Special Functions

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1 % Family 5 visual chapter. Included by the monograph and the visual guide.
2 % Exact proof sources: research_v11/coupled_growth_proof.md,
3 % coupled_identifiability.md, and coupled_structure.md.
4 
5 \pageheading{FAMILY 5}{Two functions exchange growth}
6 For four real contractions $0<q,r,a,b<1$, define the normalized pair
7 \[
8 \boxed{\begin{aligned}
9 U'(z)&=1+U(qz)V(rz),\\
10 V'(z)&=1+U(az)V(bz),\qquad U(0)=V(0)=0.
11 \end{aligned}}
12 \]
13 Each equation combines one value of $U$ with one value of $V$. The four
14 scales specify where that exchange occurs. Both functions begin with the
15 same unit slope, while later coefficients retain the two equations' different
16 scale addresses.
17 
18 \vfig[0.96]{coupled_01_matrix}{The worked pair
19 $(q,r,a,b)=(1/5,2/5,3/5,7/10)$ has exact order one. The two displayed
20 identities give the positive growth vector $(1/2,1)^T$.}
21 
22 There is a unique entire solution pair. Both functions are odd, and every
23 odd Taylor coefficient is positive. The common order is governed by the
24 positive matrix
25 \[
26 P(s)=\begin{pmatrix}q^s&r^s\\a^s&b^s\end{pmatrix},
27 \qquad \operatorname{spr}P(\rho)=1.
28 \]
29 The eigenvector supplies the relative weights that a row sum misses. In
30 the example, the row sums at $s=1$ are $3/5$ and $13/10$. The coupled
31 system nevertheless has order exactly one in both components.
32 
33 \pageheading{FAMILY 5 / CONSTRUCTION}{The coefficients preserve both addresses}
34 Write $U(z)=\sum_{n\ge0}u_nz^{2n+1}$ and
35 $V(z)=\sum_{n\ge0}v_nz^{2n+1}$, with $u_0=v_0=1$.
36 The first equation gives the exact recursion
37 \[
38 u_n=\frac1{2n+1}\sum_{i+j=n-1}
39 q^{2i+1}r^{2j+1}u_i v_j,\qquad n\ge1.
40 \]
41 The second replaces $(q,r)$ by $(a,b)$ while retaining the same products
42 $u_i v_j$. Thus each new coefficient uses the shared history of the pair.
43 
44 \vfig[0.97]{coupled_02_functions}{For
45 $(q,r,a,b)=(1/3,2/3,1/2,3/4)$, the components separate despite their
46 common initial slope. These numerical curves use odd indices through 240.
47 The two truncations at indices 120 and 240 agree at $x=8$ to displayed
48 double precision.}
49 
50 The first distinction already appears at degree three:
51 \[
52 U(z)=z+\frac{qr}{3}z^3+O(z^5),\qquad
53 V(z)=z+\frac{ab}{3}z^3+O(z^5).
54 \]
55 Induction gives $0<u_n,v_n\le2^{-n}$, which supplies a convergent germ.
56 If both functions are known on a disk of radius $R$, their integral
57 equations extend them to radius $R/Q$, where $Q=\max(q,r,a,b)<1$.
58 Repeating this step reaches every finite disk.
59 
60 The numerical agreement above checks the plotted approximation. The later
61 certificate page supplies an exact bound on the omitted tail.
62 
63 \pageheading{FAMILY 5 / GROWTH}{One order, two exact type proportions}
64 The entire functions have the same order $\rho$, with positive lower type
65 and finite upper type. Positivity places the maximum modulus on the positive
66 real axis, so $M_U(R)=U(R)$ and $M_V(R)=V(R)$.
67 
68 \vfig[0.96]{coupled_03_growth}{The order-one example has different component
69 amplitudes. The right panel shows finite-radius slopes near one. The curves
70 use odd indices through 800. Agreement with index 400 is a numerical
71 diagnostic, and the order itself follows from the analytic proof.}
72 
73 For $F_1=U$ and $F_2=V$, define
74 \[
75 \underline\tau_i=\liminf_{R\to\infty}\frac{\log M_{F_i}(R)}{R^\rho},
76 \qquad
77 \overline\tau_i=\limsup_{R\to\infty}\frac{\log M_{F_i}(R)}{R^\rho}.
