Module 2 Lesson 5 Numerical Differentiation Backward Difference Method, Lecture notes of Engineering Mathematics

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Module 2

Lesson 5

Numerical Differentiation

Backward 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 :

=Δ𝑥 lim 0

𝑎

𝑏

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

• − = →

Backward Difference Approximation of the

First Derivative Cont.

This is a backward difference approximation as you are taking a point

backward from x. To find the value of (^) f ( ) x at i x = x , we may choose another

point (^) ' Δ x '

behind as x =^ xi − 1. This gives

( )

( ) ( )

x

f x f x f x

i i i 

−  (^) 

− 1

( ) ( )

1

1

−

−

i i

i i

x x

f x f x

where

1 Δ − = − i i x x x

x-Δx^ x

x

f(x)

Figure 2 Graphical Representation of backward difference

approximation of first derivative

Backward Difference Approximation of the First

Derivative Cont.

Example 2

Solution

( )

( ) ( )

t

t t

a t

i i

  − (^1) = 16

ti Δ t = 2

14

16 2

1

=

= −

t (^) i − = ti − t

( )

( ) ( )

a 

( ) ( )

9. 8 ( 16 ) 14 10 210016

16 2000 ln 4

4

^ −

 = = 392. 07 m/s

( ) ( )

14 2000 ln

4

4

 = = 334. 24 m/s

( )

( ) ( )

2

16 14 16

 −  a 

2

 28. 915 m/s

Example 2

The absolute relative true error is

t x

The exact value of the acceleration at from Example 1 is

( )

2

a 16 = 29. 674 m/s

t = 16 s

Backward Difference Formulas- 2

nd

derivative

❑ Start with Lagrange interpolation polynomial for f (x) based on the four

points x

i

, x

i- 1

, x

i- 2

and x

i- 3

❑ Differentiate the products in the numerators twice

❑ Substitute x = xi and consider the fact that xj - xi =(j- i)h

❑ The expression of the second derivative is then:

ℎ^2

2 )