metode numerik

Moh. Sudrajat Dwi R. / 22 12 100 044 / tugas 2

Problem 1: Diketahui P(x) = - 0.02x3 + 0.1x2 0.2x + 1.66, dan melewati titik-titik (1, 1.54), (2, 15), (3, 1.42), dan (5, 0.66).

a. Hitunglah P(4). b. Hitunglah P(4). c. Hitunglah definite integral dari P(x) pada [1, 4]. d. Hitunglah extrapolasi P(5.5). e. Tunjukan perhitungan koefisien dari P(x) dengan menggunakan titik-titik yang

terdefinisi Jawab : a. P(4) = - 0.02(4)3 + 0.1(4)2 0.2(4) + 1.66 = 1.18

b. P(4) = - 0.06(4)2 + 0.2(4) 0.2 = - 0.36

c. P(x) =

d. P(5.5) = - 0.02(5.5)3 + 0.1(5.5)2 0.2(5.5) + 1.66 = 0.2575 e. a0 + a1(1) + a2(1)

2 + a3(1)3 = 1.54

a0 + a1(2) + a2(2)2 + a3(2)

3 = 15 a0 + a1(3) + a2(3)

2 + a3(3)3 = 1.42

a0 + a1(5) + a2(5)2 + a3(5)

3 = 0.66

k Xk f(Xk) f(Xk-1,Xk) f(Xk-2,Xk-1,Xk) f(Xk-3,Xk-2,Xk-1,Xk)

0 1 1.54 - - -

1 2 15 13.46 - -

2 3 1.42 -13.58 -13.52 -

3 5 0.66 -0.76 6.41 6.643333333

P(x) = 1.54 + 13.46(x-1) - 13.52(x-1)(x-2) + 6.643333333(x-1)(x-2)(x-5)

Problem 2: Diketahui f(x) = 2 sin(x/6)

a. Gunakan interpolasi quadratic lagrange berdasarkan titik x0 = 0, x1 = 1, dan x2 = 3 untuk melakukan pendekatan f(4) dan f(3.5)

b. Gunakan interpolasi cubic lagrange berdasarkan titik x0 = 0, x1 = 1, x2 = 3, dan x3 = 5 untuk melakukan pendekatan f(4) dan f(3.5)

Jawab : a. x0 = 0, f0 = 2 sin(0) = 0 ; x1 = 1, f1 = 2 sin(/6) = 1 ; x2 = 3, f2 = 2 sin(3/6) = 2

P

P2(4) =(40)(43)

2+

40 41

6 2 = 2

P2(3.5) =(3.50)(3.53)

2+

3.50 3.51

6 2 = 2.041666667

b. x0 = 0, f0 = 2 sin(0) = 0 ; x1 = 1, f1 = 2 sin(/6) = 1 ; x2 = 3, f2 = 2 sin(3/6) = 2 , x3 = 5, f3

= 2 sin(5/6) = 1

P3(x) = 1 3 5

01 03 05 0 +

0 3 5

10 13 15 1 +

0 1 5

30 31 35 2 +

0 1 3

50 51 53 (1)

= 1 3 5

15 0 +

0 3 5

8 1 +

0 1 5

12 2 +

0 1 3

40(1)

P3(4) = 41 43 45

15 0 +

40 43 45

8 1 +

40 41 45

12 2 +

40 41 43

40(1)

=2.2

P3(3.5) = 3.51 3.53 3.55

15 0 +

3.50 3.53 3.55

8 1 +

3.50 3.51 3.55

12 2 +

3.50 3.51 3.53

40(1)

=1.96875 Problem 3: Tentukan Lagrange Polynomial untuk melakukan pendekatan terhadap fungsi f(x) = x3

a. P1(x) dengan titik x0 = - 1 dan x1 = 0 b. P2(x) dengan titik x0 = - 1, x1 = 0, dan x2 = 1. c. P3(x) dengan titik x0 = - 1, x1 = 0, x2 = 1, dan x3 = 2

Jawab : a. x0 = -1, f0 = (-1)

3 = -1 ; x1 = 0, f1 = (0)3 = 0

P

b. x0 = -1, f0 = (-1)3 = -1 ; x1 = 0, f1 = 0 ; x2 = 1, f2 = (1)

3 = 1

P

c. x0 = -1, f0 = (-1)3 = -1 ; x1 = 0, f1 = 0 ; x2 = 1, f2 = (1)

3 = 1 ; x3 = 2 , f3 = (2)3 = 8

P

Problem 4: Diasumsikan f(x) = x1=2 dan point-point berikut. Hitunglah

K xk f(xk)

0 4.0 2.00000

1 5.0 2.23607

2 6.0 2.44949

3 7.0 2.64575

4 8.0 2.82843

a. Divide-different tabel

b. Tulis polinomial newton P1(x), P2(x), P3(x), dan P4(x). c. Hitung P3(4.5), P3(7.5), P4(4.5), dan P4(7.5).

