A second-order recurrence can be stable at its nominal coefficients yet become unstable at an allowed corner of a coefficient-uncertainty box. Checking one root plot is not a robustness proof.
The short answer is to write the unit-circle inequalities for the declared real quadratic, then minimise their margins over the actual uncertainty set. This guide proves the criterion, works original safe and unsafe boxes, and shows why the result applies to fixed coefficients rather than arbitrary switching. All numerical systems are fictional teaching models. The Z-Transform subject guide supplies the broader sequence and transform context.
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ROUTE 1 · CHAPTERS 1–7
Define and prove the criterion
State the fixed recurrence and derive the real-quadratic stability region.
ROUTE 2 · CHAPTERS 8–14
Test coefficient boxes exactly
Work nominal, failing and safe sets with analytical margin extrema.
ROUTE 3 · CHAPTERS 15–21
Change the assumptions and coefficients
Inspect boundaries, correlated sets, a transfer case and implementation limits.
ROUTE 4 · CHAPTERS 22–28
Distinguish frozen and switching systems
Construct a two-step counterexample and trace the source conditions.
ROUTE 5 · CHAPTERS 29–35
Reproduce and connect conventions
Check transformations, coefficient signs, rounding allowances and initial states.
Chapter contents
Define and prove the criterion · Chapters 1–7
1. Define the recurrence and the property
2. Separate nominal and robust statements
3. Write the real quadratic criterion
4. Prove necessity from roots inside the circle
5. Prove sufficiency for nonreal roots
Test coefficient boxes exactly · Chapters 8–14
8. Choose an original nominal model
9. Declare an unsafe uncertainty box
10. Verify the unstable corner by roots
12. Derive the box checks in general
Change the assumptions and coefficients · Chapters 15–21
15. Keep strict boundaries strict
16. Change the uncertainty geometry
17. Distinguish a deterministic set from a probability model
18. Transfer check with a new nominal quadratic
19. Check the changed-box solution
20. Distinguish pole location from transient size
21. Check the realized model rather than a reduced fraction alone
Distinguish frozen and switching systems · Chapters 22–28
22. Introduce the switching question explicitly
23. Build two individually stable transition matrices
24. Calculate the alternating two-step product
25. Keep the practical conclusion within its assumptions
26. Read the source criterion with the time convention intact
Reproduce and connect conventions · Chapters 29–35
29. Reproduce and communicate the result
30. Map the unit disk to a continuous-time half-plane
31. Check a box whose second coefficient is negative
32. Translate a coefficient-rounding allowance into a box
33. Compare short trajectories with the root conclusions
CHAPTER 1 OF 35 · Define and prove the criterion
1. Define the recurrence and the property
Consider the zero-input recurrence y[n]+a₁y[n−1]+a₂y[n−2]=0, with fixed real coefficients a₁ and a₂ and arbitrary finite initial values.
Its characteristic polynomial is p(z)=z²+a₁z+a₂. We ask whether every zero-input trajectory decays to zero. For this finite-dimensional constant-coefficient model, that occurs when both characteristic roots lie strictly inside the unit circle.
This is not a claim about every physical system with a similar measured transfer function. A nonminimal realization could contain additional modes, and an implementation could introduce quantization or other nonlinear rules. Here the quadratic explicitly represents the recurrence being tested. We will keep the model and property fixed while first changing its coefficients, then changing the assumptions about whether those coefficients stay constant over time.
CHAPTER 2 OF 35 · Define and prove the criterion
2. Separate nominal and robust statements
Nominal stability concerns one supplied coefficient pair. Robust fixed-parameter stability concerns every coefficient pair in a declared uncertainty set, with each pair held fixed for its trajectory.
A coefficient set can represent bounded measurement error, manufacturing variation or an educational perturbation, but the interpretation of the bounds must be supplied independently. Merely writing a small interval does not establish that it covers a real device’s uncertainty.
A robust result is conditional on that set. If a coefficient can vary beyond it, the conclusion may fail. If the true possible pairs are correlated and occupy only part of a box, checking the full box can be conservative. The first task is therefore to name the uncertainty model before choosing a root calculation or a stability test. A robust claim without a set has no defined collection of alternatives to certify.
