Jawaban Ujian Antek

1. Diberikan data:

x 0 0,5 1,0 1,5 2,0 2,5F(x) 1 2,319 2,910 3,845 5,720 8,595

a. Hitunglah f(1,1) dengan menggunakan polinomial interpolasi Newton dari orde 1 sampai 3. Pilih titik-titk untuk estimasi anda untuk mencapai keteletian yang baik.

b. Taksirlah kesalahan setiap prediksi nilai interpolasi.

Jawab : a.

a0=y0a0=2.910

a1 =f (X 1 , X 0)

=y1− y0x1−x0

= 3,845−2,910

x 1,5−1,0 = 1,87

F(x2,x1) = y2− y1x2−x1

= 5,720−3,845

2,0−1,5 = 3,75

F(x3,x2) = y3− y 2x3−x2

= 8.595−5,720

2,5−2,0 = 5,75

F(x4,x3) = y 4− y3x 4−x3

X YX0=1.0 Y0=2.910X1=1.5 Y1=3,845X2=2.0 Y2=5,720X3=2.5 Y3=8,595X4=0.5 Y4=2,319

= 2.319−8,595

0,5−2,5 = 3,108

a2 = f(x2,x1,x0)

= f ( x2 , x1 )−f (x 1 , x 0)

x 2−x 0

= 3.75−1.87

2,0−1,0 = 1,88

F(x3,x2,x1) = f ( x3 , x2 )−f (x2 , x1)

x 3−x 1

= 5,75−3,75

2,5−1,5 = 2

F(x4,x3,x2) =f ( x 4 , x3 )−f ( x 3 , x 2 )

x 4−x2

= 3.108−5.75

0,5−2,0 = 1,76

a3= f(x3,x2,x1,x0)

=f ( x 4 , x3 , x2 )−f ( x2 , x1 , x0 )

x 3−x 0

= 2−1,882,5−1,0 = 0,08

F(x4,x3,x2,x1) =f ( x 4 , x3 , x2 )−f ( x3 , x 2, x1 )

x 4−x1

= 1.76−20,5−1,5 = 0,24

a4 =f ( x 4 , x3 , x2 , x1 , x0 )

x 4−x 0

= 0.24−0,08

0,5−1,0 = -0,32

Interpolasi Newton Orde I

P1= a0+a1(x-x0)

= 2,910+1,87(1,1-1,0) = 3,097

Interpolasi Newton Orde II

P2= a0+a1(x-x0)+a2(x-x0) (x-x1)

= 3,097+(1,1-1,0)(1,1-1,5) = 3,2018

Interpolasi Newton Orde III

P3= P2+a3(x-x0) (x-x1) (x-x2)

= 3,2018+ (1,1-1,0)(1,1-1,5)(1,1-2,0) = 3,02468

b.Taksirlah kesalahan setiap prediksi nilai interpolasi

Interpolasi Newton Orde I

R1=f ( x 2 , x 1, x 0 ) (x−x0 ) ( x−x 1 )

= 1,88(1,1-1,0)(1,1-1,5) = -0,0752

Interpolasi Newton Orde II

R2=f ( x 3 , x 2 , x 1 , x 0 ) ( x−x0 ) ( x−x 1 ) ( x−x 2 )

= 0.08(1,1-1,0)(1,1-1,5)(1,1-2,0) = 2,88x10−3

Interpolasi Newton Orde III

R3=f ( x 4 , x 3 , x 2 , x 1 ) ( x−x 0 ) ( x−x1 ) ( x−x 2 )(x−x3)

= 1,12(1,1-1,0)(1,1-1,5)(1,1-2,0)(1,1-2,5) = 0,056448

2. Evaluasi integral ∫ f (x¿)dx¿dari data tabulasi berikut dengan aturan

trapesium/simpson.

A X1 X2 X3 X4 X5/bX 0 0,1 0,2 0,3 0,4 0,5

F(x) 1 8 5 ⁴ 5 8

Untuk aturan Simpson 1/3

I=(b−a ) F ( x0 )+4 ∑

i=1,3,5

n−1

F ( xi )+2 ∑i=2,4,6

n−2

F ( x1 )+F ( xn )

3 n

Jumlah segmen = 5

I=(0,5−0 )1+( 4 ) (8+3 )+2 (5+5 )+8

3. 5

I=72,515

= 48,3

Untuk aturan Trapesium

I=(b−a)f ( x 0 )+2∑

i=1

n−1

F ( xi )+F(xn)

2n

I=(0,5−0)1+2 (8+5+3+5 )+8

2.5

I=0,5+50

10= 5,05

3. Lakukan analisa komputasi serupa, tetapi gunakan integrasi Romberg untuk mengevaluasi integral. Gunakan kriteria henti sebesar Ɛs = 0,75%

