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Metode Numeric
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Nama : SATRIA ADI PRASETYA
NIM : A410090156
Kelas : VD
TUGAS METODE NUMERIK (LANJUTAN)
1. Tentukan rumus pendekatan turunan kedua beserta order kesalahannya dengan metode:
a. Forward Difference order-2, O(h^2)
b. Backward Difference order-2, O(h^2)
c. Central Difference order-4, O(h^4)
Penyelesaian:
a. Forward Difference order-2, O(h^2)
f ( x 0+h )=f ( x 0 )+h f I ( x0 )+ h2 f II ( x0 )2!
+h3 f III ( x0 )
3 !+
h4 f IV (x 0)4 !
+… x8
f ( x 0+2h )=f ( x0 )+2hf I ( x0 )+ 4 h2 f II ( x0 )2 !
+8 h3 f III ( x0 )
3 !+
16 h4 f IV ( x 0 )4 !
+…
8 f ( x 0+h )=8 f ( x 0 )+8 hf I ( x 0 )+8h2 f II (x 0 )
2 !+8
h3 f III (x 0 )3!
+8h4 f IV ( x 0 )
4 !+…
f ( x 0+2h )=f ( x0 )+2hf I ( x0 )+ 4 h2 f II ( x0 )2 !
+8 h3 f III ( x0 )
3 !+
16 h4 f IV ( x 0 )4 !
+…
8 f ( x 0+h )−f ( x 0+2 h )=7 f (x 0 )+6 hf I ( x0 )+ 4 h2 f II ( x0 )2 !
−8 h4 f IV ( x0 )
4 !+…
4 h2 f II ( x 0 )2!
=8 f ( x 0+h )−7 f ( x0 )−f ( x 0+2 h )−6hf I ( x 0 )+ 8 h4 f IV ( x 0 )4 !
+…
2 h2 f II ( x0 )=8 f ( x 0+h )−7 f ( x0 )−f ( x 0+2h )−6 hf I ( x0 )+ 8 h4 f IV ( x0 )4 !
+…
f II ( x 0 )=8 f ( x 0+h )−7 f ( x0 )−f ( x 0+2 h )−6hf I ( x 0 )2h2 +
h2 f IV ( x0 )6
+…
ϑ (h2)=h2 f IV ( x0 )6
b. Backward Difference order-2, O(h^2)
f ( x 0−h )=f (x 0 )−hf I ( x 0 )+ h2 f II (x0)2!
−h3 f III(x 0)
3!+
h4 f IV (x 0)4 !
+… x8
f ( x 0−2h )=f ( x 0 )−2hf I (x 0 )+ 4 h2 f II (x0)2 !
−8 h3 f III(x 0)
3!+
16 h4 f IV (x 0)4 !
+…
f ( x 0−2h )=f ( x 0 )−2 hf I (x 0 )+ 4 h2 f II (x0)2 !
−8 h3 f III(x 0)
3!+
16 h4 f I ( x0)4 !
+…
8 f ( x 0−h )=8 f (x 0 )−8hf I ( x 0 )+8h2 f II (x0)
2!−8
h3 f III(x 0)3 !
+8h4 f IV (x0)
4 !+…
f ( x 0−2h )−8 f ( x 0−h )=−7 f (x 0 )+6hf I ( x0 )−4h2 f II ( x0 )
2!+8
h4 f IV (x 0 )4 !
+…
4h2 f II ( x 0 )
2 !=8 f ( x 0−h )−7 f ( x 0 )−f ( x0−2h )+6hf I (x 0 )+8
h4 f IV ( x 0 )4 !
+…
2 h2 f II ( x0 )=8 f ( x 0−h )−7 f ( x 0 )−f ( x0−2h )+6 hf I ( x 0 )+8h4 f IV ( x 0 )
4 !+…
f II ( x 0 )=8 f ( x 0−h )−7 f ( x 0 )−f ( x0−2 h )+6 hf I ( x 0 )2 h2 +
h2 f IV ( x 0 )6
+…
ϑ (h¿¿2)=h2 f IV (x 0 )
6¿
c. Central Difference order-4, O(h^4)
f ( x 0+h )=f ( x 0 )+hf I ( x 0 )+ h2 f II(x 0)2 !
