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1 \pageheading{RESOLUTION ATLAS\; /\; WEAK TAPS}{A small change can alter the global order}
2 A feedback channel may be almost invisible on a fixed disk and still determine
3 the eventual growth of the function. Consider the normalized entire solution
4 \[
5 F_\varepsilon'(z)=1+(1-\varepsilon)F_\varepsilon(z/4)^2
6  +\varepsilon F_\varepsilon(3z/4)^2,
7 \qquad F_\varepsilon(0)=0.
8 \]
9 Here $0\le\varepsilon\le1$. At $\varepsilon=0$, its entire order is $1/2$.
10 At every positive weight, the order becomes
11 $\rho=\log2/\log(4/3)\approx2.40942084$.
12 
13 \vfig{resolution_01_weak_tap}{Left: divide the compact difference by the tap
14 weight to compare its shape. Right: the proved order jumps at zero. The dashed
15 curve is an analytic envelope. The function curves use 100-digit arithmetic
16 and have a normalized omitted-series bound below $5.26\times10^{-23}$ on
17 $0\le R\le1$. Floating rounding is separate from that tail bound.}
18 
19 \claim{The unit disk comes with an exact error certificate}
20 Let $f_{3/4}=F_1$, with $f_{3/4}(z)=\sum_{n\ge0}d_nz^{2n+1}$.
21 Positive coefficients give
22 \[
23 \sup_{|z|\le R}|F_\varepsilon(z)-F_0(z)|
24 \le\varepsilon\sum_{n\ge1}nd_nR^{2n+1}.
25 \]
26 A rational coefficient sum and a positive-tail estimate prove that, on the
27 unit disk, the multiplier is less than $0.241890267$.
28 For $\varepsilon=10^{-12}$, the entire disk therefore changes by less than
29 $2.41890267\times10^{-13}$.
30 
31 The observation window and the global invariant answer different questions.
32 Exact measurements can distinguish the functions. With an absolute error
33 allowance at least half the displayed bound, midpoint data can fit both.
34 The order concerns the limit through arbitrarily large radii, where the
35 active faster channel eventually matters.
36 
37 \pageheading{RESOLUTION ATLAS\; /\; OBSERVATION RADIUS}{How far is far enough to guarantee separation?}
38 The growth theorem supplies positive lower type for the weak-tap function
39 and finite upper type for the slow function. For $0<\varepsilon\le1$ and $R\ge e$,
40 \[
41 \log F_\varepsilon(R)\ge c_\varepsilon R^\rho,
42 \qquad \log F_0(R)\le C_0\sqrt R,
43 \]
44 \[
45 c_\varepsilon=\frac{\log2}{(4/\varepsilon+8/3)^\rho},
46 \qquad C_0=e^{-1/2}(1+22\log2).
47 \]
48 
49 \vfig{resolution_02_growth_resolution}{The left curve is a sufficient
50 radius obtained from conservative comparison constants. The right panel
51 shows why the lower bound for the perturbed function eventually exceeds the
52 upper bound for the slow one. These are bounds on logarithmic growth, rather
53 than numerical evaluations at the large plotted radii.}
54 
55 \claim{An explicit sufficient radius}
56 Put
57 \[
58 R_2(\varepsilon)=\max\!\left\{
59  e,\left(\frac{2C_0}{c_\varepsilon}\right)^{1/(\rho-1/2)},
60  \left(\frac{2\log2}{c_\varepsilon}\right)^{1/\rho}\right\}.
61 \]
62 Then $F_\varepsilon(R)\ge2F_0(R)$ for every $R\ge R_2(\varepsilon)$.
63 The formula makes $c_\varepsilon R^\rho$ at least both
64 $2C_0\sqrt R$ and $2\log2$, and therefore at least their half-sum.
65 Exponentiating gives the factor of two.
66 
67 For $\varepsilon=10^{-6}$, the formula is approximately
68 $1.237\times10^9$. This large radius is a guaranteed sufficient choice.
69 It is not the first radius at which a useful difference appears, and it does
70 not establish failure of every observation at smaller radii.
