Download Numerical Differentiation and Taylor Series Expansion in Engineering and more Lecture notes Engineering Mathematics in PDF only on Docsity!

CE 50 P- 2

Numerical Solutions to Engineering Problems

Module 2

Lesson 4

Numerical Differentiation

Forward Difference Method

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

Introduction

➢ Integration and differentiation are closely linked processes. They are, in

fact, inversely related.

➢ Types of functions to be differentiated or integrated:

1. Simple polynomial, exponential, trigonometric ➔ analytically
2. Complex function ➔ numerically
3. Tabulated function of experimental data ➔ numerically

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

• − = →

Taylor Series Expansion

➢ Non-elementary functions such as trigonometric,

exponential, and others are expressed in an approximate

fashion using Taylor series when their values, derivatives,

and integrals are computed.

➢ Any smooth function can be approximated as a polynomial.

Taylor series provides a means to predict the value of a

function at one point in terms of the function value and its

derivatives at another point.

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

1 st order approximation

2 nd order approximation

Numerical Application of Taylor Series

Finally

where

ξ is a value of x that lies somewhere between x i and x i+ 1 .

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

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.

Forward difference approximation from Taylor series

Taylor’s theorem says that if you know the value of a function '^ f 'at a point

i x^ and all its derivatives at that point, provided the derivatives are

continuous between i

(^) x and xi + 1 , then

( ) ( ) ( )( )

( )

( − ) + 



+ =^ +  + − + +

2 1 1 1

2!

i i

i

i i i i i x x

f x

f x f x f x x x

Substituting for convenience Δ x = xi + 1 − xi

( ) ( ) ( )

( ) ( ) + 



• = +  +

2 1 Δ 2!

Δ x

f x f x f x f x x

i i i i

( )

( ) ( ) ( ) (  )+ 

 − 

−  (^) =

x

f x

x

f x f x f x

i i i i 2!

1

( )

( ) ( )

( x )

x

f x f x f x

i i i +  

−  (^) =

0

1

Forward Difference Approximation

( )

( ) ( )

x

f x x f x

x

f x

Δ

Δ

Δ 0

lim + −

→

 =

For a finite '^ Δ x '

( )

( ) ( )

x

f x x f x

f x



+  −

 

x x+Δx

f(x)

Figure 1 Graphical Representation of forward difference approximation of first derivative.

Graphical Representation Of Forward

Difference Approximation

Example 1

The velocity of a rocket is given by

( ) 9. 8 , 0 30 14 10 2100

14 10 2000 ln 4

4

−    

  



 −

 = t t t

 t

where (^) ' ν ' is given in m/s and (^) ' t ' is given in seconds.

a) Use forward difference approximation of the first derivative of to

calculate the acceleration at. Use a step size of.

b) Find the exact value of the acceleration of the rocket.

c) Calculate the absolute relative true error for part (b).

ν ( ) t

t = 16 s Δ t = 2 s

Example 1 Cont.

Solution

( )

( ) ( )

t

t t

a t

i i i



−



 + 1 

ti = 16

Δ t = 2

18

16 2

1

=

= +

t (^) i + = ti + t

( )

( ) ( )

2

18 16 16

 −  a 

Example 1 Cont.

( ) ( )

9. 8 ( 18 ) 14 10 210018

14 10 18 2000 ln 4

4

^ − 

  



 −

  =

= 453. 02 m/s

( )

( )

9. 8 ( 16 )

14 10 210016

14 10 16 2000 ln 4

4

−  



 



 −

  =

= 392. 07 m/s

Hence

( )

( ) ( )

2

18 16 16

 −  a 

2

453. 02 − 392. 07 

2  30. 474 m/s

Example 1 Cont.

The exact value of a (^16 ) can be calculated by differentiating

( ) t t

t 9. 8 14 10 2100

14 10 2000 ln 4

4

^ − 

  



 −



 = as ( )  ν ( ) t 

dt

d

a t =

b)

Knowing that  ( ) t

t dt

d 1 ln = and 2

1 1

dt t t

d = −  

  



( ) 9. 8 14 10 2100

14 10

14 10

14 10 2100 2000 4

4

4

4

− 

 

 

 



 −

 

 

 

 





 −

dt t

t d at

( )

( )

( 2100 ) 9. 8 14 10 2100

14 10 1 14 10

14 10 2100 2000 4 2

4

4

4 − − 















 −

 − 

 

 

 





 −

t

t

t

t

200 3

4040 29. 4

− +

− −

( )

( )

200 3 ( 16 )

4040 29. 416 16 − +

− − a = = 𝟐𝟗. 𝟔𝟕𝟒m/s

𝟐

Example 1 Cont.

