Upload
jonathan-austin
View
215
Download
0
Embed Size (px)
Citation preview
NUMERICAL DIFFERENTIATIONNUMERICAL DIFFERENTIATION
The derivative of f (x) at x0 is:
h
xfhxfxf Limit
h
)( 0000
An approximation to this is:
h
xfhxfxf
)( 000
for small values of h.
Forward Difference Formula
810 . and ln)(Let xxxfFind an approximate value for 81.f
0.1 0.5877867 0.6418539 0.5406720
0.01 0.5877867 0.5933268 0.5540100
0.001 0.5877867 0.5883421 0.5554000
).( 81f ).( hf 81 h
fhf ).().( 8181 h
The exact value of 555081 .. f
Assume that a function goes through three points:
., and ,,, 221100 xfxxfxxfx
)()( xPxf
221100 xfxLxfxLxfxLxP
Lagrange Interpolating Polynomial
21202
10
12101
20
02010
21
))((
))((
))((
))((
))((
))((
xfxxxx
xxxx
xfxxxx
xxxx
xfxxxx
xxxxxP
221100 xfxLxfxLxfxLxP
)()( xPxf
21202
10
12101
20
02010
21
))((
2
))((
2
))((
2
xfxxxx
xxx
xfxxxx
xxx
xfxxxx
xxxxP
20000
000
10000
000
00000
0000
)()2()2(
)(2
)2()()(
)2(2
)2()(
)2()(2
xfhxhxxhx
hxxx
xfhxhxxhx
hxxx
xfhxxhxx
hxhxxxP
If the points are equally spaced, i.e.,
hxxhxx 20201 and
2212020 2
2
2
3xf
h
hxf
h
hxf
h
hxP
2100 432
1xfxfxf
hxP
hxfhxfxfh
xf 2432
10000
Three-point formula:
hxxhxx 0201 and If the points are equally spaced with x0 in the middle:
20000
000
10000
000
00000
0000
)()()(
)(2
)()()(
)(2
)(()(
)()(2
xfhxhxxhx
hxxx
xfhxhxxhx
hxxx
xfhxxhxx
hxhxxxP
2212020 22
0xf
h
hxf
h
hxf
hxP
hxfhxfh
xf 000 2
1
Another Three-point formula:
Alternate approach (Error estimate)
Take Taylor series expansion of f(x+h) about x:
xfh
xfh
xfhxfhxf 33
22
!32
xfh
xfh
xfhxfhxf 33
22
!32
xfh
xfh
xfh
xfhxf 32
2
!32.............. (1)
hOxfh
xfhxf
hOh
xfhxfxf
h
xfhxfxf
Forward Difference
Formula
xfh
xfh
hO 32
2
!32
xfh
xfh
xfhxfhxf 33
22
!3
8
2
422
xfh
xfh
xfhxfhxf 33
22
!3
8
2
422
xfh
xfh
xfh
xfhxf 32
2
!3
4
2
2
2
2
................. (2)
xfh
xfh
xfh
xfhxf 32
2
!32.............. (1)
xfh
xfh
xfh
xfhxf 32
2
!3
4
2
2
2
2
................. (2)
2 X Eqn. (1) – Eqn. (2)
xfh
xfh
xf
h
xfhxf
h
xfhxf
43
32
!4
6
!3
2
2
22
2
43
32
!4
6
!3
2
2
342
hOxf
xfh
xfh
xf
h
xfhxfhxf
2
2
342hOxf
h
xfhxfhxf
2
2
342hO
h
xfhxfhxfxf
h
xfhxfhxfxf
2
342
Three-point Formula
xfh
xfh
hO 43
32
2
!4
6
!3
2
Take Taylor series expansion of f(x+h) about x:
xfh
xfh
xfhxfhxf 33
22
!32Take Taylor series expansion of f(x-h) about x:
xfh
xfh
xfhxfhxf 33
22
!32
The Second Three-point Formula
Subtract one expression from another
xfh
xfh
xfhhxfhxf 66
33
!6
2
!3
22
xfh
xfh
xfhhxfhxf 66
33
!6
2
!3
22
xfh
xfh
xfh
hxfhxf 65
32
!6!32
xfh
xfh
h
hxfhxfxf 6
53
2
!6!32
2
2hO
h
hxfhxfxf
xfh
xfh
hO 65
32
2
!6!3
h
hxfhxfxf
2
Second Three-point Formula
h
xfhxfxf
Forward Difference
Formula
Summary of Errors
xfh
xfh
hO 32
2
!32Error term
h
xfhxfhxfxf
2
342
First Three-point Formula
Summary of Errors continued
xfh
xfh
hO 43
32
2
!4
6
!3
2Error term
h
hxfhxfxf
2
Second Three-point Formula
xfh
xfh
hO 65
32
2
!6!3Error term
Summary of Errors continued
Example:
xxexf
Find the approximate value of 2f
1.9 12.703199
2.0 14.778112
2.1 17.148957
2.2 19.855030
x xf
with 10.h
Using the Forward Difference formula:
0 0 0
1f x f x h f x
h
12 2 1 2
0 11
17 148957 14 7781120 1
23 708450
f f . f.