71 The exact formula carries the guarantee. Rounded figures illustrate its scale.
72 
73 \pageheading{RESOLUTION ATLAS\; /\; ARITHMETIC PRECISION}{A large excluded prime hides in a fixed window}
74 Use the squarefree arithmetic family
75 \[
76 A_Q(s)=\sum_{\substack{n\ge2\\\mu(n)^2=1\\(n,Q)=1}}
77  n^{-4}(1-n^{-2})^{2s},
78 \qquad D_p(s)=A_1(s)-A_p(s),
79 \]
80 where $p$ is prime. The difference contains exactly the squarefree atoms
81 divisible by $p$. Its largest frequency is
82 $\lambda_p=-2\log(1-p^{-2})$, and its total mass is
83 $D(4)/(p^4+1)$, with $D(4)=105/\pi^4$.
84 
85 \vfig{resolution_03_prime_precision}{Left: analytic bounds for the
86 function and three derivatives on $|s|\le5$. The connecting curves show scale,
87 while the arithmetic statement applies at primes. Right: rational bounds
88 for three particular excluded primes. The bar height is the negative
89 base-ten logarithm of the bound, not a count of recovered digits.}
90 
91 \claim{The full complex disk has a rational certificate}
92 For integer $k\ge0$,
93 \[
94 \sup_{|s|\le5}|D_p^{(k)}(s)|
95 <\frac{4(1-p^{-2})^{-10}}{3(p^4+1)}
96  \left(\frac{2}{p^2-1}\right)^k.
97 \]
98 The inequality uses $D(4)<4/3$ and
99 $\lambda_p\le2/(p^2-1)$. Every quantity on the right is rational.
100 For $k=0$, upward decimal bounds are $1.283\times10^{-8}$ at $p=101$,
101 $1.287\times10^{-12}$ at $p=1009$, and $1.330\times10^{-16}$ at $p=10007$.
102 
103 The excluded prime changes the exact inverse of the function.
104 Nevertheless, its effect becomes arbitrarily small on every fixed disk,
105 including in every fixed derivative order. A bounded error model therefore
106 needs an arithmetic parameter bound if it is to promise uniform recovery.
107 
108 \pageheading{RESOLUTION ATLAS\; /\; NEGATIVE DEPTH}{Seeing the prime can make subtraction harder}
109 At negative real depth, the difference grows. The prime atom alone supplies
110 a lower bound, while the entire removed shelf supplies an upper bound:
111 \[
112 p^{-4}e^{\lambda_pT}\le D_p(-T)
113 \le\frac{D(4)}{p^4+1}e^{\lambda_pT},\qquad T\ge0.
114 \]
115 Choose a tolerance $\eta>D(4)/(p^4+1)$. Its unique crossing occurs between
116 \[
117 T_- =\frac{\log\!\bigl(\eta(p^4+1)/D(4)\bigr)}{\lambda_p},
118 \qquad
119 T_+ =\frac{\log(\eta p^4)}{\lambda_p}.
120 \]
121 
122 \vfig{resolution_04_prime_depth}{At tolerance $10^{-6}$, the left panel
123 shows the analytic crossing brackets after rescaling each row by $T_+$.
124 The right panel shows a lower bound on the decimal cancellation scale when
125 subtracting full functions. Displayed endpoints and labels are ordinary
126 floating evaluations of exact formulas.}
127 
128 For $p=101$, the difference is no larger than $10^{-6}$ through
129 $T_-=23307.789\ldots$, and at least that large by
130 $T_+=23690.516\ldots$. The decimals illustrate the formulas and are not
131 rounded interval certificates.
132 
133 \claim{The direct positive difference and subtraction have different costs}
134 The $n=2$ atom in the full function proves
135 \[
136 \frac{A_1(-T)}{D_p(-T)}
137 \ge\frac{p^4+1}{16D(4)}e^{(\lambda_2-\lambda_p)T}.
138 \]
139 At $T_+$ for $p=101$, the base-ten logarithm of this lower bound is about
140 5924. Subtracting independently approximated full functions can therefore
141 amplify relative errors enormously. Evaluating the known positive difference
142 series is a different procedure. These conditioning figures impose no
143 universal precision requirement on every possible recovery algorithm.
