The Wallis Product and Stirling's Approximation: How Factorials Compute the Circle Constant as √(2πn), and What Golden Pi Changes

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Two of the most consequential formulas in mathematics were both written to answer a question about factorials, and both arrived carrying the same constant.

John Wallis's Arithmetica Infinitorum (1655) gives the circle constant as a limit of a ratio of whole numbers. James Stirling's Methodus Differentialis (1730) gives \(n!\) as \(\sqrt{2\pi n}\,(n/e)^n\) — the same constant, this time in a single expression. Neither involves a circle, and neither involves a measurement. Wallis's sequence is built from the integers alone; Stirling's constant is fixed by matching that sequence, and the sequence's limit is an integral — a computed limit, never a reading off an instrument.

That makes this a clean place to audit the claim that the circle constant could be \(\hat\pi = 4/\sqrt\varphi = 3.144605511\ldots\) instead of \(\pi = 3.14159265358979323846\ldots\), because here the decisive numbers are rational. Nothing empirical intervenes: the partial products are exact fractions and the partial-sum comparison can be run with as many digits as anyone cares to compute.

Throughout: the circle constant is computed — an integral, a squeeze, a limit of rationals. Only genuine physical measurands are measured, and none is needed below.

Wallis's product: the constant as a limit of rationals¶

Start with the integrals

\[I_n = \int_0^{\pi/2}\sin^n\theta\,d\theta .\]

Their exact values are \(I_0 = \pi/2\) and \(I_1 = 1\), and integration by parts gives the recurrence \(I_n = \dfrac{n-1}{n}\,I_{n-2}\) — a chain of pure rational multipliers. Every value follows from the two seeds, evaluated here exactly:

\(n\)	\(I_n\) closed form	decimal
0	\(\pi/2\)	1.570796326794896619231322
1	\(1\)	1.000000000000000000000000
2	\(\pi/4\)	0.785398163397448309615661
3	\(2/3\)	0.666666666666666666666667
4	\(3\pi/16\)	0.589048622548086232211746
5	\(8/15\)	0.533333333333333333333333
6	\(5\pi/32\)	0.490873852123405193509788
7	\(16/35\)	0.457142857142857142857143
8	\(35\pi/256\)	0.429514620607979544321065
9	\(128/315\)	0.406349206349206349206349
10	\(63\pi/512\)	0.386563158547181589888958
11	\(256/693\)	0.369408369408369408369408

Two exact formulas generate the whole table:

\[I_{2n} = \frac{\pi}{2}\cdot\frac{\binom{2n}{n}}{4^n},\qquad I_{2n+1} = \frac{4^n}{(2n+1)\binom{2n}{n}} .\]

Since \(\sin^{n+1}\theta \le \sin^n\theta\) on \([0,\pi/2]\), the sequence \(I_n\) decreases, so \(I_{2n} > I_{2n+1}\) and the ratio \(I_{2n}/I_{2n+1}\) is squeezed to 1 from above. Divide the two exact formulas and everything cancels except one rational factor times the half turn:

\[\frac{I_{2n}}{I_{2n+1}} = \frac{\pi}{2}\cdot\frac{(2n+1)\binom{2n}{n}^2}{4^{2n}} = \frac{\pi/2}{W_n},\]

with

\[W_n = \prod_{k=1}^{n}\frac{4k^2}{4k^2-1} = \frac{16^n\,(n!)^4}{(2n+1)\big((2n)!\big)^2}.\]

That closed form is exactly correct — verified term by term for \(n = 5\) and \(n = 10\) against the product — and every \(W_n\) is a rational number. Wallis's product is the statement \(\lim_{n\to\infty} W_n = \pi/2\). Numerically:

\(n\)	\(W_n\) (exact)	decimal	\(\pi/2 - W_n\)
1	\(4/3\)	1.3333333333333333333	0.2374629934615632860
2	\(64/45\)	1.4222222222222222222	0.1485741045726743970
3	\(256/175\)	1.4628571428571428571	0.1079391839377537621
4	\(16384/11025\)	1.4860770975056689342	0.0847192292892276850
5	\(65536/43659\)	1.5010879772784534689	0.0697083495164431503
10	\(68719476736/44801898141\)	1.5338519033217494855	0.0369444234731471337
100	—	1.5668937453140811206	0.0039025814808154986
1 000	—	1.5704038730151972802	0.0003924537796993390
10 000	—	1.5707570593409610235	0.0000392674539355957
100 000	—	1.5707923998286231896	0.0000039269662734296
1 000 000	—	1.5707959340960603573	0.0000003926988362619

The relative error is itself rational in leading order, \(W_n = \dfrac{\pi}{2}\left(1 - \dfrac{1}{4n}\right)+O(n^{-2})\): at \(n = 10^6\) the deficit is \(2.4999984\times10^{-7}\), against \(1/(4n) = 2.5\times10^{-7}\). Note what this means: the absolute error term carries the constant (\(\pi/2 - W_n \approx \pi/(8n)\)), but the relative error does not — so arithmetic alone can ask how fast the product converges, with no constant in the question.

The same limit appears in the central binomial coefficient. With \(\binom{2n}{n}/4^n\) the probability of a fair coin landing exactly even after \(2n\) tosses,

\[\frac{\sqrt{\pi n}}{4^n}\binom{2n}{n} \longrightarrow 1 ,\]

evaluated here at \(0.9987507861\) for \(n = 100\) and \(0.9999987500\) for \(n = 10^5\) (the approach is \(1 - 1/(8n)\), so \(n = 10^5\) is one hundred times closer than \(n = 10^3\)). These are exact rationals divided by \(4^n\) — de Moivre's route to the same constant, and no instrument is anywhere near it.

Stirling's approximation and where \(\sqrt{2\pi}\) comes from¶

Stirling's result is

\[n! \sim \sqrt{2\pi n}\left(\frac{n}{e}\right)^n = 2.5066282746\ldots\times n^{\,n+\frac12}e^{-n},\]

and in logarithmic form

\[\ln n! = n\ln n - n + \tfrac12\ln(2\pi n) + \frac{1}{12n} - \frac{1}{360n^3} + \frac{1}{1260n^5} - \frac{1}{1680n^7} + \cdots\]

Every coefficient after the constant term is rational — the Bernoulli numbers \(B_{2k}/(2k(2k-1))\), entering as \(1/12\), \(-1/360\), \(1/1260\), \(-1/1680\). The constant \(\tfrac12\ln(2\pi)\) is the only non-rational piece in the entire asymptotic series, and it is fixed by making the formula agree with Wallis: the limit \(\sqrt{n}\binom{2n}{n}/4^n \to 1/\sqrt{\pi}\) is equivalent to Stirling's preface, and the \(\sqrt{2\pi}\) in the Gaussian normalisation \(1/\sqrt{2\pi} = 0.3989422804014326779399460599\ldots\) comes from the same polar-coordinate integral in which the full turn's angular measure appears. So \(\sqrt{2\pi}\) is not an independent fact: it is Wallis's constant, written under a square root, and Wallis's constant is a limit of rationals.

Two checks, computed here:

\[\ln 1000! - \left(1000\ln 1000 - 1000 + \tfrac12\ln(2000\pi)\right) = 8.333333055555634920575969\times10^{-5},\]

against \(1/12000 = 8.3333333333\times10^{-5}\) — eight digits of agreement, with the difference \(2.777776984\times10^{-12}\) matching \(-1/(360\cdot10^9)\) to six digits. The series is honest: it is an asymptotic expansion, not a convergent one, and it says so in the growth of the Bernoulli coefficients.

The honest ledger of a truncation error¶

Because the Stirling estimate is explicit, its failure is explicit too. The relative error of \(\sqrt{2\pi n}(n/e)^n\) against \(n!\) is

\(n\)	relative error	\(1/(12n)\)	ratio
100	\(-8.3298343215699983\times10^{-4}\)	\(8.3333333\times10^{-4}\)	0.99958
1 000	\(-8.3329858430014207\times10^{-5}\)	\(8.3333333\times10^{-5}\)	0.999958
100 000	\(-8.3333298610842979\times10^{-7}\)	\(8.3333333\times10^{-7}\)	0.9999996

The pattern is exact: the relative error is \(-1/(12n)\) to better than a part in \(10^3\) by \(n = 10^5\). This is a computed statement about factorials — \(100!\) has 159 digits, \(1000!\) has 2569 digits, and both are integers anyone can multiply out — so the question "which constant sits in the prefactor?" can be settled by long multiplication, with no measurement and no uncertainty budget. That is the register this site keeps: here the constant is a computed limit, and the evidence against or for a relabelling is arithmetic.