+h3 f III(x 0)
3 !+
h4 f IV (x0)4 !
+h5 f V (x 0)
5 !+
h6 f VI( x0)6 !
+…
f ( x 0−h )=f (x 0 )−hf I ( x 0 )+ h2 f II (x0)2!
−h3 f III(x 0)
3!+
h4 f IV (x 0)4 !
−h5 f V (x0)
5 !+
h6 f VI (x 0)6 !
+…
f ( x 0+h )+ f ( x0−h )=2 f ( x0 )+2h2 f II (x 0)
2 !+2
h4 f IV (x 0)4 !
+2h6 f VI (x 0)
6 !+…
f ( x 0+2 h )=f ( x0 )+2hf I ( x0 )+ 4 h2 f II (x 0)2!
+8 h3 f III(x 0)
3 !+
16 h4 f IV (x0)4 !
+32 h5 f V (x0)
5 !+…
f ( x 0−2h )=f ( x 0 )−2 hf I (x 0 )+ 4 h2 f II (x0)2 !
−8 h3 f III(x 0)
3!+
16 h4 f IV (x 0)4 !
−32 h5 f V (x 0)
5!+…
f ( x 0+2h )+ f (x 0−2h )=2 f ( x0 )+8h2 f II ( x 0 )
2 !+32
h4 f IV ( x 0 )4 !
+128h6 f VI ( x 0 )
6 !+…
f ( x 0+h )+ f ( x0−h )=2 f ( x0 )+2h2 f II(x 0)
2 !+2
h4 f IV (x 0)4 !
+2h6 f VI (x 0)
6 !+… x16
f ( x 0+2h )+ f (x 0−2h )=2 f ( x0 )+8h2 f II ( x 0 )
2 !+32
h4 f IV ( x 0 )4 !
+128h6 f VI ( x 0 )
6 !+…
16 f ( x 0+h )+16 f ( x0−h )=32 f (x 0 )+32h2 f II (x0)
2!+32
h4 f IV (x 0)4 !
+32h6 f VI (x0)
6!+…
f ( x 0+2h )+ f (x 0−2h )−16 f ( x 0+h )−16 f ( x 0−h )=−30 f ( x 0 )−24h2 f II(x 0)
2 !+96
h6 f VI (x0)6!
+…
24h2 f II(x 0)
2 !=−f ( x 0+2h )−30 f ( x 0 )−f ( x0−2h )+96
h6 f VI (x 0)6 !
+…
12 h2 f II (x 0 )=−f ( x 0+2 h )−30 f ( x 0 )−f ( x0−2 h )+96h6 f V I (x 0)
6!+…
f II ( x 0 )=−f ( x 0+2 h )−30 f (x 0 )−f ( x0−2 h )12 h2 +
h4 f VI (x 0)90
+…
ϑ (h4 )=h4 f VI (x 0)90
2. Tentukan rumus pendekatan turunan ketiga beserta order kesalahannya dengan
metode Forward Difference, Backward Difference, dan Central Difference (Petunjuk:
ingat p>=n+1)
Penyelesaian:
a. Turunan 3 Forward Difference dengan titik x0, x0+h, x0+2h
f ( x 0+h )=f ( x 0 )+hf I ( x 0 )+ h2 f II(x 0)2 !
+h3 f III(x 0)
3 !+
h4 f IV (x0)4 !
+…
f ( x 0+2h )=f ( x0 )+2hf I ( x0 )+ 4 h2 f II(x 0)2!
+8h3 f III(x 0)
3 !+
16 h4 f IV (x0)4 !
+…
4 f ( x 0+h )=4 f (x 0 )+4 hf I ( x 0 )+4h2 f II (x0)
2!+4
h3 f III(x 0)3 !
