant digits, hence fractions are not included. Negative numbers are stored using the

2’s complement.

Example.

Represent -13 in binary form

Solution

, the extra zero to the left of the binary number is to indicate that

the number is negative.

Note: If the left most digit is 1, the number is positive.

Converting the number into 1’s complement.

Inverting each digit and adding 1 to the last bit we have

20

10010

+ 1

_____________

10011

The left most bit of a binary number which is used to indicate the sign is called the sign

bit. Thus we have n – 0 bits to represent the number. Thus a 16 bit word can contain

numbers -215 to 215 -1 (-32768 to 32767).

Floating Point Representation

Large numbers and fractional numbers are stored and processed in exponential form.

These numbers have an embedded decimal point and are called floating point numbers or

real numbers. A floating point number is characterized by 4 parameters, the base, the

number of digits t and the exponent range (m, M).

Definition1

A floating point number is a number represented in the form

………………………………………….(1)

Where are integers and satisfy .

The exponent e is such that . The fractional part is called the

Mantissa and it lies between +1 and -1.

The following numbers have been expressed in floating point form

1) 0 =

21

2)

3)

For floating point numbers the memory location is divided into 3 fields(part) as shown

below.

Typically, floating numbers use a field width of 32bits where 24 bits are used for the

mantissa, 7bits for the exponent.

Exercise

1) Convert the following numbers to floating numbers

a) 0.00596

b) 65.7451

c) -486.8

Normalization

The shifting of the decimal point to the left of the most significant digit is called

Normalization and the numbers represented in normalized form are called Normalized

Floating Point numbers.

S

I G

N

E X P O N E N T

MA

NT

ISS

A

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Example: Given 0.00011101 x 107, shifting the fractional part 3 places to the left results

in the number 0.11101 x 104 which is the normalized floating point number.

Floating Point Arithmetic

Addition

If x + y = z, let the fractional parts and the exponents be fx, fy & fz and Ex, Ey and Ez,

then the addition algorithm is

i. Set Ez = the largest of Ex and Ey. Suppose Ex Ey, then Ez = Ex.

ii. Shift Fy to the right by Ex – Ey places to make the exponent of Fx and Fy the

same.

iii. Set Fz = Fx + Fy.

iv. Normalize the sum and express as .

Example

Add 0964572E2 and 0.586351E5.

Solution

Question

Add 0.735816E4 and 0.635742E4

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Subtraction

Subtract 0.994576E-3 from 0.999658E-3

Solution

Multiplication

Algorithm

i. Multiply the fractional part:

ii. Add the exponents:

iii. Normalize if necessary.

Example

Multiply out 0.200000E4 and 0.400000E-2

Solution

Division

Algorithm

iv. Divide the fractional part:

v. Subtract the exponents:

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vi. Normalize if necessary.

Example

Divide 0.876543E-5 by 0.200000E-3

Solution

Exercise

1. Add 0.500000E1 and 0.100000E-7

2. Multiply 0.350000E40 by 0.500000E70

3. Divide 0.875000E-18 by 0.200000E95

4. Subtract 0.499998 from 0.500000

5. let , show that

a. (x + y) + z = x + (y + z)

b.

c.

Definition2

A non-zero floating point number (1) is in normal form if the value of the mantissa lies in

the interval or in the interval .

25

Definition3

A non-zero floating point number (1) is in t-digit mantissa standard from if it is

normalized and its mantissa consists of exactly t digits.

Is a number x has the representation in the form

………………….(2)

Then the floating point number f(x) in t-digit mantissa standard form can be obtained in

the following ways.

1. Chopping

Here we neglect in (2) and obtain

……………………………

(3)

2. Rounding

Here the fractional part in (2) is written as

if and the first digits

are taken to write the floating point number.

Example

Find the sum of 0.123x103 and 0.456x102 and write the results in 3digit mantissa.

Solution

26

= 0.168 x 103 for Chopping

= 0.169 x 103for rounding.

Week 3

APPROXIMATION AND ERRORS.

A computer has a finite word length and so only a fixed number of digits are stored and

used during computation. This means during storing exact decimal numbers in its

converted form in the computer memory, an error is introduced. This error is machine

dependent and is called machine epsilon. After the computation is over, the result in the

machine form (with base ) is again converted to decimal form understandable to the

users and some more error may be introduced at this stage.

