Paper IXReading edition

Congruential reciprocal packets

Reading tools
Reading libraryFind statementsInspect Lean
Switch work

Saved reading

Read side by side

Start reading

Resources & source

Original TeX with line links · Download TeX

The wording and mathematics follow the retained manuscript. This reading format does not add a proof review or extend Lean coverage.

Skip contents and start reading

D.4 Congruential reciprocal packets

D.4.1 Residue sections of the exponential series

For integers \(N\geq1\) and \(0\leq j<N\), define

\[ {\mathcal E}_{N,j}(t) = \sum_{n\geq0} \frac{t^{Nn+j}}{(Nn+j)!}. \]

The forced factor \(t^j\) may be removed without changing the residue address.

Historical definition D.20: Congruential reciprocal packet

Define

\[ \boxed{ {\mathcal C}_{N,j}(t) = \frac{t^j/j!}{{\mathcal E}_{N,j}(t)} = \sum_{n\geq0} C_{N,j}(n) \frac{t^{Nn}}{(Nn)!}. } \]
Historical theorem D.21: Triangular inverse system

The coefficients satisfy \(C_{N,j}(0)=1\) and

\[ \boxed{ \sum_{q=0}^{n} \binom{Nn+j}{Nq} C_{N,j}(q) = \delta_{n0}. } \]

For \(n\geq1\),

\[ \boxed{ C_{N,j}(n) = - \frac{ \sum_{q=0}^{n-1} \binom{Nn+j}{Nq}C_{N,j}(q) }{ \binom{Nn+j}{Nn} }. } \]
Proof

Multiply \({\mathcal C}_{N,j}(t)\) by \({\mathcal E}_{N,j}(t)\) and compare the coefficient of \(t^{Nn+j}/(Nn+j)!\). The last summand is the diagonal term, which is nonzero over \({\mathbb Q}\).

Historical proposition D.22: A denominator bound

Let

\[ D_{N,j}(n) = \prod_{r=1}^{n} \binom{Nr+j}{Nr}, \qquad D_{N,j}(0)=1. \]

Then

\[ \boxed{ D_{N,j}(n)C_{N,j}(n) \in {\mathbb Z}. } \]

For every prime \(p\),

\[ {v_p}(C_{N,j}(n)) \geq - \sum_{r=1}^{n} {v_p}\binom{Nr+j}{Nr}. \]
Proof

Use induction on \(n\) in the recurrence of Historical theorem D.21. Every earlier denominator divides \(D_{N,j}(n-1)\), while the new diagonal denominator is the final factor of \(D_{N,j}(n)\). The valuation statement follows.

Historical example D.23: Euler and Bernoulli boundaries

For \((N,j)=(2,0)\),

\[ {\mathcal C}_{2,0}(t)=\operatorname{sech}t, \]

so

\[ C_{2,0}(n)=E_{2n}. \]

For \((N,j)=(2,1)\),

\[ {\mathcal C}_{2,1}(t)=\frac{t}{\sinh t}, \]

and

\[ C_{2,1}(n) = 2\left(1-2^{2n-1}\right)B_{2n}. \]

The usual convention \(B_1=-1/2\) is used. Euler and Bernoulli number conventions are reviewed in [Comtet1974], [Stanley2012].

D.4.2 Standard windows

Fix a prime \(p\), an integer \(m\geq1\), and an address \(0\leq j<p\). Put

\[ N=mp. \]

For arbitrary \(m\), let \(\widetilde e_n^{(m,p)}\in{\mathbb F}_p\) be the unique sequence satisfying

\[ \boxed{ \sum_{q=0}^{n} \binom{mn}{mq} \widetilde e_q^{(m,p)} = \delta_{n0}. } \]

When \(m\mid p-1\), this sequence agrees with the compressed Hurwitz inverse in Historical proposition D.11.

Historical standard fact D.24: Lucas congruence

If \(a=a_0+a_1p+\cdots\) and \(b=b_0+b_1p+\cdots\) are base \(p\) expansions, then

\[ \binom ab \equiv \prod_r\binom{a_r}{b_r} \pmod p. \]

This is the classical theorem of Lucas [Lucas1878a].

Historical theorem D.25: Standard window reduction

For \(0\leq q\leq n\),

\[ \boxed{ \binom{mpn+j}{mpq} \equiv \binom{mn}{mq} \pmod p. } \]

The diagonal coefficient is a unit modulo \(p\):

\[ \binom{mpn+j}{mpn} \equiv1 \pmod p. \]
Proof

Write

\[ mpn+j=p(mn)+j, \qquad mpq=p(mq). \]

Lucas reduction separates the lowest base \(p\) digit and gives

\[ \binom{mpn+j}{mpq} \equiv \binom j0 \binom{mn}{mq} \pmod p. \]

The diagonal case is obtained by setting \(q=n\).

Historical corollary D.26: Standard window integrality

Every coefficient \(C_{mp,j}(n)\) belongs to \({\mathbb Z_p}\).

Proof

The recurrence in Historical theorem D.21 starts with a \(p\) integral constant term. The binomial coefficients are integers, and every diagonal term is a \(p\) adic unit by Historical theorem D.25. Induction gives the result.

