Catalan Constant: source and evidence

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1 \chapter{A recurrence for rational approximants}
2 \label{cb_ch_recurrence}
3 
4 The elementary series gives rational approximations to $G$ term by term.
5 Zudilin's recurrence gives a different family whose errors decrease much
6 faster \cite{Zudilin2003}. Its coefficients and initial values are rational,
7 so every finite calculation can be performed exactly. The recurrence is
8 also a useful structural object: a two dimensional transfer matrix carries
9 each pair of adjacent values to the next, and a determinant tracks the
10 independence of its two distinguished solutions.
11 
12 We first define the sequences as rational algebraic objects. We then state
13 the external theorem identifying their limit with Catalan's constant and
14 their exponential rates. This separates what follows from the recurrence
15 alone from what requires its analytic interpretation.
16 
17 \section{The equation and its initial data}
18 
19 For integer $n\geq0$, set
20 \begin{align}
21  p(n)&=20n^2-8n+1,\label{cb_eq_p_polynomial}\\
22  q(n)&=3520n^6+5632n^5+2064n^4-384n^3-156n^2+16n+7.
23  \label{cb_eq_q_polynomial}
24 \end{align}
25 For $n\geq1$, abbreviate the leading and trailing coefficients by
26 \begin{equation}\label{cb_eq_a_c_def}
27  a_n=(2n+1)^2(2n+2)^2p(n),\qquad
28  c_n=(2n-1)^2(2n)^2p(n+1).
29 \end{equation}
30 The recurrence is
31 \begin{equation}\label{cb_eq_zudilin_recurrence}
32  a_nw_{n+1}-q(n)w_n-c_nw_{n-1}=0,\qquad n\geq1.
33 \end{equation}
34 The signs of the last two terms are part of the normalization. Solving
35 forward gives $w_{n+1}=(q(n)w_n+c_nw_{n-1})/a_n$.
36 
37 \begin{definition}[The distinguished rational sequences]\label{cb_def_uv}
38 Let $(u_n)_{n\geq0}$ and $(v_n)_{n\geq0}$ satisfy
39 \eqref{cb_eq_zudilin_recurrence}, with
40 \begin{equation}\label{cb_eq_uv_initial}
41  (u_0,u_1)=\left(1,\frac74\right),\qquad
42  (v_0,v_1)=\left(0,\frac{13}{8}\right).
43 \end{equation}
44 Their rational approximants are
45 \begin{equation}\label{cb_eq_r_def}
46  r_n=\frac{v_n}{u_n}.
47 \end{equation}
48 \end{definition}
49 
50 The notation $a_n,c_n$ in this chapter and the next denotes recurrence
51 coefficients. The later source notation $a_n,b_n$ is translated explicitly
52 when it occurs, and the cleared packet coefficients $A_{n,m},B_{n,m}$ have
53 two indices. These are different objects.
54 
55 The forward equation starts at $n=1$. Substituting $n=0$ into its polynomial
56 expression would remove the trailing term and impose $4w_1=7w_0$. Although
57 the initial values of $u$ happen to satisfy that extra equation, those of
58 $v$ do not. It is therefore not a valid additional recurrence condition
59 for the pair. At $n=1$, both $w_0$ and $w_1$ are already specified and
60 $w_2$ is the first value to be computed.
61 
62 \section{Existence, uniqueness and positivity}
63 
64 \begin{proposition}[Well defined rational solutions]\label{cb_prop_forward}
65 Every initial pair $(w_0,w_1)\in\Q^2$ determines exactly one rational
66 sequence satisfying \eqref{cb_eq_zudilin_recurrence}. In particular the
67 sequences in Definition~\ref{cb_def_uv} exist uniquely. Moreover,
68 \[
69  u_n>0\quad(n\geq0),\qquad v_n>0\quad(n\geq1),\qquad v_0=0.
70 \]
71 \end{proposition}
72 
73 \begin{proof}
74 Completing the square gives
75 \[
76  p(n)=20\left(n-\frac15\right)^2+\frac15>0.