CHAPTER 3 OF 35 · Define and prove the criterion
3. Write the real quadratic criterion
For p(z)=z²+a₁z+a₂ with real coefficients, both roots lie strictly inside the unit circle exactly when |a₂|<1, 1+a₁+a₂>0 and 1−a₁+a₂>0.
The last two expressions are p(1) and p(−1). They detect boundaries where a real root reaches plus or minus one. The product of the two roots is a₂, so its magnitude supplies another essential constraint.
The criterion is the second-order form of a Jury or Schur–Cohn stability test. It is a discrete-time unit-circle condition, not a continuous-time left-half-plane rule. We will prove it for this real monic quadratic before applying it to uncertainty. That proof also explains why checking only the root product or only one endpoint polynomial value is insufficient.
CHAPTER 4 OF 35 · Define and prove the criterion
4. Prove necessity from roots inside the circle
Let the roots be r₁ and r₂. Their product is a₂, so |r₁r₂|<1 if both magnitudes are below one.
If the roots are real, each lies between minus one and one. Then p(1)=(1−r₁)(1−r₂)>0 and p(−1)=(−1−r₁)(−1−r₂)>0 because the factors in each product have matching signs.
If the roots are nonreal, real coefficients make them a conjugate pair. The endpoint values become |1−r₁|² and|−1−r₁|², which are strictly positive. Thus every stable real quadratic satisfies all three inequalities.
This direction is a useful initial check but not the whole theorem. A collection of necessary conditions is not automatically sufficient. The next two cases establish that no unstable root configuration can satisfy the full set under the stated real-quadratic assumptions.
CHAPTER 5 OF 35 · Define and prove the criterion
5. Prove sufficiency for nonreal roots
For nonreal conjugate roots r and its conjugate, a₂=r times its conjugate=|r|². The condition |a₂|<1 therefore implies |r|<1 directly.
Both roots have the same magnitude, so the product condition controls them together in this case. The endpoint expressions are automatically positive for a nonreal pair, consistent with the full criterion.
Do not extend that shortcut to arbitrary real roots. One real root can have magnitude greater than one while the other is small enough that their product remains below one. For example, roots 1.2 and 0.2 have product 0.24, yet the first root is unstable. The endpoint conditions are what exclude such real-root arrangements. Splitting the proof into the real and nonreal cases makes the roles of the conditions visible rather than treating the inequalities as an unexplained checklist.
CHAPTER 6 OF 35 · Define and prove the criterion
6. Prove sufficiency for real roots
When p(1)>0, the two real roots lie on the same side of one, except for excluded equality. When p(−1)>0, they also lie on the same side of minus one.
Together these sign conditions allow both roots inside(−1,1), both above one, or both below minus one. A configuration straddling either endpoint makes the corresponding product negative.
If both roots are above one or both below minus one, their product exceeds one, contradicting |a₂|<1. The only remaining possibility is that both lie strictly between minus one and one.
This completes the proof for a real monic quadratic. Repeated roots inside the interval cause no exception: terms involving n times r^n still decay when |r|<1. A root exactly on the unit circle fails the strict condition and generally prevents decay to zero, so boundary cases must remain separate.
CHAPTER 7 OF 35 · Define and prove the criterion
7. Interpret the stability region in coefficient space
The inequalities can also be written −1<a₂<1 and −(1+a₂)<a₁<1+a₂. The allowed coefficient pairs form an open triangular region.
At a₂=−1, the lower geometry closes to a point; at a₂=1, the top boundary is excluded. The sloping edges correspond to p(1)=0 or p(−1)=0.
This geometric picture helps with uncertainty sets because the relevant margins are linear in the coefficients. We are checking whether an entire allowed set fits within this open region, not merely whether its centre is inside.
The coefficient-space distance to a boundary is not automatically the same as a root’s Euclidean distance to the unit circle or a bound on transient amplification. Those are different quantities. Use the triangle to assess the declared stability property, then derive any additional performance or numerical-sensitivity statement separately.