Historical corollary D.27: Address independence modulo the prime

Let \(\widetilde e_n^{(m,p)}\) be the reduced standard inverse above. Then

\[ \boxed{ C_{mp,j}(n) \equiv \widetilde e_n^{(m,p)} \pmod p } \]

for every \(0\leq j<p\).

Proof

Modulo \(p\), the recurrence in Historical theorem D.21 becomes

\[ \sum_{q=0}^{n} \binom{mn}{mq} C_{mp,j}(q) = \delta_{n0}. \]

This is the defining recurrence for \(\widetilde e_n^{(m,p)}\). The diagonal is one, so the solution is unique.

Historical theorem D.28: Uniform standard window period

Assume that \(p\) is odd and \(m\mid p-1\). Put

\[ h=\frac{p-1}{m}. \]

Then for every standard window address \(0\leq j<p\),

\[ \boxed{ C_{mp,j}(n+h) \equiv C_{mp,j}(n) \pmod p \qquad(n\geq1). } \]
Proof

By Historical corollary D.27, every address reduces to the same reduced standard inverse. Under the divisibility assumption, that inverse agrees with the compressed Hurwitz inverse by Historical proposition D.11. Its positive tail has period \(h\) by Historical corollary D.12.

Historical corollary D.29: Rational residual generating function

Under the hypotheses of Historical theorem D.28, for every standard address \(j\),

\[ \sum_{n\geq0} \left(C_{mp,j}(n)\bmod p\right)z^n = 1+ \frac{ \sum_{r=1}^{h}\widetilde e_r^{(m,p)}z^r }{1-z^h}. \]
Proof

Combine Historical corollary D.27 and Historical corollary D.13.

Historical theorem D.30: Fourier invariance of the standard window operator

Let \(p\) be odd, let \(m\mid p-1\), and let \(h=(p-1)/m\). Define

\[ (T_{m,p}f)_n = \sum_{q=0}^{n}\binom{mn}{mq}f_q \quad\text{in }{\mathbb F}_p. \]

Let \(V_h\) be the vector space of sequences satisfying

\[ f_{n+h}=f_n \qquad(n\geq1). \]

Then

\[ \boxed{T_{m,p}(V_h)=V_h.} \]

The restriction of \(T_{m,p}\) to \(V_h\) is an automorphism.

Proof

Choose \(d_0=f_h\) and \(d_n=f_n\) for \(n\geq1\). Then \(d\) is fully \(h\) periodic and \(f=d+c\delta_0\) for \(c=f_0-f_h\). Since \(T_{m,p}\delta_0\) is the constant sequence one, it is enough to treat \(d\). Expand

\[ d_q=\sum_{\chi\in\mu_h}c_\chi\chi^q. \]

The \(m\)th power map on \({\mathbb F}_p^\times\) has image \(\mu_h\). Choose \(y\) with \(y^m=\chi\). The root filter gives

\[ T_{m,p}(\chi^\bullet)_n = \frac1m \sum_{\zeta\in\mu_m} (1+\zeta y)^{mn}. \]

If \(1+\zeta y=0\), the corresponding term vanishes for positive \(n\). Otherwise

\[ \left((1+\zeta y)^m\right)^h = (1+\zeta y)^{p-1} =1. \]

Every summand is therefore \(h\) periodic for positive \(n\). Thus \(T_{m,p}\) preserves \(V_h\). It is unit lower triangular on every finite initial segment. Its restriction to the finite dimensional space determined by \(f_0,f_1,\ldots,f_h\) is injective and hence bijective.

Historical corollary D.31: A second proof of the standard window period

The reduced inverse \(e^{(m,p)}\) and every standard address satisfy the period in Historical theorem D.28.

Proof

The sequence \(\delta=(1,0,0,\ldots)\) belongs to \(V_h\). Its inverse image under the automorphism in Historical theorem D.30 also belongs to \(V_h\). Address independence modulo \(p\) gives the statement for every standard address.

New reconstruction. Finite field Fourier transfer for a standard window. Each character becomes a finite sum of positive index modes with period dividing (h ).
New reconstruction. Finite field Fourier transfer for a standard window. Each character becomes a finite sum of positive index modes with period dividing \(h\).

View full-size figure (new tab)

New reconstruction. The standard window matrix for p=7, m=2, and 0 less than or equal j<7. Reduction removes the address and retains diagonal one.
New reconstruction. The standard window matrix for \(p=7\), \(m=2\), and \(0\le j<7\). Reduction removes the address and retains diagonal one.

View full-size figure (new tab)

Historical remark D.32

The theorem gives a period modulo \(p\). A period modulo higher powers requires additional control of the triangular inverse and cannot be inferred from the finite field argument alone.

Source and conversion record

Source SHA-256: 87ee9bd6c3df86066ea44a9db069d8cf75a1b673a7dffb8dfbbebb7dfe1b68c4

Unsupported prose commands stop the build. The mathematical expression sequence is checked against the source, with manuscript macros expanded for display.

Conversion records

Notation in Paper IX

Look up a symbol without leaving this chapter. Select text before opening help to search for it.

Search this work’s notation.

    Symbols can have different meanings in different passages. Read each result in its own context.

    Open the full notation index