Differentiation

\[\frac{\partial f}{\partial x}\_- = \lim _ {x \to x_0} \frac{f(x) - f(x - h)}{h}\] \[\frac{\partial f}{\partial x}\_+ = \lim _ {x \to x_0} \frac{f(x + h) - f(x)}{h}\]

\[\frac{\partial f}{\partial x}\_- = \frac{\partial f}{\partial x}\_+\]

then

\[\frac{\partial f}{\partial x} = \frac{\partial f}{\partial x}\_- = \frac{\partial f}{\partial x}\_+\]

import numpy as np

def test_np_gradient():
    xx = np.linspace(-0.1, np.pi + 0.1, 100)
    yy = np.sin(xx)
    dyy = np.cos(xx)
    dyy_approx = np.gradient(yy, xx)
    error = dyy_approx - dyy
    mean_abs_error = np.mean(np.abs(error))
    print(f'{mean_abs_error}')

test_np_gradient()

0.00015176750916622188

import sympy

def test_sympy():
  x, k = sympy.symbols('x k')
  f = sympy.sin(x**2 + k)
  df = sympy.diff(f, x)
  print(f'{f = }')
  print(f'{df = }')

test_sympy()

f = sin(k + x**2)
df = 2*x*cos(k + x**2)

Use the most accurate three-point formula to determine each missing entry in the following table.

\(x\)	\(f (x)\)	\(f' (x)\)
1.1	9.025013	17.76970
1.2	11.02318	22.19364
1.3	13.46374	27.10735
1.4	16.44465	32.51085

x = [1.1, 1.2, 1.3, 1.4]
h = (x[-1] - x[0]) / (len(x) - 1)
f = [9.025013, 11.02318, 13.46374, 16.44465]
startpoint = lambda h: (-3 * f[0] + 4 * f[1] - f[2]) / 2 / h
print(startpoint(h))
midpoint = lambda i, h: (-f[i - 1] + f[i + 1]) / 2 / h
print(midpoint(1, h))
print(midpoint(2, h))
endpoint = lambda h: (f[-3] - 4 * f[-2] + 3 * f[-1]) / 2 / h
print(endpoint(h))

76970500000003
193635000000015
107350000000014
510850000000055

Integration

\[\int_a^b f(x) \, dx = \lim_{n \to \infty} \sum_{i = 1}^n h f(a + ih)\]

where

\[h = \frac{b - a}{n}\]

Use the Midpoint rule to approximate the following integral.

\[\int_1^{1.5} x^2 \ln x \, dx\]

import math

x = (1 + 1.5) / 2
print((1.5 - 1) * x**2 * math.log(x))

0.1743308994642264