CHAPTER 8 OF 35 · Test coefficient boxes exactly
8. Choose an original nominal model
Let a₁=−1.4 and a₂=0.5. The roots are 0.7±0.1i, whose common magnitude is the square root of 0.5, approximately 0.7071.
The three margins are 1−|a₂|=0.5, p(1)=1−1.4+0.5=0.1, and p(−1)=1+1.4+0.5=2.9. All are positive, so the nominal recurrence is stable.
The small p(1) margin identifies a nearby coefficient boundary even though the nominal roots themselves look comfortably inside the unit circle. This is not a contradiction: moving coefficients and moving roots are related nonlinearly.
Keep the exact coefficient convention alongside the numbers. Some software writes a recurrence with the feedback signs on the other side of the equation. A sign conversion performed incorrectly can test a different polynomial. Reconstruct the characteristic equation from the actual recurrence before interpreting any returned roots.
CHAPTER 9 OF 35 · Test coefficient boxes exactly
9. Declare an unsafe uncertainty box
Allow |δ₁|≤0.08 and |δ₂|≤0.04 around the nominal coefficients, with a₁=−1.4+δ₁ and a₂=0.5+δ₂.
The p(1) expression is 0.1+δ₁+δ₂. Its minimum over the independent box is 0.1−0.08−0.04=−0.02. Therefore the box is not entirely inside the stable region.
The critical corner is a₁=−1.48, a₂=0.46. Both coefficient changes are within their declared bounds. A check that tests only the nominal roots would miss it.
The word independent here describes the allowed Cartesian product of coefficient values, not a probability distribution. No uniform probability or chance of failure has been assigned to the box. The result is a worst-case fixed-parameter statement: at least one allowed pair is unstable. A probability of encountering that pair would require additional information about how coefficients are distributed.
CHAPTER 10 OF 35 · Test coefficient boxes exactly
10. Verify the unstable corner by roots
At the identified corner, p(z)=z²−1.48z+0.46. Its roots are approximately 1.035972972 and 0.444027028.
The larger root exceeds one, agreeing with p(1)=−0.02. A zero-input initial condition containing that mode grows rather than decays.
This independent root calculation checks the inequality result. It is not necessary to solve a quadratic at every point of the uncertainty box once the exact criterion and margin extrema are established. It is still useful to verify a concrete failing witness and preserve it in the evidence record.
The corner result also clarifies the meaning of “small uncertainty.” The changes 0.08 and 0.04 can look numerically small, but they are large relative to the nominal p(1) margin 0.1. Relevant scale is set by the condition being tested, not by how many decimal places a perturbation occupies in a coefficient table.
Now use |δ₁|≤0.04 and |δ₂|≤0.03 around the same nominal pair. The smallest p(1) value is 0.1−0.04−0.03=0.03, still positive.
The smallest p(−1) is 2.9−0.04−0.03=2.83. The largest possible |a₂| is 0.53, below one. Thus every fixed coefficient pair in this smaller box gives a stable recurrence.
All three checks are necessary for the robust conclusion. A favourable p(1) margin alone would not rule out a different boundary failure. Here the extrema are easy because the expressions are affine and the a₂ interval stays positive.
This proof does not claim a particular settling time, maximum state magnitude or acceptable response to an external command. It establishes the stated decay property for every frozen coefficient pair in the supplied box. Performance requirements need their own quantities and bounds.
| Coefficient set | Smallest p(1) | Smallest p(−1) | Largest magnitude of a₂ |
|---|---|---|---|
| Nominal pair | 0.1 | 2.9 | 0.5 |
| Box with radii 0.08 and 0.04 | −0.02 | 2.78 | 0.54 |
| Box with radii 0.04 and 0.03 | 0.03 | 2.83 | 0.53 |
CHAPTER 12 OF 35 · Test coefficient boxes exactly
12. Derive the box checks in general
Let a₁=c₁+δ₁ and a₂=c₂+δ₂, with |δ₁|≤r₁ and |δ₂|≤r₂. Assume nonnegative radii.
The exact minima are 1+c₁+c₂−r₁−r₂ for p(1), and 1−c₁+c₂−r₁−r₂ for p(−1). The maximum of |a₂| over its interval is |c₂|+r₂.
