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Bohodir Jalolov

Taqribiy sonlar. Ularning absolyut va nisbiy xatosi. Qiymatga ega bo‘lgan raqam. To‘g‘ri ishoralar soni doc

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1-LABORATORIYA ISHI Mavzu: Taqribiy sonlar. Ularning absolyut va nisbiy xatosi. Qiymatga ega bo’lgan raqam. To’g’ri ishoralar soni . Differensiallanadigan funksiyaning y = f ( x 1 , x 2 … , x n ) absolyut xatosi argumentlarning x1,x2… ,xn deyarli kichik xato bilan chiqariladigan Δ x 1 , Δ x 2 , … Δ x n o’cham bilan baholanadi. ∆ y = ∑ i = 1n ❑ | ∂ f ∂ x i | ⋅ Δ x i (1) Bundan funksiyaning nisbiy xatosi δ y = Δ y y ∗ 100 % ( 2 ) kabi bo’ladi. Masalaning qo’yilishi: Quyidagi funksiyaning absolyut va nisbiy xatoligini aniqlang. y= a+b √(c− d)∗m a=16.342 ±0.0001 ;b= 2.5 ±0,003 ; c= 38.17 ±0,002 d=9.14 ±0,005 n=3.6 ±0,04 m= 9.14 ±0,04 Yechish: Demak: a=16.342 ;b= 2.5 ;c=38.17 ;d=9.14 ;n=3.6 ;m=9.14 ∆ a = 0.001 ; ∆ b = 0.003 ; ∆ c = 0.002 ∆ d = 0.002 ∆ n = 0.04 ∆ m = 0.005 (1) formulaga asosan funksiyaning absolyut xatosini topamiz. ∆ y = ∂ y ∂ a ⋅ Δ a + ∂ y ∂ b ⋅ Δ b + ∂ y ∂ c ⋅ Δ c + ∂ y ∂ d ⋅ Δ d + ∂ y ∂ m ⋅ m = ¿ 1 √( 38.17 − 9.14 ) ∗ 9.14 0.001 + 1 √( 38.17 − 9.14 ) ∗ 9.14 0.003 − 16.342 + 2.5 2 ∗ √ 9.14 ∗ ( 38.17 − 9.14 ) 3 2 ∙ 0.002 + 16.342 + 2.5 2 ∗√ 9.14 ∗ ( 38.17 − 9.14 ) 3 2 ∙ 0.002 + 16.342 + 2.5 2 ∗ 9.14 3 2 ( 38.17 − 9.14 ) 1 2 ∙ 0.005 = 0.000003 Endi esa funksiyaning qiymatini hisoblaymiz. y= a+b √(c− d)∗m = 16.342 +2.5 √(38.17 − 9.14 )∗9.14 =16.28004 Bunda funksiyaning nisbiy xatosi (2) formulaga asosan: δ y = Δ y y ∙ 100 % = 0.00156142 1.15673 ∙ 100 % = 0.0000018 ≈ 0.1 % kabi bo’ladi. Dastur kodi: #include <iostream> #include <cmath> #include <iomanip> double del_x ( double a, double b, double c, double d, double m, double del_a, double del_b, double del_c, double del_d, double del_m) { return std :: abs ((b * del_a) / ( 2 * sqrt ((c - d) * m) * (a + b))) + std :: abs ((-a * del_b) / ( 2 * sqrt ((c - d) * m) * (a + b))) + std :: abs (((a + b) * del_c) / ( 2 * pow ((c - d) * m, 1.5 ))) + std :: abs (((a + b) * del_d) / ( 2 * sqrt ((c - d) * m))) + std :: abs (((a + b) * del_m) / ( 2 * sqrt ((c - d) * m) * m)); } double x ( double a, double b, double c, double d, double m) { return (a + b) / sqrt ((c - d) * m); } double nisbiy ( double del_x, double x) { return (del_x / x) * 100 ; } int main () { double a = 16.342 ; double b = 2.5 ; double c = 38.17 ; double