numerical analysis- math 292 -lec 5- numerical differentiations & intergations
DESCRIPTION
numericalTRANSCRIPT
-
Numerical Analysis (MATH 292)
Lecture 5- Numerical Integration and Differentiation
1- Numerical Integration
Newton - Cotes formula
b b
na a
I f x dx f x dx Where fn(x) is a polynomial of the form
1 20 1 2 ...n
n nf x a a x a x a x Where n is order of the polynomial
The Trapezoidal rule:
.2
f b f aI b a
Error in Trapezoidal rule
True error: Et = Itrue - Iapprox
-
True percent relative error: 100true approx
t
true
I I
I
Simple application of the trapezoidal rule
2 3 4 50.2 25 200 675 900 400f x x x x x x From a = 0 to b = 0.8
The exact value of the integral is 1.640533
Solution:
0 0.2
0.8 0.2323
f
f
0.17282
f b f aI b a
1.640553 0.1728 1.46775tE
-
1.46775
100 89.5%1.640533
t
In actual situations, we would have no knowledge of the true value. Therefore, an
approximate error estimate is required
3
2
1''
12aE f b a
n
where
''''
b
af x dx
fb a
In the example above:
2 3'' 400 4050 108000 8000f x x x x
0.82 3
0'' 400 4050 108000 8000
''0.8 0
60
b
af x dx x x x dx
fb a
-
3 3
2
1 1'' 60 0.8 2.56
12 12aE f b a
n
Multiple application of the trapezoidal rule
1
0
1
22
n
i n
i
b aI f x f x f x
n
Example:
Use the trapezoidal rule to obtain an approximate value of the integral
2
315
1
d xwhere n
x
Solution:
x 1 1.2 1.4 1.6 1.8 2
3
1
1x 0.707107 0.605449 0.516811 0.442981 0.382583 0.33333
-
2 1
[0.707107 2(0.605449 0.516811 0.44298110
0.382583) 0.33333] 0.493609
b
af x dx
Simpsons 1/3 Rule
1 2
0
1,3,5 2,4,6
4 23
n n
i i n
i i
b aI f x f x f x f x
n
Example:
Use the Simpsons 1/3 rule to obtain an approximate value for the integral from 0
to 0.8 (n=2). 2 3 4 50.2 25 200 675 900 400f x x x x x x
0
1
2
0.2
0.4 2.456
0.8 0.232
f x f a
f x f
f x f
-
0.8
0.2 4 2.456 0.232 1.3674672 3
I
Recall that 1.640533trueI
1.640533 1.367467 0.2730667tE
16.6%t Error in Simpsons rule:
5
4
4180a
b aE f
n
where 4f is the average fourth derivative
0.8
2
4 024 900 120 400
24000.8 0
x x dxf
5
0.82400 0.273
2880a
-
Example:
Use Simpsons 1/3 rule with n=4 to estimate the integral
2 3 4 50.2 25 200 675 900 400f x x x x x x
Solution:
0 4 23
n
b aI f x odd even f x
n
x 0 0.2 0.4 0.6 0.8
y = f(x) 0.2 1.288 2.456 3.464 0.232
n = 4 0.8
0.24
h
0.2
0.2 4 1.288 3.464 2 2.456 0.232 1.6234673
I
5
4
0.82400 0.017067
180 4aE
1.04%t
-
Simpsons 3/8 Rule: In the similar manner to the derivation of the trapezoid and Simpsons 1/3 rule, a third order Lagrange polynomial can be fit to four points and integrate
3b b
a aI f x dx f x dx
To yield
0 1 2 33
3 38
I h f x f x f x f x
Where b a
hn
Use Simpsons 3/8 to integrate the same function for n = 3
2 3 4 50.2 25 200 675 900 400f x x x x x x
0.8 0
0.26673
b ah
n
0 0.2 0.5333 3.487177
0.2667 1.432724 0.8 0.232
f f
f f
-
0.2 3 1.432724 3.487177 0.2320.8 1.519170
8I
5
2400 1.640533 1.51917 0.1213636480
a
b aE
7.4%t
For n = 5, use the same function to integrate the first two segments using Simpsons 1/3 and the last 3 segments using Simpsons 3/8
0.80.16
5h
0 0.2f 0.16 1.296919f 0.32 1.743393f 0.48 3.186015f 0.64 3.181929f 0.8 0.323f
0.16 0.32 0.48 0.64 0.80
Simpsons 1/3 rule
Simpsons 3/8 rule
-
The integration for the first two-segments is obtained using Simpsons 1/3 rule.
0.16
0.2 4 1.296919 1.743393 0.38032373
I
For the last 3 segments, the Simpsons 3/8 rule can be used to obtain
3
0.16 1.743393 3 3.186015 3.181929 0.323 1.2647548
I
The total integral
0.3803237 1.264754 1.645077I
0.00454383tE 0.28 %t
Errors in numerical integration:
t t appE I I
100true app
t
true
I I
I
-
Approximate Error Estimate:
1) Trapezoidal rule
3
2''
12a
b aE f
n
2) Simpsons 1/3 rule
5
4
4180a
b aE f
n
3) Simpsons 3/8 rule
5
4
480a
b aE f
n
-
Numerical Differentiation
Mathematically, the derivative represents the rate of change of a dependent variable with
respect to an independent variable.