+4h4 f IV (x0)
4 !+…
-
f ( x 0+2 h )−4 f (x 0+h )=−3 f ( x 0 )−2 hf I (x 0 )+ 4 h3 f III(x 0)3 !
+12h4 f IV (x 0)
4 !+…
4 h3 f III (x0)3 !
= f (x 0+2 h )+3 f (x 0 )−4 f (x 0+h )+2hf I ( x0 )−12 h4 f IV (x 0)4 !
+…
2h3 f III (x 0)3
=f (x 0+2 h )+3 f (x 0 )−4 f (x 0+h )+2 hf I ( x0 )−12 h4 f IV (x 0)4 !
+…
f III ( x 0 )=3 f ( x 0+2 h )+9 f ( x 0 )−12 f ( x 0+h )+6 hf I ( x0 )2 h3 −3
h f IV (x0)4 !
+…
ϑ (h )=−3 h f IV (x 0)4
b. Turunan 3 Backward Difference dengan titik x0-2h, x0-h, x0
f ( x 0−h )=f (x 0 )−hf I ( x 0 )+ h2 f II (x0)2!
−h3 f III(x 0)
3!+
h4 f IV (x 0)4 !
+…
f ( x 0−2h )=f ( x 0 )−2 hf I (x 0 )+ 4 h2 f II (x0)2 !
−8 h3 f III(x 0)
3!+
16 h4 f IV (x 0)4 !
+…
4 f ( x 0−h )=4 f ( x0 )−4 hf I ( x 0 )+4h2 f II (x 0)
2 !−4
h3 f III (x0)3 !
+4h4 f IV (x 0)
4 !+…
f ( x 0−2h )−4 f ( x0−h )=−3 f ( x0 )+2 hf I ( x0 )−4h3 f III (x0)
3 !+12
h4 f IV (x 0)4 !
+…
4h3 f III (x0)
3 !=4 f ( x 0−h )−3 f ( x 0 )−f ( x0−2h )+12
h4 f IV (x 0)4 !
+…
2h3 f III (x 0)3
=4 f ( x 0−h )−3 f ( x 0 )−f ( x0−2 h )+12h4 f IV (x 0)
4 !+…
2 h3 f III ( x0 )=12 f ( x0−h )−9 f ( x 0 )−3 f ( x 0−2 h )+36h4 f IV (x 0)
4 !+…
f III ( x 0 )=12 f ( x0−h )−9 f ( x 0 )−3 f ( x 0−2 h )2h3 +
3 h f IV (x 0)4
+…
ϑh=3 h f IV (x0)
4
c. Turunan 3 Central Differance dengan titik x0-h,x0,x0+h
f ( x 0+h )=f ( x 0 )+hf I ( x 0 )+ h2 f II(x 0)2 !
+h3 f III(x 0)
3 !+
h4 f IV (x0)4 !
+h5 f V (x 0)
5 !+…
f ( x 0−h )=f (x 0 )−hf I ( x 0 )+ h2 f II (x0)2!
−h3 f III(x 0)
3!+
h4 f IV (x 0)4 !
−h5 f V (x0)
5 !+…
f ( x 0+h )−f ( x 0−h )=2 hf I ( x 0 )+2h3 f III (x0)
3 !+2
h5 f V (x 0)5 !
+…
2h3 f III (x 0)
3 !=f (x 0+h )−2 hf I ( x 0 )−f ( x0−h )−2
h5 f V (x0)5 !
+…
h3 f III(x 0)3 !
= f ( x0+h )−2 hf I (x 0 )−f ( x0−h )−2h5 f V (x 0)
5 !+…
h3 f III ( x 0 )=3 f (x 0+h )−6 hf I (x 0 )−3 f ( x0−h )−6h5 f V (x 0)
5 !+…
f III ( x 0 )=3 f ( x 0+h )−6 hf I ( x 0 )−3 f ( x 0−h )h3 −6
h5 f V (x 0)5 !
