Paper IV visual guide · Visual 9 of 26 · COEFFICIENT RECONSTRUCTION

A reciprocal is a triangular solve

Traditional monograph: Theorem 7.3 (p. 14), Theorem 7.6 (p. 15).

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Start in the reduced variable with

\[ \Delta_2(w)=1-\frac w2+\frac{w^2}{24}-\frac{w^3}{720}+\cdots. \]

Write \(1/\Delta_2(w)=\sum b_nw^n\). Matching the product with \(1\) gives a unit lower-triangular system. At each row there is exactly one new unknown.

A lower-triangular denominator matrix times the reciprocal coefficient vector equals the unit-series vector.
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Reading the drawing

Read the matrix equation from left to right: denominator convolution, the unknown coefficient vector, and the unit series. Each matrix row introduces one new coefficient. The values shown are ordinary reduced-variable coefficients. The final line separately converts them to block exponential-generating-function coefficients.

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denominator convolution
Row / columnColumn 0Column 1Column 2Column 3
Row 0\(1\)\(0\)\(0\)\(0\)
Row 1\(- \frac{1}{2}\)\(1\)\(0\)\(0\)
Row 2\(\frac{1}{24}\)\(- \frac{1}{2}\)\(1\)\(0\)
Row 3\(- \frac{1}{720}\)\(\frac{1}{24}\)\(- \frac{1}{2}\)\(1\)
\(b\)
IndexCoefficient
0\(1\)
1\(\frac{1}{2}\)
2\(\frac{5}{24}\)
3\(\frac{61}{720}\)
  • \(=\)
unit series
IndexCoefficient
0\(1\)
1\(0\)
2\(0\)
3\(0\)

\(b_1=\frac12\)

\(b_2=\frac14-\frac1{24}\)

\(b_3=\frac{61}{720}\)

  • \(P_{2,n}(1)=(2n)!b_n:\qquad 1,\ 1,\ 5,\ 61\)

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The first three equations are

\[ b_1-\tfrac12=0,\qquad b_2-\tfrac12b_1+\tfrac1{24}=0, \qquad b_3-\tfrac12b_2+\tfrac1{24}b_1-\tfrac1{720}=0. \]

They yield \(b_1=1/2\), \(b_2=5/24\), and \(b_3=61/720\). Multiplication by \((2n)!\) produces the secant coefficients \(1,5,61\).

Why this is an exact finite calculation

To compute the coefficient of \(w^n\), only denominator coefficients through \(w^n\) are needed. Later terms of the entire series cannot affect it. A finite series inverse is therefore exact at its declared truncation order, even though evaluating that truncated series is only an approximation to the full analytic function away from the origin.

Read this as

A unit constant term makes coefficient recovery triangular. It does not by itself specify the radius of analytic convergence.

Evidence note from the source edition

The checks compare an independent binomial-series expansion, the parameter recurrence, and the logarithmic Newton recurrence through block index six. The table uses ordinary coefficients, with factorial conversion shown separately.

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