interpolare lagrange
TRANSCRIPT
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05.04.1
Chapter 05.04Lagrangian Interpolation More ExamplesCivil Engineering
Example 1
To maximize a catch of bass in a lake, it is suggested to throw the line to the depth of the
thermocline. The characteristic feature of this area is the sudden change in temperature. We
are given the temperature vs. depth data for a lake in Table 1.
Table 1 Temperature vs. depth for a lake.
Temperature, C T Depth, mz
19.1 0
19.1 1
19 2
18.8 3
18.7 4
18.3 5
18.2 6
17.6 7
11.7 8
9.9 9
9.1 10
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05.04.2 Chapter 05.0
Figure 1 Temperature vs. depth of a lake.
Using the given data, we see the largest change in temperature is between m8z and
m7z . Determine the value of the temperature at m5.7z using a first order
Lagrange polynomial.
SolutionFor first order Lagrange polynomial interpolation (also called linear interpolation), the
temperature is given by
1
0
)()()(i
ii zTzLzT
)()()()( 1100 zTzLzTzL
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Lagrange Method of Interpolation-More Examples: Civil Engineering 05.04.3
Figure 2 Linear interpolation.
Since we want to find the temperature at m5.7z , we need to choose the two data points
that are closest to m5.7z that also bracket m5.7z to evaluate it. The two points are
80 z and 71 z .
Then
7.11,8 00 zTz 6.17,7 11 zTz
gives
1
00 0
0 )(
jj j
j
zzzzzL
10
1
zz
zz
1
10 1
1 )(
jj j
j
zz
zzzL
01
0
zz
zz
Hence
)()()( 101
0
0
10
1 zTzz
zzzT
zz
zzzT
78),6.17(87
8)7.11(
78
7
z
zz
)6.17(87
85.7)7.11(
78
75.7)5.7(
T
(x0,y0)
(x1,y1)
1(x)
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05.04.4 Chapter 05.0
)6.17(5.0)7.11(5.0
C65.14
You can see that 5.0)(0 zL and 5.0)(1 zL are like weightages given to the temperatures
at m8z and m7z to calculate the temperature at m5.7z .
Example 2
To maximize a catch of bass in a lake, it is suggested to throw the line to the depth of the
thermocline. The characteristic feature of this area is the sudden change in temperature. We
are given the temperature vs. depth data for a lake in Table 2.
Table 2 Temperature vs. depth for a lake.
Temperature, C T Depth, mz
19.1 0
19.1 1
19 218.8 3
18.7 4
18.3 5
18.2 6
17.6 7
11.7 8
9.9 9
9.1 10
Using the given data, we see the largest change in temperature is between m8z and
m7z . Determine the value of the temperature at m5.7z using a second order
Lagrange polynomial. Find the absolute relative approximate error for the second order
polynomial approximation.
Solution
For second order Lagrange polynomial interpolation (also called quadratic interpolation), the
temperature is given by
2
0
)()()(i
ii zTzLzT
)()()()()()( 221100 zTzLzTzLzTzL
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Lagrange Method of Interpolation-More Examples: Civil Engineering 05.04.5
Figure 3 Quadratic interpolation.
Since we want to find the temperature at ,5.7z we need to choose the three data points
that are closest to 5.7z that also bracket 5.7z to evaluate it. These three points are,90 z 81 z and 72 z . (Choosing the three points as 80 z , 71 z and 62 z
is equally valid.)
Then
9.9,9 00 zTz 7.11,8 11 zTz
6.17,7 22
zTz gives
2
00 0
0 )(
jj j
j
zz
zzzL
20
2
10
1
zz
zz
zz
zz
2
10 1
1 )(
jj j
j
zz
zzzL
21
2
01
0
zzzz
zzzz
2
20 2
2 )(
jj j
j
zz
zzzL
(x0,y0)
(x1,y1) (x2,y2)
2(x)
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05.04.6 Chapter 05.0
12
1
02
0
zz
zz
zz
zz
Hence
20
2
12
1
02
01
21
2
01
00
20
2
10
1 ),()()()(
zzz
zT
zz
zz
zz
zzzT
zz
zz
zz
zzzT
zz
zz
zz
zzzT
)6.17()87)(97(
)85.7)(95.7(
)7.11()78)(98(
)75.7)(95.7()9.9(
)79)(89(
)75.7)(85.7()5.7(
T
)6.17)(375.0()7.11)(75.0()9.9)(125.0(
C138.14
The absolute relative approximate error a obtained between the results from the first and
second order polynomial is
100138.14
65.14138.14
a
%6251.3
Example 3
To maximize a catch of bass in a lake, it is suggested to throw the line to the depth of the
thermocline. The characteristic feature of this area is the sudden change in temperature. Weare given the temperature vs. depth data for a lake in Table 3.