+…
f III ( x 0 )=3 f ( x 0+h )−6 hf I ( x 0 )−3 f ( x 0−h )h3 −
h2 f V (x 0)20
+…
ϑ (h2)=−h2 f V ( x0)20
Tugas Komputasi/Praktikum:
Tentukan nilai pendekatan turunan kedua dan ketiga untuk fungsi f(x)=exp(x)-x^3+1
pada xo=1 dengan h=0.1, 0.05, dan 0.025 dengan rumus yang diperoleh pada soal no.
1 dan 2. Selanjutnya lakukan perbandingan error dengan menggunakan order
kesalahan untuk menghitung errornya.
Penyelesaian:
Untuk soal no 1
>> syms x;
>> f=exp(x)-x^3+1;
>> x0=1;
>> h1=0.1;
>> h2=0.05;
>> h3=0.025;
>> f1=diff(f);
>> f2=diff(f1);
>> f3=diff(f2);
>> f4=diff(f3);
>> f5=diff(f4);
>> f6=diff(f5);
>> a1=x0+h1;
>> a2=x0+h2;
>> a3=x0+h3;
>> b1=x0+2*h1;
>> b2=x0+2*h2;
>> b3=x0+2*h3;
>> c1=x0-h1;
>> c2=x0-h2;
>> c3=x0-h3;
>> d1=x0-2*h1;
>> d2=x0-2*h2;
>> d3=x0-2*h3;
>> fx0=subs(f,x0);
>> f1x0=subs(f1,x0);
>> f4x0=subs(f4,x0);
>> f6x0=subs(f6,x0);
>> fa1=subs(f,a1);
>> fa2=subs(f,a2);
>> fa3=subs(f,a3);
>> fb1=subs(f,b1);
>> fb2=subs(f,b2);
>> fb3=subs(f,b3);
>> fc1=subs(f,c1);
>> fc2=subs(f,c2);
>> fc3=subs(f,c3);
>> fd1=subs(f,d1);
>> fd2=subs(f,d2);
>> fd3=subs(f,d3);
>> %forward difference
>> ffd1=(8*fa1-7*fx0-fb1-6*h1*f1x0)/2*h1.^2 %untuk h1
ffd1 =
-3.2865e-004
>> ffd2=(8*fa2-7*fx0-fb2-6*h2*f1x0)/2*h2.^2 %untuk h2
ffd2 =
-2.0518e-005
>> ffd3=(8*fa3-7*fx0-fb3-6*h3*f1x0)/2*h3.^2 %untuk h3
ffd3 =
-1.2820e-006
>> %backward difference
>> fbd1=(8*fc1-7*fx0-fd1+6*h1*f1x0)/2*h1.^2 %untuk h1
fbd1 =
-3.2860e-004
>> fbd2=(8*fc2-7*fx0-fd2+6*h2*f1x0)/2*h2.^2 %untuk h2
fbd2 =
-2.0518e-005
>> fbd3=(8*fc3-7*fx0-fd3+6*h3*f1x0)/2*h3.^2 %untuk h3
fbd3 =
-1.2820e-006
>> %central difference
>> fcd1=(-fb1-30*fx0-fd1)/12*h1.^2 %untuk h1
fcd1 =
-0.0724
>> fcd2=(-fb2-30*fx0-fd2)/12*h2.^2 %untuk h2
fcd2 =
-0.0181
>> fcd3=(-fb3-30*fx0-fd3)/12*h3.^2 %untuk h3
fcd3 =
-0.0045
>> %forward error dengan order kesalahan
>> O1=(h1.^2*f4x0)/6 %untuk h1
O1 =
0.0045
>> O2=(h2.^2*f4x0)/6 %untuk h2
O2 =
0.0011
>> O3=(h3.^2*f4x0)/6 %untuk h3
O3 =
2.8315e-004
>> %backward error dengan order kesalahan
>> O1=(h1.^2*f4x0)/6 %untuk h1
O1 =
0.0045
O2=(h2.^2*f4x0)/6 %untuk h2
O2 =
0.0011
