3-interpolasi1
TRANSCRIPT
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METODE NUMERIK
INTERPOLASI
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Tujuan
Interpolasi berguna untuk menaksir harga-harga tengah antara titik data yang sudah tepat. Interpolasi mempunyai orde atau derajat.
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Macam Interpolasi
Interpolasi Beda Terbagi Newton Interpolasi Lagrange Interpolasi Spline
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Macam Interpolasi Beda Terbagi Newton
Interpolasi Linier
Derajat/orde 1 memerlukan 2 titik
x f(x)
1 4,5
2 7.6
3 9.8
4 11.2
Berapa f(x = 1,325) = ?Memerlukan 2 titik awal :
x = 1x = 2
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Macam Interpolasi Beda Terbagi Newton
Interpolasi Kuadratik
Derajat/orde 2 memerlukan 3 titik
x = 1 f(x = 1) = . . . .
x = 2 f(x = 2) = . . . .
x = 3 f(x = 3) = . . . .f (x = 1,325) = ?
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Macam Interpolasi Beda Terbagi Newton
Interpolasi Kubik
Derajat/orde 3 memerlukan 4 titik
… Interpolasi derajat/orde ke-n
memerlukan n+1 titik
Semakin tinggi orde yang digunakan untuk interpolasi hasilnya akan semakin baik (teliti).
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Interpolasi Linier
Cara: menghubungkan 2 titik dengan sebuah garis lurus
Pendekatan formulasi interpolasi linier sama dengan persamaan garis lurus.
0
01
0101 xx
xxxfxf
xfxf
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Interpolasi Linier
Prosentase kesalahan pola interpolasi linier :
narnyaHarga_sebenarnyaHarga_sebeganl_perhitunHarga_hasi
εt
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Interpolasi Linier (Ex.1)
Diketahui suatu nilai tabel distribusi ‘Student t’ sebagai berikut :
t5% = 2,015
t2,5% = 2,571
Berapa t4% = ?
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Interpolasi Linier (Ex.1)
Penyelesaian
x0 = 5 f(x0) = 2,015
x1 = 2,5 f(x1) = 2,571 x = 4 f(x) = ?Dilakukan pendekatan dengan orde 1 :
0
01
0101 xx
xxxfxf
xfxf
237,22374,2
5455,2015,2571,2
015,2
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Interpolasi Linier (Ex.2)
Diketahui:log 3 = 0,4771213log 5 = 0,698700
Harga sebenarnya: log (4,5) = 0,6532125 (kalkulator).
Harga yang dihitung dengan interpolasi: log (4,5) = 0,6435078
%49,1%100
6532125,06532125,06435078,0
t
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Interpolasi Linier
Pendekatan interpolasi dengan derajat 1, pada kenyataannya sama dengan mendekati suatu harga tertentu melalui garis lurus.
Untuk memperbaiki kondisi tersebut dilakukan sebuah interpolasi dengan membuat garis yang menghubungkan titik yaitu melalui orde 2, orde 3, orde 4, dst, yang sering juga disebut interpolasi kuadratik, kubik, dst.
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Interpolasi Kuadratik
Interpolasi orde 2 sering disebut sebagai interpolasi kuadratik, memerlukan 3 titik data.
