ECE 3040 Lecture 21: Numerical Differentiation

ECE 3040

Lecture 21: Numerical Differentiation

? Prof. Mohamad Hassoun

This lecture covers the following topics:

?

?

?

?

Introduction

Numerical Differentiation formulas

Richardson extrapolation

Matlab numerical differentiation function: diff

? Matlab polynomial and symbolic differentiation: polyder and diff

Introduction

The differentiation of a function has many engineering applications, from finding

slopes (rate of change) to solving optimization problems to differential equations

that model electric circuits and mechanical systems. Given a function ?(?), we

denote its first derivative with respect to its independent variable ? by the symbol

(devised by the German Gottfried Leibniz, 1646-1716),

??(?)

??

. Alternative symbols

for the first derivative are: ?��(?) (attributed to the French Joseph-Louis Lagrange,

1736-1813) and ?? (?) (attributed to the English Isaac Newton, 1642-1726). Recall

(from your calculus course) that the first derivative of an analytic function ?(?) is

an analytic function expressed as the limit

?? (? ) = ? �� (? ) =

??(?)

??

?(? + ?? ) ? ?(?)

= lim

= lim

??��0 ??

??��0

??

??

The derivative of a function at a specific point ?? ,

??(? ) = ? �� (?? ) =

??(?)

??(?? )

?(?? + ?? ) ? ?(?? )

|

=

= lim

??��0

?? ?=??

??

??

is a numeric value that specifies the slope of the function at the point (?? , ?(?? )), as

depicted in the following figures.

Recall that the straight line that passes through the point (?? , ?(?? )), and has a slope

? �� (?? ), is a tangent line to ?(?).

The second derivative represents the derivative of the first derivative and is denoted

by

??(?) �� ? ��� (? ) ��

? ??(?)

?2 ?(?)

(

)��

?? ??

?? 2

The second derivative measures the rate of change of the first derivative and is

commonly referred to as the curvature. From physics, if ?(?) describes the position

of a moving object then ?(?) =

is the object��s acceleration.

??(?)

??

is the velocity of the object, and ?(?) =

?2 ?(?)

?? 2

Numerical Differentiation Formulas

We have already derived few basic numerical differentiation formulas back in

Lecture 10 in connection with truncation error. The following is a repeat (for

convenience) of such derivation. Starting from the Taylor series expansion of the

function ?(?) at a point ? = ?? and evaluating the series at the point ? = ??+1 ,

close to ?? , we obtain the (forward) Taylor expansion,

? ��� (?? )

? (3) (?? )

3

? (??+1 ) = ?(?? ) + ?? ?? ) + ?

+?

+?

2!

3!

��(

2

(1)

where ? = ??+1 ? ?? is referred to as the step size. We may now solve Eqn. (1) for

the derivative at ?? to obtain

? (??+1 ) ? ? (?? )

? ��� (?? )

? ?? ) =

??

??

?

2!

��(

(2)

or

? �� (?? ) =

?(??+1 ) ? ?(?? )

? ?(?)

?

Here, ?(?) means that the remainder of the series is of order ?. That is, if the terms

in the second and higher derivatives are dropped then the truncation error would be

proportional to the step size, ?.

The last expression should not be strange to us. In fact, if we take the limit of that

expression as ? approaches zero we arrive at the definition of the derivative [note

that ?(?) approaches zero as h approaches zero]:

? �� (?? ) = lim

?��0

? (??+1 ) ? ? (?? )

?(?? + ?) ? ?(?? )

? lim ?(?) = lim

?��0

?��0

?

?

Using a finite (but small) ?, results in the forward finite-difference formula for

numerically approximating the first derivative,

? �� (?? ) ?

? (??+1 ) ? ? (?? )

?

Based on Eqn. (2), the truncation error for the above formula is proportional to,

?���(?). This means that the formula is exact for linear functions.

The convergence of the forward difference method is of order ?, and is relatively

slow; i.e., a very, very small (positive) ? is required in order to achieve high

accuracy.

Alternatively, the Taylor series can be applied backwards with ???1 < ?? as

follows,

? ��� (?? )

? (3) (?? )

3

?(???1 ) = ?(?? ) + (???1 ? ?? )? ?? ) + (???1 ??? )

+ (???1 ??? )

2!

3!

+?

��(

2

The above expression can be rewritten, with positive ? = ?? ? ???1 , as

? ��� (?? )

? (3) (?? )

3

? (???1 ) = ?(?? ) ? ?? ?? ) + ?

??

+?

2!

3!

��(

2

(3)

from which one can derive the order ?(?) backward finite-difference formula for

computing the first derivative,

? (?? ) ? ? (???1 )

��( )

? ?? ?

?

A third way to approximate the derivative is to subtract the backward Taylor series

[Eqn. (3)] from the forward Taylor series [Eqn. (1)] to yield

? (3) (?? )

?(??+1 ) = ? (???1 ) + 2?? ?? ) + 2?

+?

3!

��(

3

which can be solved for the first derivative as

?(??+1 ) ? ? (???1 )

? (3) (?? )

2

? (?? ) =

??

+?

2?

6

��

or

? �� (?? ) =

?(??+1 ) ? ?(???1 )

? ?(?2 )

2?

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