ir taikome (5.3) formules:a = 5·100,11−15·30,05 = 49,805·55−15 2 50,0= 0, 996,b = 55·30,05−15·100,11 = 151,105·55−15 2 50,0= 3, 022.Taigi gauname prognozės funkciją y(x) = 0, 996x + 3, 022 ir apskaičiuojamey 6 = y(6) = 0, 996 · 6 + 3, 022 = 8, 998 ≈ 9, 0.5.2 užduotis savarankiškam darbui. Pagal pateiktus stebėjimųduomenis raskite funkcijos y(x) = ax + b parametrus a, b ir apskaičiuokitey (x 0 ).Užduotis y (x 0 )5.2.1y(3, 5) = 4, 01, y(3, 8) = 4, 64, y(4, 2) = 5, 37y(4, 5) = 6, 02, y(4, 9) = 6, 71, y(5, 3) = 7, 55y(3, 0)5.2.2y(3, 5) = 3, 99, y(3, 8) = 4, 62, y(4, 2) = 5, 38y(4, 5) = 6, 03, y(4, 9) = 6, 68, y(5, 3) = 7, 52y(3, 0)5.2.3y(3, 5) = 3, 99, y(3, 8) = 4, 59, y(4, 2) = 5, 37y(4, 5) = 6, 01, y(4, 9) = 6, 69, y(5, 3) = 7, 57y(3, 0)5.2.4y(3, 5) = 4, 02, y(3, 8) = 4, 58, y(4, 2) = 5, 39y(4, 5) = 6, 03, y(4, 9) = 6, 70, y(5, 3) = 7, 54y(3, 0)5.2.5y(3, 5) = 4, 02, y(3, 8) = 4, 57, y(4, 2) = 5, 41y(4, 5) = 5, 97, y(4, 9) = 6, 65, y(5, 3) = 7, 50y(3, 0)5.2.6y(3, 5) = 4, 01, y(3, 8) = 4, 64, y(4, 2) = 5, 37y(4, 5) = 6, 02, y(4, 9) = 6, 71, y(5, 3) = 7, 55y(6, 0)5.2.7y(3, 5) = 3, 99, y(3, 8) = 4, 62, y(4, 2) = 5, 38y(4, 5) = 6, 03, y(4, 9) = 6, 68, y(5, 3) = 7, 52y(6, 0)5.2.8y(3, 5) = 3, 99, y(3, 8) = 4, 59, y(4, 2) = 5, 37y(4, 5) = 6, 01, y(4, 9) = 6, 69, y(5, 3) = 7, 57y(6, 0)5.2.9y(3, 5) = 4, 02, y(3, 8) = 4, 58, y(4, 2) = 5, 39y(4, 5) = 6, 03, y(4, 9) = 6, 70, y(5, 3) = 7, 54y(6, 0)5.2.10y(3, 5) = 4, 02, y(3, 8) = 4, 57, y(4, 2) = 5, 41y(4, 5) = 5, 97, y(4, 9) = 6, 65, y(5, 3) = 7, 50y(6, 0)5.2.11y(1, 0) = 2, 02, y(1, 5) = 1, 47, y(1, 8) = 1, 17y(2, 3) = 0, 67, y(2, 6) = 0, 39, y(2, 9) = 0, 11y(3, 0)5.2.12y(1, 3) = 4, 39, y(1, 4) = 2, 51, y(1, 7) = 2, 20y(1, 4) = 1, 63, y(2, 5) = 0, 73, y(2, 8) = 0, 31y(0, 3)5.2.13y(1, 0) = 1, 99, y(1, 5) = 1, 51, y(1, 8) = 1, 20y(2, 3) = 0, 63, y(2, 6) = 0, 37, y(2, 9) = 0, 13y(3, 0)84
Užduotis y (x 0 )5.2.14y(1, 0) = 1, 98, y(1, 5) = 1, 52, y(1, 8) = 1, 19y(2, 3) = 0, 66, y(2, 6) = 0, 38, y(2, 9) = 0, 12y(3, 0)5.2.15y(1, 0) = 2, 00, y(1, 5) = 1, 51, y(1, 8) = 1, 21y(2, 3) = 0, 64, y(2, 6) = 0, 36, y(2, 9) = 0, 14y(3, 0)5.2.16y(3, 5) = 4, 02, y(3, 8) = 4, 57, y(4, 2) = 5, 40y(4, 5) = 5, 96, y(4, 9) = 6, 66, y(5, 3) = 7, 49y(6, 0)5.2.17y(1, 0) = 2, 02, y(1, 5) = 1, 47, y(1, 8) = 1, 17y(2, 3) = 0, 67, y(2, 6) = 0, 39, y(2, 9) = 0, 11y(0, 5)5.2.18y(1, 0) = 1, 99, y(1, 5) = 1, 51, y(1, 8) = 1, 20y(2, 3) = 0, 63, y(2, 6) = 0, 37, y(2, 9) = 0, 13y(0, 5)5.2.19y(1, 0) = 1, 98, y(1, 5) = 1, 52, y(1, 8) = 1, 19y(2, 3) = 0, 66, y(2, 6) = 0, 38, y(2, 9) = 0, 12y(0, 5)5.2.20y(1, 0) = 2, 00, y(1, 5) = 1, 51, y(1, 8) = 1, 21y(2, 3) = 0, 64, y(2, 6) = 0, 36, y(2, 9) = 0, 14y(0, 5)5.1.5 Mažiausių kvadratų metodo apibendrinimasNagrinėsime dviejų kintamųjų x ir y funkcijąu(x, y) = x α y β . (5.4)5.4 pavyzdys. Žinomos kelios (5.4) funkcijos reikšmėsx 10 15 20 30 35y 20 30 15 25 10u(x, y) 14, 9 21, 8 12, 8 20, 4 9, 9Apskaičiuokime funkcijos parametrų α ir β reikšmes.Sprendimas. Apskaičiuojame funkcijos u(x, y) logaritmusir sudarome lentelęln u(x, y) = α ln x + β ln y (5.5)ln x 2, 3026 2, 7081 2, 9957 3, 4012 3, 5553ln y 2, 9957 3, 4012 2, 7081 3, 2189 2, 3026ln u 2, 7014 3, 0819 2, 5494 3, 0155 2, 292585