Bentuk polinomial orde ini adalah :
f2(x) = a0 + a1x + a2x2
dengan mengambil:
a0 = b0 – b1x0 + b2x0x1
a1 = b1 – b2x0 + b2x1
a2 = b2
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Interpolasi Kuadratik
Sehingga
f2(x) = b0 + b1(x-x0) + b2(x-x0)(x-x1)
dengan
Pendekatan dengan kelengkungan
Pendekatan dengan garis linier
01202
01
01
12
12
2
0101
011
00
,,
,
xxxfxx
xxxfxf
xxxfxf
b
xxfxxxfxf
b
xfb
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Interpolasi Kubik
f3(x) = b0 + b1(x-x0) + b2(x-x0)(x-x1) + b3(x-x0)(x-x1)(x-x2)
dengan:
012303
0121233
01202
01
01
12
12
02
01122
0101
011
00
,,,],,[],,[
,,],[],[
,
xxxxfxx
xxxfxxxfb
xxxfxx
xxxfxf
xxxfxf
xxxxfxxf
b
xxfxxxfxf
b
xfb
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Interpolasi Beda Terbagi Newton
Secara umum:
f1(x) = b0 + b1(x-x0)
f2(x) = b0 + b1(x-x0) + b2(x-x0)(x-x1)
f3(x) = b0 + b1(x-x0) + b2(x-x0)(x-x1) +
b3(x-x0)(x-x1)(x-x2)…
fn(x) = b0 + b1(x-x0) + b2(x-x0)(x-x1) +
b3(x-x0)(x-x1)(x-x2) + … +
bn(x-x1)(x-x2)…(x-xn-1)
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Interpolasi Beda Terbagi Newton
Dengan: b0 = f(x0)
b1 = f[x1, x0]
b2 = f[x2, x1, x0]
… bn = f[xn, xn-1, xn-2, . . . ., x0]
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Interpolasi Beda Terbagi Newton (Ex.)
Hitung nilai tabel distribusi ‘Student t’ pada derajat bebas dengan = 4%, jika diketahui:
t10% = 1,476 t2,5% = 2,571
t5% = 2,015t1% = 3,365
dengan interpolasi Newton orde 2 dan orde 3!
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Interpolasi Beda Terbagi Newton (Ex.)
Interpolasi Newton Orde 2: butuh 3 titik x0 = 5 f(x0) = 2,015
x1 = 2,5 f(x1) = 2,571
x2 = 1 f(x2) = 3,365 b0 = f(x0) = 2,015
02
01
01
12
12
2 xxxxxfxf
xxxfxf
b
222,0
55,2015,2571,2
01
011
xxxfxf
b
077,051
55,2015,2571,2
5,21571,2365,3
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Interpolasi Beda Terbagi Newton (Ex.)
f2(x) = b0 + b1(x-x0) + b2(x-x0)(x-x1)
= 2,015 + (-0,222) (4-5) +
0,077 (4-5)(4-2,5)
= 2,121
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Interpolasi Beda Terbagi Newton (Ex.)
Interpolasi Newton Orde 3: butuh 4 titik x0 = 5 f(x0) = 2,015
x1 = 2,5 f(x1) = 2,571
x2 = 1 f(x2) = 3,365
x3 = 10 f(x3) = 1,476
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Interpolasi Beda Terbagi Newton (Ex.)
b0 = f(x0) = 2,015
b1 = -0,222 f[x1,x0]
b2 = 0,077 f[x2,x1,x0]
007,05
077,0043,0510
077,05,210
5,21571,2365,3
110365,3476,1
3
b
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Interpolasi Beda Terbagi Newton (Ex.)
f3(x) = b0 + b1(x-x0) + b2(x-x0)(x-x1) +
b3(x-x0)(x-x1)(x-x2)
= 2,015 + (-0,222)(4-5) +
0,077 (4-5)(4-2,5) +
(-0,007)(4-5)(4-2,5)(4-1)
= 2,015 + 0,222 + 0,1155 + 0,0315
= 2,153
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Kesalahan Interpolasi Beda Terbagi Newton
Rn = |f[xn+1,xn,xn-1,…,x0](x-x0)(x-x1)…(x-xn)|
Menghitung R1
Perlu 3 titik (karena ada xn+1)
R1 = |f[x2,x1,x0](x-x0)(x-x1)|
Menghitung R2
Perlu 4 titik sebagai harga awal
R2 = |f[x3,x2,x1,x0](x-x0)(x-x1)(x-x2)|
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Kesalahan Interpolasi Beda Terbagi Newton (Ex.)
