1519 differentiation-integration-02

NUMERICAL DIFFERENTIATION AND INTEGRATION
ENGR 351
Numerical Methods for Engineers
Southern Illinois University Carbondale
College of Engineering
Dr. L.R. Chevalier
Dr. B.A. DeVantier

Copyright © 2003 by Lizette R. Chevalier
Permission is granted to students at Southern Illinois University at Carbondale
to make one copy of this material for use in the class ENGR 351, Numerical
Methods for Engineers. No other permission is granted.
All other rights are reserved. No part of this publication may be reproduced,
stored in a retrieval system, or transmitted, in any form or by any means,
electronic, mechanical, photocopying, recording, or otherwise, without
the prior written permission of the copyright owner.

Specific Study Objectives
• Understand the derivation of the
Newton-Cotes formulas
• Recognize that the trapezoidal and
Simpson’s 1/3 and 3/8 rules represent
the areas of 1st, 2nd, and 3rd order
polynomials
• Be able to choose the “best” among
these formulas for any particular
problem

• Recognize the difference between open
and closed integration formulas
• Understand the theoretical basis of
Richardson extrapolation and how it is
applied in the Romberg integration
algorithm and for numerical
differentiation

• Recognize why both Romberg integration and
Gauss quadrature have utility when
integrating equations (as opposed to tabular
or discrete data).
• Understand basic finite difference
approximations
• Understand the application of high-accuracy
numerical-differentiation.
• Recognize data error on the processes of
integration and differentiation.

Numerical Differentiation and
Integration
• Calculus is the mathematics of change.
• Engineers must continuously deal with
systems and processes that change, making
calculus an essential tool of our profession.
• At the heart of calculus are the related
mathematical concepts of differentiation and
integration.

Differentiation
• Dictionary definition of differentiate - “to mark
off by differences, distinguish; ..to perceive
the difference in or between”
• Mathematical definition of derivative - rate of
change of a dependent variable with respect
to an independent variable
( ) ( )
x
xfxxf
x
y ii
∆
−∆+
=
∆
∆

( ) ( )∆
∆
∆
∆
y
x
f x x f x
x
i i
=
+ −
( )f xi
( )f x xi + ∆
∆y
∆x
f(x)
x

Integration
• The inverse process of differentiation
• Dictionary definition of integrate - “to bring
together, as parts, into a whole; to unite; to
indicate the total amount”
• Mathematically, it is the total value or
summation of f(x)dx over a range of x. In fact
the integration symbol is actually a stylized
capital S intended to signify the connection
between integration and summation.

Mathematical Background
d
dx
x
d
dx
e
d
dx
x
d
dx
a
d
dx
x
d
dx
x
d
dx
x
if u and v are functions of x
d
dx
u
d
dx
uv
x
x
n
n
sin ? ?
cos ? ?
tan ? ?
ln ?
? ( ) ?
= =
= =
= =
=
= =

udv
u du
a dx
dx
x
e dx
n
bx
ax
=
=
=
=
=
∫
∫
∫
∫
∫
?
?
?
?
?
Mathematical Background

Overview
• Newton-Cotes Integration Formulas
• Trapezoidal rule
• Simpson’s Rules
• Unequal Segments
• Open Integration
• Integration of Equations
• Romberg Integration
• Gauss Quadrature
• Improper Integrals

Overview
• Numerical Differentiation
• High accuracy formulas
• Richardson’s extrapolation
• Unequal spaced data
• Uncertain data
• Applied problems

Newton-Cotes Integration
• Common numerical integration scheme
• Based on the strategy of replacing a complicated
function or tabulated data with some approximating
function that is easy to integrate
( ) ( )
( )
I f x dx f x dx
f x a a x a x
a
b
n
a
b
n n
n
= ≅
= + + +
∫ ∫
0 1 ....