Therefore the entire closed box is strictly stable for fixed coefficients exactly when |c₂|+r₂<1, 1+c₁+c₂>r₁+r₂ and 1−c₁+c₂>r₁+r₂.
These are not conservative replacements for this particular real quadratic and rectangular uncertainty set: they are the exact extrema of its necessary-and-sufficient inequalities. The result would need reconsideration for a higher-order polynomial, complex coefficients or a different uncertainty geometry. State those restrictions rather than treating a convenient three-line test as a universal robust-control theorem.
CHAPTER 13 OF 35 · Test coefficient boxes exactly
13. Explain why vertices suffice here
An affine function on a rectangular box attains its minimum at a vertex. The p(1) and p(−1) margins are affine, while the two inequalities a₂<1 and −a₂<1 are also affine.
Checking all four box vertices against the quadratic criterion therefore certifies every interior point and edge. The direct extremum formulas are a compact version of that vertex check.
This argument depends on the stability region’s linear description for a real monic quadratic. It is not a general licence to check only vertices of every uncertain polynomial family. A higher-order stability region can have a more complicated geometry.
A useful learner check is to identify the exact step that justified vertex sufficiency: the inequalities are affine in the uncertain coefficients and jointly characterize the whole target region. Removing either feature can invalidate the shortcut. The property of the uncertainty set and the property of the stability criterion must be considered together.
CHAPTER 14 OF 35 · Test coefficient boxes exactly
14. Distinguish a grid search from an exact box proof
A dense grid can reveal unstable examples, but a passing finite grid does not by itself cover every point of a continuous uncertainty box. A narrow failing region could lie between tested points.
For the present quadratic, the analytical extrema solve that coverage problem exactly. A numerical grid remains a useful implementation check: compare its root classifications with the inequality result and examine points close to the boundaries.
Record grid spacing, floating-point tolerance and how equality was treated. A root whose displayed magnitude rounds to 1.0000 is not automatically proven inside or outside. Use exact or higher-precision coefficient calculations when the conclusion depends on a small margin.
The evidence roles differ. A failing point is a witness that a universal claim is false. A passing collection of points needs a coverage argument before it becomes a universal proof. Our affine-margin derivation supplies that argument for the stated family.
CHAPTER 15 OF 35 · Change the assumptions and coefficients
15. Keep strict boundaries strict
If p(1)=0, then z=1 is a characteristic root. The corresponding constant mode need not decay. If p(−1)=0, then z=−1 is a root and an alternating mode can persist.
If a₂=1 and the remaining conditions place a nonreal conjugate pair on the circle, its magnitude is one. Repeated unit-circle roots can introduce additional growth factors.
Therefore replacing strict inequalities by nonstrict ones changes the property being certified. A numerical tolerance used for classification should be reported as such, rather than silently redefining asymptotic decay.
A boundary can still be useful for finding an uncertainty radius or visualising a transition. The last allowed radius at which a root touches the circle is a threshold, not a member of the strictly stable closed uncertainty set. Keep that distinction when rounding a reported radius or choosing a practical engineering margin.
CHAPTER 16 OF 35 · Change the assumptions and coefficients
16. Change the uncertainty geometry
Consider a correlated line a₁=−1.4+t and a₂=0.5−t for |t|≤0.08. Now p(1)=0.1 exactly, because the coefficient changes cancel in that expression.
The other endpoint margin is p(−1)=2.9−2t, whose minimum is 2.74. The a₂ range is 0.42 to 0.58, safely within(−1,1). Every fixed pair on the line is stable.
Its enclosing independent box has radii 0.08 and 0.08. That larger box permits combinations absent from the line and has p(1) minimum −0.06, so it fails a robust box test.
There is no contradiction. The box test correctly rejects the larger set, while the line proof certifies the actual correlated set. If correlations are independently established, retaining them can avoid unnecessary conservatism. If they are only hoped for, using them as a guarantee would be unsupported.