O3=(h3.^2*f4x0)/6 %untuk h3
O3 =
2.8315e-004
>> %central error dengan order kesalahan
>> O1=(h1.^4*f6x0)/90 %untuk h1
O1 =
3.0203e-006
>> O2=(h2.^4*f6x0)/90 %untuk h2
O2 =
1.8877e-007
>> O3=(h3.^4*f6x0)/90 %untuk h3
O3 =
1.1798e-008
Untuk yang no 2
>> syms x;
>> f=exp(x)-x.^3+1;
>> x0=1;
>> h1=0.1;
>> h2=0.05;
>> h3=0.025;
>> f1=diff(f);
>> f2=diff(f1);
>> f3=diff(f2);
>> f4=diff(f3);
>> f5=diff(f4);
>> a1=x0+h1;
>> a2=x0+h2;
>> a3=x0+h3;
>> b1=x0-h1;
>> b2=x0-h2;
>> b3=x0-h3;
>> c1=x0+2*h1;
>> c2=x0+2*h2;
>> c3=x0+2*h3;
>> d1=x0-2*h1;
>> d2=x0-2*h2;
>> d3=x0-2*h3;
>> fx0=subs(f,x0);
>> fa1=subs(f,a1);
>> fa2=subs(f,a2);
>> fa3=subs(f,a3);
>> fb1=subs(f,b1);
>> fb2=subs(f,b2);
>> fb3=subs(f,b3);
>> fc1=subs(f,c1);
>> fc2=subs(f,c2);
>> fc3=subs(f,c3);
>> fd1=subs(f,d1);
>> fd2=subs(f,d2);
>> fd3=subs(f,d3);
>> f1x0=subs(f1,x0);
>> f4x0=subs(f4,x0);
>> f5x0=subs(f5,x0);
>> %untuk forward
>> ffd1=(3*fc1+9*fx0-12*fa1+6*h1*f1x0)/2*h1.^3 %untuk h1
ffd1 =
-3.0680e-006
>> ffd2=(3*fc2+9*fx0-12*fa2+6*h2*f1x0)/2*h2.^3 %untuk h2
ffd2 =
-4.9646e-008
>> ffd3=(3*fc3+9*fx0-12*fa3+6*h3*f1x0)/2*h3.^3 %untuk h3
ffd3 =
-7.8861e-010
>> %untuk backward
>> fbd1=(12*fb1-9*fx0-3*fd1)/2*h1.^3 %untuk h1
fbd1 =
8.1039e-005
>> fbd2=(12*fb2-9*fx0-3*fd2)/2*h2.^3 %untuk h2
fbd2 =
5.2294e-006
>> fbd3=(12*fb3-9*fx0-3*fd3)/2*h3.^3 %untuk h3
fbd3 =
3.2932e-007
>> % untuk central
>> fcd1=(3*fa1-6*h1*f1x0-3*fb1)/h1.^3 %untuk h1
fcd1 =
-3.2804
>> fcd2=(3*fa2-6*h2*f1x0-3*fb2)/h2.^3 %untuk h2
fcd2 =
-3.2814
>> fcd3=(3*fa3-6*h3*f1x0-3*fb3)/h3.^3 %untuk h3
fcd3 =
-3.2816
>> % forward error dengan order kesalahan
>> O1=-(3*h1*f4x0)/4 %untuk h1
O1 =
-0.2039
>> O2=-(3*h2*f4x0)/4 %untuk h2
O2 =
-0.1019
>> O3=-(3*h3*f4x0)/4 %untuk h3
O3 =
-0.0510
>> %backward error dengan order kesalahan
>> O1=(3*h1*f4x0)/4 %untuk h1
O1 =
0.2039
>> O2=(3*h2*f4x0)/4 %untuk h2
O2 =
0.1019
>> O3=(3*h3*f4x0)/4 %untuk h3
O3 =
0.0510
>> %central error dengan order kesalahan
>> O1=-(h1.^2*f5x0)/20 %untuk h1
O1 =
-0.0014
>> O2=-(h2.^2*f5x0)/20 %untuk h2
O2 =
-3.3979e-004
>> O3=-(h3.^2*f5x0)/20 %untuk h3
O3 =
-8.4946e-005