The mathematical definition of the derivative begins with a difference approximation
( ) ( )i if x x f xy
x x
If x is allowed to approach zero, the difference becomes a derivative
0
( ) ( )lim i ix
f x x f xdy
dx x
The derivative dy
dx is the slope of the tangent to the curve ( )y f x at
ix .
-
Taylor Series:
Taylor series is of great value in the study of numerical methods.
By using the Taylor method we can predict a function value at one point in terms of the
function value and its derivatives at another point.
2
1 1 1
( )
1
''' ...
2!
!
i
i i i i i i i
nni
i i n
f xf x f x f x x x x x
f xx x R
n
where nR is the reminder term.
It is usually convenient to simplify the Taylor series by defining a step size 1i ih x x
Let 1i ix x h , then
( )2
1
'''
2! !
n
ni i
i i i n
f x f xf x f x f x h h h R
n
-
Finite divided difference approximation of derivative
First derivative
Taylor series can be solved for
1
1
( )
2
1
' ( ) (1)
where ( )
''and
2! !
i i
i
n
ni i
n
f x f xf x O h
h
RO h
h
f x f xR h h R
n
1( ) ( )i if x f x is the first forward difference and h is called the step size ( length of
interval)
-
There are three ways to approximate the first derivative form Taylor series
1- Forward difference: 1' i ii
f x f xf x
h
2- Backward difference: 1' i ii
f x f xf x
h
3- Centered difference: 1 1'
2
i ii
f x f xf x
h
Example:
Use forward, backward and centered difference approximation to estimate the first
derivative of
4 3 20.1 0.15 0.5 0.25 1.2f x x x x x
at 0.5x using a step size 0.5h . Repeat the computation using 0.25h , compute the
true percent relative error
-
Solution:
Exact first derivative 3 2' 0.4 0.45 0.25f x x x x ' 0.5 0.9125f
For h = 0.5
1 1
1 1
0 1.2
0.5 0.925
1.0 0.2
i i
i i
i i
x f x
x f x
x f x
Forward divided difference:
0.2 0.925
' 1.45 58.9%0.5
i tf x
Backward divided difference:
0.925 1.2
' 0.55 39.7%0.5
i tf x
Centered divided difference:
0.2 1.2
' 1.0 9.6%1.0
i tf x
-
For h = 0.25
1 1
1 1
0.25 1.10351
0.5 0.925
0.75 0.6363281
i i
i i
i i
x f x
x f x
x f x
Forward divided difference:
' 0.25 1.155 26.5%tf
Backward divided difference:
' 0.5 0.714 21.7%tf
Centered divided difference:
' 0.75 0.934 2.4%tf
For both step sizes, the centered difference approximation is more accurate than the other.
-
Finite difference of higher derivative:
Taylor Series expansion can be used to derive numerical estimates of higher derivatives.
To do this, we write a forward Taylor expansion of 2( )if x in terms of ( )if x
2
2
''' 2 2 ...
2!
i
i i i
f xf x f x f x h h (2)
The Taylor expansion of 1( )if x is
2
1
''' ...
2!
i
i i i
f xf x f x f x h h (3)
Multiply equation (3) by 2 and subtract from equation (2)
22 12 '' ...i i i if x f x f x f x h
2 122
'' i i iif x f x f x
f x O hh
(4)
Equation (4) is called the second forward finite divided difference.
-
High-accuracy differentiation Formulas
High-accuracy forward divided-difference
We have
1 '''
2
i i i
i
f x f x f xf x h
h
(5)
Substitution of equation (4) into equation (5), gives
1 2 1
2
2 1
2' .
2
4 3
2
i i i i i
i
i i i
f x f x f x f x f x hf x
h h
f x f x f x
h
High-accuracy backward divided-difference
1 23 4( )
2
i i i
i
f x f x f xf x
h
-
High-accuracy centered divided-difference
2 1 1 28 8( )
12
i i i i
i
f x f x f x f xf x
h
Example:
Find the first derivative of 4 3 20.1 0.15 0.5 0.25 1.2f x x x x x at 0.5x , use high accuracy differentiation formulas with 0.25h
Solution:
2
1
1
2
0 0 1.2
0.25 0.25 1.103515
0.5 0.5 0.925
0.75 0.75 0.636328
1.00 1.0 0.2
i
i
i
i
i
x f
x f
x f
x f
x f
-
Forward difference:
0.2 4 0.636328 3 0.925' 0.5 0.859375
2 0.25f
5.82 %t
Backward difference:
3 0.925 4 1.103515 1.2' 0.5 0.878125
2 0.25f
3.77 %t
Centered difference:
0.2 8 0.636328 8 1.103515 1.2
' 0.5 0.912512
fh
0 %t