78 \]
79 If $P(\rho)w=w$ with $w>0$, then both type vectors have precisely these
80 Perron proportions:
81 \[
82 \boxed{\underline{\boldsymbol\tau}=\underline\tau\,w,
83 \qquad\overline{\boldsymbol\tau}=\overline\tau\,w,
84 \qquad0<\underline\tau\le\overline\tau<\infty.}
85 \]
86 For the worked pair, both the lower type and the upper type of $V$ are
87 twice the corresponding type of $U$. Either both ordinary type limits
88 exist or neither does. Equality of the lower and upper types remains open.
89 The finite-radius slope picture does not decide that question.
90 
91 \pageheading{FAMILY 5 / PARAMETER SPACE}{Growth follows a spectral surface}
92 The order is the unique positive solution of
93 \[
94 \boxed{(1-q^\rho)(1-b^\rho)=(ar)^\rho.}
95 \]
96 The left side increases from zero to one as $\rho$ increases. The right
97 side decreases from one to zero. Their unique crossing determines the
98 growth exponent without fitting a truncated Taylor curve.
99 
100 \vfig[0.80]{coupled_04_landscape}{Numerical solutions of the proved
101 characteristic equation, with $r=2/5$ and $a=3/5$. The highlighted pair
102 $(q,b)=(1/5,7/10)$ has order exactly one by rational arithmetic.}
103 
104 The order depends on the self scales $q,b$ and on the product $ar$ of
105 the cross scales. Holding $q,b,ar$ fixed preserves the order, while
106 changing the cross scales separately generally changes the functions and
107 their type proportions:
108 \[
109 \frac{w_1}{w_2}=\frac{r^\rho}{1-q^\rho}
110 =\frac{1-b^\rho}{a^\rho}.
111 \]
112 Consequently, equal order does not determine the family member. It fixes
113 one growth exponent within a larger parameter space. The coefficients
114 carry enough additional information to recover all four ordered scales
115 away from the reduction surfaces described below.
116 
117 \pageheading{FAMILY 5 / PRECISION}{A coefficient list with a certified tail}
118 A finite Taylor list becomes an exact approximation certificate when it
119 is paired with a bound on a larger disk. Let $F$ be either component,
120 and let $F_N$ retain the odd indices $0,\ldots,N$.
121 
122 \vfig[0.95]{coupled_05_certification}{Each curve is computed from an exact
123 rational upper bound for the same worked pair. Increasing the coefficient
124 count certifies a smaller uniform error on the stated complex disk.
125 The logarithmic plotting coordinates are rounded.}
126 
127 Put $Q=\max(q,r,a,b)$. One elementary majorant starts with
128 $B(x)=x/(1-x^2)$ for $0\le x<1$ and works outward through the finite
129 contraction ladder using
130 \[
131 B(x)=x\bigl(1+B(Qx)^2\bigr).
132 \]
133 The integral equations show that this bounds both components on the disk
134 $|z|\le x$. For $S>R\ge0$, positivity then gives
135 \[
136 \boxed{\max_{|z|\le R}|F(z)-F_N(z)|
137 \le B(S)\left(\frac RS\right)^{2N+3}.}
138 \]
139 The exponent is the first omitted odd degree. Every later term has an
140 even smaller ratio between its values at radii $R$ and $S$.
141 For the worked pair, $N=30$, $R=1$ and $S=2$ give an error below
142 $1.600\times10^{-13}$ for both functions. The laboratory returns
143 rational interval endpoints, so the certificate survives decimal rounding.
144 
145 \pageheading{FAMILY 5 / SEQUENCES}{The phase calculus transports the pair}
146 Use the derivative normalization
147 $A_n=(2n+1)!u_n$, $B_n=(2n+1)!v_n$. Its recurrence has integer
148 binomial weights:
149 \[
150 A_n=\sum_{i+j=n-1}\binom{2n}{2i+1}
151 q^{2i+1}r^{2j+1}A_iB_j,\qquad n\ge1.