The error = True value – Approximate value

The relative error = …………………………………(5)

Absolute error = …………………………(6)

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Types of Errors

1) Inherent Error

these are errors present in the data supplied to the computer. They are present in the

statement of the problem before solution. They are of two types.

a) Data Errors

These arise when data is obtained by some experimental means and are therefore

of limited accuracy and precision.

b) Conversion Errors

These arise due to the limitation of the computer to store the data exactly after

converting from one number system to another.

2) Numerical Errors

They are introduced during the process of implementation of a numerical method.

They are of two forms;

a) Round-off Errors

They occur when a fixed number of digits are used to represent exact numbers.

The round off error is the quantity e which must be added to the finite

representation of a computed number in order to make it the true representation of

that digit. Thus, if x is the computed number given by

, then

thye relative error (5) for t-digit mantissa standard form representation of x

becomes

28

Thus the bound on the relative error of a floating point number is reduced by half

when rounding is used than chopping.

b) Chopping Errors

The extra digits are dropped. For instance if we use a computer with a fixed word

length of 4 digits, then the number 42.79893 can be expressed as follows

This can be expressed in the general form as

True x =

Where fx - Mantissa, d - length of the mantissa, E – exponent.

SOLUTION OF NON-LINEAR EQUATIONS

Consider equations of the form

………………………………………………….. 2.1

Where is a polynomial of degree in

x.

Definition

29

A number is a solution of f(x) = 0, if f() . Such a solution is called a root or zero of

f(x) = 0.

Geometrically, a root of the equation 2.1 is the value of x at which the graph of y = f(x)

intersects the x – axis. Equation 2.1 can be solved using the following two methods.

i. Direct Methods

These methods give the exact value of the roots in a finite number of steps.

ii. Iterative methods

These methods are based on the idea of successive approximation, that is starting

with one or more initial approximations to the root, we obtain a sequence of

approximations or iterates (xk) which in the limit converges to the root.

Definition

A sequence of iterates (xk) is said to converge to the root if

If xk, xk-1…., xk-m+1 are m approximations to the root then a multipoint iteration method is

defined as

.

The function is called the multipoint iteration function. To get an appropriate initial

approximation it might be necessary to sketch the curve so that any value in the

neighbourhood of the point the curve crosses the x axis may be taken as an initial

approximation.

Alternatively, we can use the Intermediate Value theorem stated below.

Theorem

30

If f(x) is a continuous function on some initial interval and f(a)f(b) < 0 then the

equation f(x) = 0 has atleast one real root or an odd number of roots in the interval (a,b).

Example

Obtain an initial approximation to a root of the equation f(x) = cosx – xex = 0

Solution

We prepare a table of the values of the function f(x) for various values of x.

x 1 0.5 1 1.5 2

f(x) 1 0.0532 -2.1780 -6.6518 -15.1942

From the table f(x) = 0, has at least one root in the interval (0.5, 1).

The exact root correct to 10d.p is 0.5177573637.

Bisection Method

This method is based on the repeated application of the intermediate value theorem.

Suppose the root f(x) = 0 lies in the interval I0 = (a0, b0), then we determine

and denote by I1, the new subinterval which satisfies the IVT.

We then bisect I1 and get a subinterval I2 which satisfies the IVT. Continuing this way we

obtain a sequence of nested sets of such intervals such that each

subinterval contains the root.

Suppose we repeat the bisection q times, then we obtain either the root or the subinterval

Iq of length which contains the root.

31

We take the midpoint of the last subinterval as the desired approximation to the root. If

the permissible error is , then the approximate number of iterations required is obtained

from the relation

Example

Perform five iterations of the bisection method to obtain the smallest positive root of the

equation f(x) = x3 – 5x + 1 = 0

Solution

Since f(0) = 1 > 0 and f(1) = -3 < 0, then the smallest root lies in the interval (0, 1).

Taking a0 = 0 and b0 = 1 we get,

F(m1) = -1.375 and f(a0)f(m1)<0.

