The Islamic University of Gaza

Faculty of Engineering

Civil Engineering Department

Numerical Analysis

ECIV 3306

Chapter 4

1

Truncation Errors and Taylor Series

Introduction

Truncation errors

• Result when approximations are used to represent exact mathematical procedure

• For example:

2

3

Taylor Series - Definition

• Mathematical Formulation used widely in numerical methods to express functions in an approximate fashion……. Taylor Series.

• It is of great value in the study of numerical methods.

• It provides means to predict a functional value at one point in terms of:

- the function value - its derivatives at another point

Taylor’s Theorem

Where:

ii xxh 1

1)1(

)!1(

)(

n

n

n hn

fR

4

n

ni

nii

iiiR

n

hxfhxfhxfhxfxfxf

!

)(.......

!3

)(

!2

)('')()()(

)(3)3(2'

1

General Expression

Rn is the remainder term to account

for all terms from n+1 to infinity.

And is a value of x that lies somewhere between xi and xi+1

Taylor’s Theorem

)()(1 ii

xfxf

!2

)('')()()(

2'

1

hxfhxfxfxf i

iii

hxfxfxf iii)()()( '

1

5

Zero- order approximation: only true if xi+1 and xi are very close to each other. First- order approximation: in form of a straight line

Second- order approximation:

Any smooth function can be approximated as a polynomial

Taylor’s Theorem - Remainder Term

Remainder Term: What is ξ ?

h

Rf o)(' oii

Rxfxf

)()(1

6

If Zero- order approximation:

Taylor Series - Example Use zero-order to fourth-order Taylor series expansions to

approximate the function. f(x)= -0.1x4 – 0.15x3 – 0.5x2 – 0.25x +1.2

From xi = 0 with h =1. Predict the function’s value at xi+1 =1.

Solution f(xi)= f(0)= 1.2 , f(xi+1)= f(1) = 0.2 ………exact solution

• Zero- order approx. (n=0) f(xi+1)=1.2

Et = 0.2 – 1.2 = -1.0

• First- order approx. (n=1) f(xi+1)= 0.95 f(x)= -0.4x3 – 0.45x2 – x – 0.25, f ’(0)= -0.25 f( xi+1)= 1.2- 0.25h = 0.95 Et = 0.2 - 0.95 = -0.75

)()(1 ii

xfxf

hxfxfxf iii)()()( '

1

7

Taylor Series - Example• Second- order approximation (n=2) f(xi+1)= 0.45

f ’’(x) = -1.2 x2 – 0.9x -1 , f ’’(0)= -1

f( xi+1)= 1.2 - 0.25h - 0.5 h2 = 0.45

Et = 0.2 – 0.45 = -0.25

• Third-order approximation (n=3) f(xi+1)= 0.3

f( xi+1)= 1.2 - 0.25h - 0.5 h2 – 0.15h3 = 0.3

Et = 0.2 – 0.3 = -0.1

!2

)('')()()(

2'

1

hxfhxfxfxf i

iii

!3

)(

!2

)('')()()(

3(3)2'

1

hxfhxfhxfxfxf ii

iii

8

Taylor Series - Example

• Fourth-order approximation (n = 4) f(xi+1)= 0.2

f( xi+1)= 1.2 - 0.25h - 0.5 h2 – 0.15h3 – 0.1h 4= 0.2

Et = 0.2 – 0.2 = 0

The remainder term (R4) = 0

because the fifth derivative of the fourth-order polynomial is zero.

!4

)(

!3

)(

!2

)('')()()(

4)4(3)3(2'

1

hxfhxfhxfhxfxfxf iii

iii

5)5(

4 !5

)(h

fR

9

10

Approximation using Taylor Series Expansion

The nth-order Approximation

Taylor Series

• In General, the n-th order Taylor Series will be exact for n-th order polynomial.

• For other differentiable and continuous functions, such as exponentials and sinusoids, a finite number of terms will not yield an exact estimate. Each additional term will contribute some improvement.

(see example 4.2)

11

Taylor Series

• Truncation error is decreased by addition of terms to the Taylor series.

• If h is sufficiently small, only a few terms may be required to obtain an approximation close enough to the actual value for practical purposes.

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