CHAPTER 17 OF 35 · Change the assumptions and coefficients
17. Distinguish a deterministic set from a probability model
An interval bound describes allowed coefficient values. It does not assign probabilities to them. A random coefficient model can support probabilities of different outcomes, but it introduces an additional distributional assumption.
Monte Carlo draws from a chosen distribution can estimate how often a criterion fails under that distribution. They do not prove every coefficient pair in its support is stable, especially when rare or boundary regions matter.
Conversely, a robust box failure does not say failure is common. It identifies at least one allowed pair with the undesired property. The scientific or engineering meaning depends on whether the uncertainty set actually represents the system and how the values arise.
A report should identify which question it answers: stability of the nominal pair, every fixed pair in a set, or a probability under a stated model. The same collection of coefficients can be analysed in all three ways, but their conclusions are not interchangeable.
CHAPTER 18 OF 35 · Change the assumptions and coefficients
18. Transfer check with a new nominal quadratic
Use p(z)=z²−0.8z+0.2. Its nominal roots are 0.4±0.2i, with magnitude the square root of 0.2.
Allow radii r₁=0.15 and r₂=0.10. Before reading the result, calculate the worst p(1), worst p(−1) and largest |a₂|. Decide whether every fixed pair in the box is stable.
Then scale both radii by a common factor t≥0 and find the limiting t for strict stability. Identify which boundary is reached first and give its coefficient pair.
The changed centre and radii require rebuilding the margins rather than reusing the earlier nominal value 0.1. Explain which algebraic condition controls the answer. A successful transfer predicts the limiting boundary before solving a collection of root equations or running a grid search.
CHAPTER 19 OF 35 · Change the assumptions and coefficients
19. Check the changed-box solution
The nominal p(1) margin is 0.4 and p(−1) margin is 2.0. Subtracting 0.15+0.10 gives minima 0.15 and 1.75. The largest |a₂| is 0.30. Thus the declared box is strictly stable for every fixed pair.
When the radii become 0.15t and 0.10t, p(1) remains positive exactly while 0.4−0.25t>0, or t<1.6. The a₂ and p(−1) constraints permit t<8, so they do not limit this first threshold.
At t=1.6, the critical corner is a₁=−1.04 and a₂=0.04. Its polynomial has roots 1 and 0.04. The unit root confirms that the threshold itself is excluded from the strictly stable closed box.
This example shows how a robustness radius can be derived exactly within the model. It should not be interpreted as a real manufacturing tolerance unless the coefficient uncertainty and the required property have been independently justified.
CHAPTER 20 OF 35 · Change the assumptions and coefficients
20. Distinguish pole location from transient size
Roots inside the unit circle guarantee decay for the fixed recurrence, but they do not specify every transient amplitude from every initial condition. Closely spaced roots, repeated roots and the chosen state coordinates can affect finite-time behaviour.
For a repeated root r with |r|<1, solutions can contain n r^n. That term eventually decays, but its early values need not decrease monotonically. A stable recurrence can therefore have a transient that exceeds a practical limit.
If a task requires a bound on output magnitude, settling time or sensitivity, derive that bound using the actual initial-state and input assumptions. Do not convert a positive Jury margin into an unproved performance guarantee. The present coefficient-space test establishes root location. Its usefulness increases when that specific conclusion is kept separate from other engineering requirements rather than being advertised as a certificate for every desirable behaviour.
CHAPTER 21 OF 35 · Change the assumptions and coefficients
21. Check the realized model rather than a reduced fraction alone
Our quadratic is the characteristic polynomial of the stated recurrence. If it instead came from an unreduced or reduced transfer-function description, the relationship to internal modes would need checking.
The existing hidden-mode cancellation guide shows why a stable selected external map can coexist with an unstable internal state. A stability test applied to a reduced denominator does not automatically assess a different, larger realization.
Likewise, the rounded-recurrence guide explains how quantization changes a linear update into a nonlinear one. The exact quadratic criterion here does not replace that arithmetic analysis. Identify the equation and property represented by the polynomial before importing the root test into an implementation claim.
CHAPTER 22 OF 35 · Distinguish frozen and switching systems
22. Introduce the switching question explicitly
So far every allowed coefficient pair was held fixed over time. Suppose instead that a₁ can change at each update while a₂ remains 0.5. A different mathematical problem results.