Thus the root lies in the interval (0, 0.5). The sequence of interval are given in the table

below.

k ak-1 bk-1 mk f(mk)

1 0 1 0.5 < 0

2 0 0.5 0.25 < 0

3 0 0.25 0.125 < 0

4 0.125 0.25 0.1875 < 0

5 0.1875 0.25 0.21875 < 0

Hence the root lies in (0.1875, 0.21875)

x = 0.203125

32

ITERATIVE METHODS BASED ON 1ST DEGREE EQUATIONS

Suppose f(x) = a0x + a1 = 0 …………………………………………………….2.1,

is a 1st degree approximation of f(x) then the solution of 2.1 is given by

…………………………………………………………3.1

Where a0 = 0 and a1 are arbitrary parameters.

SECANT METHOD

If xk-1 and xk are two approximations to the root, then to get a0 and a1 we use the

conditions

, where

, where

………………………………………………………..4.1

Which implies ……………………………………..5.1

This is called the Secant or the Chord Method.

33

THE REGULA – FALSI METHOD

If the approximations in the secant method are such that f(x) f(xk-1) < 0, then the method

is known as the Regula-Falsi method.

Example

Use the Secant and Regula-Falsi methods to determine the root of the equation

Cos x – xex = 0

Solution

Taking the initial approximations as x0 = 0, x1 = 1, we have for the secant method

F(0) = 1

F(1) = cos 1 – e = -2.17797523

F(x2) = f(0.3146653378) = 0.519871175

The computed results are as follows.

k xk+1 f(xk+1)

2 0.44678144 0.519871

3 0.5317058606 -0.429311 x 10-1

4 0.5169044676 0.259276 x 10-2

5 0.5977477653 0.301119 x 10-4

6 0.5177573708 -0.215132 x 10-7

7 0.5177573637 0.178663 x 10-12

34

8 0.517757367 0.222045 x 10-15

Hence the root is 0.5178 (4d.p)

The table for Regula- Falsi method is

k xk+1 f(xk+1)

1 0.3146653378 0.519871

2 0.4467281446 0.203545

3 0.4940153366 0.708023 x 10-1

4 0.5099461404 0.236077 x 10-1

.

.

.

.

10 0.5177478783 0.288554 x 10-4

.

.

20 0.5177573636 0.396288 x 10-9

x = 0.5178 (4d.p)

Note: The Secant methods converges faster than the Regula-Falsi method.

THE NEWTON-RAPHSON METHOD

Suppose f(xk) = a0xk + a1, then f`(xk) = a0 …………………………….6.1

a1 = - xk f`(xk) - f(xk)

From 3.1, we have

35

which is the Newton-Raphson Method.

Example

Apply Newton-Raphson Method to determine a root of the equation

f(x) = Cos x – xex = 0 such that f(x*) < 10-8 where x* is the approximation to the root.

Solution

f(x) = Cos x – xex = 0

Since the roots lie in we set xo = 1 and obtain the table below.

k xk f(xk)

0 1

1 0.653079406 -4.606 x 10-1

2 0.531343367 -4.180 x 10-2

3 0.517909913 -4.641 x 10-4

4 0.51775738 -5.926 x 10-7

5 0.517757363 -2.910 x 10-10

X 0.5178

36

POLYNOMIAL INTERPOLATION

Suppose that we have a table of numerical values of a function

x x0 x1 ……… xn

y y0 y1 ……… yn

Table 1.1

We are faced with three problems

i. To find a simple and convenient formulas that reproduces the given points

exactly.

ii. We assume, given the table of numerical values it is contaminated by errors. We

therefore seek a formula that represents the data(approximately) and if possible

filter out the errors.

iii. We need to find a function f that is simple to evaluate and produce a reasonable

approximation to g(which might be given) with full machine precision.

In all these problems, a simple function p can be obtained that represents or approximates

the given table / function f. the representation p can always be taken to be a polynomial.

Uses of Polynomials

i. Reconstruct the function f(x) when it is not given explicitly and only the values of

f(x) and / or its certain order derivatives at a set of points called nodes, tabular

points or arguments are known.

ii. To represent the function f(x) by the interpolating polynomial p(x) so that

common operations such as determination of roots, differentiations and

integrations may be performed using p(x).

37

Definition

A polynomial p(x) is called interpolating polynomial if the values of p(x) and / or its

certain order derivatives coincide with those of f(x) and /or its same order derivatives at

one or more tabular points. Its represented as

,

where n is a non negative integer and a0, a1,….. an are constants.