d = 9.14 ; double m = 9.14 ; double del_a = 0.001 ; double del_b = 0.003 ; double del_c = 0.002 ; double del_d = 0.002 ; double del_m = 0.005 ; double absolut_xato = del_x(a, b, c, d, m, del_a, del_b, del_c, del_d, del_m); double funksiya_qiymati = x(a, b, c, d, m); double nisbiy_xato = nisbiy(absolut_xato, funksiya_qiymati); std :: cout << "Funksiyaning qiymati: " << funksiya_qiymati << std :: endl ; std :: cout << "Absolut xatolik: " << absolut_xato << std :: endl ; std :: cout << "Nisbiy xatolik: " << std ::fixed << std ::setprecision( 1 ) << nisbiy_xato << " %" << std :: endl ; return 0 ; } Dastur natijasi: Chiziqlimas tenglamalarni yechishning sonli usullari. Nyuton, oddiy iteratsiya, vatarlar usullari. Nyuton usuli def f(x): return x**2 - 4 def f_prime(x): h = 1e-6 return (f(x + h) - f(x)) / h def newton_method(f, f_prime, x0, tol=1e-6, max_iter=100): x = x0 for i in range(max_iter): fx = f(x) if abs(fx) < tol: break fp = f_prime(x) if fp == 0: break x = x - fx / fp return x x0 = 2.0 root = newton_method(f, f_prime, x0) print("Natija:", root) Vatarlar usuli # Sonli usul def sonli_usul(n): if n == 0: return 1 else: return n * sonli_usul(n-1) # Vatarlar usuli def vatarlar_usuli(n): result = 1 for i in range(1, n+1): result *= i return result n = int(input("Ixtiyoriy son kiriting: ")) print(f"{n} ning sonli usuli: {sonli_usul(n)}") print(f"{n} ning vatarlar usuli: {vatarlar_usuli(n)}") Oddiy iteratisiya usuli def iteratsiya_method(a, b, c): epsilon = 0.0001 x0 = 0 while True: x1 = (b * x0**2 + c) / a if abs(x1 - x0) < epsilon: break x0 = x1 print("Iteratsiya usuli natijasi:", x1) a = float(input("a ni kiriting: ")) b = float(input("b ni kiriting: ")) c = float(input("c ni kiriting: ") iteratsiya_method(a, b, c) Oddiy itereatsiya usuli x 3 − 3 x 2 + 12 x − 12 = 0 f 1( x ) = x 3 − 3 x 2 f2(x)=12 x− 12 x -1 0 1 y -28 -12 -2 x0≤12 x n + 1 = x n − f ( x n ) f I ( x n ) f I ( x 0 ) = 3 x 2 − 6 x + 12 x1= x0− f(x0) fI(x0) ≈7,5 x2= x1− f(x1) fI(x1) ≈6,3 | x n − x n − 1 ≤ ε ε ≤ 10 − 4 | x ≈ 1,22 Vatar usuli f ( a) < 0 x n + 1 = x n − f ( x n ) f ( b ) − f ( x n ) ∗ ( b − x n ) f ( a) = − 12 f ( x 1 ) = − 12 − − 2316 − 4632 ∗ ( − 12 − 12 ) = − 1,99x -1 0 1 2 f1(x) -4 0 -2 -4 f 2 ( x ) -24 -12 0 12 x n + 1 = x n − f( x n ) f ( a ) − f ( x n ) ∗ ( a − x n ) = 1,2 x≈0,79 Nyuton usuli [ a , b ] = [ 1,2 ] a + b 2 = 1,5 f(a)∗ f(c)<0f(1)∗ f(1,5 )<0 f ( b) ∗ f ( c) < 0 f ( 2 ) ∗ f ( 1,5 ) < 0 f(1)=-2 f(2)=8 