77 \]
78 Thus $a_n,c_n>0$ for $n\geq1$. Also, for such $n$,
79 \[
80  2064n^4-384n^3-156n^2\geq1524n^4>0,
81 \]
82 because $n^3\leq n^4$ and $n^2\leq n^4$. All the remaining terms of $q(n)$
83 are positive, so $q(n)>0$. Division by $a_n$ therefore makes forward
84 iteration possible. It preserves rationality and determines the next term
85 uniquely. Induction proves existence and uniqueness at every index.
86 
87 The initial $u$ values are positive, and each subsequent value is a positive
88 linear combination of its two predecessors. For $v$, the initial $v_0$ is
89 zero and $v_1>0$. The first forward step gives $v_2>0$, and the same
90 induction applies thereafter. Consequently the quotient $r_n$ is defined
91 at every index.
92 \end{proof}
93 
94 These conclusions concern an infinite sequence but require only the
95 recurrence and a finite induction rule. They do not require a numerical
96 run through a large terminal index. The computer checks test whether an
97 implementation follows the rule correctly.
98 
99 The solution space also has a concrete basis. For any rational solution
100 $w$, put
101 \[
102  x=w_0,\qquad y=\frac8{13}\left(w_1-\frac74w_0\right).
103 \]
104 The sequence $xu_n+yv_n$ has the same initial pair as $w_n$. By uniqueness,
105 \begin{equation}\label{cb_eq_solution_basis}
106  w_n=xu_n+yv_n\qquad(n\geq0).
107 \end{equation}
108 Thus the entire space of rational solutions is two dimensional. The factor
109 $13/8$ in the initial data is nonzero, so there is no degeneracy in this
110 basis calculation. The same argument works for real initial values,
111 yielding the two dimensional real solution space.
112 
113 \begin{figure}[tbp]
114 \centering\includegraphics[width=0.93\linewidth]{../visual/figures/09_recurrence_structure.pdf}
115 \caption{Two preceding values determine the next rational value. The
116 forward rule begins at $n=1$ and uses the stated initial pair. Here and in
117 subsequent recurrence diagrams, $A_n=a_n$, $Q_n=q(n)$ and $C_n=c_n$.}
118 \label{cb_fig_recurrence}
119 \end{figure}
120 
121 \section{Two complete forward steps}
122 
123 The first steps expose all constants and signs. At $n=1$,
124 \[
125  p(1)=13,\quad p(2)=65,\quad q(1)=10699,\qquad
126  a_1=1872,\quad c_1=260.
127 \]
128 For $u$, the recurrence gives
129 \[
130  u_2=\frac{10699(7/4)+260}{1872}
131      =\frac{75933/4}{1872}=\frac{649}{64}.
132 \]
133 For $v$, its zero initial term removes the trailing contribution:
134 \[
135  v_2=\frac{10699(13/8)+260\cdot0}{1872}
136      =\frac{139087/8}{1872}=\frac{10699}{1152}.
137 \]
138 Therefore the second rational approximant is
139 \[
140  r_2=\frac{10699}{11682}.
141 \]
142 
143 At $n=2$ the coefficients become
144 \[
145  p(2)=65,\quad p(3)=157,\quad q(2)=434871,\qquad
146  a_2=58500,\quad c_2=22608.
147 \]
148 Using the fractions just obtained,
149 \begin{align*}
150  u_3
151  &=\frac{434871(649/64)+22608(7/4)}{58500}\\
152  &=\frac{284763375/64}{58500}=\frac{19471}{256},\\[3pt]
153  v_3
154  &=\frac{434871(10699/1152)+22608(13/8)}{58500}\\
155  &=\frac{521667445/128}{58500}=\frac{8025653}{115200}.
156 \end{align*}
157 It follows that $r_3=8025653/8761950$. All these values were also checked
158 against the recovered exact rational script. A floating point calculation
159 is unnecessary for the equality at any stage.
160 
161 The first ratios are $r_0=0$ and $r_1=13/14$. The subsequent values lie
162 near the decimal for $G$, but proximity is only a first observation.