TAYLOR’S INTERPOLATION

If the polynomial p(x) is written as the Taylor’s expansion for the function f(x) at a point

in the form

………4.1

Then p(x) may be regarded as an interpolating polynomial of degree n, satisfying the

condition.

, k = 0, 1,…….., n. ………………………………………4.2

The term

, x0 < < x……………………………4.3

Which has been neglected in 4.1 is called the remainder or the truncation error.

The number of terms to be included in 4.1 may be determined by the acceptable error. If

this error > 0 and the series in truncated at the term f(n)(x0), then

…………………………………………4.4

38

Where

Example1

Obtain a polynomial approximation p(x) to f(x) = e-x using the Taylor’s expansion about

x0 = 0 and determine

i. X when the error in p(x) obtained from the 1st four terms only is to be less than 10-

6 after rounding.

ii. They number of terms in the approximation to find results correct to 10-10 for

.

Solution

i. From f(x) = e-x, we have

from 4.1,

and from 4.4,

where

x = 120 x 10-7

< 0.058856619

x < 0.06

39

ii. From 4.1 we obtain

, x = 1, M4 = 1

n ≥ 14

LINEAR INTERPOLATION

From Table 1.1, we assume that the xi’s form a set of n +1 distinct points. The table

represents n+1 points in the Cartesian plane and we want to find a polynomial curve that

passes through all the points. Thus we seek to determine a polynomial that is defined for

all x and takes on the corresponding values of yi for each of the n+1 distinct xi’s in this

table. A polynomial p for which p(xi) = yi when is said to interpolate the table.

The points xi are called nodes.

Consider the 1st and simple case, n = 0. Here a constant function solves the problem i.e

the polynomial p of degree 0 defined by p(x) = y0 reproduces the one none table.

For n = 1, since a straight line can be passes through two pints, a linear function is

capable of solving the problem. Explicitly, the polynomial is defined by

…………………………………………..4.5

is of 1st degree and reproduces the table. In this case p(x0) = y0 and p(x1) = y1.

This p is used for linear interpolation.

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Example 1

Find the polynomial of least degree that interpolates the table below.

x 1.4 1.25

y 3.7 3.9

Solution

By definition

Example 2

Find a polynomial of degree 2 or less to approximate f(x) = sin x near x0 = 0 and use this

polynomial to approximate sin(0.1).

Solution

The equation of this approximating polynomial is

To determine the constants, we let

x0 = 0

P(0) = ao, f(0) = sin 0 = 0 ao = 0

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and a1 = 1

and a2 = 0

and

The approximating polynomial of degree 3 is given by

0.099833

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LAGRANGE’S INTERPOLATION

Suppose we wish to interpolate arbitrary functions at a set of nodes x0, x1, ………xn. We

first define a system of n+1 special polynomial of degree n known as cardinal

polynomials. These are denoted by lo, l1,……ln

and

at x = x0, L0(x0) = 1, L1(x0) = 0

x = x1, L0(x1) = 0, L1(x1) = 1

Once these are available, we can interpolate any function f by the Lagrange form of the

interpolating polynomial

where

The formula indicates that li(x) is the product of n linear factors

Example

Determine the 2nd degree interpolating polynomial for using the nodes x0=2,

x1=2.5, and x2=4.0.

Solution

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……………………………(1)

, ,

and the coefficients are given by

Substituting in the polynomial (1), we have

= 0.05003x2 – 0.425185x + 1.15025

P2(x) 0.05x2 – 0.425x + 1.15

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Example 2

Find the Lagrange’s interpolating polynomial to interpolate the table below.

x 1

f(x) 2 -1 7

Solution

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NEWTON’S DIVIDED DIFFERENCE INTERPOLATION

This is in the form

……….4.6

Where and is called the 1st divided difference of f(x)

relative to x0 and x1.

Equation 4.6 may be written as

…………………………………………………….4.7

Equation 4.6 and 4.7 are the linear Newtonian interpolating polynomial with divided

differences.

TRUNCATED ERROR BOUNDS

The polynomial p(x) coincides with the function f(x) at x0 and x1 and it deviates at all

other points in the interval (x0, x1) as shown below.

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This deviation is called the truncation error and may be written as

E(f,x) = f(x) – p(x).

