I'm trying to calculate the weights for a finite-difference approximation of the first derivative f′(x)f'(x)f′(x) using the Taylor series expansion. I'm solving for weights a,b,c,d,ea, b, c, d, ea,b,c,d,e such that: af(x+2Δx)+bf(x+Δx)+cf(x)+df(x−Δx)+ef(x−2Δx)a f(x+2\Delta x) + b f(x+\Delta x) + c f(x) + d f(x-\Delta x) + e f(x-2\Delta x)af(x+2Δx)+bf(x+Δx)+cf(x)+df(x−Δx)+ef(x−2Δx) approximates f′(x)f'(x)f′(x). Here's what I did:  I used the Taylor series expansion for f(x±kΔx)f(x \pm k\Delta x)f(x±kΔx), where k=1,2k = 1, 2k=1,2.  I built a system of linear equations to enforce the following conditions:  Coefficients of f(x): a+b+c+d+e=0  Coefficients of f′(x): 2a+b−d−2e=1  Coefficients of f′′(x): 4a+b+d+4e=0  Coefficients of f(3)(x): 8a+b−d−8e=0  Coefficients of f(4)(x): 16a+b+d+16e=0    I implemented the matrix equation A⋅z=bA \cdot z = bA⋅z=b in Python: import numpy as np A = np.array([ [1, 1, 1, 1, 1], # Coefficients of f(x) [2, 1, 0, -1, -2], # Coefficients of f'(x) [4, 1, 0, 1, 4], # Coefficients of f''(x) [8, 1, 0, -1, 8], # Coefficients of f'''(x) [16, 1, 0, 1, 16] # Coefficients of f''''(x) ]) b = np.array([0, 1, 0, 0, 0]) # Targeting the first derivative z = np.linalg.solve(A, b) print(z)   The Issue: The output I'm getting is: [0.25,0,-0,0,-0.25]  However, the expected weights for the first derivative should be something like:   [-1/12,2/3,0,-2/3,1/12]   What I Tried:  Double-checked the coefficients in matrix A to ensure they match the Taylor series expansion.  Verified that the right-hand side vector b=[0,1,0,0,0] is correct for approximating f′(x).   Despite this, the weights are incorrect. Am I missing something in the matrix setup or the Python implementation?  Expected Behavior: I want the solution to match the theoretical weights for a central finite difference approximation of the first derivative f′(x) using five points.  
You have a typo in the 4th line of the matrix A, the last element should be -8 instead of 8: A = np.array([ [1, 1, 1, 1, 1], # Coefficients of f(x) [2, 1, 0, -1, -2], # Coefficients of f'(x) [4, 1, 0, 1, 4], # Coefficients of f''(x) [8, 1, 0, -1, -8], # Coefficients of f'''(x) [16, 1, 0, 1, 16] # Coefficients of f''''(x) ]) 