163 Chapter~\ref{cb_ch_wronskian} proves exact nested brackets and an error
164 sign at every index. The external identification below then makes those
165 brackets enclosures of Catalan's constant itself.
166 
167 \section{The transfer matrix and its determinant}
168 
169 Normalize the recurrence by putting
170 \[
171  \alpha_n=\frac{q(n)}{a_n},\qquad
172  \beta_n=\frac{c_n}{a_n}\qquad(n\geq1).
173 \]
174 Both quantities are positive rationals. For the adjacent state vector
175 $X_n(w)=(w_n,w_{n-1})^{\mathsf T}$, define
176 \begin{equation}\label{cb_eq_transfer}
177  M_n=\begin{pmatrix}\alpha_n&\beta_n\\1&0\end{pmatrix}.
178  \qquad X_{n+1}(w)=M_nX_n(w).
179 \end{equation}
180 Hence
181 \[
182  X_{N+1}(w)=M_NM_{N-1}\cdots M_1X_1(w)\qquad(N\geq1).
183 \]
184 The order of multiplication follows the temporal order: $M_1$ acts first.
185 This is a product of changing matrices, not a power of one fixed matrix.
186 
187 The determinant is
188 \[
189  \det M_n=-\beta_n=-\frac{c_n}{a_n}<0.
190 \]
191 Every transfer is invertible, with
192 \[
193  M_n^{-1}=\begin{pmatrix}0&1\\1/\beta_n&-\alpha_n/\beta_n\end{pmatrix}.
194 \]
195 An adjacent pair at index $n+1$ therefore determines the preceding pair at
196 index $n$, within the stated recurrence domain. This does not extend the
197 equation through $n=0$, where the trailing coefficient would vanish.
198 
199 The two distinguished solutions can be collected into the fundamental
200 matrix
201 \[
202  \mathcal F_n=
203  \begin{pmatrix}u_n&v_n\\u_{n-1}&v_{n-1}\end{pmatrix},\qquad n\geq1.
204 \]
205 It satisfies $\mathcal F_{n+1}=M_n\mathcal F_n$. Its initial determinant is
206 \[
207  \det\mathcal F_1=
208  \det\begin{pmatrix}7/4&13/8\\1&0\end{pmatrix}=-\frac{13}{8}.
209 \]
210 Multiplicativity of determinants now gives
211 \[
212  \det\mathcal F_{N+1}
213  =-\frac{13}{8}\prod_{n=1}^N(-\beta_n)\ne0.
214 \]
215 Thus adjacent values of the two solutions remain independent at every
216 index. Each transfer reverses the determinant sign. The closed evaluation
217 of this product is the Wronskian theorem in
218 Chapter~\ref{cb_ch_wronskian}, where the factors telescope and yield exact
219 quotient increments. Here the matrix description identifies the mechanism
220 without duplicating that full proof.
221 
222 \section{Continuants and the finite fraction structure}
223 
224 The transfer product can also be written using continuants. For integers
225 $1\leq r\leq s$, let $K_{r,s}$ be the determinant of the tridiagonal matrix
226 whose diagonal is $\alpha_r,\ldots,\alpha_s$, lower adjacent diagonal is
227 one, and upper adjacent diagonal is
228 $-\beta_{r+1},\ldots,-\beta_s$. For instance,
229 \[
230  K_{r,r}=\alpha_r,\qquad
231  K_{r,r+1}=\det
232  \begin{pmatrix}\alpha_r&-\beta_{r+1}\\1&\alpha_{r+1}\end{pmatrix}
233  =\alpha_r\alpha_{r+1}+\beta_{r+1}.
234 \]
235 Set $K_{r,r-1}=1$ for the empty determinant. Expansion at the last row gives
236 \begin{equation}\label{cb_eq_continuant}
237  K_{r,s}=\alpha_sK_{r,s-1}+\beta_sK_{r,s-2}
238  \qquad(s\geq r+1).
