Numerical Methods

FE Fundamentals of Engineering Exam

Track your progress as you read

Finished sections get a check mark.

1 Introduction to the FE Exam

About the FE Exam FE Exam Study Strategies Navigating the FE Reference Handbook

2 Mathematics (8–12 Questions)

Analytic Geometry and Trigonometry Single-Variable Calculus Differential Equations Linear Algebra and Matrix Operations Numerical Methods

3 Probability and Statistics (6–9 Questions)

Probability Fundamentals Probability Distributions Statistical Estimation and Hypothesis Testing

4 Chemistry (5–8 Questions)

Chemical Reactions and Stoichiometry Acids, Bases, and pH Oxidation-Reduction and Corrosion

5 Ethics and Professional Practice (5–8 Questions)

Engineering Ethics Fundamentals

6 Engineering Economics (6–9 Questions)

Time Value of Money Economic Analysis Methods

7 Statics (9–14 Questions)

Force Systems and Equilibrium Trusses, Frames, and Machines Centroids and Moments of Inertia Static Friction

8 Dynamics (9–14 Questions)

Particle Kinematics Kinetics: Force, Mass, and Acceleration Work, Energy, Impulse, and Momentum Rigid-Body Rotation and Mechanical Vibrations

9 Strength of Materials (9–14 Questions)

Stress and Strain Fundamentals Beam Analysis: Shear and Moment Diagrams Bending Stress and Torsion Stress Transformation and Material Failure

10 Materials Science (6–9 Questions)

Material Properties and Selection Phase Diagrams and Alloy Systems

11 Fluid Mechanics (12–18 Questions)

Fluid Properties and Statics Fluid Dynamics and Bernoulli's Equation Pumps, Turbines, and Flow Measurement

12 Thermodynamics and Heat Transfer (9–14 Questions)

Laws of Thermodynamics and Properties Thermodynamic Cycles and Applications Heat Transfer: Conduction, Convection, and Radiation

13 Basic Electrical Engineering (6–9 Questions)

Electrical Fundamentals and DC Circuits AC Circuits and Three-Phase Power

14 Safety, Health, and Environment (6–9 Questions)

Industrial Safety and Hazard Management

15 Instrumentation and Controls (4–6 Questions)

Sensors, Data Acquisition, and Control Systems

16 Computational Tools and Computer Science

Spreadsheets, Structured Programming, and Flowcharts

Key Takeaways Explain This Topic AI

Newton-Raphson iterates x_{n+1} = x_n − f(x_n)/f′(x_n) with fast (quadratic) convergence but needs f′ and a good start.

Bisection is guaranteed to converge when f(a)·f(b) < 0, halving the error each step: error ≤ (b−a)/2ⁿ.

Trapezoidal rule has O(h²) error; Simpson's 1/3 rule has O(h⁴) error and needs an even number of intervals.

Euler's method (O(h) per step) and Runge-Kutta 4 (O(h⁴)) advance ODE solutions numerically from y_{n+1} = y_n + h·f.

Truncation error comes from approximating infinite processes; round-off error comes from finite machine precision.

On multiply/divide, the result keeps as many significant figures as the least-precise input.

Last updated: June 2026

Numerical methods give approximate answers when a closed-form solution is impractical. The FE tests which method applies, its convergence/error order, and a single hand iteration — rarely a long computation. Know the trade-offs: guaranteed-but-slow versus fast-but-fragile.

Root-Finding
Bisection Method

If f(a) and f(b) have opposite signs (f(a)·f(b) < 0), the Intermediate Value Theorem guarantees a root in [a, b]. Repeatedly take the midpoint c = (a+b)/2 and keep the half-interval that still brackets the sign change.

Convergence: guaranteed but slow — the bracket (and thus the error bound) halves each step: error ≤ (b−a)/2ⁿ. It does not require f to be differentiable.

Newton-Raphson Method

x_{n+1} = x_n − f(x_n)/f′(x_n). Fast (quadratic) convergence near a root, but it requires the derivative, can diverge from a poor start, and fails if f′(x_n) = 0.

Worked example — Newton-Raphson. Find a root of f(x) = x² − 4 starting at x₀ = 3, with f′(x) = 2x. First step: x₁ = 3 − (3²−4)/(2·3) = 3 − 5/6 = 3 − 0.833 = 2.167 . A second step gives x₂ = 2.167 − (2.167²−4)/(2·2.167) = 2.167 − 0.696/4.334 ≈ 2.006, rapidly approaching the exact root x = 2.

