16. Split reciprocal coefficients into cyclic packets
All examples · Glossary: formal reciprocal · Papers, proofs, Lean, and code
This example introduces the coefficient operations behind Paper IV. Work over the rational numbers and seek the formal reciprocal of \(A(z)=1+z+z^2\).
Solve one coefficient at a time
Write \(B(z)=\sum_{n\ge0}b_nz^n\) and require \(A(z)B(z)=1\). Comparing coefficients gives \(b_0=1\), \(b_1+b_0=0\), and, for \(n\ge2\),
\[b_n+b_{n-1}+b_{n-2}=0.
\]
The first six calculations are
| Degree | Calculation | Coefficient |
|---|---|---|
| 0 | Constant product must be one | \(b_0=1\) |
| 1 | \(b_1=-b_0\) | \(-1\) |
| 2 | \(b_2=-b_1-b_0\) | \(0\) |
| 3 | \(b_3=-b_2-b_1\) | \(1\) |
| 4 | \(b_4=-b_3-b_2\) | \(-1\) |
| 5 | \(b_5=-b_4-b_3\) | \(0\) |
Thus the truncation through degree five is \(B_{<6}(z)=1-z+z^3-z^4\).
Check exactly what the truncation proves
Multiplying the two polynomials gives
\[(1+z+z^2)(1-z+z^3-z^4)=1-z^6.
\]
Every positive-degree coefficient below degree six vanishes. The product equals one modulo \(z^6\), but the degree-six discrepancy is \(-1\). The full formal identity follows instead from
\[\frac{1}{1+z+z^2}=\frac{1-z}{1-z^3}
=(1-z)\sum_{k\ge0}z^{3k}.
\]
The last expression can also be evaluated as a convergent complex series when \(|z|<1\). The finite polynomial calculation alone did not establish an identity at arbitrary complex inputs.
Select coefficients by residue class
Keep exponents congruent to \(0\), \(1\) and \(2\) modulo three:
\[P_0B_{<6}=1+z^3,\qquad P_1B_{<6}=-z-z^4,\qquad P_2B_{<6}=0.
\]
Their sum recovers the truncation. Each selector is idempotent, and distinct selectors annihilate each other because they keep disjoint exponent classes.
Over the complex numbers, let \(\omega\) be a primitive cube root of unity. The same selection has the root-filter formula
\[P_rf(z)=\frac13\sum_{k=0}^{2}\omega^{-rk}f(\omega^kz).
\]
On a monomial \(z^n\), the multiplier is one when \(n-r\) is divisible by three and zero otherwise, using \(1+\omega+\omega^2=0\).
Try it
Compute \(b_6,b_7,b_8\) and check the next truncation.
Answer: They are \(1,-1,0\). Hence \(B_{<9}=B_{<6}+z^6-z^7\) and \(A(z)B_{<9}(z)=1-z^9\). The coefficient pattern repeats in blocks of three.
Papers, proofs, Lean, and code
Terms in this example: Coefficient · Power series · Truncation · Formal reciprocal · Coefficient convolution · Root of unity
This coefficient model introduces cyclic projection and reciprocal recursion in Paper IV. Its finite product check does not address the paper’s analytic growth or spectral-zero results.
Paper IV: Cyclic Analytic Functions, Reciprocal Families, and Spectral Zero Data
Read · Written proofs · Lean coverage · Code and setup · Evidence
Source passages:
Related library guide and code examples. The guide covers related constructions and their software interfaces.
Python arithmetic check for this example: cyclic_reciprocals(). This checks the displayed rational, modular or finite-set calculations exactly.
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