Questions: Numerical Stability and Conditioning

Question 1

A programmer evaluates p(x) = x² − 4x + 4 near x = 2 and gets a wildly inaccurate answer. They rewrite it as p(x) = (x − 2)² and the result is accurate. The underlying mathematical function is identical. What does this illustrate?

• A. The problem was ill-conditioned — the answer is inherently sensitive to small input changes near x = 2
• B. The algorithm choice determines numerical stability even for a well-conditioned problem
• C. Floating-point arithmetic is always unreliable for polynomial evaluation
• D. The rewritten form changes the mathematical function being computed

Answer: (B)

This is the central example of numerical stability: the same well-conditioned function can be computed accurately or inaccurately depending on algorithm choice. Near x = 2, p(x) = x² − 4x + 4 subtracts two nearly equal large numbers, causing catastrophic cancellation. The factored form (x − 2)² avoids this entirely. The problem is not ill-conditioned — small changes in x near 2 produce small changes in p(x). The instability is entirely an artifact of the expanded algorithm, not the problem itself.

Question 2

An algorithm is called 'backward stable' if it:

• A. Produces an output error smaller than machine epsilon
• B. Computes the exact answer to a slightly perturbed version of the original input
• C. Never amplifies rounding errors beyond the precision of the inputs
• D. Always converges to the correct answer given enough iterations

Answer: (B)

Backward stability is a specific, technical concept: the algorithm computes the exact answer to a *slightly perturbed* problem. This separates algorithm error (backward error: how much did the input have to change?) from problem sensitivity (conditioning: how much does the output change for that input perturbation?). If the backward error is small and the problem is well-conditioned, the output is close to the true answer. Gaussian elimination with partial pivoting and QR decomposition are backward stable. The definition does not require output error smaller than machine epsilon, convergence, or bounded amplification in absolute terms.

Question 3

A stable algorithm cannot save you from an ill-conditioned problem — if the problem amplifies small input errors into large output errors, no choice of algorithm can prevent inaccuracy.

• T. True
• F. False

Answer: True

True. Conditioning is a property of the mathematical problem itself, independent of any algorithm. An ill-conditioned problem has a high condition number: small relative perturbations in input produce large relative perturbations in output. Even if an algorithm is backward stable (introducing only tiny backward error), the problem's own amplification factor magnifies that error into a large output error. Stability and conditioning are independent axes — both must be favorable for accurate results.

Question 4

A numerically stable algorithm is one that produces the mathematically correct answer for nearly every possible input.

• T. True
• F. False

Answer: False

False. Stability does not mean correctness for all inputs — it means small perturbations in inputs produce proportionally small changes in outputs, and the algorithm does not amplify errors beyond what the problem requires. An algorithm can be stable yet produce large output errors when the *problem itself* is ill-conditioned (a large condition number). Stability is about controlling algorithmic error amplification relative to problem sensitivity, not about achieving exact answers.

Question 5

Why is it useful to distinguish conditioning (a property of the problem) from stability (a property of the algorithm)? What does each concept tell you?

Model answer: Conditioning tells you whether the problem is inherently sensitive to input perturbations — if the true answer changes a lot when inputs change slightly, the problem is ill-conditioned and no algorithm can produce reliable results. Stability tells you whether the algorithm avoids unnecessarily amplifying rounding errors introduced during computation. The distinction matters because it diagnoses the source of numerical failure: if you compute an inaccurate result, is it because the problem is fundamentally sensitive (ill-conditioned) or because the algorithm is introducing avoidable error (unstable)? Only the second cause can be fixed by redesigning the algorithm.

The practical payoff is diagnostic: when a numerical result seems wrong, the first question is 'is this ill-conditioned or is the algorithm bad?' For an ill-conditioned problem, the answer is inherently unreliable — you need more data, a reformulation, or to accept limited accuracy. For a stable problem with a bad algorithm, switching to a backward-stable method (pivoted Gaussian elimination, QR decomposition) can recover accuracy. Without the distinction, all numerical failures look the same and the fix is unclear.