A small change can alter the global order
A feedback channel may be almost invisible on a fixed disk and still determine the eventual growth of the function. Consider the normalized entire solution
Here \(0\le\varepsilon\le1\). At \(\varepsilon=0\), its entire order is \(1/2\). At every positive weight, the order becomes \(\rho=\log2/\log(4/3)\approx2.40942084\).
The unit disk comes with an exact error certificate
Let \(f_{3/4}=F_1\), with \(f_{3/4}(z)=\sum_{n\ge0}d_nz^{2n+1}\). Positive coefficients give
A rational coefficient sum and a positive-tail estimate prove that, on the unit disk, the multiplier is less than \(0.241890267\). For \(\varepsilon=10^{-12}\), the entire disk therefore changes by less than \(2.41890267\times10^{-13}\).
The observation window and the global invariant answer different questions. Exact measurements can distinguish the functions. With an absolute error allowance at least half the displayed bound, midpoint data can fit both. The order concerns the limit through arbitrarily large radii, where the active faster channel eventually matters.
How far is far enough to guarantee separation?
The growth theorem supplies positive lower type for the weak-tap function and finite upper type for the slow function. For \(0<\varepsilon\le1\) and \(R\ge e\),
An explicit sufficient radius
Put
Then \(F_\varepsilon(R)\ge2F_0(R)\) for every \(R\ge R_2(\varepsilon)\). The formula makes \(c_\varepsilon R^\rho\) at least both \(2C_0\sqrt R\) and \(2\log2\), and therefore at least their half-sum. Exponentiating gives the factor of two.
For \(\varepsilon=10^{-6}\), the formula is approximately \(1.237\times10^9\). This large radius is a guaranteed sufficient choice. It is not the first radius at which a useful difference appears, and it does not establish failure of every observation at smaller radii. The exact formula carries the guarantee. Rounded figures illustrate its scale.
A large excluded prime hides in a fixed window
Use the squarefree arithmetic family
where \(p\) is prime. The difference contains exactly the squarefree atoms divisible by \(p\). Its largest frequency is \(\lambda_p=-2\log(1-p^{-2})\), and its total mass is \(D(4)/(p^4+1)\), with \(D(4)=105/\pi^4\).
The full complex disk has a rational certificate
For integer \(k\ge0\),
The inequality uses \(D(4)<4/3\) and \(\lambda_p\le2/(p^2-1)\). Every quantity on the right is rational. For \(k=0\), upward decimal bounds are \(1.283\times10^{-8}\) at \(p=101\), \(1.287\times10^{-12}\) at \(p=1009\), and \(1.330\times10^{-16}\) at \(p=10007\).
The excluded prime changes the exact inverse of the function. Nevertheless, its effect becomes arbitrarily small on every fixed disk, including in every fixed derivative order. A bounded error model therefore needs an arithmetic parameter bound if it is to promise uniform recovery.
Seeing the prime can make subtraction harder
At negative real depth, the difference grows. The prime atom alone supplies a lower bound, while the entire removed shelf supplies an upper bound:
Choose a tolerance \(\eta>D(4)/(p^4+1)\). Its unique crossing occurs between
For \(p=101\), the difference is no larger than \(10^{-6}\) through \(T_-=23307.789\ldots\), and at least that large by \(T_+=23690.516\ldots\). The decimals illustrate the formulas and are not rounded interval certificates.
The direct positive difference and subtraction have different costs
The \(n=2\) atom in the full function proves
At \(T_+\) for \(p=101\), the base-ten logarithm of this lower bound is about 5924. Subtracting independently approximated full functions can therefore amplify relative errors enormously. Evaluating the known positive difference series is a different procedure. These conditioning figures impose no universal precision requirement on every possible recovery algorithm.
A degree-seven witness exposes a signed defect
Let \(x_n=(1-n^{-2})^2\), and change the prime-two sieve strength to \(\lambda_2=-1-\delta\). For a real polynomial \(f\), the associated quadratic form is
Every even squarefree atom has negative mass. At \(\delta=10^{-12}\), the polynomial below certifies that this is detectable at finite degree.
The certificate uses rational arithmetic throughout
Take \(S=\{1,3,5,7\}\) and define
Its degree is seven. All remaining positive nodes begin at \(n=11\). With \(C=756482048/20777538825\) and \(r=3856/102487\), a complete tail bound gives
The inequality is certified as a comparison of exact rational numbers. The first four zeros remove the dominant positive contributions. The factor \((1-x)^3\) suppresses the remaining cluster near one. The construction exhibits a detecting degree, without asserting that degree seven is the smallest possible choice.
Passing every coarse test can still miss the defect
The previous witness would be less informative without a converse guarantee at lower resolution. For the same defect \(\delta=10^{-12}\), every nonzero polynomial of degree at most four has strictly positive quadratic form. This quantifies the information that the coarse test actually misses.
Interpolation controls the whole polynomial space
Choose the odd squarefree labels \(n_i=1,3,5,7,11\), write \(y_i=x_{n_i}\), and put
Lagrange interpolation and weighted Cauchy–Schwarz show that, for every polynomial \(f\) of degree at most four,
Exact rational comparison gives \(12/K_4>7.9790\times10^{-11}>10^{-12}\). A nonzero polynomial cannot vanish at all five selected nodes, so the right side is strictly positive. Thus the order-five Hankel matrix is positive definite, while the degree-seven witness makes the order-eight matrix fail positive semidefiniteness.
The complete atlas contains the general interpolation and detector bounds. Its reproducible check records 985 exact assertions and 16 floating algebra checks. The proofs establish the universal statements. The tests verify the stated finite certificates and implementations.