. ...
Using the 1st Three-point formula:
032310.22
855030.19
148957.174778112.1432.0
1
)2.2()1.2(4)2(31.02
12
ffff
hxfhxfxfh
xf 2432
10000
Using the 2nd Three-point formula:
228790.22
703199.12148957.172.0
1
)9.1()1.2(1.02
12
fff
The exact value of 22.167168 :is 2f
hxfhxfh
xf 000 2
1
Comparison of the results with h = 0.1
Formula Error
Forward Difference 23.708450 1.541282
1st Three-point 22.032310 0.134858
2nd Three-point 22.228790 0.061622
2f
The exact value of 2f is 22.167168
Second-order Derivative
xfh
xfh
xfhxfhxf 33
22
!32
xfh
xfh
xfhxfhxf 33
22
!32
Add these two equations.
xfh
xfh
xfhxfhxf 44
22
!4
2
2
22
2 4
2 42 22
2 4!
h hf x h f x f x h f x f x
xfh
xfh
hxfxfhxf 42
22 !4
22
xfh
h
hxfxfhxfxf 4
2
22
!4
22
2
2 2
h
hxfxfhxfxf
NUMERICAL INTEGRATIONNUMERICAL INTEGRATION
b
a
dxxf )( area under the curve f(x) between
. to bxax
In many cases a mathematical expression for f(x) is unknown and in some cases even if f(x) is known its complex form makes it difficult to perform the integration.
y
xx 0 = a x 0 = b
f(a)
f(b)f(x)
y
xx 0 = a x 0 = b
f(a)
f(b)
f(x)
Area of the trapezoid
The length of the two parallel sides of the trapezoid are: f(a) and f(b)
The height is b-a
)()(
)()()(
bfafh
bfafab
dxxfb
a
2
2
Simpson’s Rule:
2
0
2
0
x
x
x
x
dxxPdxxf )()(
hxxhxx 20201 and
2 2
0 0
2
0
2
0
1 20
0 1 0 2
0 21
1 0 1 2
0 12
2 0 2 1
x x
x x
x
x
x
x
( x x )( x x )P x dx f x dx
( x x )( x x )
( x x )( x x )f x dx
( x x )( x x )
( x x )( x x )f x dx
( x x )( x x )
)()(
)()(
210 43
2
0
2
0
xfxfxfh
dxxPdxxfx
x
x
x
y
xx 0 = a x 0 = b
f(a)
f(b)
f(x)
y
xx 0 = a x 0 = b
f(a)
f(b)
f(x)
y
xx 0 = a x 0 = b
f(a)
f(b)
f(x)
y
xx 0 = a x 0 = b
f(a)
f(b)
f(x)
Composite Numerical Integration
Riemann Sum
The area under the curve is subdivided into n subintervals. Each subinterval is treated as a rectangle. The area of all subintervals are added to determine the area under the curve.
There are several variations of Riemann sum as applied to composite integration.
In Left Riemann sum, the left-side sample of the function is used as the height of the individual rectangle.
a b
Left Riemann Sumy
xx
1
2
3 2
1i
x b a / n
x a
x a x
x a x
x a i x
1
nb
iai
f x dx f x x
In Right Riemann sum, the right-side sample of the function is used as the height of the individual rectangle.
a b
Right Riemann Sumy
x
x
1
2
3
2
3
i
x b a / n
x a x
x a x
x a x
x a i x
1
nb
iai
f x dx f x x
In the Midpoint Rule, the sample at the middle of the subinterval is used as the height of the individual rectangle.
a b
Midpoint Ruley
xx
1
2
3
2 1 1 2
2 2 1 2
2 3 1 2
2 1 2i
x b a / n
x a x /
x a x /
x a x /
x a i x /
1
nb
iai
f x dx f x x
Composite Trapezoidal Rule:
Divide the interval into n subintervals and apply Trapezoidal Rule in each subinterval.
)()()( bfxfafh
dxxfn
kk
b
a
1
1
22
where
nkkhaxn
abh k , ... , , ,for and 210
by dividing the interval into 20 subintervals.
Find π
sin0
(x)dx
2021020
20
20
....., , , , ,π
π
kk
khax
n
abh
n
k
9958861
2020
40
22
19
1
1
10
.
πsinπ
sinsinπ
)()()sin(π
k
n
kk
k
bfxfafh
dxx
Composite Simpson’s Rule:
Divide the interval into n subintervals and apply Simpson’s Rule on each consecutive pair of subinterval. Note that n must be even.
)()(
)()(
/
)/(
bfxf
xfafh
dxxf
n
kk
n
kk
b
a
2
112
12
12
4
23
nkkhaxn
abh k , ... , , ,for and 210
where
by dividing the interval into 20 subintervals.
Find π
sin0
(x)dx
2021020
2020
....., , , , ,π
π
kk
khax
n
abhn
k
0000062
20
124
20
220
6010
1
9
10
.
)πsin(π)(
sin
πsinsin
π)sin(
π
k
k
k
kdxx