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Contracted nonlinear feedback

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Contracted nonlinear feedback

Two scales inside the function

For real \(0<q,r<1\), define the normalized odd solution

\[ \boxed{T_{q,r}'(z)=1+T_{q,r}(qz)T_{q,r}(rz),\qquad T_{q,r}(0)=0.} \]

The contractions are part of the defining feedback. The function is symmetric in \(q,r\), but unequal scales retain left/right address information.

Computed Taylor curves. Each entire curve uses terms through z 321 ; positive-coefficient Cauchy estimates bound the omitted tail on each displayed interval. The black dashed curve is ordinary tangent.
Computed Taylor curves. Each entire curve uses terms through \(z^{321}\); positive-coefficient Cauchy estimates bound the omitted tail on each displayed interval. The black dashed curve is ordinary tangent.

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For \(T_{q,r}(z)=\sum_{n\ge0}c_nz^{2n+1}\), \(c_0=1\), coefficient matching gives

\[ c_n=\frac{1}{2n+1}\sum_{i=0}^{n-1} q^{2i+1}r^{2(n-1-i)+1}c_ic_{n-1-i},\qquad n\ge1. \]

In particular,

\[ T_{q,r}(z)=z+\frac{qr}{3}z^3+ \frac{q^2r^2(q^2+r^2)}{15}z^5+O(z^7). \]

The coefficients are positive and bounded by tangent coefficients, giving a convergent germ. If the solution is known on \(|z|<R\), its integral equation extends it to \(|z|<R/\max(q,r)\). Repetition proves that it is entire.

The parameters can be recovered, up to swapping, from \(c_1,c_2\). Put \(p=3c_1\) and \(u=15c_2/p^2\). Then \(q,r\) are the roots of \(x^2-\sqrt{u+2p}\,x+p=0\).

A scale-balance atlas

\[ \boxed{\rho(T_{q,r})=\rho\quad\text{where}\quad q^\rho+r^\rho=1.} \]

Here \(M_f(R)=\max_{|z|=R}|f(z)|\) and \(\rho(f)=\limsup_{R\to\infty}\log\log M_f(R)/\log R\) is the order of an entire function. Moreover \(\log M_{T}(R)=O(R^\rho)\): the upper type is finite.

The displayed values numerically solve the proved characteristic equation. Colors saturate at order 7; the upper corner contains larger orders. The dashed diagonal is the previously known one-scale family.
The displayed values numerically solve the proved characteristic equation. Colors saturate at order 7; the upper corner contains larger orders. The dashed diagonal is the previously known one-scale family.

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ScalesOrderInterpretation
\(q=r=1/4\)\(1/2\)Slow entire growth
\(q+r=1\)\(1\)Many distinct functions at one order
\(q=r=1/\sqrt2\)\(2\)Quadratic logarithmic growth scale
\(q=r=3/4\)\(\log2/\log(4/3)\)Approximately \(2.4094208397\)
Why this equation appears

Nonnegative coefficients give \(M(R)=T_{q,r}(R)\). The integral equation implies

\[ M(R)\le R[1+M(qR)M(rR)], \]
\[ M(R)\ge(1-\alpha)R M(\alpha qR)M(\alpha rR),\quad0<\alpha<1. \]

Taking logarithms turns product feedback into a sum over contracted scales. Comparison with \(CR^\rho\) proves the upper bound; comparison with \(cR^{\rho_\alpha}\), where \((\alpha q)^{\rho_\alpha}+ (\alpha r)^{\rho_\alpha}=1\), proves the lower bound. Letting \(\alpha\uparrow1\) gives the same order. The complete interval-induction proof is given in Appendix A.

Positive type and tap continuity

More than a rescaled one-scale seed

Nonrescaling theorem

If \(q\ne r\), then \(T_{q,r}\) cannot equal \(A f_s(Bz)\) for nonzero constants \(A,B\) and any real \(0<s<1\), where \(f_s'=1+f_s(sz)^2\) is the known family.