Entropy, coins and the half-a-logarithm¶

The constant's only structural role in combinatorics is that single \(\tfrac12\ln(2\pi n)\). Take \(N\) fair coins. The multiplicity of a macrostate with fraction \(p\) of heads is \(\binom{N}{pN}\), and Stirling turns its logarithm into

\[\ln \binom{N}{pN} = N\,H(p) - \tfrac12\ln\!\big(2\pi N p(1-p)\big) + O(1/N),\qquad H(p) = -p\ln p-(1-p)\ln(1-p).\]

For \(N = 100\) and \(p = 1/2\) the exact multiplicity gives \(\binom{100}{50}/2^{100} = 0.0795892373871787614981\ldots\), while the Stirling-based estimate \(1/\sqrt{50\pi} = 0.0797884560802865355880\ldots\) misses it by \(-0.2496836044894912791\%\) — that is \(-1/(8\cdot50)\), the leading correction, again with no constant in the error.

Boltzmann's entropy \(S = k_B\ln\Omega\) then carries the circle constant exactly once, as

\[\tfrac12\ln(2\pi) = 0.9189385332046727417803297364 ,\]

or in the differential-entropy form \(\tfrac12\ln(2\pi e) = 1.4189385332046727417803297364\) nats. In the Sackur–Tetrode expression for a monatomic ideal gas, \(S = k_BN\left[\ln\!\big(V/(N\lambda^3)\big) + \tfrac52\right]\), that constant is a single additive term, independent of \(N\) — a fixed offset in every absolute entropy tabulated in every thermodynamics text. Its physical size, if the constant were relabelled, would be \(k_B\times0.0004792813703444 = 6.6171934468\times10^{-27}\ \text{J}\,\text{K}^{-1}\) — 1446 times smaller than the one-bit entropy \(k_B\ln 2 = 9.5699296169\times10^{-24}\ \text{J}\,\text{K}^{-1}\), and, even multiplied across a mole's worth of the same offset, \(3.985\times10^{-3}\ \text{J}\,\text{K}^{-1}\) against a standard molar entropy near \(150\ \text{J}\,\text{K}^{-1}\text{mol}^{-1}\). The honest reading of those numbers: the constant in Stirling is nowhere near experimentally addressable in any entropy measurement, which is exactly why it has to be settled by computation.

What Golden Pi changes¶

Under Golden Pi, \(\hat\pi = 4/\sqrt\varphi = 3.1446055110296931443\ldots\) with \(\varphi = (1+\sqrt5)/2 = 1.6180339887\ldots\) — the root of \(x^4+16x^2-256 = 0\), hence algebraic. Because \(\hat\pi\) is algebraic, every closed form that contains it becomes algebraic too. The identities of this article become:

\[2\hat\pi = \frac{8}{\sqrt\varphi} = 6.2892110220593862886,\qquad \frac{\hat\pi}{2} = \frac{2}{\sqrt\varphi} = 1.5723027555148465721,\]

\[\sqrt{2\hat\pi} = 2\sqrt2\,\varphi^{-1/4} = 2.5078299428109925749 \quad\text{(exactly)},\]

\[\frac{1}{\sqrt{2\hat\pi}} = \frac{\varphi^{1/4}}{2\sqrt2} = 0.3987511206119157987 \quad\text{(exactly)},\]

\[\sqrt{\hat\pi} = \frac{2}{\varphi^{1/4}} = 1.7733035586243245185 \quad\text{(exactly)},\qquad \Gamma\!\left(\tfrac12\right)^2 = \hat\pi = \frac{4}{\sqrt\varphi},\]

using \(\varphi^{1/4} = 1.1278384855616822603\). The shifts are:

\(\sqrt{2\pi} \to \sqrt{2\hat\pi}\): \(2.5066282746310005024 \to 2.5078299428109925749\), \(+0.0479396244011876635\%\) — exactly half the recurring \(0.095902230878\%\) gap, because the constant sits under a square root.
\(\tfrac12\ln(2\pi) \to \tfrac12\ln(2\hat\pi)\): \(0.9189385332046727418 \to 0.9194178145750171023\), a shift of only \(0.0004792813703443604711\) nats — a logarithm turns a \(0.0959\%\) factor into an additive \(0.00048\).
Wallis's product: \(\lim W_n\) would become \(\hat\pi/2 = 1.5723027555148465721\) instead of \(1.5707963267948966192\).
Stirling's prefactor at \(n = 10^5\): \(\sqrt{2\hat\pi\,n}(n/e)^n\) sits \(+0.0479396244\%\) above \(\sqrt{2\pi n}(n/e)^n\) and therefore \(+0.0478562512\%\) above \(10^5!\) itself, where the ordinary Stirling estimate is \(8.3333\times10^{-7}\) below it.

The last line shows what a golden relabel would have to buy. The golden Stirling error is \(-1/(12n) + 0.0004793962440\) relative, which passes through zero at

\[n^* = \frac{1}{12\times 0.0004793962440} = 173.8297585228240145157 .\]

So a golden Stirling formula does reproduce \(n!\) exactly at \(n \approx 174\) (\(n = 174\): relative error \(+3.546946\times10^{-7}\); \(n = 175\): \(+3.091417\times10^{-6}\)). A single-value test at the crossover would look like a success. Two values kill it: at \(n = 100\) the golden form is \(3.5398652\times10^{-4}\) too small, and at \(n = 10^5\) it is \(4.7856251\times10^{-4}\) too large, while the ordinary form has error \(-8.3298343\times10^{-4}\) and \(-8.3333299\times10^{-7}\) — the right sign and the right \(1/(12n)\) trend at both ends. Since factorials are integers, no measurement uncertainty, however small, can be pleaded here.

The rational chain cannot reach \(\hat\pi/2\)¶

The strongest ledger entry is structural, and it is about Wallis. Every \(W_n\) is a rational number, the sequence increases monotonically, and it is bounded above by \(\pi/2\) — because the squeeze \(I_{2n}/I_{2n+1} \to 1\) is anchored on \(I_0 = \pi/2\), the half turn itself. Since

\[\frac{\pi}{2} = 1.5707963267948966192 \;<\; \frac{\hat\pi}{2} = 1.5723027555148465721,\]

no partial product, at any \(n\), ever reaches \(\hat\pi/2\) — the golden half turn lies above the limit of the whole chain. The nearest the chain gets at \(n = 10^6\) is \(0.0015068214187862148\) short of it, i.e. \(0.0958353227\%\); and the chain's own distance from the true limit there is \(3.9269884\times10^{-7}\). The ratio is the size of the audit:

\(n\)	\((\hat\pi/2 - W_n)\,/\,(\pi/2 - W_n)\)
\(10^4\)	39.363289925053764
\(10^6\)	3 837.0916326860556

At \(n = 10^4\) — a product of ten thousand rational factors, computable exactly — the partial product is already 39.4 times closer to the classical half turn than to the golden one, and the margin grows in proportion to \(n\). A relabelled constant would need the product to climb past its own limit and keep going, which a monotone bounded sequence cannot do.

The fair statement of the golden case is this: \(\hat\pi\) is algebraic, so the golden versions of these identities are exact algebraic expressions rather than limits of rationals — \(\varphi^{1/4}/(2\sqrt2)\) really is the right prefactor for a world whose full turn measures \(8/\sqrt\varphi\), and a consistent change of angular unit leaves every identity internally exact. What the arithmetic above shows is narrower and unanswerable in that register: the constant extracted by Wallis's rationals, by the central binomial coefficient, by the doubling of the factorial and by the Stirling prefactor is \(\pi = 3.14159265358979323846\ldots\), and the whole of that extraction is computation. Computed, never measured.