239 \end{equation}
240 The plus sign comes from the minus sign on the upper adjacent diagonal.
241 This convention matches the positive trailing term in the recurrence.
242 All these continuants are positive rationals.
243 
244 \begin{proposition}[Initial data through continuants]\label{cb_prop_continuant}
245 For any solution $w$ and every $n\geq2$,
246 \begin{equation}\label{cb_eq_continuant_solution}
247  w_n=K_{1,n-1}w_1+\beta_1K_{2,n-1}w_0.
248 \end{equation}
249 \end{proposition}
250 
251 \begin{proof}
252 At $n=2$, the formula is
253 $w_2=\alpha_1w_1+\beta_1w_0$, using the empty determinant for $K_{2,1}$.
254 At $n=3$, direct iteration gives
255 \[
256  w_3=(\alpha_2\alpha_1+\beta_2)w_1+\alpha_2\beta_1w_0,
257 \]
258 which agrees with the formula. For later indices, both coefficients in
259 \eqref{cb_eq_continuant_solution} satisfy the same recurrence by
260 \eqref{cb_eq_continuant}. Induction finishes the proof.
261 \end{proof}
262 
263 This expression separates the dependence on the two initial values from
264 the intervening recurrence coefficients. It is useful when checking a
265 different normalization or a shifted index: the same two continuant
266 coefficients must act on the corresponding initial pair.
267 
268 For a positive solution, ratios of consecutive values give another finite
269 description. If $x_n=w_n/w_{n-1}$ is defined, then
270 \[
271  x_{n+1}=\alpha_n+\frac{\beta_n}{x_n}.
272 \]
273 For $u$, every such ratio is defined. The first two iterations are
274 \[
275  \frac{u_2}{u_1}=\alpha_1+\frac{\beta_1}{7/4},\qquad
276  \frac{u_3}{u_2}=
277  \alpha_2+\frac{\beta_2}{\alpha_1+\beta_1/(7/4)}.
278 \]
279 Repeated substitution yields a finite generalized continued fraction.
280 For $v$, one cannot begin with $v_1/v_0$, because $v_0=0$. Its ratio
281 iteration instead begins at $v_2/v_1=\alpha_1$. These formulas are exact
282 finite identities. Establishing a limit for an infinite continued fraction
283 requires an additional argument and is not implied merely by writing the
284 finite nested fractions.
285 
286 \section{The limiting characteristic equation}
287 
288 The leading terms of the coefficient polynomials are
289 \[
290  a_n=320n^6+O(n^5),\qquad
291  q(n)=3520n^6+O(n^5),\qquad
292  c_n=320n^6+O(n^5).
293 \]
294 Consequently $\alpha_n\to11$, $\beta_n\to1$, and
295 \[
296  M_n\longrightarrow M_\infty=\begin{pmatrix}11&1\\1&0\end{pmatrix}.
297 \]
298 The characteristic polynomial of this limiting matrix is
299 \begin{equation}\label{cb_eq_characteristic}
300  \lambda^2-11\lambda-1.
301 \end{equation}
302 Its two roots are
303 \begin{equation}\label{cb_eq_characteristic_roots}
304  \lambda_+=\frac{11+5\sqrt5}{2}=\phi^5,\qquad
305  \lambda_-=\frac{11-5\sqrt5}{2}=-\phi^{-5}.
306 \end{equation}
307 Here $\phi=(1+\sqrt5)/2$. The identity $\phi^5=5\phi+3$ follows by
308 repeatedly using $\phi^2=\phi+1$. The roots have product $-1$, as required
309 by the limiting determinant, and very different magnitudes.
310 
311 The limiting polynomial explains the scales that appear in the analytic
312 theorem, but calculating its roots is not a substitute for that theorem.
313 The original transfer matrices vary with $n$. One must also identify which
314 solution or linear combination occupies which asymptotic scale. In
315 particular, numerical agreement between the ratio and $G$ cannot supply
316 the identification of the decaying combination $u_nG-v_n$.
