Error in Quadrature Formula | Trapezoidal & Simpson's Rule

A Taylor polynomial approximates a function (such as trigonometric, exponential, or logarithmic functions) using a finite sum of polynomial terms derived from the function's derivatives at a specific point, while the remainder term (Restglied) represents the error between the actual function and its polynomial approximation, which increases as you move away from the development point.

Taylor series expand functions as infinite polynomials around a point x₀: f(x) = Σ f⁽ⁿ⁾(x₀)(x-x₀)ⁿ/n! + Rₙ(x). The Lagrange remainder Rₙ(x) = f⁽ⁿ⁺¹⁾(ξ)(x-x₀)ⁿ⁺¹/(n+1)! provides error estimation. When x₀ = 0, this becomes the Maclaurin series. For f(x) = √(2+x) at x₀ = -1, the second-order polynomial is P₂(x) = 1 + (1/2)(x+1) - (1/8)(x+1)². To approximate √0.9, set x = -1.1, giving √0.9 ≈ 0.94875. The error bound uses the Lagrange remainder: |R₂(-1.1)| ≤ (3/8)(1/0.9^(5/2))|(-1.1+1)|³ ≈ 0.0016267.

This comprehensive lesson covers Taylor series expansions with Lagrange form of remainder. The Taylor series for a function f(x) that is (n+1) times differentiable is: f(x) = f(a) + f'(a)(x-a) + f''(a)(x-a)²/2! + ... + f⁽ⁿ⁾(a)(x-a)ⁿ/n! + Rₙ(x), where Rₙ(x) = f⁽ⁿ⁺¹⁾(c)(x-a)ⁿ⁺¹/(n+1)!. The lesson demonstrates finding the Lagrange remainder for f(x) = 1 - x^(5/2) at x = 1, yielding c = 9/25. It also derives Taylor series for log(1+x) about x = a: log(1+x) = log(1+a) + (x-a)/(1+a) - (x-a)²/(2(1+a)²) + ..., and for e^x: e^x = e^a + e^a(x-a) + e^a(x-a)²/2! + ..., with Maclaurin series obtained by setting a = 0.

This segment covers Taylor Series, Maclaurin Series, and error term. Taylor's Theorem states that if f(x) and its derivatives up to order n-1 are continuous on [a, a+h], then f(a+h) = f(a) + h·f'(a) + h²/2!·f''(a) + ... + h^(n-1)/(n-1)!·f^(n-1)(a) + R_n, where R_n is the remainder term. Maclaurin Series is a special case of Taylor's Series where the expansion is centered at x = 0: f(x) = f(0) + x·f'(0) + x²/2!·f''(0) + ... + x^(n-1)/(n-1)!·f^(n-1)(0) + R_n. Common Maclaurin series include: e^x = 1 + x + x²/2! + x³/3! + ...; sin(x) = x - x³/3! + x⁵/5! - ...; cos(x) = 1 - x²/2! + x⁴/4! - ... The remainder term R_n = h^n/n!·f^(n)(a + θh) for some θ between 0 and 1. For the Taylor Series to be exact, R_n must be zero, which occurs when n approaches infinity for certain functions like e^x, sin(x), and cos(x).

The Taylor series expansion formula is: f(b) = f(a) + (b-a)f'(a) + (b-a)²/2! f''(a) + (b-a)³/3! f'''(a) + ... + (b-a)ⁿ/n! f⁽ⁿ⁾(a) + Rₙ. Lagrange's form of remainder is: Rₙ = (b-a)ⁿ⁺¹/(n+1)! f⁽ⁿ⁺¹⁾(c), where c is some point in (a,b). This formula allows approximation of a function using its derivatives at a single point, with the remainder term providing the error bound.

Numerical integration approximates definite integrals using discrete methods when analytical solutions are difficult; the Trapezoidal Rule uses straight-line approximations between points, Simpson's 1/3 Rule applies parabolic approximations over pairs of intervals, and Simpson's 3/8 Rule extends this to cubic approximations over three intervals, each requiring specific conditions for optimal application.

Numerical integration approximates definite integrals using discrete function values. The Trapezoidal Rule formula is: ∫a^b f(x)dx = h/2 * [y0 + 2(y1 + y2 + ... + yn-1) + yn], where h = (b-a)/n. Simpson's 1/3 Rule formula is: ∫a^b f(x)dx = h/3 * [y0 + 4(y1 + y3 + ... + yn-1) + 2(y2 + y4 + ... + yn-2) + yn], requiring n to be even. Simpson's 3/8 Rule formula is: ∫a^b f(x)dx = 3h/8 * [y0 + 3(y1 + y2 + ... + yn-1) + 2(y3 + y6 + ...) + yn], requiring n to be a multiple of 3. The step size h is calculated as (b-a)/n, where n is the number of intervals.