F= ₀∫30 200 (z/5+z) e-2.z/30 dz

F = ₀∫3 200 (3/5+3) e-2.3/30

= 61,40

F = ₀∫6 200 (6/5+6) e-2.6/30

= 73,13

F = ₀∫9 200 (9/5+9) e-2.9/30

= 70,56

F = ₀∫12 200 (12/5+12) e-2.12/30

= 63,43

F = ₀∫15 200 (15/5+15) e-2.15/30

= 55,18

F = ₀∫18 200 (18/5+18) e-2.18/30

= 47,14

F = ₀∫21 200 (21/5+21) e-2.21/30

= 39,83

F = ₀∫25 200 (25/5+25) e-2.25/30

= 33,42

F = ₀∫27 200 (27/5+27) e-2.27/30

= 27,89

F = ₀∫30 200 (30/5+30) e-2.30/30

= 23,20

f ( x 1 )= ∫0

1.764706

2001.764706

5+1.764706e

−2( 1.764706 )30 dz=¿ 46.38312

Z (z/5+z) e−2 z /30 f(z)1.764706 200 0.26087 0.88901 46.383123.529412 200 0.413793 0.790338 65.407315.294118 200 0.514286 0.702619 72.269337.058824 200 0.585366 0.624635 73.127978.823529 200 0.638298 0.555306 70.8901810.58824 200 0.679245 0.493673 67.0649812.35294 200 0.711864 0.43888 62.484614.11765 200 0.738462 0.390169 57.6248915.88235 200 0.760563 0.346864 52.7623617.64706 200 0.779221 0.308365 48.0569119.41176 200 0.795181 0.27414 43.5981121.17647 200 0.808989 0.243713 39.4321922.94118 200 0.821053 0.216663 35.5783624.70588 200 0.831683 0.192616 32.0390326.47059 200 0.841121 0.171237 28.8062528.23529 200 0.849558 0.152231 25.86588

30 200 0.857143 0.135335 23.20033

f ( x 2 )= ∫0

3.529412

2003.529412

5+3.529412e

−2 (3.529412 )30 dz=¿ 65.40731

f ( x 3 )= ∫0

5.294118

2005.294118

5+5.294118e

−2( 5.294118)30 dz=¿ 72.26933

f ( x 4 )= ∫0

7.058824

2007.058824

5+7.058824e

−2 (7.058824 )30 dz=¿ 73.12797

f ( x 5 )= ∫0

8.823529

2008.823529

5+8.823529e

−2 (8.823529 )30 dz=¿ 70.89018

f ( x 6 )= ∫0

10.58824

20010.58824

5+10.58824e

−2 (10.58824 )30 dz=¿ 67.06498

f ( x 7 )= ∫0

12.35294

20012.35294

5+12.35294e−2 (12.35294 )

30 dz=¿ 62.4846

f ( x 8 )= ∫0

14.11765

20014.11765

5+14.11765e

−2 (14.11765)30 dz=¿ 57.62489

f ( x 9 )= ∫0

15.88235

20015.88235

5+15.88235e

−2(15.88235)30 dz=¿ 52.76236

f ( x 10 )= ∫0

17.64706

2005.294118

5+5.294118e

−2 (17.64706 )30 dz=¿ 48.05691

f ( x 11)= ∫0

19.41176

20019.41176

5+19.41176e

−2 (19.41176 )30 dz=¿ 43.59811

f ( x 12 )= ∫0

21.17647

20021.17647

5+21.17647e

−2 (21.17647 )30 dz=¿ 39.43219

f ( x 13 )= ∫0

22.94118

20022.94118

5+22.94118e

−2 (22.94118)30 dz=¿ 35.57836

f ( x 14 )= ∫0

24.70588

20024.70588

5+24.70588e

−2 (24.70588 )30 dz=¿ 32.03903

f ( x 15 )= ∫0

26.47059

20026.47059

5+26.47059e

−2( 26.47059)30 dz=¿ 28.80625

f ( x 16 )= ∫0

28.23529

20028.23529

5+28.23529e

−2 (28.23529 )30 dz=¿25.86588

f ( x 17 )=∫0

30

20030

5+30e

−2( 30)30 dz=¿ 23.20033

I=b−a2. n

f ( x0 )+2¿17)+f ( x 16 )

I=30/34(0+2(46.38312+65.40731+72.26933+73.12797+70.89018+67.06498+62.4846+57.

62489+ 2.76236+48.05691+43.59811+39.43219+35.57836+32.03903+28.80625)

I= 1469.985237

ε t = 1480.6-1469.985237

= 10.614763

εs = 10.614763/1480.6 (100%) = 0.7169231%