Rolle’s Theorem

If g(x) is continuous on the interval [a,b] and if g(a) = 0, g(b) = 0, then there is at least

one point inside (a,b) for which g`() = 0.

If , , x1 is fixed, then for this x we define a function g(t) as

…………………………..4.8

g(t) = 0 for t = x0, x1 and x.

Differentiating 4.8 twice with respect to x we get

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, since P’’(t) = 0 by Rolle’s Theorem. …..4.9

Solving 4.9 we have

……………………………………4.10

Where min(x0, x1, x) < < max (x0, x1, x)

the truncation error in linear interpolation is given by

………………………………………4.11

The bound for f’’(x) in [x0,x1] i.e , then

…………………………..4.12

The max value of occurs at x = and 4.12 becomes

………………………………………………….4.13

For fully spaced nodal points xi = a + ih, i = 0, 1, ……n and , the max truncation

error using the linear interpolating polynomial P(x) is less than a given > 0, we have

………………………………………….4.14

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Example1

Using sin(0.1) = 0.09983 and sin(0.2) = 0.19867, find an approximate value of sin (0.15)

by Lagrange’s interpolation, obtain a bound on the truncation error.

Solution

= 0.049915 + 0.099335

= 0.14925

The truncation error is

=

Where ,

The maximum value of is sin 0.2 = 0.19867,

Thus

= (0.00125)(0.19867)

0.00025

Example2

Determine the stepsize h to be used in the tabulation of f(x) = sin x in the interval [1,3] so

that the linear interpolation will be correct to 4 decimal places.

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Solution

and f(x) = sin x, f’(x) = cos x, f’’(x) = -sin x

,

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WEEK 7 and 8

1) Least square approximation to find an approximation to a function f(x) given in

tabular form.

2) Evaluate errors of interpolating polynomials.

3) Use divided differences to determine the explicit representation of an interpolating

polynomial.

APPROXIMATION

The existence of a polynomial function P(x) which approximates any continous function

f(x) on a finite interval is guaranteed by the Weierstrass approximation Theorem.

Weierstrass approximation Theorem

If the function f(x) is continuous on a finite interval , then given any ε > 0, there

exists a n = n(ε) and a polynomial P(x) of degree n such that

In order to determine an approximation to f(x) we assume an expression of the form

……………………(1)

Where (i = 0, 1, ………………, n) are independent

functions and the parameter ci (i = 0, 1, ………………, n) are constants to be determined.

The function are called Coordinate Functions which for polynomial approximations

are considered as , (i = 0, 1, ………………, n)

The error of the approximation ε is defined as

……………………….(2)

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Where is a well defined norm. the aim of approximation is to determine the

coefficient ci (i = 0, 1, ………………, n) inorder that the error is made as small as

possible.

For different types of norm, we obtain different types of approximations. Once a

particular norm is fixed the function which make these error smallest according to this

norm is called the “Best Approximation”.

The most commonly used norms are;

Norm For Discrete Data

a) Lp Norm is defined as

P ≥ 1 ………………………….(3)

b) Euclidean Norm

……………………………………….(4)

written as and is called the square norm.

c) Uniform Norm

……………………………………………(5)

Which is a particular case of (3) for P → ∞.

Norm for Continuous Data

If the function f(x) is continuous in the interval , then the norm is

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a) , P ≥ 1 ………………………….(6)

Where4 w(x) > 0 and is the weighted function.

b) For p = 2, we have the Euclidean or square norm

………………………………………….(7)

c) For p = ∞, we have the uniform norm.

…………………………………………………….(8)

When we use the Euclidean norm we obtain the Least Square Approximation and when

uniform norm is used we get the Uniform Approximation.

Least Square Approximation

Least Square Approximations are most commonly used approximations for

approximating a function f(x) which may be given in tabular form or known explicitly

over a given interval. In this case we use the Euclidean norm (4). The best approximation

in the least square sense is defined as that for which the constants Ci (i = 0, 1……….., n )

are determined so that the aggregate of the weighted function w(x) over a given interval

is as small as possible for w(x) > 0.

For functions whose values are given at n + 1 points x0, x1, …..xn, we have

………………..(9)

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For functions which are continuous over the interval

…………………..(10)

Where the coordinate , i = 0, 1, ………., n. and the weight function w(x) = 1.