Jawab : a .

k Xk f(Xk) f(Xk-

1,Xk)

f(Xk-2,Xk-1,Xk) f(Xk-3,Xk-2,Xk-1,Xk) f(Xk-4,Xk-3,Xk-2,Xk-1,Xk)

0 4.0 2.00000 - - - -

1 5.0 2.23607 0.23607 - - -

2 6.0 2.44949 0.21342 -0.011325 - -

3 7.0 2.64575 0.19626 -0.00858 0.000915 -

4 8.0 2.82843 0.18268 -0.00679 0.000596666 0.000079583

b .

P4(x) = 2 + 0.23607(x-4) -0.011325(x-4)(x-5)+0.000915(x-4)(x-5)(x-6)+0.000079583(x-4)(x5)(x-6)(x-7) P3(x) = 2 + 0.23607(x-4) -0.011325(x-4)(x-5)+0.000915(x-4)(x-5)(x-6) P2(x) = 2 + 0.23607(x-4) -0.011325(x-4)(x-5) P1(x) = 2 + 0.23607(x-4)

c . P3(4.5) = 2 + 0.23607(4.5 - 4) -0.011325(4.5 - 4)(4.5 - 5)+0.000915(4.5 - 4)(4.5 - 5)(4.5 - 6) =2.121209375

P3(7.5) = 2 + 0.23607(7.5 - 4) -0.011325(7.5 - 4)(7.5 - 5)+0.000915(7.5 - 4)(7.5 - 5)(7.5 - 6) =2.793160625

P4(4.5) = 2 + 0.23607(4.5 - 4) -0.011325(4.5 - 4)(4.5 - 5)+0.000915(4.5 - 4)(4.5 - 5)(4.5 - 6) +0.000079583(4.5 - 4)(4.5 - 5)(4.5 - 6)(4.5 - 7) = 2.121209375 0.000018359

=2.121191016 P4(7.5) = 2 + 0.23607(7.5 - 4) -0.011325(7.5 - 4)(7.5 - 5)+0.000915(7.5 - 4)(7.5 - 5)(7.5 - 6) +0.000079583(7.5 - 4)(7.5 - 5)(7.5 - 6)(7.5 - 7) = 2.793160625 + 0.000522263

=2.7936828888

Problem 5: Diasumsikan f(x) = 3 sin2(x/6) dan point-point berikut. Hitunglah

k xk f(xk)

0 0.0 0.00

1 1.0 0.75

2 2.0 2.25

3 3.0 3.00

4 4.0 2.25

(a) Divide-different tabel (b) Tulis polinomial newton P1(x), P2(x), P3(x), dan P4(x). (c) Hitung P3(1.5), P3(3.5), P4(1.5), dan P4(3.5) Jawab : a .

k Xk f(Xk) f(Xk-

1,Xk)


0 1.0 0.00 - - - -

1 2.0 0.75 0.75 - - -

2 3.0 2.25 1.5 0.225 - -

3 4.0 3.00 0.75 -0.375 -0.2 -

4 5.0 2.25 -0.75 -0.75 -0.125 -0.0125

b . P4(x) = 0 + 0.75(x-1) +0.225(x-1)(x-2)-0.2(x-1)(x-2)(x-3)-0.0125(x-1)(x-2)(x-3)(x-4) P3(x) = 0 + 0.75(x-1) +0.225(x-1)(x-2)-0.2(x-1)(x-2)(x-3) P2(x) = 0 + 0.75(x-1) +0.225(x-1)(x-2) P1(x)= 0 + 0.75(x-1)

c . P3(1.5) = 0 + 0.75(1.5-1) +0.225(1.5-1)(1.5-2)-0.2(1.5-1)(1.5-2)(1.5-3) =0.24375

P3(3.5) = 0 + 0.75(3.5-1) +0.225(3.5-1)(3.5-2)-0.2(3.5-1)(3.5-2)(3.5-3) =2.34375

P4(1.5) = 0 + 0.75(1.5-1) +0.225(1.5-1)(1.5-2) - 0.2(1.5-1)(1.5-2)(1.5-3) - 0.0125(1.5-

1)(1.52)(1.5-3)(1.5-4) = 0.24375 + 0.01171875

=0.25546875 P4(3.5) = 0 + 0.75(3.5-1) +0.225(3.5-1)(3.5-2)-0.2(3.5-1)(3.5-2)(3.5-3)-0.0125(3.5-

1)(3.52)(3.5-3)(3.5-4) = 2.34375 + 0.01171875

=2.35546875

Problem 6: Diasumsikan f(x) = e-x dan point-point berikut. Hitunglah

k xk f(xk)

