NUMERICAL ANALYSIS

Lecture -2

Muhammad Rafiq

Assistant Professor University of Central Punjab

Lahore Pakistan

ERROR ANALYSIS

ERROR

The difference in exact and approximate values is error.

Suppose exact value of a number be = x

Approximate exact value of a number be = x

Error = x- x

Absolute error=|x- x|

Relative error=A.E/|x|

=|x- x|/|x|

Percentage relative error=R.E100%

ERROR IN ADDITION

Suppose exact value of a number be= x1

Suppose exact value of another number be= x2

Suppose approximate value of a number be= x1

Suppose approximate value of another number be= x2

Error in first number = e1=x1-x1

Error in first number=e2=x2-x2

Let Z be the sum of two numbers=Z=x1+x2

Let Z be the sum of approximate values=Z=x1 +x2

Error=ez=Z-Z

= (x1+x2) - (x1+x2)

= x1+x2- x1-x2

=(x1-x1)+(x2-x2)

ez=e1+e2

Absolute error=|ez|=|e1+e2||e1|+|e2|

And |ez||e1|+|e2|

ERROR IN SUBTRACTION

Suppose exact value of a number be= x1

Suppose exact value of another number be= x2

Suppose approximate value of a number be= x1

Suppose approximate value of another number be= x2

Error in first number = e1=x1-x1

Error in first number=e2=x2-x2

Let Z be the difference of two numbers=Z=x1-x2

Let Z be the difference of approximate values=Z=x1-x2

Error=ez=Z-Z

= (x1-x2 )- (x1-x2)

= x1-x2- x1+x2

=(x1-x1)-(x2-x2)

ez=e1-e2

Absolute error=|ez|=|e1-e2||e1|+|e2|

And |ez| |e1|+|e2|

Thus for n numbers eZ = e1 e2 e3 en

|ez| |e1|+|e2|+|e3|++|en|

Note:

If a number is approximated to n-decimal places, then absolute

error in it is given by

A.E 1/2 10-n

For Example =1.73 e=1/2 10-2

=1.7320 e=1/2 10-4

EXAMPLE 1

If numbers in 0.3062-0.25026+2.51342 are rounded. Estimate

the maximum absolute error and relative error. Find also the

range in which true answers lies.

Solution:

Let x1=0.3062 e1=1/210-4

x2=0.25026 e2=1/210-5

x3=2.51342 e3=1/210-5

Let Z=x1-x2+x3

=0.3062-0.25026+2.51342

= 2.56936

Absolute error 1/210-4+1/210-5+1/210-5

1/210-5(10+1+1)

1/210-5(12)

610-5

0.610-4

Maximum Absolute error=0.610-4

Relative error=A.E/|Z|

=0.610-4

/2.56936

=2.335210-5

The range in which true answer lies

Z-A.E Z Z+A.E

2.56936-0.610-4 Z 2.56936+0.610-4

2.56930 Z 2.56942.

(Correct to 3d.p)

EXAMPLE 2

The values of x1 and x2 are estimated as x1=4.57+e1 and

x2=8.84+e2 where |e1|

Exercise

Q # 1

If numbers in 5.2163+1.35202-1.41265+2.14169 are rounded.

Estimate the maximum absolute error and relative error. Find

also the range in which true answers lies.

Q # 2

If numbers in 3.123-2.5169+2.234-2.01267+2.1006 are

rounded. Estimate the maximum absolute error and relative

error. Find also the range in which true answers lies.