Newton-Cotes Integration
• Common numerical integration scheme
• Based on the strategy of replacing a complicated
function or tabulated data with some approximating
function that is easy to integrate
( ) ( )
( )
I f x dx f x dx
f x a a x a x
a
b
n
a
b
n n
n
= ≅
= + + +
∫ ∫
0 1 .... fn(x) is an nth order
polynomial

0
1
2
3
4
5
0 5 10
x
f(x)
0
1
2
3
4
5
0 5 10
x
f(x)
The approximation of an integral by the area under
- a first order polynomial
- a second order polynomial

We can also approximated the integral by using a
series of polynomials applied piece wise.
0
1
2
3
4
5
0 5 10
x
f(x)
0
1
2
3
4
5
0 5 10
x
f(x)
The approximation of an integral by the area under
- a first order polynomial
- a second order polynomial

0
1
2
3
4
5
0 1 2 3 4 5 6 7 8 9 10
x
f(x)
An approximation of an integral by the area under
straight line segments.

Newton-Cotes Formulas
• Closed form - data is at the beginning and
end of the limits of integration
• Open form - integration limits extend
beyond the range of data.
0
1
2
3
4
5
0 1 2 3 4 5 6 7 8 9 10
x
f(x)
0
1
2
3
4
5
0 1 2 3 4 5 6 7 8 9 10
x
f(x)

Trapezoidal Rule
• First of the Newton-Cotes closed
integration formulas
• Corresponds to the case where the
polynomial is a first order
( ) ( )
( )
I f x dx f x dx
f x a a x
a
b
a
b
n
= ≅
= +
∫ ∫ 1
0 1

( ) ( )
( )
I f x dx f x dx
f x a a x
a
b
a
b
n
= ≅
= +
∫ ∫ 1
0 1
A straight line can be represented as:
( ) ( )
( ) ( )
( )f x f a
f b f a
b a
x a1 = +
−
−
−
Trapezoidal Rule

( ) ( )
( )
( ) ( )
( )
I f x dx f x dx
f a
f b f a
b a
x a dx
a
b
a
b
a
b
= ≅
= +
−
−
−
∫ ∫
∫
1
Integrate this equation. Results in the
trape zo idalrule .
( )
( ) ( )I b a
f a f b
≅ −
+
2
Trapezoidal Rule

( ) ( ) ( )
2
bfaf
abI
+
−≅
Recall the formula for computing the area of a
trapezoid:
height x (average of the bases)
height
base
base
Trapezoidal Rule

The concept is the same but the trapezoid
is on its side.
height
base
base
widthheight
height
( ) ( ) ( )
2
bfaf
abI
+
−≅
Trapezoidal Rule

Error of the Trapezoidal Rule
( )( )E f b a
where a b
t = − −
< <
1
12
3
'' ξ
ξ
This indicates that is the function being integrated is linear, the
trapezoidal rule will be exact.
Otherwise, for section with second and higher order derivatives (that is
with curvature) error can occur.
A reasonable estimate of x is the average value of b and a

Multiple Application of the
Trapezoidal Rule
• Improve the accuracy by dividing the integration
interval into a number of smaller segments
• Apply the method to each segment
• Resulting equations are called multiple-application or
composite integration formulas

( ) ( ) ( ) ( ) ( ) ( )
I f x dx f x dx f x dx
I h
f x f x
h
f x f x
h
f x f x
x
x
x
x
x
x
n n
n
n
= + + +
≅
+
+
+
+
+
∫ ∫ ∫
−
−
( ) ( ) ( )
0
1
1
2
1
0 1 1 2 1
2 2 2


where there are n+ 1 equally spaced base points.
Trapezoidal Rule

( ) ( ) ( ) ( ) ( ) ( )
( )
( ) ( ) ( )
I h
f x f x
h
f x f x
h
f x f x
I b a
f x f x f x
n
x
x
x
x
x
x
n n
i n
i
n
n
n
= + + +
≅
+
+
+
+
+
≅ −
+ +
∫ ∫ ∫
∑
−
−
=
−
( ) ( ) ( )
0
1
1
2
1
0 1 1 2 1
0
1
1
2 2 2
2
2