The frozen-coefficient polynomials can all be stable while products of their state-transition matrices grow. A root test on each individual matrix does not automatically control arbitrary switching products.
This distinction is important enough to demonstrate rather than leave as a generic warning. We will use two coefficient pairs well inside the same quadratic stability region and calculate their alternating two-step transition.
The purpose is not to claim that coefficients in a real device switch in this manner. It is to identify an assumption behind the robust fixed-parameter result. If a task includes time-varying coefficients, its required stability notion and admissible switching rule must be stated and analysed separately.
CHAPTER 23 OF 35 · Distinguish frozen and switching systems
23. Build two individually stable transition matrices
Represent the recurrence state as x[n]=(y[n],y[n−1])ᵀ. The next state is A(a₁)x[n], where A(a₁)=[[-a₁,−0.5],[1,0]].
Choose a₁=−1.4 for A₊=[[1.4,−0.5],[1,0]] and a₁=+1.4 for A₋=[[-1.4,−0.5],[1,0]]. Each fixed quadratic z²±1.4z+0.5 has conjugate roots of magnitude the square root of 0.5.
In fact every frozen pair with a₁ between −1.4 and 1.4 and a₂=0.5 passes the criterion: both endpoint margins are at least 0.1, and |a₂|=0.5.
Thus this entire fixed-parameter interval is stable in the previously proved sense. Any failure under an alternating update cannot be blamed on choosing an individually unstable member. It must arise from the time-varying composition of the allowed transitions, which is precisely the new assumption being tested.
CHAPTER 24 OF 35 · Distinguish frozen and switching systems
24. Calculate the alternating two-step product
Apply A₊ first and A₋ second. The two-step transition is A₋A₊=[[-2.46,0.7],[1.4,−0.5]]. Its trace is −2.96 and determinant 0.25.
Its eigenvalues solve ξ²+2.96ξ+0.25=0. One is approximately −2.872982412, with magnitude greater than one. An initial state with a component along that eigenvector grows under repeated two-step products.
This proves that individually stable frozen matrices do not guarantee decay under the specified alternating switching rule. The fixed-coefficient criterion was correct; the stronger time-varying claim was never among its premises.
The calculation is an original exact matrix-product example using decimal rational coefficients. It should be reproduced from the two stated matrices and their order. Reversing multiplication represents a different time sequence, and treating a sequence of matrices as though it were one constant average matrix can hide the behaviour the example was constructed to reveal.
CHAPTER 25 OF 35 · Distinguish frozen and switching systems
25. Keep the practical conclusion within its assumptions
If coefficients are uncertain but remain fixed during a trajectory, the box or correlated-set tests above address that model. If they vary with time, additional tools or restrictions are needed to control products of transitions.
A common contraction bound in one norm, a suitable common Lyapunov argument or a restricted switching analysis can provide sufficient results in appropriate settings. This guide does not claim to construct those for every uncertain family.
The useful next step is to identify which physical or computational mechanism changes the coefficients and on what timescale. A fitted interval alone does not answer that question. Stable nominal roots, stable frozen alternatives and stable arbitrary switching are three different statements. The original safe box and alternating counterexample make their logical separation concrete, allowing a reader to choose the correct question before reaching for a more advanced theorem.
CHAPTER 26 OF 35 · Distinguish frozen and switching systems
26. Read the source criterion with the time convention intact
Kamran Iqbal’s Stability of Sampled-Data Systems presents the unit-circle setting and second-order Jury conditions. The coefficient boxes, correlated line and switching product in this article are independently constructed teaching calculations.
The existing Routh–Hurwitz guide develops continuous-time polynomial stability and explicitly distinguishes the discrete-time criterion. Do not replace our unit-circle test with a left-half-plane sign rule without a justified transformation of the model.
When comparing sources, check whether their polynomial is monic and whether coefficient indices run from highest or lowest degree. Translate the actual equation before copying a condition. A familiar theorem name does not guarantee that the same symbols or feedback signs are being used.