317 
318 \begin{theorem}[External recurrence identification and rates]\label{cb_thm_source_rates}
319 For the exact normalization \eqref{cb_eq_uv_initial}, Zudilin's theorem and
320 its stated asymptotic rates give
321 \begin{align}
322  \lim_{n\to\infty}\frac{v_n}{u_n}&=G,\label{cb_eq_source_limit}\\
323  \lim_{n\to\infty}u_n^{1/n}&=\phi^5,\label{cb_eq_source_u_rate}\\
324  \lim_{n\to\infty}|u_nG-v_n|^{1/n}&=\phi^{-5}.
325  \label{cb_eq_source_error_rate}
326 \end{align}
327 These are external mathematical inputs, recorded precisely in source
328 contracts LC01 and LC02 \cite{Zudilin2003}.
329 \end{theorem}
330 
331 The source limit is about the sequences just defined, not a differently
332 scaled initial pair. Multiplying both sequences by the same nonzero
333 constant would leave their quotient unchanged, but scaling only one
334 sequence would change it. The rational initial data and the source
335 dictionary therefore belong to the statement of the theorem.
336 
337 Chapter~\ref{cb_ch_wronskian} gives a recurrence only proof that the ratios
338 have a limit, with strict alternating brackets. It then invokes
339 \eqref{cb_eq_source_limit} to name that limit $G$. This division of labor
340 keeps the exact recurrence reasoning visible while acknowledging the
341 external analytic identification.
342 
343 \begin{figure}[tbp]
344 \centering\includegraphics[width=0.92\linewidth]{../visual/figures/11_recurrence_growth.pdf}
345 \caption{The growing coefficient scale corresponds to the larger
346 characteristic root. Finite plotted values illustrate the behavior. The
347 universal limiting rate is a separate source theorem.}
348 \label{cb_fig_recurrence_growth}
349 \end{figure}
350 
351 \section{Normalized error and the arithmetic question}
352 
353 An immediate consequence of the two source rates is
354 \begin{equation}\label{cb_eq_ratio_error_rate}
355  \lim_{n\to\infty}|G-r_n|^{1/n}
356  =\lim_{n\to\infty}\frac{|u_nG-v_n|^{1/n}}{u_n^{1/n}}
357  =\phi^{-10}.
358 \end{equation}
359 Thus the ratio error has a smaller exponential scale than the unnormalized
360 linear form. The distinction comes from division by the growing
361 coefficient $u_n$. Also, since $r_n\to G>0$, one has
362 $r_n^{1/n}\to1$ and consequently $v_n^{1/n}\to\phi^5$. Both individual
363 sequences grow, while their specified linear combination decays.
364 
365 The scaled coefficient used in the packet construction has rate
366 \[
367  \lim_{n\to\infty}(16^nu_n)^{1/n}
368  =16\phi^5=88+40\sqrt5.
369 \]
370 Chapter~\ref{cb_ch_integrality} proves that this scaled coefficient is an
371 integer at every index using a separate half binomial identity. Its growth
372 rate and its integrality are distinct mathematical facts.
373 
374 \begin{figure}[tbp]
375 \centering\includegraphics[width=0.92\linewidth]{../visual/figures/12_approximant_error.pdf}
376 \caption{The normalized rational error decreases rapidly. Exact interval
377 computations support the finite plotted values. The asymptotic rate
378 depends on the identified source solution, and irrationality requires
379 denominator control in addition.}\label{cb_fig_ratio_error}
380 \end{figure}
381 
382 The approximants therefore do much more than the original alternating
383 partial sums in numerical accuracy. Their rapid convergence still does
384 not classify the limit as irrational. The fractions $u_n$ and $v_n$ carry
385 denominators, and the denominator of their ratio may grow rapidly too.
386 To apply the integer lattice criterion, one must turn the error into a
387 nonzero integer linear form and show that its size tends to zero after
388 all necessary clearing. The recurrence supplies the coefficients and the
389 analytic theorem supplies the error scale. The subsequent determinant,
390 integrality and packet chapters examine the additional arithmetic needed
391 to connect these facts to that criterion.

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