We can group terms to express a general form
}
}
width average height
Trapezoidal Rule

( )
( ) ( ) ( )
I b a
f x f x f x
n
i n
i
n
≅ −
+ +
=
−
∑0
1
1
2
2
}
}width average height
The average height represents a weighted average
of the function values
Note that the interior points are given twice the weight
of the two end points
( )
E
b a
n
fa = −
−
3
2
12
''
Trapezoidal Rule

Example
Evaluate the following integral using the trapezoidal rule
and h = 0.1
( )
( ) ( ) ( )
I b a
f x f x f x
n
i n
i
n
≅ −
+ +
=
−
∑0
1
1
2
2
I e dxx
= ∫
2
1
1 6.
h
b a
n
=
−
0
5
10
15
20
25
30
0 0.3 0.6 0.9 1.2 1.5 1.8
x
f(x)

Simpson’s 1/3 Rule
function is a second order polynomial
( ) ( )
( )
I f x dx f x dx
f x a a x a x
a
b
a
b
n
= ≅
= + +
∫ ∫ 2
0 1 2
2

• Designate a and b as x0 and x2, and
estimate f2(x) as a second order
Lagrange polynomial
( ) ( )
( )( )
( )( )
( )
I f x dx f x dx
x x x x
x x x x
f x dx
a
b
a
b
x
x
= ≅
=
− −
− −
+






∫ ∫
∫
2
1 2
0 1 0 2
0
0
2
.......

• After integration and algebraic
manipulation, we get the following
equations
( ) ( ) ( )[ ]
( )
( ) ( ) ( )
I
h
f x f x f x
b a
f x f x f x
≅ + +
≅ −
+ +
3
4
4
6
0 1 2
0 1 2
}
}
width average height

Error
( )( )E f b a
where a b
t = − −
< <
1
12
3
'' ξ
ξ
Single application of Trapezoidal Rule.
Single application of Simpson’s 1/3 Rule
( )( )E f b at = − −
1
2880
4 5( )
ξ

Multiple Application of Simpson’s
1/3 Rule
( )
( ) ( ) ( ) ( )
( ) ( )
I b a
f x f x f x f x
n
E
b a
n
f
x
x
x
x
x
x
i j n
j
n
i
n
a
n
n
= + + +
≅ −
+ + +
= −
−
∫ ∫ ∫
∑∑
−
=
−
=
−
( ) ( ) ( )
, , .., , ..
0
1
1
2
1
0
2 4 6
2
1 3 5
1
5
4
4
4 2
3
180


( )
( ) ( ) ( ) ( )
I b a
f x f x f x f x
n
i j n
j
n
i
n
≅ −
+ + +
=
−
=
−
∑∑0
2 4 6
2
1 3 5
1
4 2
3
, , .., , ..
The odd points represent the middle term for each
application. Hence carry the weight 4.
The even points are common to adjacent
applications and are counted twice.
f(x)
x
i=1 (odd)
weight of 4
i=2 (even)
weight of 2

function is a third order polynomial
( ) ( )
( )
( ) ( ) ( ) ( )[ ]3210
3
3
2
210
3
33
8
3
xfxfxfxf
h
I
xaxaxaaxf
dxxfdxxfI
n
b
a
b
a
+++≅
+++=
≅= ∫∫

Integration of Unequal
Segments
• Experimental and field study data is often
unevenly spaced
• In previous equations we grouped the term
(i.e. hi) which represented segment width.
( )
( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( )
I b a
f x f x f x
n
I h
f x f x
h
f x f x
h
f x f x
i n
i
n
n n
≅ −
+ +
≅
+
+
+
+
+
=
−
−
∑0
1
1
0 1 1 2 1
2
2
2 2 2


Segments
• We should also consider alternately using
higher order equations if we can find data in
consecutively even segments
trapezoidal
rule