CHAPTER 27 OF 35 · Distinguish frozen and switching systems
27. Common misconceptions to repair
A stable centre proves a stable uncertainty box: the original failing corner disproves it. A small decimal coefficient change must be harmless: the relevant margin can be smaller than the combined change.
A full box is always the correct uncertainty set: the correlated-line example shows why that can add impossible combinations. Stable roots guarantee acceptable transient performance: root location alone supplies no task-specific amplitude limit.
Every allowed frozen matrix is stable, so arbitrary switching is stable: the alternating two-step eigenvalue disproves that conclusion. A grid with many passing points is a complete proof: it needs a coverage argument.
The repair is to write the recurrence, property, uncertainty set and time behaviour of the coefficients before calculating. Then state exactly which inequality or transition product supports the conclusion. This sequence preserves the benefit of a compact stability test without allowing it to answer a different, stronger question by implication.
CHAPTER 28 OF 35 · Distinguish frozen and switching systems
28. Questions readers often ask
Are the three inequalities necessary and sufficient?
Yes, for the stated real monic second-order constant-coefficient polynomial and strict unit-circle root location. Other model classes need their own conditions.
Must I calculate roots at every box point?
No. The affine margin extrema give an exact box test here. A few independent root calculations are useful checks and failing witnesses.
Does equality count as stable decay?
Not for the asymptotic-decay property used here. A unit root is on the excluded boundary; distinguish it from strict interior stability.
Does the box assign a probability of failure?
No. It supplies a set of allowed values. A probability calculation requires a separate distribution over them.
Does the result cover changing coefficients?
Only if a separate argument establishes the required time-varying property. The frozen-parameter proof alone does not, as the switching example shows.
CHAPTER 29 OF 35 · Reproduce and connect conventions
29. Reproduce and communicate the result
First derive the quadratic criterion from its real and conjugate-root cases. Calculate the original nominal margins, then verify the failing corner of the larger box and the positive margins of the smaller one.
Next work the changed centre and scaled-radius exercise. Check the root at the limiting boundary independently. Finally multiply the two switching matrices and compare their individual eigenvalues with the two-step product’s eigenvalues.
A clear report states the polynomial convention, allowed coefficient set, whether coefficients remain fixed, and the property certified. It then gives the worst margins or an explicit failing witness. This is more reproducible than a statement that a root plot looked safe or that a dense numerical grid found no problem under an unstated uncertainty model.
CHAPTER 30 OF 35 · Reproduce and connect conventions
30. Map the unit disk to a continuous-time half-plane
The substitution z=(1+s)/(1−s) maps the open left half of the s-plane into the open unit disk. Indeed, |1+s|<|1−s| exactly when the real part of s is negative.
Multiplying p((1+s)/(1−s)) by (1−s)² gives the transformed quadratic (1−a₁+a₂)s²+2(1−a₂)s+(1+a₁+a₂). Its three coefficients display the same margins in another form.
For a real second-order continuous-time polynomial with positive leading coefficient, positive remaining coefficients give roots in the open left half-plane. Here those conditions require p(−1)>0, a₂<1 and p(1)>0. Adding the two endpoint inequalities also gives a₂>−1, recovering the unit-circle criterion.
This is a consistency route between two conventions, not permission to apply a continuous-time test directly to the original z polynomial. The variable transformation is the step that changes the stability boundary. Keep it explicit, including the excluded unit-circle endpoint cases, when connecting the discrete and continuous interpretations.
CHAPTER 31 OF 35 · Reproduce and connect conventions
31. Check a box whose second coefficient is negative
Take a new centre c₁=0.2, c₂=−0.3 and radii r₁=0.1, r₂=0.2. The maximum |a₂| is |−0.3|+0.2=0.5, not c₂+r₂=−0.1.
The endpoint minima are 1+0.2−0.3−0.1−0.2=0.6 and 1−0.2−0.3−0.1−0.2=0.2. Both remain positive, so this box is strictly stable for every fixed pair.
The p(−1) minimum occurs at a₁=0.3 and a₂=−0.5. Checking that corner gives p(−1)=0.2 directly. This example tests the absolute-value part of the general formula, which the original positive a₂ centre could not expose as clearly.