The absolute relative true error is

100 TrueValue

True Value-ApproximateValue  t = x

100

29. 674

30. 674 30. 474 x

−

= 2. 6967 %

Finite Difference Approximation of Higher Derivatives

One can use Taylor series to approximate a higher order derivative.

For example, to approximate f^ ( )^ x , the Taylor series for

( ) ( ) ( )( )

( ) ( )

( ) ( ) + 





• = +  +

2 3 2 2 Δ 3!

2 Δ 2!

2 Δ x

f x x

f x f x f x f x x

i i i i i

where xi^ + 2 = xi +^2 Δ x

( ) ( ) ( )( )

( ) ( )

( ) ( ) 

2 3 1 2! 3!

x

f x x

f x f x f x f x x

i i i i i 

  +



• = +   +

where x x x

i i

Δ

1

= −

−

Finite Difference Approximation of Higher Derivatives

Subtracting 2 times equation ( 4 ) from equation ( 3 ) gives

( ) ( ) ( ) ( )( ) ( )( ) 

2 3 f xi 2 2 f xi 1 f xi f xi Δ x f xi Δ x − =− +  + 

( )

( ) ( ) ( )

( )

− ( )( )+ 

− +  (^) =

f x x x

f x f x f x f x i

i i i i Δ Δ

2

2

2 1

( )

( ) ( ) ( )

( )

( x )

x

f x f x f x f x

i i i i^0 Δ Δ

2

2

2 1

− +  (^) 

• • ( 5 )

Example 2

The velocity of a rocket is given by

( ) 9. 8 , 0 30 14 10 2100

14 10 2000 ln 4

4

−    

  



 −

 = t t t

 t

Use forward difference approximation of the second derivative of to

calculate the “pull” or “jerk” at. Use a step size of.

ν ( ) t

t = 16 s Δ t = 2 s

(^23) http://numericalmethods.eng.usf.edu

Example 2

Solution ( )

( ) ( ) ( )

( )

2

t

t t t

j t

i i i i



− +



 +  +  = 16 ti

18

16 2

1

=

= +

t (^) i + = ti + t (^ )

( )

20

16 2 2

2 2

=

= +

t (^) i + = ti +  t

( )

( ) ( ) ( )

( )

2 2

20 2 18 16 16

 −  +  j 

 t = 2

( ) ( )

9. 8 ( 20 ) 14 10 210020

14 10 20 2000 ln 4

4

−  

  



 −

  = = 517. 35 m/s

( ) ( )

9. 8 ( 18 ) 14 10 210018

14 10 18 2000 ln 4

4

−  

  



 −

  = = 453. 02 m / s

( ) ( )

9. 8 ( 16 ) 14 10 210016

14 10 16 2000 ln 4

4

−  

  



 −

  = = 392. 07 m/s

( )

( )

4

517. 35 2 453. 02 392. 07 16

− + j 

3  0. 84515 m/s

(^24) http://numericalmethods.eng.usf.edu

Example 2

The exact value of j (^16 ) can be calculated by differentiating

( ) t t

t 9. 8 14 10 2100

14 10 2000 ln 4

4

^ − 

  



 −

  =

twice as ( )  ν ( ) t  dt

d

a t = and ( )^  a^ ( )^ t 

dt

d

j t =

Knowing that  ( ) t

t dt

d 1 (^) ln = and 2

1 1

dt t t

d =−  

  



( ) 9. 8 14 10 2100

14 10

14 10

14 10 2100 2000 4

4

4

4

− 

 

 

 



 −

 

 

 

 





 −

dt t

t d at

t

t

200 3

4040 29. 4

− +

− − ( ) (^) = ( )

( 2100 ) 9. 8 14 10 2100

14 10 1 14 10

14 10 2100 2000 4 2

4

4

4

− − 















 −

 − 

















 −

t

t

(^25) http://numericalmethods.eng.usf.edu

Example 2

( ) (^)  ( )

2 ( 200 3 )

18000

t

a t dt

d j t

− +

=

= ( )

3

2

1. 77909 m/s

[ 200 3 ( 16 )]

18000 16

=

− +

j =

The absolute relative true error is

100

1. 77909

2. 77909 0. 84515 

−  t = = 8. 4797 %

Similarly it can be shown that