Segments
trapezoidal
rule
1/3
rule

Segments
trapezoidal
rule
1/3
rule
3/8
rule

Segments
trapezoidal
rule
1/3
rule
3/8
rule
trapezoidal
rule

Example
Integrate the following using the trapezoidal rule, Simpson’s
1/3 Rule, a multiple application of the trapezoidal rule with
n=2 and Simpson’s 3/8 Rule. Compare results with the
analytical solution.
xe dxx2
0
4
∫
0
2000
4000
6000
8000
10000
12000
14000
0 1 2 3 4
x
f(x)

Integration of Equations
• Integration of analytical as opposed to
tabular functions
• Romberg Integration
• Richardson’s Extrapolation
• Romberg Integration Algorithm
• Gauss Quadrature
• Improper Integrals

Richardson’s Extrapolation
• Use two estimates of an integral to compute a third
more accurate approximation
• The estimate and error associated with a multiple
application trapezoidal rule can be represented
generally as:
• I = I(h) + E(h)
• where I is the exact value of the integral
• I(h) is the approximation from an n-segment
application
• E(h) is the truncation error
• h is the step size (b-a)/n

Make two separate estimates using step sizes
of h1 and h2 .
I(h1) + E(h1) = I(h2) + E(h2)
Recall the error of the multiple-application of the trapezoidal
rule
E
b a
h f= −
−
12
2
''
Assume that is constant regardless of the step sizef ''
( )
( )
E h
E h
h
h
1
2
1
2
2
2
≅

( )
( )
( ) ( )
E h
E h
h
h
E h E h
h
h
1
2
1
2
2
2
1 2
1
2
2
≅
≅






Substitute into previous equation:
I(h1) + E(h1) = I(h2) + E(h2)
( )
( ) ( )
E h
I h I h
h
h
2
1 2
1
2
2
1
=
−
−







Thus we have developed an estimate of the truncation
error in terms of the integral estimates and their step
sizes. This estimate can then be substituted into:
I = I(h2) + E(h2)
to yield an improved estimate of the integral:
( ) ( ) ( )[ ]I I h
h
h
I h I h≅ +


 

 −










−2
1
2
2 2 1
1
1
( )
( ) ( )
E h
I h I h
h
h
2
1 2
1
2
2
1
=
−
−







( ) ( ) ( )[ ]I I h
h
h
I h I h≅ +


 

 −










−2
1
2
2 2 1
1
1
What is the equation for the special case
where the interval is halved?
i.e. h2 = h1 / 2

( ) ( ) ( )[ ]
( ) ( )
h
h
h
h
I I h I h I h
collecting terms
I I h I h
1
2
2
2
2 2 2 1
2 1
2
2
1
2 1
4
3
1
3


 

 = 

 

 =
≅ +
−



 −
≅ −

Example
Use Richardson’s extrapolation to evaluate:
xe dxx2
0
4
∫
0
2000
4000
6000
8000
10000
12000
14000
0 1 2 3 4
x
f(x)

( ) ( )
( ) ( )
( ) ( )
I I h I h
I I h I h
I I h I h
I
I I
m l
m l
j k
k
j k j k
k
≅ −
≅ −
≅ −
≅
−
−
−
+ − −
−
4
3
1
3
16
15
1
15
64
63
1
63
4
4 1
2 1
1
1 1 1
1,
, ,
We can continue to improve the estimate by successive
halving of the step size to yield a general formula:
k = 2; j = 1
Romberg Integration
Note:
the subscripts
m and lrefer to
m o re and le ss
accurate estimates

Gauss Quadrature
f(x)
x
f(x)
x
Extend the area
under the straight
line

Method of Undetermined
Coefficients
Recall the trapezoidal rule
( )
( ) ( )
I b a
f a f b
≅ −
+
2
This can also be expressed as
( ) ( )I c f a c f b≅ +0 1
where the c’s are constant
Before analyzing
this method,
answer this
question.
What are two
functions that
should be evaluated
exactly
by the trapezoidal
rule?

