Transcript Chapter 7 Differentiation and Integration

Chapter 7 Differentiation and
Integration
• Finite-difference differentiation
df ( x ) f ( x  x )  f ( x )
forward difference

dx
x
df ( x ) f ( x )  f ( x  x )
backward difference

dx
x
df ( x ) f ( x  x )  f ( x  x )
two - step method

dx
2x
7- 1
Example: Evaporation Rates
Table: Saturation Vapor Pressure (es) in
mm Hg as a Function of Temperature (T) in °C
T(°C)
20
21
22
23
24
25
es(mm Hg)
17.53
18.65
19.82
21.05
22.37
23.75
7- 2
The slope of the saturation vapor pressure curve at
22°C (3 methods) :
des es (23)  es (22) 21.05  19.82
forward


 1.23 mm Hg/C
dT
23  22
1
des es (22)  es (21) 19.82  18.65
backward


 1.17 mm Hg/C
dT
22  21
1
des es (23)  es (21) 21.05  18.65
two - step


 1.20 mm Hg/C
dT
23  21
2
The true value is 1.20 mm Hg/°C, so the two-step
method provides the most accurate estimate.
7- 3
Differentiation Using a Finitedifference Table
Example: Finite-difference Table for Specific
Enthalpy (h) in Btu/lb and Temperature (T) in ºF
T
h
800
1305
Δh
Δ2h
Δ3h
Δ4h
155
1000
1460
-30
125
1200
1585
25
-5
120
1400
1705
-20
5
0
120
1600
1825
7- 4
• For example, at a temperature of 1200 ºF, the
forward, backward, and two-step methods yield:
h 1705  1585 120
cp 


 0.6 Btu/lb/ F
T 1400  1200 200
cp 
h 1585  1460 125


 0.625 Btu/lb/ F
T 1200  1000 200
h 1705  1460 245
cp 


 0.6125 Btu/lb/ F
T 1400  1000 400
• The rate of change of cp at T= 1200 ºF is
2h
5
2



0
.
025
Btu/lb/(

F)
T 2 200
7- 5
Differentiating an Interpolating Polynomial
f ( x)  bn xn  bn1xn1  bn2 xn2  ... b1x  b0
The derivative:
df ( x)
 nbn x n 1  (n  1)bn 1 x n  2  ...  b1
dx
Gregory-Newton interpolation polynomial:
f(x)  a1  a2 ( x  x1 )  a3 ( x  x1 )( x  x2 ) 
a4 ( x  x1 )( x  x2 )( x  x3 )  
It is more difficult to evaluate the derivative:
df ( x )
 a2  a3[( x  x2 )  ( x  x1 )]  a4 [( x  x2 )( x  x3 )
dx
 ( x  x1 )( x  x3 )  ( x  x1 )( x  x2 )]  
7- 6
Differentiation Using Taylor Series
Expansion
df ( x )
d 2 f ( x ) ( x )2 d 3 f ( x ) ( x )3
f ( x  x )  f ( x ) 
x 

 ...
2
3
dx
dx
2!
dx
3!
T runcatingtermswith thesecondderivativeand higher derivatives :
df ( x )
f ( x  x )  f ( x ) 
x
dx
T hen,we get
df ( x ) f ( x  x )  f ( x )
f ' ( x) 

dx
x
It is thesame as theforwardfinite- difference.
7- 7
• The second-order approximation:
df ( x ) f ( x  x )  f ( x ) d 2 f ( x ) x


dx
x
dx2 2!
Withforward difference:
d 2 f ( x ) f ' ( x  x )  f ' ( x )

2
dx
x
f ( x  2x )  2 f ( x  x )  f ( x )

( x ) 2
The second-order approximation of the first
derivative with forward difference:
df ( x )  f ( x  2x )  4 f ( x  x )  3 f ( x )

dx
2x
7- 8
• The second-order approximation of the second
derivative with forward difference:
d 2 f ( x)  f ( x  3x)  4 f ( x  2x)  5 f ( x  x)  2 f ( x)

