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Pertemuan 6 Metnum 2011 Bilqis

Pertemuan 6 Metnum 2011 Bilqis. PENCOCOKAN KURVA: INTERPOLASI. INTERPOLASI.

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Pertemuan 6 Metnum 2011 Bilqis

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  1. Pertemuan6Metnum2011Bilqis

  2. PENCOCOKAN KURVA:INTERPOLASI KomNum

  3. INTERPOLASI Jika pada materi pencocokan kurva sebelumnya anda diminta menaksir bentuk fungsi melalui sederetan data, maka sekarang kita diminta untuk mengestimasi nilai fungsif(x) di antara beberapa nilai fungsi yang diketahui (tanpa mengetahui bentuk fungsi yang menghasilkannya).

  4. INTERPOLASI • Untuk menaksir harga tengahan diantara titik-titik data yang telah ada : • Polinomial Newton • Polinomial Lagrange • Interpolasi Newton-Gregory

  5. Beda Hingga Tabel Beda Diagonal Maju s x f(x) ∆f(x) ∆2f(x) ∆3f(x) ∆4f(x) … 0 x0 f0 ∆f0 1 x1 f1 ∆2f0 ∆f1 ∆3f0 2 x2 f2 ∆2f1 ∆4f0 ∆f2 ∆3f1 . . . . . . . . . ∆3fn-4 n-2 xn-2 fn-2 ∆2fn-3 ∆4fn-4 ∆fn-2 ∆3fn-3 n-1 xn-1 fn-1 ∆2fn-2 ∆fn-1 n xn fn

  6. Interpolasi Newton-Gregory Newton-Gregory Forward (NGF) s(s-1) s(s-1)(s-2) s(s-1)(s-2)…(s-n+1) f(xs) = f0 + s∆f0 + ∆2f0 + ∆3f0 + … + ∆nf0 2! 3! n! xs – x0 s = dengan h = ∆x h Newton-Gregory Backward (NGB) s(s+1) s(s+1)(s+2) s(s+1)(s+2)…(s+n-1) f(xs) = f0 + s∆f-1 + ∆2f-2 + ∆3f-3 + … + ∆nf-n 2! 3! n! xs – x0 s = dengan h = ∆x h

  7. Interpolasi Newton-Gregory contoh : carilah nilai f(xs) untuk xs = 1,03, jika diketahui fungsi tsb menghasilkan nilai2 sbb : x 1,0 1,3 1,6 1,9 2,2 2,5 2,8 f(x) 1,449 2,060 2,645 3,216 3,779 4,338 4,898 Langkah 1  mencari nilai beda x f(x) ∆f(x) ∆2f(x) ∆3f(x) ∆4f(x) ∆5f(x) ∆6f(x) 1,0 1,449 0,611 1,3 2,060 -0,026 0,585 0,012 1,6 2,645 -0,014 -0,006 0,571 0,006 0,004 1,9 3,216 -0,008 -0,002 -0,001 0,563 0,004 0,003 2,2 3,779 -0,004 0,001 0,559 0,005 2,5 4,338 0,001 0,560 2,8 4,898 1,03 ada di dekat titik awal. shg NGF lebih cocok digunakan.

  8. Interpolasi Newton-Gregory Langkah 2  mencari nilai s (lebar interval) s = (1,03 – 1) / (1,3 – 1) = 0,1 Langkah 3  mencari nilai f(xs) 0,1 (0,1 – 1) 0,1 (0,1 – 1)(0,1 – 2) f(1,03) = 1,449 + 0,1(0,611) + . -0,026 + . 0,012 2! 3! 0,1 (0,1 – 1)(0,1 – 2)(0,1 – 3) 0,1 (0,1 – 1)(0,1 – 2)(0,1 – 3)(0,1 – 4) + . -0,006 + . 0,004 4! 5! 0,1 (0,1 – 1)(0,1 – 2)(0,1 – 3)(0,1 – 4)(0,1 – 5) + . -0,001 6! = 1,5118136

  9. Interpolasi Newton-Gregory contoh : carilah nilai f(xs) untuk xs = 2,67, jika diketahui fungsi tsb menghasilkan nilai2 sbb : x 1,0 1,3 1,6 1,9 2,2 2,5 2,8 f(x) 1,449 2,060 2,645 3,216 3,779 4,338 4,898 Langkah 1  mencari nilai beda x f(x) ∆f(x) ∆2f(x) ∆3f(x) ∆4f(x) ∆5f(x) ∆6f(x) 1,0 1,449 0,611 1,3 2,060 -0,026 0,585 0,012 1,6 2,645 -0,014 -0,006 0,571 0,006 0,004 1,9 3,216 -0,008 -0,002 -0,001 0,563 0,004 0,003 2,2 3,779 -0,004 0,001 0,559 0,005 2,5 4,338 0,001 0,560 2,8 4,898 2,67 ada di dekat titik akhir. Jadi NGB adalah pilihan terbaik.

  10. Interpolasi Newton-Gregory Langkah 2  mencari nilai s (lebar interval) s = (2,67 – 2,8) / (1,3 – 1) = -0,43333 Langkah 3  mencari nilai f(xs) -0,433 (-0,433 + 1) -0,433 (-0,433 + 1)(-0,433 + 2) f(1,03) = 4,898 + -0,433(0,560) + . 0,001 + . 0,005 2! 3! 0,433 (0,433 + 1)(0,433 + 2)(0,433 + 3) + . 0,001 4! 0,433 (0,433 + 1)(0,433 + 2)(0,433 + 3)(0,433 + 4) + . 0,003 5! 0,433 (0,433 + 1)(0,433 + 2)(0,433 + 3)(0,433 + 4)(0,433 + 5) + . -0,001 6! = 4,654783

  11. Interpolasi Newton-Gregory Coba cari nilai f(x) ketika x = 2.0 Dengan • NGF • NGB

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