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Truncation Errors and Taylor Series

Introduction

Truncation errors

Result when approximations are used to represent exact mathematical procedure

For example:

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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

Taylors Theorem

Where:

ii xxh 1

1)1(

)!1(

)(

n

n

n hn

fR

4

n

n

i

n

iiiii

Rn

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

Taylors Theorem

)()(1 ii

xfxf

!2

)('')()()(

2'

1

hxfhxfxfxf iiii

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

Taylors Theorem - Remainder Term

Remainder Term: What is ?

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 functions 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

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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 iiii

!3

)(

!2

)('')()()(

3(3)2'

1

hxfhxfhxfxfxf iiiii

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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 iiiiii

5)5(

4!5

)(h

fR

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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)

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Example 4.2

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Effect of non-linearity

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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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Effect of step size

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