f(3/2)=2,6 Chiziqli algebraik tenglamalar sistemasini (ChATS) yechishning sonli usullari. Gauss, oddiy iteratsiyalar, Zeydel usullari 1. Biz tenglamalarni matritsa ko'rinishida yozamiz: 1.7 10 -1.3 2.1 | 3.1 3.1 1.7 -2.1 5.4 | 2.1 3.3 -7.7 4.4 -5.1 | 1.9 10 -20.1 20.4 1.7 | 1.8 2. Birinchi qattni ikkinchi qatrga 1.7 va uchinchi qatrga 3.3 qo'shib qo'yamiz: 1.7 10 -1.3 2.1 | 3.1 0 -15.3 -6 0.95 | -3.9 0 25.78 7.33 -11.82| -7.07 10 -20.1 20.4 1.7 | 1.8 3. Ikkinchi qattni uchinchi qatrga 25.78 bo'lib 10 qo'shib qo'yamiz: 1.7 10 -1.3 2.1 | 3.1 0 -15.3 -6 0.95 | -3.9 0 0 160.39 -122.85 | -49.89 10 -20.1 20.4 1.7 | 1.8 4. Keyingi qobriqni 160.39 ga bo'lip birinchi va uchinchi qatrlarga ko'paytiramiz: 1.7 10 -1.3 2.1 | 3.1 0 -15.3 -6 0.95 | -3.9 0 0 160.39 -122.85 | -49.89 0 -37.91 115.75 -19.77 | -29.33 5. Oxirgi qatrdan birinchi va uchinchi qatrlarni 115.75 ga bo'lib ko'paytiramiz: 1.7 10 -1.3 2.1 | 3.1 0 -15.3 -6 0.95 | -3.9 0 0 160.39 -122.85 | -49.89 0 0 118.18 -33.74 | -16.92 6. Bizning yechimiz: x = 0.25 y = 0.254902 z = -0.306122 d = -0.122449 Gaus usuli import numpy as np def gauss_elimination(A, B): n = len(B) for i in range(n): max_index = i max_value = abs(A[i][i]) for j in range(i+1, n): if abs(A[j][i]) > max_value: max_index = j max_value = abs(A[j][i]) A[[i, max_index]] = A[[max_index, i]] B[[i, max_index]] = B[[max_index, i]] for j in range(i+1, n): factor = A[j][i] / A[i][i] B[j] -= factor * B[i] A[j] -= factor * A[i] X = np.zeros(n) for i in range(n-1, -1, -1): X[i] = (B[i] - np.dot(A[i,i+1:], X[i+1:])) / A[i][i] return X A = np.array([[2.0, 1.0], [5.0, 3.0]]) B = np.array([4.0, 10.0]) X = gauss_elimination(A, B) print("Natija:") print(X) Zeydel usuli import numpy as np def zeydel_usuli(A, b, x0, eps=1e-5, max_iter=1000): n = len(b) x = np.copy(x0) x_new = np.copy(x) for _ in range(max_iter): for i in range(n): x_new[i] = b[i] for j in range(n): if j != i: x_new[i] -= A[i, j] * x_new[j] x_new[i] /= A[i, i] if np.linalg.norm(x_new - x) < eps: return x_new x = np.copy(x_new) return x A = np.array([[10, 2, 1], [1, 5, 1], [2, 3, 10]]) b = np.array([7, -8, 6]) x0 = np.array([0, 0, 0]) print("Zeydel usuli:") print(zeydel_usuli(A, b, x0)) x = (3.1 - 10*y + 1.3*z - 2.1*d) y = (2.1 - 3.1*x + 2.1*z - 5.4*d) z = (1.9 - 3.3*x + 5.1*d) d = (1.8 - 10*x - 20.1*d) k x 1 (k) x 2 (k) x 3 (k) x 4 (k) 0.3,1 2,1 1,9 1,8 …… ……. ………. …… …… ……. …… ……. …… import numpy as np def chiziqli_ustun(A, b, eps=1e-5, max_iter=1000): n = len(b) x = np.zeros(n) for _ in range(max_iter): x_new = np.copy(x) for i in range(n): x_new[i] = (b[i] - np.dot(A[i, :i], x_new[:i]) - np.dot(A[i, i+1:], x[i+1:])) / A[i, i] if np.linalg.norm(x_new - x) < eps: return x_new x = x_new return x A = np.array([[10, 2, 1], [1, 5, 1], [2, 3, 10]]) b = np.array([7, -8, 6]) print("Chiziqli ustunlar:") print(chiziqli_ustun(A, b)) Nochiziqli tenglamalar sistamasini yechish usullari. Oddiy iteratsiyalari, Nyuton usullari. Oddiy iteratisya usuli x = 0 y = 0 1. x(n+1) = - sin(y(n)) - 0.4 2. y(n+1) = cos(x(n)+1) / 2 x(1) = -sin(0) - 0.4 = -0.4 y(1) = cos(1)/2 ≈ 0.417 import numpy as np def oddiy_iteratsiyalar(A, b, x0, eps=1e-5, max_iter=1000): n = len(b) x = np.copy(x0) for _ in range(max_iter): x_new = np.zeros(n) for i in range(n): x_new[i] = b[i] for j in range(n): if j != i: x_new[i] -= A[i, j] * x[j] x_new[i] /= A[i, i] if np.linalg.norm(x_new - x) < eps: return x_new x = x_new return x A = np.array([[10, 2, 1], [1, 5, 1], [2, 3, 10]]) b = np.array([7, -8, 6]) x0 = np.array([0, 0, 0]) print("Oddiy iteratsiyalar usuli:") print(oddiy_iteratsiyalar(A, b, x0)) Nyuton usuli x=-0,8 y=0,4 f(x,y) = 2*y - cos(x+1) - 0 g(x,y) = x + sin(y) + 0.4 FI(x,y)= cos (x+1) 2G I ( x , y ) = arcsin ( − x − 0.4 ) F I ( x , y ) = − cos ( y + 1 ) 2 G ( x , y ) = − sin ( y ) − 0.4 J ( x , y ) = | cos ( x + 1 ) 2 − cos ( y + 1 ) 2 arcsin ( − x − 0.4 ) − sin ( y ) − 0.4 | = − 0,45 import numpy as np def f(x): return np.array([ x[0]**2 + x[1]**2 + x[2]**2 - 2*x[0] - 3*x[1] + x[2] + 6, x[0]**2 + 4*x[1]**2 + x[2]**2 - 3*x[0] - x[1] + 2*x[2] - 7, x[0]**2 + x[1]**2 + 9*x[2]**2 + x[0] - x[2] + 6 ]) def df(x): return np.array([ [2*x[0] - 2, 2*x[1] - 3, 2*x[2] + 1], [2*x[0] - 3, 8*x[1] - 1, 2*x[2] + 2], [2*x[0] + 1, 2*x[1], 18*x[2] - 1] ]) def newton_method(f, df, x0, eps=1e-5, max_iter=1000): x = np.copy(x0) for _ in range(max_iter): delta_x = np.linalg.solve(df(x), -f(x)) x = x + delta_x if np.linalg.norm(delta_x) < eps: return x return x x0 = np.array([0, 0, 0]) solution = newton_method(f, df, x0) print("Nyuton usuli:") print(solution) MATRISANING XOS SON VA XOS VEKTORI MASALASI. DANILEVSKIY VA KRILOV USULLARI Danilevskiy usuli ( 1 2 3 7 − 1 5 2 3 − 3 ) = ¿ m ( x ) = det ( A − αl ) m(x)= det ( 1− x 2 3 7 1− x 5 2 3 3− x)=¿m ( x ) = ( − 1 ) 3 ∗ ( 1 − x ) ∗ ( − x 2 + 4 ∗ x − 8 ) − 2 ∗ ( 7 ∗ ( − 3 − x ) − 5 ∗ 2 ) + 3 ∗ ( 7 ∗ 3 − 5 ∗ 2 ) = − x 3 + 6 x 2 − 5 x + 33 m(x)=− 