The two cases that should be evaluated exactly
by the trapezoidal rule: 1) y = constant
2) a straight line
f(x)
x
y = 1
(b-a)/2-(b-a)/2
f(x)
x
y = x
(b-a)/2
-(b-a)/2

Thus, the following equalities should hold.
( ) ( )
( )
( )
( )
( )
I c f a c f b
c c dx
c
b a
c
b a
xdx
b a
b a
b a
b a
≅ +
+ ≅
−
−




 +
−




 ≅
−
−
−
−
−
−
∫
∫
0 1
0 1
2
2
0 1
2
2
1
2 2
FOR y=1
since f(a) = f(b) =1
FOR y =x
since f(a) = x =-(b-a)/2
and
f(b) = x =(b-a)/2

Evaluating both integrals
c c b a
c
b a
c
b
0 1
0 1
2
1
2
0
+ = −
−
−
+
−
=
For y = 1
For y = x
Now we have two equations and two unknowns, c0 and c1.
Solving simultaneously, we get :
c0 = c1 = (b-a)/2
Substitute this back into:
( ) ( )I c f a c f b≅ +0 1

( )
( ) ( )
I b a
f a f b
≅ −
+
2
We get the equivalent of the trapezoidal rule.
DERIVATION OF THE TWO-POINT
GAUSS-LEGENDRE FORMULA
( ) ( )I c f x c f x≅ +0 0 1 1
Lets raise the level of sophistication by:
- considering two points between -1 and 1
- i.e. “open integration”

f(x)
x-1 x0 x1 1
Previously ,we assumed that the equation
fit the integrals of a constant and linear function.
Extend the reasoning by assuming that it also
fits the integral of a parabolic and a cubic
function.

( ) ( )
( ) ( )
( ) ( )
( ) ( )
c f x c f x dx
c f x c f x xdx
c f x c f x x dx
c f x c f x x dx
0 0 1 1
1
1
0 0 1 1
1
1
0 0 1 1
2
1
1
0 0 1 1
3
1
1
1 2
0
2 3
0
+ = =
+ = =
+ = =
+ = =
−
−
−
−
∫
∫
∫
∫
/
We now have four
equations and four
unknowns
c0 c1 x0 and x1
What equations are
you solving?

( ) ( ) ( ) ( )
( ) ( )
( ) ( )
( ) ( )
c f x c f x dx c c
c f x c f x xdx c x c x
c f x c f x x dx c x c x
c f x c f x x dx c x c x
0 0 1 1
1
1
0 1
0 0 1 1
1
1
0 0 1 1
0 0 1 1
2
1
1
0 0
2
1 1
2
0 0 1 1
3
1
1
0 0
3
1 1
3
1 2 1 1 2
0 0
2 3 2 3
0 0
+ = = + =
+ = = + =
+ = = + =
+ = = + =
−
−
−
−
∫
∫
∫
∫
`
/ /
Solve these equations simultaneously
f(xi) is either 1, xi, xi
2
or xi
3

c c
x
x
I f f
0 1
0
1
1
1
3
1
3
1
3
1
3
= =
=
−
=
≅
−




 +






This results in the following
The interesting result is that the integral can be
estimated by the simple addition of the function values
at −1
3
1
3
and

A simple change in variables can be use to translate
other limits.
Assume that the new variable xd is related to the
original variable x in a linear fashion.
x = a0 + a1xd
Let the lower limit x = a correspond to xd = -1 and the
upper limit x=b correspond to xd=1
a = a0 + a1(-1) b = a0 + a1(1)
What if we aren’t integrating from –1 to 1?

a = a0 + a1(-1) b = a0 + a1(1)
SOLVE THESE EQUATIONS
SIMULTANEOUSLY
a
b a
a
b a
0 1
2 2
=
+
=
−
( ) ( )x a a x
b a b a x
d
d
= + =
+ + −
0 1
2
substitute

( ) ( )x a a x
b a b a x
dx
b a
dx
d
d
d
= + =
+ + −
=
−
0 1
2
2
These equations are substituted for x and dx
respectively.
Let’s do an example to appreciate the theory
behind this numerical method.

