Mata kuliah : K0624 - Metode Numerik II
Tahun
: 2010
Pertemuan 3
Differensiasi
DIFFERENSIASI NUMERIK
f  x0  
Limit
h 0
f  x0  h  f ( x0 )
h
f  x0  h  f ( x0 )
f  x0  
h
Forward Difference Formula
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f  x  h  f  x 
 f  x   O h
h
f  x  h  f  x 
f  x  
 O h
h
f  x  h  f  x 
f  x  
h
Forward Difference Formula
h 2 
h2 3 
Oh  f  x  
f x  
2
3!
DIFFERENSIASI NUMERIK
 f ( x  2h)  4 f ( x  h)  3 f ( x )
f ( x) 
h
f ( x )  f ( x  h)
f ( x) 
h
Backward Difference Formula
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Backward Difference
11 July 2017
Metode Numerik II
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DIFFERENSIASI NUMERIK
3 f ( x )  4 f ( x  h )  f ( x  2h )
f ( x) 
2h
1
f  x0    f  x0  h  f  x0  h
2h
Central Difference Formula
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2h 3  3 
2h 6  6 
f  x  h  f  x  h  2hf  x  
f x 
f x  
3!
6!
f  x  h  f  x  h
h2 3 
h5 6 
 f  x  
f x  
f x  
2h
3!
6!
f  x  h  f  x  h h2 3 
h5 6 
f  x  

f x  
f x  
2h
3!
6!
f  x  h  f  x  h
f  x  
 O h2
2h
 
 
2
5
h
h
O h2  
f 3   x  
f 6   x   
3!
6!
Misalkan
f(x) = ln x dan x0 = 1.8
f 1.8
Hitung
0.1
0.5877867 0.6418539
f (1.8  h)  f (1.8)
h
0.5406720
0.01
0.5877867 0.5933268
0.5540100
0.001 0.5877867 0.5883421
0.5554000
h
f (1.8)
f (1.8  h)
Nilai eksak adalah
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f 1.8  0.55 5
9
Example:
Bila
f  x   xe
x
Hitung
x
f x
1.9
12.703199
2.0
14.778112
2.1
17.148957
2.2
19.855030
f 2
dengan h  0.1
Gunakan Forward Difference formula:
1
f   x0    f  x0  h   f  x0 
h
1
f   2 
f  2.1  f  2 

0.1
1

17.148957  14.778112
0.1
 23.708450
Gunakan untuk tiga titik:
1
 3 f  x0   4 f  x0  h  f  x0  2h
f  x0  
2h
1
 3 f ( 2)  4 f ( 2.1)  f ( 2.2)
f 2  
2  0 .1
1
  3  14.778112  4  17.148957

0.2
 19.855030 
 22.032310
Central difference formula:
1
 f  x0  h  f  x0  h
f  x0  
2h
1
 f ( 2.1)  f (1.9)
f 2  
2  0 .1
1
 17.148957  12.703199

0.2
 22.228790
Nilai Exact
f’(x) = 22.167168

Perbandingan hasil dengan h = 0.1
Nilai exact dari
f   2  adalah 22.167168
Rumus
f   2
Error
Forward Difference
23.708450
1.541282
Tiga Titik
22.032310
0.134858
Central
22.228790
0.061622
Turunan order dua
h2 2 
h3 3 
f  x  h  f  x   hf  x  
f x 
f x  
22
3!3
h 2 
h 3 
f  x  h  f  x   hf  x  
f x 
f x  
2
3!
Dari kedua persamaan di atas diperoleh
2
4
2h  2 
2h  4 
f  x  h  f  x  h  2 f  x  
f x 
f x  
2
4!
2
4
2h  2 
2h  4 
f  x  h  2 f  x   f  x  h 
f x 
f x  
2
4!
2
f  x  h  2 f  x   f  x  h
2
h
2 
4 


x  

f
x

f
2
h
4!
2






f
x

h

2
f
x

f
x

h
2
h
4 
x  
f 2   x  

f
2
h
4!
f
2 
f  x  h  2 f  x   f  x  h
x 
h2
Contoh :
Hitung f’(0.5) . h = 0.5
nilai exact
forwad difference
backward difference
central difference
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Metode Numerik II
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Contoh
11 July 2017
f(x) = e-x sin(x)
f '(x) = e-x cos(x) - e-x sin(x)
f '(1) = - 0.110794 , h = 0.5
f '(1) = ( f(1+0.5) – f(1-0.5))/1 = - 0.0682151
Metode Numerik II
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TURUNAN ORDER TINGGI
(1) dan (2) dijumlahkan
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Forward dan Backward dari order O(h2)
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Forward dan Backward dari order O(h2)
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Contoh
11 July 2017
f(x) = e-x sin(x)
f '' (x) = e-x cos(x) - e-x sin(x)
f '(1) = - 0.110794 , h = 0.2
f '(1) = ( f(1+0.2) – f(1-0.2))/(0.04) = - 0.10401
Metode Numerik II
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Contoh
11 July 2017
f(x) = e-x sin(x)
f '' (x) = e-x cos(x) - e-x sin(x)
f '(1) = - 0.110794 , h = 0.1
f '(1) = ( f(1+0.1) – f(1-0.1))/(0.01) = - 0.110794
Metode Numerik II
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Contoh
11 July 2017
f(x) = e-x sin(x)
f '' (x) = e-x cos(x) - e-x sin(x)
f '(1) = - 0.110794 , h = 0.05
f '(1) = ( f(1+0.05) – f(1-0.05))/(0.1) = - 0.11087
Metode Numerik II
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TERIMA KASIH
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