0 4.0 1.00000

1 5.0 0.36788

2 6.0 0.13534

3 7.0 0.04979

4 8.0 0.01832

(a) Divide-different tabel (b) Tulis polinomial newton P1(x), P2(x), P3(x), dan P4(x). (c) Hitung P3(0.5), P3(1.5), P4(0.5), dan P4(1.5) Jawab: a .

k Xk f(Xk) f(Xk-

1,Xk)


0 4.0 1.00000 - - - -

1 5.0 0.3678 -0.6322 - - -

2 6.0 0.13534 -0.23246 0.43233 - -

3 7.0 0.04979 -0.08555 0.073455 -0.119625 -

4 8.0 0.01832 -0.03147 0.02704 -0.015471666 0.104153334

b .

P1(x) = 1 - 0.6322(x-4)

P2(x) = 1 - 0.6322(x-4) + 0.43233(x-4)(x-5) P3(x) = 1 - 0.6322(x-4) + 0.43233(x-4)(x-5)-0.119625(x-4)(x-5)(x-6) P4(x) = 1 - 0.6322(x-4) + 0.43233(x-4)(x-5)-0.119625(x-4)(x-5)(x-6)+0.104153334(x-4)(x-

5)(x-6)(x-7) c .

P3(0.5) = 1 - 0.6322(0.5-4) + 0.43233(0.5-4)(0.5-5)-0.119625(0.5-4)(0.5-5)(0.5-6) =20.38441313

P3(1.5) = 1 - 0.6322(1.5-4) + 0.43233(1.5-4)(1.5-5)-0.119625(1.5-4)(1.5-5)(1.5-6) =11.07362188

P4(0.5) = 1 - 0.6322(0.5-4) + 0.43233(0.5-4)(0.5-5)-0.119625(0.5-4)(0.5-5)(0.5-6) + 0.104153334(0.5-4)(0.5-5)(0.5-6)(0.5-7)

= 20.38441313 + 58.64483663 =79.02924976 P4(1.5) = 1 - 0.6322(1.5-4) + 0.43233(1.5-4)(1.5-5)-0.119625(1.5-4)(1.5-5)(1.5-6) +

0.104153334(1.5-4)(1.5-5)(1.5-6)(1.5-7) = 11.07362188 + 22.55570639 =33.62932827

Problem 7: Termistor digunakan untuk mengukur temperatur tubuh. Termistor berbasikan

pada perubahan resistansi pada suatu material. Suatu manufactur memberikan Anda suatu

data percobaan hasil pengukuran suhu dan hambatan dari termistor sebagai berikut :

R(ohm) T(C)

1101.0 25.113

911.3 30.131

636.0 40.120

451.1 50.128

(a) Tentukan persamaan linear spline (b) Persamaan kalibrasi yang dilakukan oleh manufaktur adalah

[ln R] + a2[ln R]2

tentukan persamaan quadratic spline untuk persamaan diatas. a . Quadratic spline Ax + B

Jawab : f(x) = f(x0) + f(x1)-f(x0)

(x x0) x0 x x1

x1-x0

f(x1) + f(x2)-f(x1)

(x x0) x1 x x2

x2-x1

f(x2) + f(x3)-f(x2)

(x x0) x2 x x3

x3-x2

= 25.113 + (30.131 25.113)/(911.3-1101) x (x-1101) 1101 x 911.3

= 30.131 + (40.120 30.131)/(636 911.3) x ( x 911.3) 911.3 x 636

= 40.120 + (50.128 40.120 )/(451.1 636) x ( x 636) 636 x 451.1

25.113 + (x 1101)x(-0.026452293) 1101 x 911.3 30.131 + (x 911.3)x(-0.036284053) 911.3 x 636 40.120 + (x 636)x(-0.054126554) 636 x 451.1

b. .