2
dx
(x) 2
• The first-order and second-order approximation of
the first derivative with backward difference:
df ( x) f ( x)  f ( x  x)

dx
x
df ( x) 3 f ( x)  4 f ( x  x)  f ( x  2x)

dx
2x
7- 9
• The first-order and second-order approximation of
the second derivative with backward difference:
d 2 f ( x) f ( x)  2 f ( x  x)  f ( x  2x)

2
dx
(x) 2
d 2 f ( x) 2 f ( x)  5 f ( x  x)  4 f ( x  2x)  f ( x  3x)

2
dx
(x) 2
• How to derive
(for reference only)
f ( x  3x )  f ( x )  f ' ( x )  3x  f ' ' ( x )  92 ( x )2
f ( x  2x )  f ( x )  f ' ( x )  2x  f ' ' ( x )  2( x )2
f ( x  x )  f ( x )  f ' ( x )  x  f ' ' ( x )  12 ( x ) 2
1
[  f ( x  3x )  4 f ( x  2x )  5 f ( x  x )  2 f ( x  x )]
( x ) 2
1
2
9

[

f
(
x
)

3

xf
'
(
x
)

(

x
)
f ' ' ( x)
2
( x ) 2
 4 f ( x )  8xf ' ( x )  8( x ) 2 f ' ' ( x )
- 5 f ( x )  5xf ' ( x )  25 ( x )2 f ' ' ( x )  2 f ( x )]
 f ' ' ( x)
7- 10
• The first-order and second-order approximation of
the first derivative with the two-step method:
df ( x) f ( x  x)  f ( x  x)

dx
2x
df ( x)  f ( x  2x)  8 f ( x  x)  8 f ( x  x)  f ( x  2x)

dx
12x
• The first-order and second-order approximation of
second derivative with the two-step method:
d 2 f ( x) f ( x  x)  2 f ( x)  f ( x  x)

dx2
(x) 2
d 2 f ( x)  f ( x  2x)  16 f ( x  x)  30 f ( x)  16 f ( x  x)  f ( x  2x)

dx2
12(x) 2
7- 11
Example: Evaporation Rates
Second-order with forward, backward, two-step:
df ( x)  es (24)  4es (23)  3es (22)

dx
2(1)
 22.37  4(21.05)  3(19.82)

 1.185 mm Hg/C
2(1)
df ( x) 3es (22)  4es (21)  es (20)

dx
2(1)
3(19.82)  4(18.65)  17.53

 1.195 mm Hg/ C
2(1)
df ( x)  es (24)  8es (23)  8es (21)  es (20)

dx
12(1)
 (22.37)  8(21.05)  8(18.65)  (17.53)

 1.19667mm Hg/ C
12(1)
The true value at T = 22ºC is 1.2 mm Hg/ºC
7- 12
Numerical Integration
• The area under the curve f(x) between x=a and x=b:
b
integral  f ( x)dx
a
• Example: the volume rate of flow (Q) of water in a channel
or through a pipe is the integral of the velocity (V) and the
incremental area (dA):
Q   VdA
7- 13
Interpolation Formula Approach
The Gregory-Newton interpolation polynomial:
f ( x )  a1  a2 ( x  x1 )  a3 ( x  x1 )( x  x2 )  ... 
an 1 ( x  x1 )( x  x2 )...(x  xn )
After a1 , a2 , an 1 are determined, we can rearrange
it toform an nth - order polynomial:
f ( x )  b1 x n  b2 x n 1  ...  bn 1
T hen,it can be integratedanalytically as follows:
b1 n 1 b2 n
 f ( x )dx  n  1 x  n x  ...  bn1 x  c
7- 14
Trapezoidal Rule

xn
x1
f ( xi 1 )  f ( xi )
f ( x)dx   ( xi 1  xi )
2
i 1
n 1
7- 15
• Another way to get the trapezoidal formula
The linear polynomial passing the data points:
f ( xi 1 )  f ( xi )
f ( x)  f ( xi ) 
( x  xi )
xi 1  xi
Integrating f ( x ) between two pointsa and b, we get
b
b  a (b  a )
f
(
x
)
dx