The necessary condition for equation (9) and (10) to have a minimum value is that

, i = 0, 1, ………., n

This gives a system of n + 1 linear equations in n + 1 unknowns C0, C1…….Cn.

These equations are called the Normal equations.

The normal equations for (9) and (10) are respectively

……………………………(12)

and

, j = 0, 1, 2…….n ….........(13)

Example

Obtain a linear polynomial approximation to the function f(x) = x3 on interval using

the least square approximation with W(x) = 1.

Solution

This equation is continuous hence we use a linear polynomial

P(x) = C0 + C1(x) , where C0 and C1 are arbitrary constants.

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Using equation (10)

The necessary and sufficient conditions for a minimum to occur is given by equation (11)

Taking the partial derivatives of equation (1) above we have

……………………………………….(2)

Equation (2) is known as the Normal equations.

Solving the normal equations simultaneously, we get

and

The polynomial will be

Example 2

Obtain the least square polynomial approximation of degree 1 and 2 for f(x) = on the

interval .

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Solution

W(x) = 1.

a) For n = 1

………………………………(1)

The necessary and sufficient conditions for a minimum

……………………………….(a)

…………………………….(b)

Solving Simultaneously (a) and (b) we have

and

b) For n = 2

…………………….(2)

The least square polynomial is of the form

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…………………………(a)

……………………….(b)

………………………(c)

Rearranging (a), (b) and (c), we have

Solving the above equations simultaneously, we have

, and

TCHEBYSHEV POLYNOMIALS

57

The Tchebyshev polynomial of the 1st kind Tn(x) is defined on the interval are

given by

…………………….(1)

and satisfies the differential equations

……………………………(2)

One independent solution of (2) gives the Tn(x) and the second independent solution

gives Chebyshev’s polynomial of the 2nd kind Un(x).

…………………………(3)

The Tchebyshev polynomial Tn(x) satisfies the recurrence relation

……………………………..(4)

From equation (4)

Thus we have

Thus

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Properties of Tchebyshev polynomial

1. Tn(x) is a polynomial of degree n and if n is even Tn(x) is an even polynomial and if n

is odd Tn(x) is odd polynomial.

2. Tn(x) has n simple zeros , on the interval .

3. Tn(x) assumes extreme values at n + 1 points and the

extreme value at xk is (-1)k.

4.

5. If Pn(x) is any polynomial of degree n with leading coefficient unity (monomial) and

is the monic Chebyshev polynomial.

6. Tn(x) are orthogonal with respect to the weight function

Example

Using the Chebyshev polynomial, obtain the least squares approximation of second

degree for on .

Solution

59

.

We have

The required approximation is

60

61

WEEK 9 and 10

1) State Chebychev oscillation theorem

2) Chebychev’s polynomial to find an approximation to a function f(x) given in tabular

form.

Theorem(CHEBYSHEV EQUIOSCILLATION)

Let f(x) be continuous on and Pn(x) be the best uniform polynomial approximation.

Denote and . Then tere are at least n + 2

points , where

i.

ii. ……………………(1)

Example

Obtain the Chebychev’s linear polynomial approximation to the function on

.

Solution

P(x) = C0 + C1(x)

and x0 = 0 and x1 = and x2 = 1.

Using (1) we have

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Thus we have

which gives

, and

The Chebychev’s linear approximation is given by

Question

Obtain the Chebychev’s polynomial approximation of second degree to the function

on .

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CHOICE OF THE METHOD

1) Understand and use the concept of Extrapolation to the limit.

We have considered various forms of the approximating polynomials. The interpolating

polynomials and the function f(x) agree in value at a set of points xi, i = 0, 1, ….., n

belonging to an interval . We use this interpolating polynomial to determine the

value of f(x) at any point . If , we call it an interpolating problem,

otherwise it is called an extrapolation problem. Normally interpolating problem produce

better results. The same interpolating polynomial can be used for obtaining roots of f(x),

certain order derivatives of f(x) at tabular or non-tabular points, integral of f(x) between

known limits or for any other operation desired on f(x).

For a given set of tabular data, the Lagrange interpolating polynomial can be easily

obtained. In many applications, the exact number of tabular points to be used to achieve a

certain degree of accuracy is not readily known.

WEEK 13 AND 14

i) Students hand in the assignment

ii) Students take sit in course examination.

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