Numerical Differentiation and Integration

• Numerical differentiation and integration covered
• Newton-Cotes Integration Formulas are introduced, including the trapezoidal rule and Simpson's rules

Numerical Differentiation

• Finds the rate of change $f'(x)$ of a function $f(x)$

• The formula for the numerical differentiation is:

  • $f'(x) = \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}$

• Derivatives are approximated using a small $\Delta x$

• The formula for the forward difference approximation:

  • $f'(x) \approx \frac{f(x + \Delta x) - f(x)}{\Delta x}$

• The formula for the backward difference approximation:

  • $f'(x) \approx \frac{f(x) - f(x - \Delta x)}{\Delta x}$

• The formula for the central difference approximation:

  • $f'(x) \approx \frac{f(x + \Delta x) - f(x - \Delta x)}{2 \Delta x}$

Example of derivative of $f(x) = e^x$ at $x = 2$ where $\Delta x = 0.1$:

• The true solution is $f'(x) = e^x$, so $f'(2) = e^2 = 7.389056099$
• Forward difference approximation:
  • $= \frac{e^{2.1} - e^2}{0.1} = 7.707426066$
  • Absolute error $ = |7.707426066 - 7.389056099| = 0.318369967$
  • Relative error $ = |\frac{0.318369967}{7.389056099}| = 0.0430869$
• Backward difference approximation:
  • $\frac{e^{2} - e^{1.9}}{0.1} = 7.081880265$
  • Absolute error $ = |7.081880265 - 7.389056099| = 0.307175834$
  • Relative error $ = |\frac{0.307175834}{7.389056099}| = 0.0415713$
• Central difference approximation
  • $\frac{e^{2.1} - e^{1.9}}{0.2} = 7.394653166$
  • Absolute error $ = |7.394653166 - 7.389056099| = 0.005597067$
  • Relative error $ = |\frac{0.005597067}{7.389056099}| = 0.0007575$
• $f''(x)$ second derivative approximation:
  • $f''(x) \approx \frac{f'(x + \Delta x) - f'(x - \Delta x)}{2 \Delta x}$
• Using central difference approximation for $f'(x + \Delta x)$ and $f'(x - \Delta x)$:
  • $f''(x) \approx \frac{f(x + 2\Delta x) - 2f(x) + f(x - 2\Delta x)}{(2 \Delta x)^2}$

Numerical Integration

• Finds the area under the curve of $f(x)$
• Numerical integration approximates a definite integral using numerical methods, dividing the interval $[a, b]$ into subintervals to approximate the area in each

Newton-Cotes Integration Formulas

• Approximates the integrand $f(x)$ with a polynomial
• Approximates the integral of $f(x)$ with the integral of the polynomial
• The most common are the trapezoidal rule and Simpson's rule

Trapezoidal Rule

• Approximates the integral of $f(x)$ by dividing the area under the curve into trapezoids
• The area of each trapezoid is:
  • $A_i = \frac{f(x_i) + f(x_{i+1})}{2} \Delta x$
• The integral is approximated by the sum of the areas of the trapezoids using:
  • $\int_a^b f(x) dx \approx \frac{\Delta x}{2} [f(x_1) + 2f(x_2) + 2f(x_3) +... + 2f(x_{n-1}) + f(x_n)]$

Example using the trapezoidal rule of $f(x) = x^2$ from $a = 0$ to $b = 2$ using $n = 4$ subintervals

• The width of each subinterval is $\Delta x = \frac{b - a}{n} = \frac{2 - 0}{4} = 0.5$
• The $x$ values are $x_1 = 0, x_2 = 0.5, x_3 = 1, x_4 = 1.5, x_5 = 2$
• $\int_0^2 x^2 dx \approx \frac{0.5}{2} [0^2 + 2(0.5)^2 + 2(1)^2 + 2(1.5)^2 + 2^2] = 2.75$
• The true solution is $\int_0^2 x^2 dx = \frac{x^3}{3} |_0^2 = \frac{8}{3} = 2.666666667$

Simpson's Rules

• Approximates the integrand $f(x)$ with quadratic polynomials
• Simpson's 1/3 rule approximates the integral of $f(x)$ by dividing the area under the curve into parabolas, using the formula:
  • $\int_a^b f(x) dx \approx \frac{\Delta x}{3} [f(x_1) + 4f(x_2) + 2f(x_3) + 4f(x_4) +... + 2f(x_{n-1}) + 4f(x_n) + f(x_{n+1})]$
• $n$ must be even for Simpson's 1/3 rule
• Simpson's 3/8 rule approximates the integral of $f(x)$ by dividing the area under the curve into cubic polynomials, using the formula:
  • $\int_a^b f(x) dx \approx \frac{3\Delta x}{8} [f(x_1) + 3f(x_2) + 3f(x_3) + 2f(x_4) + 3f(x_5) + 3f(x_6) + 2f(x_7) +... + f(x_n)]$
• $n$ must be a multiple of 3 for Simpson's 3/8 rule

Example of Simpson's 1/3 rule with $f(x) = x^2$ from $a = 0$ to $b = 2$ using $n = 4$ subintervals

• The width of each subinterval is $\Delta x = \frac{b - a}{n} = \frac{2 - 0}{4} = 0.5$
• The $x$ values are $x_1 = 0, x_2 = 0.5, x_3 = 1, x_4 = 1.5, x_5 = 2$
• $\int_0^2 x^2 dx \approx \frac{0.5}{3} [0^2 + 4(0.5)^2 + 2(1)^2 + 4(1.5)^2 + 2^2] = 2.666666667$
• The true solution is $\int_0^2 x^2 dx = \frac{x^3}{3} |_0^2 = \frac{8}{3} = 2.666666667$