Example
Estimate the following using two-point Gauss Legendre:
xe dxx2
0
4
∫
0
2000
4000
6000
8000
10000
12000
14000
0 1 2 3 4
x
f(x)

Higher-Point Formulas
( ) ( ) ( )I c f x c f x c f xn n≅ + + + − −0 0 1 1 1 1
For two point, we determined that c0 = c1= 1
For three point:
c0 = 0.556 x0=-0.775
c1= 0.889 x1=0.0
c2= 0.556 x2=0.775

Higher-Point Formulas
Your text goes on to provide additional weighting
factors (ci’s) and function arguments (xi’s)
in Table 8.5 p. 593

Numerical Differentiation
• Forward finite divided difference
• Backward finite divided difference
• Center finite divided difference
• All based on the Taylor Series
( ) ( ) ( )
( )
.........
!2
''
' 2
1 +++=+ h
xf
hxfxfxf i
iii

Forward Finite Difference
( ) ( ) ( )
( )
( )
( ) ( )
( )
( )
( ) ( ) ( ) ( )21
1
2
1
2
''
'
'
.........
!2
''
'
hOh
xf
h
xfxf
xf
hO
h
xfxf
xf
h
xf
hxfxfxf
iii
i
ii
i
i
iii
+−
−
=
+
−
=
+++=
+
+
+

Forward Divided Difference
( )
( ) ( )
( ) ( )f x
f x f x
x x
O x x
f
h
O hi
i i
i i
i i
i
' =
−
−
+ − = ++
+
+
1
1
1
∆
f(x)
x
(xi, yi)
What is derivative at this
point?

( )
( ) ( )
( ) ( )f x
f x f x
x x
O x x
f
h
O hi
i i
i i
i i
i
' =
−
−
+ − = ++
+
+
1
1
1
∆
f(x)
x
(xi, yi)
(x i+1,y i+1)
Determine a second point
base on Dx (h)

( )
( ) ( )
( ) ( )f x
f x f x
x x
O x x
f
h
O hi
i i
i i
i i
i
' =
−
−
+ − = ++
+
+
1
1
1
∆
f(x)
x
(xi, yi)
(x i+1,y i+1)
estimateHow does
this compare
to the actual
first derivative
at xi?

( )
( ) ( )
( ) ( )f x
f x f x
x x
O x x
f
h
O hi
i i
i i
i i
i
' =
−
−
+ − = ++
+
+
1
1
1
∆
f(x)
x
(xi, yi)
(x i+1,y i+1)
estimate
actual

( ) ( )f x
f
h
O hi
i
' = +
∆
first forward divided difference
Error is proportional to
the step size
O(h2
) error is proportional to the square of the step size
O(h3
) error is proportional to the cube of the step size

f(x)
x
estimate
actual
(xi,yi)
(xi-1,yi-1)

( ) ( ) ( )
( )
( ) ( ) ( )
( )
( )
( ) ( )
f x f x f x h
f x
h
f x f x f x h
f x
h
f x
f x f x
h
f
h
i i i
i
i i i
i
i
i i i
+
−
−
= + + +
= − + +
=
−
=
∇
1
2
1
2
1
2
2
'
''
!
......
'
''
!
.....
'
Backward Difference Approximation of the
First Derivative
Expand the Taylor series backwards
The error is still O(h)