The Trapezoidal Rule formula: ∫f(x)dx from a to b = (h/2)[y₀ + 2y₁ + 2y₂ + ... + 2yₙ₋₁ + yₙ], where: (1) h = (b-a)/n is the interval width, (2) y₀, y₁, y₂, ... are function values at equally spaced points, (3) Coefficients are 1, 2, 2, ..., 2, 1. The instructor explains this rule approximates the area under the curve using trapezoids.

The trapezoidal rule is a numerical integration method that estimates the definite integral (area under a curve) by dividing the interval [a, b] into n subintervals of width Δx = (b-a)/n, then calculating the sum: Tₙ = (Δx/2)[f(x₀) + 2f(x₁) + 2f(x₂) + ... + 2f(xₙ₋₁) + f(xₙ)], where all interior points are multiplied by 2 but the first and last points are not; unlike Riemann sums which use only n out of n+1 points, the trapezoidal rule uses all n+1 points to create trapezoids instead of rectangles, providing a more accurate approximation of the area under the curve.

Simpson's 1/3 Rule is a numerical integration method that approximates definite integrals by fitting a quadratic polynomial (degree 2) through three points: the initial point, midpoint, and final point. The basic formula is: I ≈ (h/3) × [f(x₀) + 4f(x₁) + f(x₂)], where h is the step size between points. For repeated application over multiple intervals, the formula becomes: I ≈ (h/3) × [f(x₀) + 4∑f(x_odd) + 2∑f(x_even) + f(xₙ)], where n is the number of intervals (must be even). This method reduces error compared to the trapezoidal rule by better approximating curved functions, though accuracy depends on the specific function being integrated.

Reducing the step size h in Euler's method decreases the error, as smaller steps cause the polygonal approximation to follow the true solution curve more closely. Larger step sizes result in greater deviation from the actual solution because the linear segments cannot capture rapid changes in the curve. This trade-off between step size and accuracy is fundamental to numerical methods—smaller steps improve accuracy but increase computational cost. Euler's method is classified as a first-order numerical method because its global error is proportional to the step size h (E ∝ Ch for some constant C). This classification refers specifically to the power of h in the error expression, not to the order of the differential equation being solved. The mantra 'have the step size' captures the principle that reducing h reduces error proportionally.

This comprehensive section covers the foundational principles of numerical analysis for solving differential equations. The two fundamental challenges are: (1) The 'omnipresence of infinity' - equations on intervals contain infinitely many equations with infinitely many unknowns. (2) The complexity of the derivative operator - unlike algebraic equations, derivatives cannot be easily isolated. The solution involves discretization: selecting finite mesh points xᵢ = i·h where h = 1/(n+1), transforming infinite problems into finite ones. Taylor's formula enables finite difference approximation: u''(x) = [u(x-h) - 2u(x) + u(x+h)]/h² + O(h²). Big-O notation describes error: f(h) is O(hⁿ) if |f(h)| ≤ C|h|ⁿ. The key insight is the trade-off between accuracy and solvability - neglecting error terms transforms unsolvable exact equations into solvable approximate equations. Substituting into the differential equation yields a tridiagonal linear system that can be solved efficiently, demonstrating how numerical methods transform differential equations into computationally tractable algebraic systems.

f(n) = O(g(n)) means there exist positive constants c and n0 such that f(n) ≤ c*g(n) for all n ≥ n0. This means f(n) grows no faster than g(n) asymptotically.

This section provides a comprehensive explanation of asymptotic notation, focusing on Big-O and Omega. Asymptotic notation is a mathematical tool for representing algorithm complexity. Big-O (O) provides an upper bound (worst case), Omega (Ω) provides a lower bound (best case), and Theta (θ) provides a tight bound (average case). Big-O notation means f(n) = O(g(n)) if there exist positive constants c and n₀ such that f(n) ≤ c × g(n) for all n ≥ n₀. The instructor demonstrates proof using f(n) = 3n + 2 and g(n) = n. To prove f(n) = O(g(n)), we need to find c and n₀ such that 3n + 2 ≤ c × n for all n ≥ n₀. The instructor shows that 3n + 2 ≤ 5n for all n ≥ 1, so c = 5 and n₀ = 1. Omega notation means f(n) = Ω(g(n)) if there exist positive constants c and n₀ such that f(n) ≥ c × g(n) for all n ≥ n₀. To prove f(n) = Ω(g(n)), we need to find c and n₀ such that 3n + 2 ≥ c × n for all n ≥ n₀. The instructor shows that 3n + 2 ≥ (1/5)n for all n ≥ 1, so c = 1/5 and n₀ = 1.

Any polynomial of degree d belongs to Θ(n^d), where d is the highest power with non-zero coefficient. For example, a polynomial of degree 3 is Θ(n³), and degree 8 is Θ(n^8). When d = 0, the polynomial is constant, belonging to Θ(1). Big-O (O) notation provides an upper bound: f(n) = O(g(n)) if there exist c and n0 such that f(n) ≤ c·g(n) for all n > n0. Unlike Big-Theta, Big-O only requires an upper bound. For example, n = O(n²) and n² + 2n = O(n²).