(3h/8)[f(x₀) + f(xₙ) + 2(sum of ordinates with non-multiple-of-3 indices) + 3(sum of ordinates with indices multiple of 3)]

(3h/8)[f(x₀) + 4f(x₁) + 4f(x₂) + 2f(x₃) + 4f(x₄) + ... + f(xₙ)]

(3h/8)[f(x₀) + f(xₙ) + 4(sum of all interior ordinates)]

(3h/8)[f(x₀) + f(xₙ) + 3(sum of ordinates with non-multiple-of-3 indices) + 2(sum of ordinates with indices multiple of 3)]

(3h/8)[f(x₀) + f(xₙ) + 4(sum of ordinates with non-multiple-of-3 indices) + 2(sum of ordinates with indices multiple of 3)]

Line 5: Explicitly zeroing U[j, i] corrupts the decomposition.

Line 2: U should be a triangular matrix initialization, not a full copy of A.

Line 1: Incorrect initialization of L as an identity matrix.

Line 3: Division by zero if U[i,i] is zero (no pivoting).

Line 4: Incorrect row update for U (should be U[j, i+1:]).

The derivative of the function must be zero 

The function must be periodic 

The function must be linear

The function must be differentiable

The absolute value of the derivative of the iteration function must be less than 1

A is symmetric positive definite.

A is strictly diagonally dominant. 

The spectral radius of the iteration matrix ρ(T)<1

A is tridiagonal.

A has no zero diagonal elements.

A x = e_i for each i 

L U x = e_i  for each i 

L y = e_i  followed by U x = e_i for each i

L y = e_i  followed by U x = y  for each i 

U y = e_i  followed by L x = y  for each i 

Line B: Row swapping syntax is incorrect 

Should check if A[max_row][col] is zero before swapping 

Line C: Should return the pivot value, not a boolean 

Line A: Should start from col+1 instead of col 

No error in the code 

h=(b-a)/2

h=(b-a)/4

h=(b-a)/5

h=(b-a)/6

h=(b-a)/3

4

3

2

6

5

def newton_backward(x, y, x_val):

    n = len(x)

    h = x[1] - x[0]

    u = (x_val - x[-1]) / h

    result = y[-1]

    for i in range(1, n):

        term = 1

        for j in range(i):

            term *= (u + j) / (j + 1)

        result += term * backward_difference(y)[i][-1]

    return result

def newton_backward(x, y, x_val):

    n = len(x)

    h = x[1] - x[0]

    u = (x_val - x[-1]) / h

    result = y[-1]

    for i in range(1, n):

        term = 1

        for j in range(i):

            term *= (u - j) / (j + 1)

        result += term * forward_difference(y)[-1][i]

    return result

def newton_backward(x, y, x_val):

    n = len(x)

    h = x[1] - x[0]

    u = (x_val - x[0]) / h

    result = y[0]

    for i in range(1, n):

        term = 1

        for j in range(i):

            term *= (u - j) / (j + 1)

        result += term * forward_difference(y)[i][0]

    return result

def newton_backward(x, y, x_val):

    n = len(x)

    h = x[1] - x[-1]

    u = (x_val - x[-1]) / h

    result = y[-1]

    for i in range(1, n):

        term = 1

        for j in range(i):

            term *= (u - j) / (j + 1)

        result += term * forward_difference(y)[-1][i]

    return result

def newton_backward(x, y, x_val):

    n = len(x)

    h = x[1] - x[0]

    u = (x_val - x[-1]) / h

    result = y[-1]

    for i in range(1, n):

        term = 1

        for j in range(i):

            term *= (u - j) / (j + 1)

        result += term * forward_difference(y)[i][0]

    return result

f(x) = sin(x)

(x) = e^x

(x) = x^2

f(x) = 5x + 3

f(x) = x³