POLITECHNIKA WARSZAWSKA

WYDZIAŁ INŻYNIERII ŚRODOWISKA

Metody numeryczne

Ćwiczenie 3

Wykonała

Weronika Hejko gr. IW

Rok 2015/2016

Scałkować w przedziale  (0,4 ⟩zapomocą wszystkich typów kwadratur oraz policzyć błędy funkcji

$$y = \frac{Nx^{N + 7} + N}{x^{7} + 7}$$

N= 6

$$y = \frac{6x^{13} + 6}{x^{7} + 7}$$

$$\int_{0}^{4}\frac{6 + 6x^{13}}{{(x}^{7} + 7)}$$

Dane

x	y
0	0.857
1	1.500
2	364.133
3	4360.047
4	24565.505

Wzory na pochodne funkcji:

$$y^{'} = \frac{6x^{6}({6x}^{13} + {91x}^{6} - 42)}{{(x^{7} + 7)}^{2}}$$

$$y" = \frac{6x^{5}({5x}^{20} - {49x}^{13} + 336x^{7} - {7644x}^{6} + 1764)}{{(x^{7} + 7)}^{3}}$$

$$y^{(3)} = \frac{12x^{4}*(10x^{27} + 511x^{20} - 1512x^{14} - 41307x^{13} + 28224x^{7} + 294294x^{6} - 30870)}{{(x^{7} + 7)}^{4}}$$

$$y^{(4)} = \frac{36x^{3}(10x^{34} + 42x^{27} + 5040x^{21} + 180075x^{20} - 223440x^{14} - 3404275x^{13} + 971376x^{7} + 6866860x^{6} - 288120)}{{(x^{7} + 7)}^{5}}$$

1. kwadratura trapezowa

h_T=4

wzór na kwadraturę

$$S\left( f \right) = \frac{1}{2}h\left( f_{0} + f_{4} \right) = \sum_{i = 0}^{n}{f\left( x_{i} \right)*A_{i}}$$

$$S\left( f \right) = \frac{1}{2}*4*\left( 0,857 + 24565,5 \right) = 14246,94$$

Błąd wzoru trapezu

$$\xi = \frac{b - a}{2} = 2$$

$$E\left( f \right) = \frac{1}{12}h^{3}f^{\left( 2 \right)}\left( \xi \right) = \frac{h^{3}}{12}\frac{6\xi^{5}({5\xi}^{20} - {49\xi}^{13} + 336\xi^{7} - {7644\xi}^{6} + 1764)}{{(\xi^{7} + 7)}^{3}} =$$

$$= \frac{1}{12}*4^{3}*\frac{6{*2}^{5}({5*2}^{20} - {49*2}^{13} + 336{*2}^{7} - {7644*2}^{6} + 1764)}{{({*2}^{7} + 7)}^{3}} = 2338,065$$

1. Kwadratura parabol

h_S= h_T/2=2

$$S\left( f \right) = \frac{1}{3}*h*\left( f_{0} + 4*f_{2} + f_{4} \right) = 26022,9$$

$$S\left( f \right) = \frac{1}{3}*2*\left( 0,857 + 4*364,13 + 364,13 \right) = 26022,9$$

Błąd wzoru paraboli

$$\xi = \frac{b - a}{2} = 2$$

$$y^{\left( 4 \right)} = \frac{36\xi^{3}\left( 10\xi^{34} + 42\xi^{27} + 5040\xi^{21} + 180075\xi^{20} - 223440\xi^{14} - 3404275\xi^{13} + 971376\xi^{7} + 6866860\xi^{6} - 288120 \right)}{\left( \xi^{7} + 7 \right)^{5}} = 2220,576$$

$$E\left( f \right) = \frac{h^{5}}{2880}f^{\left( 4 \right)}\left( \xi \right) = \frac{2^{5}}{2880}*2220,576 = 24,67$$

1. Kwadratura złożona trapezu

h=2

$$S\left( f \right) = h*\left( \frac{1}{2}f_{0} + f_{1} + \frac{1}{2}f_{4} \right)$$

$$S\left( f \right) = 2*\left( \frac{1}{2}*0,857 + 364,13 + \frac{1}{2}*364,13 \right) = 25294,63$$

Błąd wzoru złożonego trapezu

$$\xi = \frac{b - a}{2} = 2$$

$$f^{(2)} = \frac{6\xi^{5}({5\xi}^{20} - {49\xi}^{13} + 336\xi^{7} - {7644\xi}^{6} + 1764)}{{(\xi^{7} + 7)}^{3}} = 419,21$$

$$E\left( f \right) = \frac{\left( b - a \right)^{3}}{12n^{2}}f^{\left( 2 \right)}\left( \xi \right)$$

$$E\left( f \right) = \frac{\left( 4 - 0 \right)^{3}}{12{*2}^{2}}*419,21 = 558,95$$

1. Kwadratura złożonych parabol

$$h = \frac{b - a}{n} = \frac{4 - 0}{4} = 1$$

$$S\left( f \right) = \frac{h}{3}\left( f_{0} + f_{4} + 2*f_{2} + 4*\left( f_{1} + f_{3} \right) \right)$$

$$S\left( f \right) = \frac{1}{3}\left( 0,857 + 24565,505 + 2*364,13 + 4*\left( 1,5 + 4360,047 \right) \right) = 14246,94$$

Błąd wzoru złożonych parabol

$$E\left( f \right) = \frac{\left( b - a \right)^{5}}{180n^{4}}*f^{\left( 4 \right)}\left( \xi \right)$$

$$E\left( f \right) = \frac{\left( 4 - 0 \right)^{5}}{180{*4}^{4}}*2220,576 = 49,35$$