[
( f ( xi 1 )  f ( xi ))  xi 1 f ( xi )  xi f ( xi 1 )]
a
xi 1  xi
2
Let b  xi 1 and a  xi , we get

xi 1
xi
f ( x )dx 
xi 1  xi
[ f ( xi 1 )  f ( xi )]
2
7- 16
• The absolute value of the upper bound on
the error for the Trapezoidal rule is:
( xi 1  xi )3
error 
f ''(xj )
12
where f ' ' ( x j )  max f ' ' ( x )
xi  x  xi 1
7- 17
Example: Trapezoidal Rule for Integration
f ( x)  x3  5x 2  4x  3
• n=4
• The trapezoidal rule provides
2 1
3 2
43
1 f ( x)dx  2 [3  (1)]  2 [1  (3)]  2 [3  (3)]
 1  ( 2)  0  1
4
7- 18
7- 19
Example: Flow Rate
• The flow rate (Q) of an incompressible fluid is given by
the integral.
Q   VdA
in which V is the velocity and A is the area.
• For a circular pipe of radius r, the incremental area dA is
equal to 2πrdr.
r0
Q   V( 2rdr)
0
7- 20
• Table 6: The Data for Estimating the Flow Rate of a Fluid
in a Circular Piple
i
ri (ft)
Vi (fps)
1
0
10.000
2
1/12
9.722
3
1/6
8.889
4
1/4
7.500
5
1/3
5.556
6
5/12
3.056
7
1/2
0.
7- 21
• For a pipe of diameter of 1 ft
trapezoidal rule:
r 6
Q
[Vi (2π ri )  Vi 1 (2π ri 1 )]

2 i 1
Q
Q
1
) 6
12
[Vi ri  Vi 1ri 1 ]

2
i 1
2π (
π 
 1  
1
 1 
{100  9.722   9.722   8.889 
12 
 12  
 12 
 6 

1
 1    1 
 1 
 8.889   7.5   7.5   5.556 
6
 4    4 
 3 


1
 5  
 5   1 
 5.556   3.056   3.056   0 }
 3
 12  
 12   2 

 3.818ft3/sec
7- 22
Simpson’s Rule
• Simpson’s rule:

b
a
f ( x)dx 
x 
x
[ f (a)  4 f (a  x)  f (b)]
3
ba
2
where
• Simpson’s rule can only be applied when there are
an even number of subintervals:

xn
x1
xi 1  xi
f ( x)dx  
[ f ( xi )  4 f ( xi 1 )  f ( xi  2 )]
3
i 1, 3, 5
n2
7- 23
• Proof of Simpson’s Rule
Using a second-order polynomial:
f ( x)  kx2  hx  l
7- 24
Passing through the three data points:
f ( a )  ka 2  ha  l
f ( a  x )  k ( a  x ) 2  h( a  x )  l
f (b)  kb2  hb  l
Integrating thesecond- order polynomial:

b
a
(kx 2  hx  l )dx
k (b3  a 3 ) h (b 2  a 2 )