2x2+10 x+87 m(0)=−2(0)2+10 (0)+87 =87 m ( 1 ) = − 2 ( 1) 2 + 10 ( 1 ) + 87 = − 2 + 10 + 87 = 95 m(2)=−2(2)2+10 (2)+87 =−8+20 +87 =99 import numpy as np def danilevsky_method(A): n = A.shape[0] P = np.eye(n) for k in range(n-1, 0, -1): v = np.zeros(n) v[k] = 1 for j in range(k-1, -1, -1): mu = A[j, k] / A[k, k] v[j] = -mu A[j, :] = A[j, :] + mu * A[k, :] P_k = np.eye(n) - np.outer(v, P[:, k]) P = np.dot(P_k, P) return P, A # Matrisni yaratish A = np.array([ [1, 2, 3], [2, 5, 4], [3, 4, 11] ]) P, B = danilevsky_method(A) print("P:") print(P) print("Danilevski usuli:") print(B) Krilov usuli ( 1 2 3 7 − 1 5 2 3 − 3 ) = ¿ X 1 = A ∗ X 0 = A ∗ ( 1 , 1 , 1 ) ¿ ( 1 ∗ 1 + 2 ∗ 1 + 3 ∗ 1 , 7 ∗ 1 + ( − 1 ) ∗ 1 + 5 ∗ 1 , 2 ∗ 1 + 3 ∗ 1 + ( − 3 ) ∗ 1 ) = ( 1 + 2 + 3 , 7 − 1 + 5 , 2 + 3 − 3 ) = ( 6 , 11 , 2 ) import numpy as np def krilov_method(A, b): n = A.shape[0] x = np.zeros(n) x[0] = 1 for i in range(1, n): x = np.dot(A, x) x = x / np.linalg.norm(x) phi = np.dot(A, x) lambd = np.dot(x, phi) return lambd, x # Matris va b vektorini yaratish A = np.array([[2, -1], [1, 3]]) b = np.array([1, 4]) lambd, x = krilov_method(A, b) print("Xalqalar son: ", lambd) print("Xalqalar vektori: ", x) FUNKSIYALARNI INTERPOLYASIYALASH. LAGRANJ va Nyuton INTERPOLYASION FORMULASI Lagranj usuli Teng uzoklikda joylashmagan tugunlarda y funksiyaning qiymatlari berilgan. Lagranj interpolyasion formulalari yordamida ko’phad quring va х nuqtadagi qiymatini hisoblang. X Y № x 0,02 1,02316 2 0,102 0,08 1,09590 4 0,114 0,12 1,14725 6 0,125 0,17 1,21483 8 0,203 0,23 1,30120 10 0,154 0,30 1,40976 12 0,087 L 5 (x)=( x − x 1 )( x − x 2 )( x − x 3 )( x − x 4 )( x − x 5 ) ( x 0 − x 1 )( x 0 − x 2 )( x 0 − x 3 )( x 0 − x 4 )( x 0 − x 5 ) y 0 + + ( x − x 0 )( x − x 1 )( x − x 2 )( x − x 3 )( x − x 4 ) ( x 1 − x 0 )( x 1 − x 2 )( x 1 − x 3 )( x 1 − x 4 )( x 1 − x 5 ) y 1 + +(x− x1)(x− x0)(x− x3)(x− x4)(x− x5) (x2− x1)(x2− x0)(x2− x3)(x2− x4)(x2− x5) y2 + +(x− x1)(x− x2)(x− x0)(x− x4)(x− x5) (x3− x1)(x3− x2)(x3− x0)(x3− x4)(x3− x5) y3 + + ( x − x 1 )( x − x 2 )( x − x 3 )( x − x 0 )( x − x 5 ) ( x 4 − x 1 )( x 4 − x 2 )( x 4 − x 3 )( x 4 − x 0 )( x 4 − x 5 ) y 4 + + ( x − x 1 )( x − x 2 )( x − x 3 )( x − x 4 )( x − x 0 ) ( x 5 − x 1 )( x 5 − x 2 )( x 5 − x 3 )( x 5 − x 4 )( x 5 − x 0 ) y 5 import numpy as np def lagrange_interpolation(x, y, x_eval): result = 0 for i in range(len(y)): term = y[i] for j in range(len(x)): if j != i: term = term * (x_eval - x[j]) / (x[i] - x[j]) result += term return result # Input ma'lumotlarni yaratish x = np.array([1, 2, 3, 4, 5]) y = np.array([2, 3, 5, 7, 11]) # Interpolatsiya uchun qiymatni hisoblash x_eval = 3.5 result = lagrange_interpolation(x, y, x_eval) print("Interpolatsiyalangan qiymat: ", result) Nyuton usuli Teng uzoklikda joylashgan tugunlarda y funksiyaning qiymatlari berilgan. Nyutonning birinchi va ikkinchi interpolyasion formulalari yordamida ko’phad quring va х nuqtadagi qiymatini hisoblang. I X Y Δ y ❑Δ1y❑ Δ2y❑ Δ3y❑ Δ4y❑ 0 0,115 8,65729 -0,36 0,021 0,00864 -0,0155 -0,1437 1 0,120 8,29729 -0,339 0,02964 -0,00686 -0,01592 2 0,125 7,95829 -0,30936 0,02278 -0,02278 3 0,130 7,64893 -0,28658 0,02036 4 0,135 7,36235 -0,26622 5 0,140 7,09613 N 5 (x)=8,65729+ − 0,36 1 ! ∙ 0,005 ( x − 0,115 ) + 0,021 2 ! ∙ 0,005 ( x − 0,115 )( x − 0,120 ) + 0,00864 3 ! ∙ 0,005 ( x − 0,115 )( x − 0,120 )( x − 0,125 ) + − 0,0155 4 ! ∙ 0,005 ( x − 0,115 )( x − 0,120 )( x − 0,125 )( x − 0,130 ) + − 0,1437 5 ! ∙ 0,005 ( x − 0,115 )( x − 0,120 )( x − 0,125 )( x − 0,135 ) = ¿ 16.96568846-72,47830871x+1,962718117x 2 +0,4698441667x 3 -0,3686666667x 4 x=0,1315 dagi qiymatini hisoblaymiz 16.96568846-72,47830871x+1,962718117x 2 +0,4698441667x 3 -0,3686666667x 4 = > Javob:7,47 ODDIY DIFFERENSIAL TENGLAMALARNI SONLI YECHISHNING RUNGE-KUTTA USULI. Runge-kutte usulix0 y0 ∆ y k1i k 2i k 3i k 4i 0,6 0,8 0,093 0,085 0,091 0,0914 0,1128 0,7 0,9 0,104 0,098 0,104 0,1045 0,11 0,8 1 k1i= 0,1 ∗ f(0,6 ,0,8 )=0,085 k 2i = 0,1 ∗ f ( 0,65 , 0,8425 ) = 0,091 k3i= 0,1 ∗ f(0,65 ,0,8455 )= 0,0914 k 4i = 0,1 ∗ f ( 0,7 , 0,8914 ) = 0,1128 ∆ y = 1 6 ( 0,085 + 0,1128 + 2 ∗ 0,091 + 2 ∗ 0,0914 ) = 0,093 y 1 = ∆ y + y 0 = 0,09 k1i= 0,1 ∗ f(0,7 ,0,9 )=0,098 k2i=0,1 ∗ f(0,75 ,0,94 )=0,104 k3i= 0,1 ∗ f(0,75 ,0,95 )= 0,1045 k 4i = 0,1 ∗ f ( 0,8 , 1,0045 ) = 0,11 ∆ y = 1 6 ( 0,098 + 0,11 + 2 ∗ 0,104 + 2 ∗ 0,1045 ) = 0,104 y 1 = ∆ y + y 0 = 1 import numpy as np def f(x, y): return x - y def runge_kutta(x0, y0, h, N): x = x0 y = y0 for i in range(N): k1 = h * f(x, y) k2 = h * f(x + h/2, y + k1/2) k3 = h * f(x + h/2, y + k2/2) k4 = h * f(x + h, y + k3) y = y + (k1 + 2*k2 + 2*k3 + k4) / 6 x = x + h return y x0 = 0 y0 = 1 h = 0.1 N = 10 result = runge_kutta(x0, y0, h, N) print("Yechish natijasi:", result)

Taqribiy sonlar. Ularning absolyut va nisbiy xatosi. Qiymatga ega bo‘lgan raqam. To‘g‘ri ishoralar soni doc

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