152 \]
153 The second component again replaces $(q,r)$ by $(a,b)$.
154 
155 \vfig[0.91]{coupled_06_polyphase}{The full derivative array of $V$ is split
156 into three residue lanes. Applying taps $(1,-2,1)$ with the correct block
157 carries, then interleaving, agrees exactly with direct convolution.}
158 
159 These derivative arrays are polynomials with nonnegative integer
160 coefficients in the four scale variables. Their total degree is at most
161 $n(n+1)$. If every parameter denominator divides $D_0$, then
162 \[
163 D_0^{n(n+1)}A_n,\qquad D_0^{n(n+1)}B_n
164 \quad\hbox{are integers}.
165 \]
166 This uniform clearing factor makes exact rational families available to
167 integer sequence experiments. It need not be the smallest possible factor.
168 
169 Ordinary taps on a derivative array correspond to repeated integration
170 shifts of its exponential generating function. Multiplication of two
171 exponential generating functions instead requires binomial convolution.
172 The catalog preserves that distinction and marks generic integer and
173 signed scale examples as formal series outside the proved analytic domain.
174 
175 \pageheading{FAMILY 5 / REDUCTION}{Two surfaces where one component rescales to the other}
176 Set $p=qr$, $k=ab$, and use the ordinary Taylor coefficients $u_2,v_2$
177 at degree five. The exact separation quantity factors as
178 \[
179 \boxed{D=k^2u_2-p^2v_2
180 =\frac{pk ar}{15}(q^2-b^2)(qb-ar).}
181 \]
182 
183 \vfig[0.88]{coupled_07_reduction_geometry}{The two reduction surfaces are
184 $q=b$ and $qb=ar$. Their union is exactly the zero set of $D$ in the
185 positive contraction domain. Outside this union, the components cannot
186 be obtained from each other by amplitude and argument rescaling.}
187 
188 There exist nonzero complex constants $A,B$ with $V(z)=A\,U(Bz)$
189 if and only if $D=0$. On either surface, put $s=\sqrt{k/p}>0$.
190 The scalar family is defined by
191 $T_{x,y}'(z)=1+T_{x,y}(xz)T_{x,y}(yz)$, $T_{x,y}(0)=0$.
192 It gives the explicit reduction
193 \[
194 U(z)=\sqrt{s}\,T_{q,sr}(z/\sqrt{s}),\qquad
195 V(z)=s^{-1/2}T_{q,sr}(\sqrt{s}\,z).
196 \]
197 Both effective scales lie in $(0,1)$. Equality $U=V$ is more restrictive:
198 it occurs exactly when $\{q,r\}=\{a,b\}$.
199 
200 This classifies relations between the normalized components. Separately,
201 no invertible constant linear coordinate change decouples the defining
202 operator into two autonomous scalar equations on every analytic input.
203 That operator statement does not exclude nonlinear changes of variables
204 or other special-function identities.
205 
206 \pageheading{FAMILY 5 / RECOVERY}{Degree seven identifies the four scales}
207 Away from $D=0$, the six coefficients
208 $(u_1,v_1,u_2,v_2,u_3,v_3)$ uniquely recover the ordered parameters
209 $(q,r,a,b)$ in $(0,1)^4$. These are the coefficients at degrees three,
210 five and seven of the two labeled functions.
211 
212 \vfig[0.92]{coupled_08_inverse}{Cubic terms recover the two scale products.
213 Fifth-degree terms leave at most two candidates for each remaining scale.
214 The seventh-degree terms select the unique admissible candidates when
215 $D\ne0$. Repeated quadratic roots are already unique.}
216 
217 First $p=3u_1$ and $k=3v_1$. The possible values $x=q^2$ and $y=a^2$
218 satisfy
219 \[
220 p x^2-\frac{15u_2}{p}x+kp^2=0,\qquad
221 p y^2-\frac{15v_2}{k}y+k^3=0.
222 \]
223 For two distinct positive candidates $x,x'$, the predicted seventh
224 coefficients differ by
225 \[
226 u_3(x)-u_3(x')=\frac{pD}{7k^2}(x^2-x'^2)\ne0.
227 \]
228 The second component supplies the analogous separation for $y$.
229 Finally take the positive square roots and recover $r=p/q$, $b=k/a$.
230 
231 This is an exact-data inverse theorem. Near a reduction surface, small
232 $D$ can make candidate predictions close, so a uniform stable inverse
233 does not follow. The degree-seven recovery and the reduction classification
234 are proved statements. Historical priority of these particular results
235 is addressed separately by the bounded literature comparison.

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