Suppose such an equality held. Normalization gives \(AB=1\). Write \(p=qr\), \(u=q^2+r^2\), and let \(\rho\) be the common order. The first two nontrivial coefficients and the growth law force

\[ B^2s^2=p,\qquad s^2=u/2,\qquad 2s^\rho=q^\rho+r^\rho=1. \]

Consequently

\[ 2\left(\frac{q^2+r^2}{2}\right)^{\rho/2}=q^\rho+r^\rho. \]

For \(\rho\ne2\), strict convexity or concavity of \(x^{\rho/2}\) forces \(q=r\). For the exceptional case \(\rho=2\), we have \(u=1\), \(s^2=1/2\), \(B^2=2p\). The two \(z^7\) coefficients become

\[ \frac{p^3(3-p^2)}{315}\quad\text{and}\quad\frac{11p^3}{1260}. \]

Their equality forces \(p=1/2\) and again \((q-r)^2=u-2p=0\). This completes the exceptional-order case.

A worked order-one member
\[ T_{1/3,2/3}(z)=z+\frac{2}{27}z^3+ \frac{4}{2187}z^5+\frac{232}{11160261}z^7+O(z^9). \]

It has order one, and \(T_{1/3,2/3}(1)\approx1.0759239897408249\). The diagonal point \(q=r=1/2\) gives \(f_{1/2}(z)=\sqrt2\sinh(z/\sqrt2)\). Every off-diagonal member is separated from all rescaled diagonal members by the theorem above.

A combinatorial address formula

Sum over full ordered binary trees with \(n\) internal vertices. A vertex with \(i\) internal vertices in its left subtree and \(j\) in its right contributes

\[ \frac{q^{2i+1}r^{2j+1}}{2(i+j+1)+1}. \]

Multiply these factors over internal vertices, then sum over trees. Root splitting reproduces the coefficient recurrence exactly. Unequal contractions therefore record left and right path lengths separately.

This is a precise separation result from one authenticated antecedent. It does not prove that no earlier author studied the asymmetric equation.

Feedback banks and weak-channel sensitivity

Choose finitely many channels \(j=1,\ldots,J\), with \(h_j>0\), \(\sum_jh_j=1\), and \(0<q_j,r_j<1\). Define

\[ F_\Theta'(z)=1+\sum_jh_jF_\Theta(q_jz)F_\Theta(r_jz),\qquad F_\Theta(0)=0. \]

These taps act inside the nonlinear definition. With \(\mu_1=\sum_jh_jq_jr_j\) and \(\mu_2=\sum_jh_jq_jr_j(q_j^2+r_j^2)\),

\[ F_\Theta(z)=z+\frac{\mu_1}{3}z^3+\frac{\mu_1\mu_2}{15}z^5+O(z^7). \]

If \(q_j^{\rho_j}+r_j^{\rho_j}=1\), then

\[ \boxed{\rho(F_\Theta)=\max_{j\text{ active}}\rho_j.} \]

The positive weights alter local coefficients. The most expansive active channel determines the asymptotic order. A zero-weight channel is removed before taking the maximum.

A two-channel example: scales (1/4,1/4) with weight 1- varepsilon , and (3/4,3/4) with weight varepsilon . The function varies continuously on every fixed compact set, while its growth order jumps when the second channel becomes active.
A two-channel example: scales \((1/4,1/4)\) with weight \(1-\varepsilon\), and \((3/4,3/4)\) with weight \(\varepsilon\). The function varies continuously on every fixed compact set, while its growth order jumps when the second channel becomes active.

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For \(0<\varepsilon\le1\) the order is \(\log2/\log(4/3)\); at \(\varepsilon=0\) it is \(1/2\). Coefficients depend continuously on \(\varepsilon\) and are dominated by those of the entire \(f_{3/4}\). Dominated convergence of the Taylor series therefore proves uniform convergence on compact sets as \(\varepsilon\downarrow0\).

This gives a useful design distinction: finite-window similarity does not control entire growth order. For a bank with weights \(1/4,3/4\) at these scales,

\[ F(z)=z+\frac7{48}z^3+\frac{427}{30720}z^5+ \frac{176681}{188743680}z^7+O(z^9). \]

Recovering dilation parameters

Part II: Arithmetic depth and signed measures

Complex parameters, positive kernels and the boundaries of positivity.

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