Central Divided Difference Method, Lecture notes of Engineering Mathematics

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Module 2

Lesson 6

Numerical Differentiation

Central Divided Difference

Edgar M. Adina

Instructor

CE 50 P- 2

Numerical Solutions to Engineering Problems

Introduction

➢ Calculus is the mathematics of change. Because engineers must

continuously deal with systems and processes that change, calculus is

an essential tool of engineering.

➢ Standing in the heart of calculus are the mathematical concepts of

differentiation and integration :

Δ𝑦

Δ𝑥

=

𝑓(𝑥𝑖 + Δ𝑥) − 𝑓(𝑥𝑖)

Δ𝑥 𝑑𝑦

𝑑𝑥

=Δ𝑥 lim 0

𝑓(𝑥𝑖 + Δ𝑥) − 𝑓(𝑥𝑖)

Δ𝑥

𝐼 = න

𝑎

𝑏

𝑓(𝑥)𝑑𝑥

Applications

➢ Differentiation has so many engineering applications

(heat transfer, fluid dynamics, chemical reaction kinetics,

etc…)

➢ Integration is equally used in engineering (compute work

in ME, nonuniform force in SE, cross-sectional area of a

river, etc…)

Differentiation

The finite difference becomes a derivative as Δx approaches zero.

𝜟𝒚

𝜟𝒙

=

𝒇(𝒙𝒊 + 𝜟𝒙) − 𝒇(𝒙𝒊)

𝜟𝒙 x

f x x f x

dx

dy (^) i i

x (^) 





( ) ( ) lim 0

• − = →

Numerical Application of Taylor Series

If f (x) and its first n+ 1 derivatives are continuous on an

interval containing x

i+ 1

and x

i

, then:

Where the remainder R

n

is defined as:

𝑓 𝑥𝑖+ 1 ≅ 𝑓 𝑥𝑖 + 𝑓

′ 𝑥𝑖 𝑥𝑖+ 1 − 𝑥𝑖 +

𝑓

′′ 𝑥𝑖

2!

𝑥𝑖+ 1 − 𝑥𝑖

2

𝑓

3 𝑥𝑖

3!

𝑥𝑖+ 1 − 𝑥𝑖

3 +... +

𝑓

𝑛 𝑥𝑖

𝑛!

𝑥𝑖+ 1 − 𝑥𝑖

𝑛

• 𝑅𝑛

𝑅𝑛 =

𝑓

𝑛+ 1 𝜉

𝑛 + 1!

𝑥𝑖+ 1 − 𝑥𝑖

𝑛+ 1

Numerical Application of Taylor Series

➢ The series is built term by term

➢ Continuing the addition of more terms to get better

approximation we have:

𝑓 𝑥𝑖+ 1 ≅ 𝑓 𝑥𝑖

𝑓 𝑥𝑖+ 1 ≅ 𝑓 𝑥𝑖 + 𝑓

′ 𝑥𝑖 𝑥𝑖+ 1 − 𝑥𝑖

𝑓 𝑥𝑖+ 1 ≅ 𝑓 𝑥𝑖 + 𝑓

′ 𝑥𝑖 𝑥𝑖+ 1 − 𝑥𝑖 +

𝑓

′′ 𝑥𝑖

2!

𝑥𝑖+ 1 − 𝑥𝑖

2

zero order approximation

st

order approximation

nd

order approximation

Taylor Series: ξ in the Remainder Term

❑ Limitations

➢ ξ is not exactly known but lies somewhere between xi and xi+ 1

➢ To evaluate R

n

, the (n+ 1 ) derivative of f (x) has to be determined.

To do this f (x) must be known

➔ if f (x) was known there would be no need to perform the Taylor series

expansion!!!

❑ Modification

R

n

= O(h

n+ 1

) the truncation error is of the order of h

n+ 1.

(h = x

i+ 1

• x i

➢ If the error is O(h), halving the step size will halve the error.

➢ If the error is O(h

2

), halving the step size will quarter the error.

➢ In general, the truncation error is decreased by addition of more terms in the

Taylor series.

Centered Difference Formulas- 1

st derivative

Start with the 2

nd degree Taylor expansions about x for f (x+h) and f (x-h):

Subtract ( 5 ) from ( 4 )

Hence

𝑓(𝑥𝑖+ 1 ) = 𝑓(𝑥𝑖) + 𝑓′(𝑥𝑖)ℎ +

𝑓′′(𝑥𝑖)

2

ℎ

2

• 𝑂(ℎ

3 )

2

3

( 4 )

( 5 )

𝑓(𝑥𝑖+ 1 ) − 𝑓(𝑥𝑖− 1 ) = 2 𝑓′(𝑥𝑖)ℎ + 𝑂(ℎ

3 )

2

Centered Difference Formulas- 2

nd derivative

Start with the 3

rd degree Taylor expansions about x for f (x+h) and f (x-h):

Add equations ( 8 ) and ( 9 ), and solve for f ’’(x)

2

3

• 𝑂(ℎ

4 ) ( 8 )

2 −

3

• 𝑂(ℎ

4 ) (^9 )

Centered Difference Formulas- 2

nd derivative

Start with the addition between the 5

th degree Taylor expansions about x for f (x+h)

and f (x-h):

Use the step size 2 h, instead of h, in ( 10 )

Multiply equation ( 10 ) by 16 , subtract ( 11 ) from it, and solve for f ’’(x)

2

( 4 )

4

6 ) (^10 )

( 11 )

2

4

2

( 4 )

4

6

Example 1

Solution

( )

( ) ( )

t

t t

a t

i i i

• −

 1  1 ti = 16

1

t i + = ti + t

14

16 2

1

=

= −

t (^) i − = ti − t

( )

( ) ( )

2 ( ) 2

18 14 16

a 

( ) ( )

 t = 2

( ) ( )

9. 8 ( 18 ) 14 10 210018

18 2000 ln 4

4

−  

 = = 453. 02 m/s

14 2000 ln

4

4

 = = 334. 24 m/s

( )

( ) ( )

a 

2  29. 694 m/s

Example 1

The absolute relative true error is

 t =

= 0. 069157 %

The exact value of the acceleration at is

( )

2 a 16 = 29. 674 m/s

t = 16 s

Example 2

The velocity of a rocket is given by

( ) 9. 8 , 0 30 14 10 2100

2000 ln 4

4

−    

= t t t

 t

Use central difference approximation of second derivative of to

calculate the jerk at. Use a step size of.

ν ( ) t

t = 16 s Δ t^^ =^2 s

Example 2

Solution ( )

( )

2

(^1 )

t

t t t a t

i i i i 

• −

ti

1

t i + = ti + t

1

t i − = ti − t

( )

( ) ( ) ( )

( )

2

j 

 t = 2

( ) ( )

9. 8 ( 18 ) 14 10 210018

18 2000 ln 4

4

−  

 = = 453. 02 m/s

( ) ( )

9. 8 ( 16 ) 14 10 210016

16 2000 ln 4

4

−  

 = =^392.^07 m/s

( ) ( )

9. 8 ( 14 ) 14 10 210014

14 10 14 2000 ln 4

4

−  

  



 −

  = =^334.^24 m/s

( )

( ) ( ) ( )

( )

2 2

j 

( )

3  0. 77969 m/s