Berdasarkan contoh:
R1 = |f[x2,x1,x0](x-x0)(x-x1)|
= |0.077 (4-5)(4-2.5)|
= 0.1155
R2 = |f[x3,x2,x1,x0](x-x0)(x-x1)(x-x2)|
= |-0.007 (4-5)(4-2.5)(4-1)|
= 0.0315
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Interpolasi Lagrange
Interpolasi Lagrange pada dasarnya dilakukan untuk menghindari perhitungan dari differensiasi terbagi hingga (Interpolasi Newton)
Rumus:
dengan
n
iiin xfxLxf
0.
n
ijj ji
ji xx
xxxL
0
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Interpolasi Lagrange
Pendekatan orde ke-1
f1(x) = L0(x)f(x0) + L1(x)f(x1)
10
10 xx
xxxL
01
01 xx
xxxL
101
00
10
11 xf
xxxx
xfxxxx
xf
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Interpolasi Lagrange
Pendekatan orde ke-2
f2(x) = L0(x)f(x0) + L1(x)f(x1) + L2(x)f(x2)
20
2
10
1
200 xx
xxxxxx
xL
ijni
21
2
01
0
211 xx
xxxxxx
xL
ijni
12
1
02
0
222 xx
xxxxxx
xL
ijni
212
1
02
01
21
2
01
00
20
2
10
12 xf
xxxx
xxxx
xfxxxx
xxxx
xfxxxx
xxxx
xf
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Interpolasi Lagrange
Pendekatan orde ke-3
f3(x) = L0(x)f(x0) + L1(x)f(x1) + L2(x)f(x2) + L3(x)f(x3)
131
3
21
2
01
00
30
3
20
2
10
12 xf
xxxx
xxxx
xxxx
xfxxxx
xxxx
xxxx
xf
323
2
13
1
03
02
32
3
12
1
02
0 xfxxxx
xxxx
xxxx
xfxxxx
xxxx
xxxx
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Interpolasi Lagrange (Ex.)
Berapa nilai distribusi t pada = 4 %?
= 2,5 % x0 = 2,5 f(x0) = 2,571
= 5 % x1 = 5 f(x1) = 2,015
= 10 % x2 = 10 f(x2) = 1,476
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Interpolasi Lagrange (Ex.)
Pendekatan orde ke-1
f1(x) = L0(x)f(x0) + L1(x)f(x1)
101
00
10
11 xf
xxxx
xfxxxx
xf
237,2
015,25,255,24
571,255,2
54
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Interpolasi Lagrange (Ex.)
Pendekatan orde ke-2
f2(x) = L0(x)f(x0) + L1(x)f(x1) + L2(x)f(x2)
214,2
476,151054
5,2105,24
015,2105104
5,255,24
571,2105,2
10455,2
54
212
1
02
01
21
2
01
00
20
2
10
12 xf
xxxx
xxxx
xfxxxx
xxxx
xfxxxx
xxxx
xf
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Interpolasi Spline
Tujuan: penghalusan Interpolasi spline linear, kuadratik, kubik.
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Interpolasi Cubic Spline
dimana Si adalah polinomial berderajat 3:
p(xi) = di + (x-xi) ci + (x-xi)2 bi + (x-xi)3 ai, i=1,2, …, n-1
Syarat: Si(xi) = Si+1(xi), Si’(xi) = Si+1’(xi), Si’’(xi) = Si+1’’(xi)
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Interpolasi Cubic Spline
Interpolasi spline kubik menggunakan polinomial p(x) orde 3
p(x) = di + (x-xi) ci + (x-xi)2 bi + (x-xi)3 ai
Turunan pertama dan kedua p(xi) yaitu:
p’(x) = ci + 2bi (x-xi) + 3ai (x-xi)2
p”(x) = 2bi + 6ai (x-xi)
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Interpolasi Cubic Spline
Evaluasi pada titik x=xi menghasilkan:
pi = p(xi) = di
pi” = p”(xi) = 2bi
Evaluasi pada titik x=xi+1 menghasilkan:
pi = di + (xi+1-xi) ci + (xi+1-xi)2 bi + (xi+1-xi)3 ai
p(xi) = di + hi ci + hi2 bi + hi
3 ai
p”i = 2bi + 6ai (xI+1-xi)
p”(xi+1) = 2bi + 6ai hi
dimana hi = (xI+1-xi)
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Interpolasi Cubic Spline
Jadi:
di = pi
Sehingga:
2"p
b ii
i
i1ii 6h
p"p"a
6p"2hp"h
hpp
c ii1ii
i
i1ii
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Interpolasi Cubic Spline (Ex.)