Error analysis handout

The errors of the sumThe errors of product 3 E XERCISES Exercises in writing form Multiple choice exercises... The deviation is the difference between approximate value and exact value:

ERROR ANALYSIS

ELECTRONIC VERSION OF LECTURE

Dr Lê Xuân Đại

HoChiMinh City University of Technology Faculty of Applied Science, Department of Applied Mathematics

Email: ytkadai@hcmut.edu.vn

HCMC — 2018.

The errors of the sum

The errors of product

3 E XERCISES

Exercises in writing form

Multiple choice exercises

R EAL WORLD PROBLEM

D EFINITION

DEFINITION 1.1

Suppose that a is an approximation to the exact value A. The deviation is the difference between approximate value and exact value:

a − A

DEFINITION 1.2

The number a is called an approximation of exact value A, we denote it bya≈A.

DEFINITION 1.3

The number ∆ = |a − A|is called actual error

In reality, we do not know exact value A, so

we can estimate actual error by a positive

that |A − a| É ∆ a Then the number ∆a is called

Note In engineering problems, we denote

exact value using absolute error by

A = a ± ∆

Therefore, there are many ways to choose

DEFINITION 1.4

exact value Ais the number which is less

than or equal to δ a, where δ a is defined by formula

δ a= |A − a|

|A| ·

|a|·So relative error É

∆a

|a|

EXAMPLE 1.2

If the lengths of 2 tables are a = 10cm and

b = 1cm respectively, where ∆a= ∆b = 0.01cm.Then

δ a = 0.01

10 = 0.1%; δ b= 0.01

hay δ b = 10δ a Therefore, the measurement of

a is better than that of b even though ∆a = ∆b

Machine numbers are represented in the normalized decimal form

DEFINITION 1.5

The error that results from replacing a

number with its floating-point form is called

round-off error regardless of whether the

rounding or chopping method is used.

k + 1st digit is 5or greater, then 1is added to

chopped.

EXAMPLE 1.4

The irrational number π has an infinite

decimal expansion of the form

π = 3.1415926535

The four-digit floating-point form of π

chopping is 3.1416

DEFINITION 1.6

The actual error of round-off ea and the

approximation a is called round-off error So

θ ae= |a −ea|.

|ea − a| + |a − A| É θea+ ∆a = ∆ea Since θ aeÊ 0 then

increases , so we just round and chop the last results.

R OUNDING FOR INEQUALITY

EXAMPLE 1.5

If a < 13.9236,we will round up 2 numbers after decimal point and receive a < 13.93.

If b > 78.6789, we will round down 2

numbers after decimal point and receive

b > 78.67.

E RROR OF FUNCTIONS OF 2 VARIABLES

Let ∆x = |X − x| ⇒ ∆x É ∆ x

Let ∆y = |Y − y| ⇒ ∆y É ∆ y

3 u = f (x,y)is an approximation of exact

Relative error of uis less than or equal to

F UNCTION OFnVARIABLES

x i , y (i = 1 n)be exact values of variables and function, and their approximations

Relative error of y is less than or equal to

EXAMPLE 2.1

Evaluate absolute error and relative error of the volume of the sphere V = 1

6πd3, if the diameter d = 3.70cm ± 0.05cm and

2× (3.14) × (3.70)2× 0.05 = 1.088172467

So the absolute error is less than or equal to

1.0882 The relative error is less than or

|v| = 0.04105009468 ≈ 0.0411.

|16× 3.14 × 3.703|

T HE ERRORS OF THE SUM

¯= 1, (i = 1 n). Therefore, absolute error

∆y = ∆x1+ ∆x2+ + ∆x n

and relative error of y is less than or equal to

δ y= ∆y

|y|.

Relative error of y is less than or equal to

δ y = ∆y

|y| = 0.003749722 ≈ 0.0038.

|47.132 + 47.111 + 45.234|

Therefore, absolute error of y is less than or equal to

∆y = δ y |y| = M × |47.132 × 47.111 × 45.234| =

1159.250261 ≈ 1159.2503

EXERCISE 1.1

If a = 1.85 has relative error δ a = 0.12%

Evaluate absolute error of a.

SOLUTION.

δ a = ∆a

|a| ⇒ ∆a = δ a × |a| = 0.12% × 1.85 = 0.00222

numbers, we obtain that absolute error of

a É 0.0023.

EXERCISE 1.3

If the sphere has radius of R = 5 ± 0.005(m)

and π = 3.14 ± 0.002.Evaluate absolute error and relative error of the volume of the

sphere.

Solution. Considering π and R variables of

Relative error is less than or equal to

δ v= ∆v

|v| = 0.003636942 ≈ 0.0037.

|43× 3.14 × 53|

EXERCISE 1.5

Let f (x) = 3x5− 2x2+ 7 andx = 1.234 ± 0.00015.

Evaluate absolute error of f (x).

∆f = |f0(x)|.∆ x

so

∆f = |15 × (1.234)4− 4 × 1.234| × 0.00015 =

= 0.004476868 ≈ 0.0045

The absolute error of a∗ is less than or equal to

∆a∗ = ∆a + θ a∗= δ a × |a| + |a∗− a| = 0.01969118.

The absolute error of f is less than or equal to

THANK YOU FOR YOUR ATTENTION