Centered Difference Approximation of the
First Derivative
Subtract backward difference approximation
from forward Taylor series expansion
( ) ( ) ( )
( )
( )
( ) ( )
( ) ( ) ( )
( )
( )
( ) ( )
( )211
2
11
1
1
2
1
2
'
6
'''
'2
'
....
!2
''
'
hO
h
xfxf
xf
h
xf
hxfxfxf
xx
xfxf
xf
h
xf
hxfxfxf
ii
i
i
iii
ii
ii
i
i
iii
−
−
=
+++=
−
−
=
+++=
−+
−−
+
+
+


f(x)
x
estimate
actual
(xi,yi)
(xi-1,yi-1)
(xi+1,yi+1)
( )
( ) ( ) ( )211
2
' hO
h
xfxf
xf ii
i −
−
= −+

f(x)
x
f(x)
x
f(x)
x
f(x)
x
true derivative
forward
finite divided
difference approx.
backward
finite divided
difference approx.
centered
finite divided
difference approx.

Numerical Differentiation
• You should be familiar with the following
Tables in your text
• Table 7.1: Common Finite Difference
Formulas
• Table 7.2: Higher order finite difference
formulas

Richardson Extrapolation
• Two ways to improve derivative
estimates
¤ decrease step size
¤ use a higher order formula that employs
more points
• Third approach, based on Richardson
extrapolation, uses two derivatives
estimates to compute a third, more
accurate approximation

Richardson Extrapolation
( ) ( ) ( )[ ]
( ) ( )
( ) ( )
I I h
h
h
I h I h
Special case where h
h
I I h I h
In a similar fashion
D D h D h
≅ +


 

 −
−
=
≅ −
≅ −
2
1
2
2 2 1
2
1
2 1
2 1
1
1
2
4
3
1
3
4
3
1
3
For a centered difference
approximation with
O(h2
) the application of
this formula will yield
a new derivative estimate
of O(h4
)

Example
Given the following function, use Richardson’s
extrapolation to determine the derivative at 0.5.
f(x) = -0.1x4
- 0.15x3
- 0.5x2
- 0.25x +1.2
Note:
f(0) = 1.2
f(0.25) =1.1035
f(0.75) = 0.636
f(1) = 0.2

Derivatives of Unequally
Spaced Data
• Common in data from experiments or field studies
• Fit a second order Lagrange interpolating polynomial
to each set of three adjacent points, since this
polynomial does not require that the points be equi-
spaced
• Differentiate analytically
( ) ( )
( )( )
( )
( )( )
( )
( )( )
f x f x
x x x
x x x x
f x
x x x
x x x x
f x
x x x
x x x x
i
i i
i i i i
i
i i
i i i i
i
i i
i i i i
' =
− −
− −
+
− −
− −
+
− −
− −
−
+
− − +
− +
− +
+
−
+ + −
1
1
1 1 1
1 1
1 1
1
1
1 1 1
2 2
2

Derivative and Integral
Estimates for Data with Errors
• In addition to unequal spacing, the other
problem related to differentiating empirical data
is measurement error
• Differentiation amplifies error
• Integration tends to be more forgiving
• Primary approach for determining derivatives
of imprecise data is to use least squares
regression to fit a smooth, differentiable
function to the data
• In absence of other information, a lower order
polynomial regression is a good first choice

0
5 0
1 00
1 5 0
200
25 0
0 5 1 0 1 5
t
y
0
1 0
20
30
0 1 0
t
dy/dt
0
5 0
1 00
1 5 0
200
25 0
0 5 1 0 1 5
t
y
0
1 0
20
30
40
0 1 0
t
dy/dt

• Recognize why both Romberg
integration and Gauss quadrature have
utility when integrating equations (as
opposed to tabular or discrete data).
• Understand the application of high-
accuracy numerical-differentiation.
• Recognize data error on the processes
of integration and differentiation.

1519 differentiation-integration-02

Recommandé

Recommandé

Contenu connexe

Tendances

Tendances (20)

En vedette

En vedette (20)

Similaire à 1519 differentiation-integration-02

Similaire à 1519 differentiation-integration-02 (20)

Plus de Dr Fereidoun Dejahang

Plus de Dr Fereidoun Dejahang (20)

Dernier

Dernier (20)

1519 differentiation-integration-02