A sign mistake can hide behind a correct-looking interval endpoint if a program simply takes the upper endpoint as the largest magnitude. Compare both interval endpoints or use |c₂|+r₂ for the symmetric box. The quantity being maximised is magnitude, not signed coefficient value.
CHAPTER 32 OF 35 · Reproduce and connect conventions
32. Translate a coefficient-rounding allowance into a box
Suppose an independently specified implementation rounds a₁ to a grid of spacing 0.1 and a₂ to spacing 0.02, using nearest rounding. Each coefficient error has magnitude at most half its grid spacing, so a box with radii 0.05 and 0.01 contains the possible errors.
For the original centre (−1.4,0.5), the worst p(1) margin is 0.1−0.06=0.04, while the other conditions also remain positive. Thus this bounding-box model certifies the frozen rounded-coefficient recurrence.
The calculation addresses coefficient errors, not rounding of the evolving state. The latter is nonlinear and belongs to the separate rounded-recurrence analysis. If a larger error box fails, that failure also need not prove a particular deterministic rounding result is unstable: the box may include coefficient combinations that the stated conversion rule never produces.
A bound is useful only when its role is clear. Record the actual coefficient values and rounding convention when known, use a justified over-approximation when needed, and distinguish sufficient box guarantees from claims about a specific converted pair.
CHAPTER 33 OF 35 · Reproduce and connect conventions
33. Compare short trajectories with the root conclusions
Set y[−1]=0 and y[0]=1. For the nominal a₁=−1.4, a₂=0.5 recurrence, the first later values are 1.4,1.46 and 1.344. The initial rise does not contradict eventual decay.
For the failing corner a₁=−1.48, a₂=0.46, the corresponding values are 1.48,1.7304 and 1.880192. Its larger real root eventually dominates.
When the two real roots r₊ and r₋ are distinct, this startup sequence is y[n]=(r₊^(n+1)−r₋^(n+1))/(r₊−r₋). Substitution checks y[0]=1 and y[−1]=0, and the root equations verify the recurrence. For the unstable corner, the nonzero coefficient of r₊ confirms growth beyond merely observing three rising samples.
A short plot can therefore illustrate a model but cannot by itself prove asymptotic stability or instability. Use the trajectory to check timing and signs, then use the root or margin argument for the long-run conclusion. Preserve the stated initial conditions so the comparison remains reproducible.
CHAPTER 34 OF 35 · Reproduce and connect conventions
34. Normalise a nonmonic polynomial before using the test
If the supplied polynomial is b₀z²+b₁z+b₂ with b₀≠0, divide by b₀ first. The monic coefficients are a₁=b₁/b₀ and a₂=b₂/b₀.
For example, 2z²−2.8z+1 has exactly the same roots as the original nominal z²−1.4z+0.5. Testing the unnormalised constant coefficient 1 against |a₂|<1 would incorrectly place it on a boundary.
If b₀ is uncertain, dividing by it can create a correlated, nonrectangular uncertainty set in the monic coefficients. Replacing that set by a box requires its own inclusion argument. If b₀ can reach zero, the polynomial order can change, so the stated quadratic theorem no longer covers the whole original parameter set without separate cases.
The leading coefficient is therefore part of the model contract. Keep the original coefficients, their units and the normalisation step in the record, then apply the criterion to the quantities it actually defines.
CHAPTER 35 OF 35 · Reproduce and connect conventions
35. Continue through the relevant learning routes
Return to the Z-Transform subject guide for sequence and realization foundations. Use the hidden-mode guide and rounded-recurrence guide for different implementation limits.
For a different boundary on transform reasoning, use the bilateral convolution and convergence guide. It checks whether the infinite operation exists before identifying a rational product with an output sequence.
Read The Gold Standard of Education: Learning Transfers into Life for the wider learning practice of original examples, changed assumptions and explicit transfer checks. The How X Works hub connects those habits to other disciplines.
The final claim should be as precise as the proof: every fixed real coefficient pair in this declared set has roots strictly inside the unit circle, or this allowed pair provides a counterexample. Further performance and switching claims need further evidence.