Secant Method

x_{n+1} = x_n − f(x_n)·(x_n − x_{n−1})/(f(x_n) − f(x_{n−1})). It approximates f′ with a finite difference, so it needs no analytic derivative ; convergence is superlinear (order ≈ 1.618).

Choosing a Root-Finder

The FE often frames this as a trade-off: bisection is robust (cannot fail once a bracket exists) but needs many iterations; Newton-Raphson is fast but can shoot off to a different root or diverge if f′ is small near the start; the secant method is a middle ground that keeps Newton-like speed without computing a derivative. A practical engineering strategy — and a defensible FE answer — is to bracket coarsely with bisection, then switch to Newton-Raphson once you are near the root.

Numerical Integration

With n equal subintervals of width h = (b−a)/n and nodes xᵢ = a + ih:

Method Formula (composite) Error order Restriction
Trapezoidal (h/2)[f₀ + 2f₁ + … + 2f_{n−1} + f_n] O(h²) none
Simpson's 1/3 (h/3)[f₀ + 4f₁ + 2f₂ + 4f₃ + … + f_n] O(h⁴) n even
Simpson's 3/8 (3h/8)[f₀ + 3f₁ + 3f₂ + 2f₃ + …] O(h⁴) n multiple of 3

Simpson's 1/3 is far more accurate than the trapezoidal rule for smooth functions because it fits parabolas instead of straight lines.

Worked example — trapezoidal rule. Estimate ∫₀² x² dx with n = 2 (so h = 1, nodes x = 0, 1, 2; f = 0, 1, 4). Trapezoidal: (1/2)[0 + 2(1) + 4] = (1/2)(6) = 3.0. The exact value is [x³/3]₀² = 8/3 ≈ 2.667, so the trapezoidal estimate is high by ~0.33 — straight-line segments overshoot a convex curve.

Worked example — Simpson's 1/3. Same integral, same nodes: (1/3)[0 + 4(1) + 4] = (1/3)(8) = 2.667, which equals the exact value. Simpson's rule integrates any cubic (and hence x²) exactly , demonstrating its O(h⁴) accuracy advantage.

Numerical ODE Solvers

For dy/dx = f(x, y) with y(x₀) = y₀ and step size h:

Method Update Error/step
Euler y_{n+1} = y_n + h·f(x_n, y_n) O(h)
Heun (improved Euler) average of slopes at the ends O(h²)
Runge-Kutta 4 (RK4) y_{n+1} = y_n + (h/6)(k₁ + 2k₂ + 2k₃ + k₄) O(h⁴)

RK4 evaluates the slope four times per step (k₁ at the start, k₂ and k₃ at the midpoint, k₄ at the end) and is the standard workhorse for its accuracy-vs-effort balance.

Worked example — Euler's method. Solve dy/dx = x + y, y(0) = 1, with h = 0.1; find y at x = 0.1. y₁ = y₀ + h·f(x₀, y₀) = 1 + 0.1·(0 + 1) = 1.10 . The exact solution y = 2e^x − x − 1 gives y(0.1) = 1.1103, so Euler's first-order O(h) error already shows a small lag.

Error Analysis

Term Definition
Absolute error |approx − exact|
Relative error |absolute error / exact|
Percent error relative error × 100%
Truncation error from approximating infinite processes (series, derivatives) with finite steps
Round-off error from finite machine precision

Significant figures: in multiplication/division the result carries as many sig figs as the least-precise factor; in addition/subtraction align by decimal place. Worked example — relative error. If a method yields 2.167 for a true value of 2.000, the absolute error is 0.167 and the relative error is 0.167/2.000 = 0.0835 = 8.35%.

Step Size, Stability, and the Speed-Accuracy Trade

Shrinking the step size h reduces truncation error — halving h cuts a trapezoidal estimate's error by ~4× (O(h²)) and a Simpson or RK4 estimate by ~16× (O(h⁴)). But arbitrarily small h is not free: more steps mean more arithmetic and more accumulated round-off error, and some explicit ODE methods become numerically unstable if h is too large for the problem's fastest-changing mode. The practical takeaway the FE tests is that total error has two competing parts, and the optimal h balances truncation against round-off rather than driving h toward zero. When a question gives you the error order, you can predict how the error scales without re-running the computation — e.g., a method labeled O(h⁴) that already gives 1% error will give roughly 0.06% error if you halve h.