Introduction and Error Analysis - Examples & Applications

This section explores practical examples covering mathematical modeling, error definitions (true, relative, and approximate), round-off and truncation errors, Taylor Series expansions, and error propagation.

Mathematical Modeling

A mathematical model formulates physical principles into an equation. The classic example is Newton's Second Law applied to a falling parachutist.

Basic: Terminal Velocity of a Parachutist

A parachutist with a mass of 68.1 kg68.1 \text{ kg} jumps out of a stationary hot air balloon. The drag coefficient is c=12.5 kg/sc = 12.5 \text{ kg/s}. Compute velocity prior to opening the chute at t=2 st = 2 \text{ s}. Acceleration due to gravity is g=9.81 m/s2g = 9.81 \text{ m/s}^2.
The analytical solution for velocity is:
v(t)=gmc(1e(c/m)t)v(t) = \frac{gm}{c} \left(1 - e^{-(c/m)t}\right)

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Intermediate: Radioactive Decay Modeling

The rate of decay of a radioactive substance is proportional to the amount of substance present: dNdt=kN\frac{dN}{dt} = -kN. Given an initial amount N0=100 gN_0 = 100 \text{ g} and a decay constant k=0.05 day1k = 0.05 \text{ day}^{-1}, compute the amount remaining after 10 days10 \text{ days} using the exact analytical model.

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Advanced: Unforced Spring-Mass System

An unforced, undamped spring-mass system is modeled by md2xdt2+kx=0m\frac{d^2x}{dt^2} + kx = 0. Given m=2 kgm = 2 \text{ kg}, k=50 N/mk = 50 \text{ N/m}, initial displacement x(0)=0.5 mx(0) = 0.5 \text{ m}, and initial velocity v(0)=0 m/sv(0) = 0 \text{ m/s}, determine the displacement at t=1 st = 1 \text{ s}.

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Accuracy, Precision, and Significant Digits

Accuracy refers to how closely a computed or measured value agrees with the true value. Precision refers to how closely individual computed or measured values agree with each other. The concept of significant digits is crucial in numerical methods to quantify precision.

Case Study 1: Bridge Deflection Measurements

An engineering team measures the mid-span deflection of a bridge under a test load. The true theoretical deflection is 15.0 mm15.0 \text{ mm}.
Team A's measurements: 14.8 mm,14.9 mm,15.1 mm,15.0 mm14.8 \text{ mm}, 14.9 \text{ mm}, 15.1 \text{ mm}, 15.0 \text{ mm}
Team B's measurements: 16.5 mm,16.5 mm,16.6 mm,16.5 mm16.5 \text{ mm}, 16.5 \text{ mm}, 16.6 \text{ mm}, 16.5 \text{ mm}
Analyze the accuracy and precision of both teams.

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Case Study 2: Concrete Compressive Strength Testing

Three batches of concrete cylinders are tested for their 28-day compressive strength. The design target (true value) is 30 MPa30 \text{ MPa}.
Batch X: 29.8,30.2,30.1,29.9 MPa29.8, 30.2, 30.1, 29.9 \text{ MPa} Batch Y: 25.1,35.0,28.5,31.4 MPa25.1, 35.0, 28.5, 31.4 \text{ MPa} Batch Z: 34.8,35.1,35.0,34.9 MPa34.8, 35.1, 35.0, 34.9 \text{ MPa}
Evaluate the accuracy and precision.

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Error Definitions

Basic: True and Relative Error

The true value of π\pi is approximately 3.141593.14159. A student estimates π\pi as 22/73.14285722/7 \approx 3.142857. Calculate the true error and true percent relative error.

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Intermediate: True and Relative Error Calculation

The derivative of a function f(x)=0.1x40.15x30.5x20.25x+1.2f(x) = -0.1x^4 - 0.15x^3 - 0.5x^2 - 0.25x + 1.2 at x=0.5x = 0.5 is true value 0.9125-0.9125. A numerical method estimates the derivative to be 0.9025-0.9025. Calculate the true error and the true percent relative error.

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Advanced: Approximate Percent Relative Error

In an iterative numerical method to find a root, the estimate in iteration 2 is 0.50.5, and the estimate in iteration 3 is 0.56710.5671. Calculate the approximate percent relative error.

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Taylor Series and Truncation Errors

Truncation errors occur when exact mathematical formulations are replaced by approximations, such as truncating a Taylor series after a finite number of terms.

Basic: Zero-Order Taylor Series

Approximate f(x)=exf(x) = e^x at xi+1=1x_{i+1} = 1 using a zero-order Taylor series based at xi=0x_i = 0. True value is e12.71828e^1 \approx 2.71828.

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Intermediate: Second-Order Taylor Series

Approximate f(x)=exf(x) = e^x at x=1x = 1 using a second-order Taylor series based at x=0x = 0. True value is 2.718282.71828.

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Advanced: Taylor Series Approximation

Approximate the function f(x)=cos(x)f(x) = \cos(x) at xi+1=π3x_{i+1} = \frac{\pi}{3} using a zero-order and first-order Taylor series expansion based on the base point xi=π4x_i = \frac{\pi}{4}. Compute the true percent relative error for each approximation.
Given: True value f(π3)=cos(π3)=0.5f(\frac{\pi}{3}) = \cos(\frac{\pi}{3}) = 0.5 h=xi+1xi=π3π4=π12h = x_{i+1} - x_i = \frac{\pi}{3} - \frac{\pi}{4} = \frac{\pi}{12}

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Error Propagation

Error propagation deals with how errors in individual variables or measurements combine to produce error in a calculated result.

Basic: Error Propagation in Addition

Given two measurements with their associated estimated errors: x1=100±2x_1 = 100 \pm 2 x2=50±1x_2 = 50 \pm 1
Determine the maximum possible error in the sum y=x1+x2y = x_1 + x_2.

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Intermediate: Error Propagation in Multiplication

Given the sides of a rectangle w=10±0.1 mw = 10 \pm 0.1 \text{ m} and l=20±0.2 ml = 20 \pm 0.2 \text{ m}, find the maximum error in the calculated area A=w×lA = w \times l.

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Advanced: General Error Propagation Formula

The deflection of a beam is given by y=FL33EIy = \frac{FL^3}{3EI}. If the measured force is F=100 NF = 100 \text{ N} with an error of ΔF=±5 N\Delta F = \pm 5 \text{ N}, and the length is L=2 mL = 2 \text{ m} with an error of ΔL=±0.02 m\Delta L = \pm 0.02 \text{ m}. Assume EE and II are exact, where 3EI=1000 Nm23EI = 1000 \text{ N}\cdot\text{m}^2. Use the general first-order error propagation formula to estimate the maximum error in deflection.

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