Table 3 Temperature vs. depth for a lake.
Temperature, C T Depth, mz
19.1 0
19.1 1
19 2
18.8 3
18.7 4
18.3 5
18.2 6
17.6 7
11.7 8
9.9 9
9.1 10
Using the given data, we see the largest change in temperature is between m8z and
m7z .
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Lagrange Method of Interpolation-More Examples: Civil Engineering 05.04.7
a) Determine the value of the temperature at m5.7z using a third order Lagrangepolynomial. Find the absolute relative approximate error for the third order
polynomial approximation.
b) The position where the thermocline exists is given where 02
2
dz
Td. Using the
expression from part (a), what is the value of the depth at which the thermoclineexists?
Solution
a) For third order Lagrange polynomial interpolation (also called cubic interpolation), we
choose the temperature given by
3
0
)()()(i
ii zTzLzT
)()()()()()()()( 33221100 zTzLzTzLzTzLzTzL
Figure 4 Cubic interpolation.
Since we want to find the temperature at ,5.7z we need to choose the four data points
that are closest to 5.7z that also bracket 5.7z to evaluate it. The four data points are
,90 z ,81 z 72 z and 63 z .Then
9.9,9 00 zTz 7.11,8 11 zTz 6.17,7 22 zTz 2.18,6 33 zTz
y
x
3(x)
(x0,y0)
(x1,y1)
(x2,y2)
(x3,y3)
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05.04.8 Chapter 05.0
gives
3
00 0
0 )(
jj j
j
zz
zzzL
30
3
20
2
10
1
zz
zz
zz
zz
zz
zz
3
10 1
1 )(
jj j
j
zz
zzzL
31
3
21
2
01
0
zz
zz
zz
zz
zz
zz
3
20 2
2 )(
jj j
j
zz
zzzL
32
3
12
1
02
0
zzzz
zzzz
zzzz
3
30 3
3 )(
jj j
j
zz
zzzL
23
2
13
1
03
0
zz
zz
zz
zz
zz
zz
Hence
)()(
)()()(
3
23
2
13
1
03
0
2
32
3
12
1
02
0
1
31
3
21
2
01
0
0
30
3
20
2
10
1
zTzz
zz
zz
zz
zz
zzzT
zz
zz
zz
zz
zz
zz
zT
zz
zz
zz
zz
zz
zzzT
zz
zz
zz
zz
zz
zzzT
30 zzz
)2.18()76)(86)(96(
)75.7)(85.7)(95.7()6.17(
)67)(87)(97(
)65.7)(85.7)(95.7(
)7.11()68)(78)(98(
)65.7)(75.7)(95.7()9.9(
)69)(79)(89(
)65.7)(75.7)(85.7()5.7(
T
)2.18)(0625.0()6.17)(5625.0()7.11)(5625.0()9.9)(0625.0(
C725.14 The absolute relative approximate error a obtained between the results from the second
and third order polynomial is
100725.14
138.14725.14
a
%9898.3
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Lagrange Method of Interpolation-More Examples: Civil Engineering 05.04.9
b) To find the position of the thermocline, we must find the points of inflection of the third
order polynomial, given by .02
2
dz
Td
)2.18()76)(86)(96(
)7)(8)(9()6.17(
)67)(87)(97(
)6)(8)(9(
)7.11()68)(78)(98(
)6)(7)(9()9.9(
)69)(79)(89(
)6)(7)(8()(
zzzzzz
zzzzzzzT
69 z 69,5667.155.3558.2629.615 32 zzzz
69,7.41.7158.262 2 zzzdz
dT
69,4.91.712
2
zzdz
Td
Simply setting this expression equal to zero, we get
69,4.91.710 zz m5638.7z
This answer can be verified due to the fact that it falls within the specified range of the third
order polynomial and it also falls within the region of the greatest temperature change in the
collected data from the lake.