R(ohm) T(C) Ln R (Ln R)2

1101.0 25.113 7.003974137 49.05565371

911.3 30.131 6.814872152 46.44248245

636.0 40.120 6.455198563 41.66958849

451.1 50.128 6.111689044 37.35274298

1

=a0 + a1[ln R] + a2[ln R]

2

Y = 1/T a2 = a1 ; a1 = b1 ; a0 = C1

Y = a1x2 + b1x + c1 1101 x 911.3

Y = a2x2 + b1x + c2 911.3 x 636

Y = a3x2 + b1x + c3 636 x 451.1

a1[49.05565371] + b1[7.003974137] + C1 = 0.039820013

a1[46.44248245] + b1[6.814872152] + C1 = 0.03318841

a2[46.44248245] + b2[6.814872152] + C2 = 0.03318841

a2[41.66958849] + b2[6.455198563] + C2 = 0.024925224

a3[41.66958849] + b3[6.455198563] + C3 = 0.024925224

a3[37.35274298] + b3[6.111689044] + C3 = 0.01994893

Persamaan bantu

2a1[6.814872152] + b1 = 2a2[6.814872152] + b2

2a2[6.455198563] + b2 = 2a3[6.455198563] + b3

Asumsi a1 = 0 maka

b1[7.003974137] + c1 = 0.039820013

b1[6.814872152] + c1 = 0.03318841

----------------------------(dikurangkan)

b1 = 0.035091715 masuk ke b1[6.814872152] + c1 = 0.03318841

c1 = -0.205957141

a2[46.44248245] + b2[6.814872152] + c2 = 0.03318841

a2[41.66958849] + b2[6.455198563] + c2 = 0.024925224

2a2[6.814872152] + b2 = 0.035091715

Dari eliminasi gauss didapatkan

a2 = 0.084012234 b2 = -1.120140043 c2 = 3.765062887

a3[41.66958849] + b3[6.455198563] + c3= 0.024925224

a3[37.35274298] + b3[6.111689044] + c3= 0.01994893

2a3[6.455198563] + b3 = 12.71307419

Dari eliminasi gauss didapatkan

a3 = -0.145542855 b3 = 1.843507313 c3 = -5.810569678

Persamaan quadratic spline

0.035091715X - 0.205957141 1101< R < 911.3

0.084012234X2 -1.120140043X + 3.765062887 911.3< R < 636

-0.145542855X2

+1.843507313+ -5.810569678 636< R < 451.1

y = 1/T T = 1/y X = ln R Problem 8: Lanjutan dari soal no 7. Tentukan temperatur jika R = 754.8 , R = 465.77 dan

R = 956.3 dengan menggunakan (a) Persamaan quadratic spline pada poin (a) pada no 7. (b) Persamaan quadratic spline pada poin (b) pada no 7. Jawab: a .

25.113 + (956.3 1101)x(-0.026452293) = 28.9406468 1101 x 911.3 30.131 + (754.8 911.3)x(-0.036284053)= 35.80945429 911.3 x 636 40.120 + (465.77 636)x(-0.054126554) = 49.33396329 636 x 451.1

b . 0.035091715(ln 956.3)-0.205957141 = 28.669877560 0.084012234(ln 754.8)2-1.120140043(ln 754.8) + 3.765062887 = 31.771651090 -0.145542855(ln 465.77)2+1.843507313(ln 465.77) -5.810569678 = 45.738408827

Problem 9: Temukan solusi least-square y = f(x) = Ax+B untuk data berikut dan kalkulasi E2(f)

xk yk f(xk)

-2 1 1.2

-1 2 1.9

0 3 2.6

1 3 3.3

2 4 4.0

Jawab:

xk yk f(xk) (xk)2 xk.yk ek ek2

-2 1 1.2 4 4 0.2 0.04

-1 2 1.9 1 2 -0.1 0.01

0 3 2.6 0 0 -0.4 0.16

1 3 3.3 1 3 0.3 0.09

2 4 4.0 4 16 0 0

0 13 10 25 0 0.3

10A + 0B = 25 A = 2.5 Y = 2.5(x) + 2.6 0A + 5B = 13 B = 2.6

e2(f) = (0.3/5)1/2 = 0.244948974

Problem 10: Temukan solusi least-square y = f(x) = Ax+B untuk data berikut dan kalkulasi

E2(f)

xk yk f(xk)

-6 -5.3 -6.00

-2 -3.5 -2.48

0 -1.7 -1.26

2 0.2 0.32

6 4.0 3.48

Jawab: xk yk f(xk) (xk)2 xk.yk ek ek2

-6 -5.3 -6.00 36 31.8 -0.7 0.49

-2 -3.5 -2.48 4 7 1.02 1.0404

0 -1.7 -1.26 0 0 0.44 0.1936

2 0.2 0.32 4 0.4 0.12 0.0144

6 4.0 3.48 36 24 -0.52 0.2704

0 -6.3 - 80 63.2 0.36 2.0088

80A + 0B = 63.2 A = 0.79 Y = 0.79(x) 1.26 0A + 5B = -6.3 B = -1.26

e2(f) = (2.0088/5)1/2 = 0.633845407

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