 l (b  a )
3
2
Then, we can obtain the Simpson’s formula.
7- 25
• The absolute value of the upper bound on
the error for the Simpson’s rule is estimated
by
4
( xi 1  xi )5 d f ( x j )
error 
90
dx4
d 4 f (xj )
d 4 f ( x)
where
 max
4
xi  x  xi 1
dx
dx4
7- 26
Example: Flow Rate Problem
• Applying Simpson’s rule to the data of Table.6
r0
Q   V 2π rdr 
0
2π ( 121 )
3
5
 [V r  4V
r 1, 3
i i
r  Vi  2 ri  2 ]
i 1 i 1
π
1
1
1
1
 [10(0)  4(9.722)( )  8.889( )]  [8.889( )  4(7.5)( )
18
12
6
6
4
1
1
5
1
 5.556( )]  [5.556( )  4(3.056)( )  0( )]  3.927 ft3 /sec
3
3
12
2
7- 27
Romberg Integration
• Denoting the trapezoidal estimate as I01
ba
I 01 
( f (a )  f (b))
2
where a and b are the start and end of an interval.
• A second estimate I11:
ba 1
1
I11 
( f (a )  f (m)  f (b))
2 2
2
ab
ba
where m 
a
2
2
1
ba
It can be rewritten : I11  [ I 01  (b  a ) f (a 
)]
2
2
7- 28
• A third estimate I21 can be obtained using three
equally spaced intermediate points m1, m2, and m3
ba 1
1
1
1
1
I 21 
[ f (a)  f (m1 )  f (m2 )  f (m3 )  f (b)]
2 4
2
2
2
4
and it can be rewritten as


3
1
ba
ba 
I 21   I11 
f
(
a

k )

2
2 k 1
4


k

2


7- 29
• Continuing this subdividing of the interval leads to
the following recursive relationship
1
b  a 2 1
ba 
I i1   I i 1,1  i 1  f (a  i k ) for i  1, 2,...
2
2 k 1,3,5
2

i
• The general extrapolation formula in recursive
form is
I ij 
4 j 1 I i 1, j 1  I i , j 1
4 j 1  1
7- 30
• The values of Iij can be presented in the following
upper-triangular matrix form:
I01
I02
I03
I04
…
I0,N-1
I0,N
I11
I12
I13
.
…
I1,N-1
I1,N
I21
I22
.
.
…
I2,N-1
I31
.
.
.
.
.
.
.
.
IN-2,3
.
IN-1,2
I0,N+1
IN1
7- 31
Example: Romberg Method for
Integration
• For the function f(u) = ueku, the integral is
ku
e
ku
ue
 du  k 2 (ku  1)  c
• Let k = 2, we want to calculate
1

0
ue2 u du
The true value of the integral is 2.097264 (seven
significant digits).
7- 32
We have a=0, b=1.
For i  0
I 01  0.5(b  a )( f ( a )  f (b))
 0.5(1  0)(0  e 20  1  e 21 )  3.69453
For i  1
ba
)]
2
 0.5[3.69453 (1  0)(0.5  e 20.5 )]  2.52683
I11  0.5[ I 01  (b  a ) f ( a 
For i  2
ba
ba
I 21  0.5[ I11  0.5(b  a )( f ( a 
)  f (a  3
))]
4
4
 0.5[2.52683 0.5(0.25  e 20.25  0.75  e 20.75 )]
 2.20678
7- 33
For i  3
ba
3(b  a )
I 31  0.5[ I 21  0.25(b  a )( f (a 
)  f (a 
)
8
8
5(b  a )
7(b  a )
 f (a 
)  f (a 
))]
8
8
 2.12478
For i  4
ba
3(b  a )
I 41  0.5[ I 31  0.125(b  a )( f ( a 
)  f (a 
)
16
16
5(b  a )
7( b  a )
9( b  a )
 f (a 
)  f (a 
)  f (a 
)
16
16
16
11(b  a )
13(b  a )
15(b  a )
 f (a 
)  f (a 
)  f (a 
))]
16
16
16
 2.10415
7- 34
Table.7 Computations According to the Romberg Method
i
j=1
2
3
4
5
6
0
3.69453
2.13760
2.09759
2.09727
2.09726
2.09726
1
2.52683
2.10009
2.09727
2.09726
2.09726
2
2.20678
2.09745
2.09726
2.09726
3
2.12478
2.09728
2.09726
4
2.10415
2.09727
5
2.09899
7- 35
• Similarly for i = 5, I51=2.09899
• I02 is computed as following:
421 (2.52683)  3.69453
I 02 
 2.13760
2 1
4 1
• I03 is computed as following:
431 (2.10009)  2.13760
I 03 
 2.09759
31
4 1
• The Romberg method yields an exact value to six
significant digits for the integral of 2.09726
7- 36