144 
145 \pageheading{RESOLUTION ATLAS\; /\; NEGATIVE ATOMS}{A degree-seven witness exposes a signed defect}
146 Let $x_n=(1-n^{-2})^2$, and change the prime-two sieve strength to
147 $\lambda_2=-1-\delta$. For a real polynomial $f$, the associated quadratic form is
148 \[
149 Q_\delta(f)=\sum_{\substack{n\ge1\\\mu(n)^2=1\\n\text{ odd}}}
150  \frac{f(x_n)^2}{n^4}
151 -\delta\sum_{\substack{n\ge1\\\mu(n)^2=1\\n\text{ even}}}
152  \frac{f(x_n)^2}{n^4}.
153 \]
154 Every even squarefree atom has negative mass. At $\delta=10^{-12}$,
155 the polynomial below certifies that this is detectable at finite degree.
156 
157 \vfig{resolution_05_signed_detector}{The detector vanishes on the first
158 four positive shelves and equals one at the negative atom $n=2$.
159 The positive bar is an upper bound for the entire remaining positive tail.
160 The total upper bound is negative, so no omitted atom can reverse the
161 conclusion.}
162 
163 \claim{The certificate uses rational arithmetic throughout}
164 Take $S=\{1,3,5,7\}$ and define
165 \[
166 P(x)=\prod_{n\in S}\frac{x-x_n}{x_2-x_n}
167  \left(\frac{1-x}{1-x_2}\right)^3.
168 \]
169 Its degree is seven. All remaining positive nodes begin at $n=11$.
170 With
171 $C=756482048/20777538825$ and $r=3856/102487$, a complete tail bound gives
172 \[
173 Q_{10^{-12}}(P)\le-\frac{10^{-12}}{16}
174  +\frac{C^2r^6}{3\cdot7^3}
175  <-5.8845708899\times10^{-14}.
176 \]
177 The inequality is certified as a comparison of exact rational numbers.
178 The first four zeros remove the dominant positive contributions.
179 The factor $(1-x)^3$ suppresses the remaining cluster near one.
180 The construction exhibits a detecting degree, without asserting that degree
181 seven is the smallest possible choice.
182 
183 \pageheading{RESOLUTION ATLAS\; /\; ALL LOW DEGREES}{Passing every coarse test can still miss the defect}
184 The previous witness would be less informative without a converse guarantee
185 at lower resolution. For the same defect $\delta=10^{-12}$, every nonzero
186 polynomial of degree at most four has strictly positive quadratic form. This quantifies
187 the information that the coarse test actually misses.
188 
189 \vfig{resolution_06_degree_gap}{The blue boundary certifies positivity
190 for every polynomial through that degree. The orange boundary gives an
191 explicit detecting polynomial. The shaded gap remains unresolved by these
192 certificates. At $\delta=10^{-12}$, degrees through four pass, degree seven
193 fails, and degrees five and six are undecided here.}
194 
195 \claim{Interpolation controls the whole polynomial space}
196 Choose the odd squarefree labels $n_i=1,3,5,7,11$, write $y_i=x_{n_i}$,
197 and put
198 \[
199 b_i=\prod_{j\ne i}\frac{\max(y_j,1-y_j)}{|y_i-y_j|},
200 \qquad K_4=\sum_{i=0}^{4}n_i^4b_i^2.
201 \]
202 Lagrange interpolation and weighted Cauchy--Schwarz show that, for every
203 polynomial $f$ of degree at most four,
204 \[
205 Q_\delta(f)\ge
206 \left(1-\frac{\delta K_4}{12}\right)
207 \sum_{i=0}^{4}n_i^{-4}f(y_i)^2.
208 \]
209 Exact rational comparison gives $12/K_4>7.9790\times10^{-11}>10^{-12}$.
210 A nonzero polynomial cannot vanish at all five selected nodes, so the right
211 side is strictly positive. Thus the order-five Hankel matrix is positive
212 definite, while the degree-seven witness makes the order-eight matrix fail
213 positive semidefiniteness.
214 
215 The complete atlas contains the general interpolation and detector bounds.
216 Its reproducible check records 985 exact assertions and 16 floating algebra
217 checks. The proofs establish the universal statements. The tests verify the
218 stated finite certificates and implementations.

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