Thinking in Waves Episode 61 / Part XIX — Closing
Four routes land on the same number
How many times did \(\alpha\) appear? ──
and \(-1/12\) did not depend on the cutoff
Recomputing the nine numbers of Episodes 56–60 (all reproduced), we then measured \(\text{dB/oct}=6\alpha\) along four independent routes at once ── differentiation, fractional differentiation, a window’s edge, and subtracting a trend.
★★★ And we closed the open item Episode 45 left in a footnote ── \(-1/12\) comes out the same for five different cutoffs. The divergent term’s coefficient ranges over 0.443–1.832; the constant does not budge.
★★ That is precisely where Episode 60’s “running variance” parts company.
Part XIX opened as “back to physics, carrying Part XVIII’s tools.” ★ To close it, we recompute every number it produced and count how many strands of the spine ran through it.
01Recomputing Episodes 56–60
| item | recomputed | as published | verdict |
|---|---|---|---|
| Ep. 56 slope | 0.9915 | 0.9916 | OK |
| Ep. 56 intercept | −2.1690 | −2.1708 | OK |
| Ep. 57 cost of an edge [dec] | 2.2098 | 2.2098 | OK |
| Ep. 57 slope (\(\gamma{=}0.6\)) | 2.3810 | 2.3810 | OK |
| Ep. 58 rectangular dB/oct | −5.9337 | −5.9340 | OK |
| Ep. 58 Hann dB/oct | −17.8760 | −17.8760 | OK |
| Ep. 60 \(q{=}0\) dB/oct | 6.0141 | 6.0140 | OK |
| Ep. 60 \(q{=}2\) dB/oct | 18.0572 | 18.0570 | OK |
★ All nine reproduce. But one number that “agrees” still needs fixing.
| how Episode 56’s intercept was obtained | value |
|---|---|
| \(T=3,4,6,8,10,12\) (what Episode 56 used) | −2.1690 |
| \(T=4,6,8,10,12\) (dropping \(T{=}3\)) | −2.2395 |
| the analytic value derived in Episode 57 | −2.2098 |
The core of §01 ── we did it again
★★ \(T=3\) is about the same size as the 2.2 decades being lost, so it has not reached the asymptotic regime. Including it in the fit pulled the intercept shallow (−2.17). Dropping it gives −2.24, with the analytic −2.2098 in between.
★★★ So Episode 56’s “−2.1708” had four digits where the evidence supports two ── it moves by ±0.04 with the range of \(T\). The right statement is “about −2.2.”
★★★ One episode after Episode 55 wrote “0.0812 is only good to two figures,” we did exactly the same thing. Writing it in the handbook does not stop you doing it.
02★★★ Measuring \(\text{dB/oct}=6\alpha\) four independent ways
| route | \(\alpha\) | measured dB/oct | predicted \(6\alpha\) | diff | from |
|---|---|---|---|---|---|
| ① 1 difference | 1 | 6.016 | 6.021 | −0.005 | Ep. 1 |
| ① 3 differences | 3 | 18.048 | 18.062 | −0.014 | Ep. 1 |
| ② fractional, order 0.5 | 0.5 | 3.010 | 3.010 | −0.000 | Ep. 13, 41 |
| ② fractional, order 2.5 | 2.5 | 15.051 | 15.052 | −0.000 | Ep. 13, 41 |
| ③ \(\cos^0\) window sidelobes | 1 | 5.934 | 6.021 | −0.087 | Ep. 58 |
| ③ \(\cos^4\) window sidelobes | 5 | 29.873 | 30.103 | −0.230 | Ep. 58 |
| ④ detrending \(q{=}0\) | 1 | 6.014 | 6.021 | −0.006 | Ep. 60 |
| ④ detrending \(q{=}3\) | 4 | 24.078 | 24.082 | −0.004 | Ep. 60 |
The core of §02
Four entirely different operations land on the same \(6\alpha\).
★★★ Differentiation (Ep. 1), fractional differentiation (Ep. 13, 41), a window’s edge (Ep. 58), subtracting a trend (Ep. 60). The entrances are unrelated; the exit is the same.
★★ And \(\alpha\) need not be an integer (orders 0.5 and 2.5 above). Episodes 41–42’s “the order leaves the integers” is at work here too.
★ Only route ③ shows a sizeable gap (−0.09 to −0.23), which comes from the coarse way the sidelobe envelope is sampled, not from the principle.
★ This is how far one line from Episode 1 ── “differentiate a sound and it goes high” ── has travelled.
03★★★ Closing Episode 45’s open item ── \(1+2+3+\cdots=-1/12\)
Episode 45 treated the Casimir effect and, with an exponential cutoff, obtained \(\sum n\,e^{-\delta n}=1/\delta^2-1/12+O(\delta^2)\). Its footnote then said: “that this does not depend on the choice of regularisation has not been checked here.” Let us check it.
| cutoff \(f\) | \(A\) | \(\delta{=}0.02\) | \(\delta{=}0.01\) | \(\delta{=}0.005\) |
|---|---|---|---|---|
| \(e^{-x}\) (Ep. 45) | 1.000000 | −0.083332 | −0.083333 | −0.083333 |
| \(e^{-x^2}\) (Gaussian) | 0.500000 | −0.083337 | −0.083334 | −0.083334 |
| \(e^{-x^4}\) | 0.443113 | −0.083333 | −0.083333 | −0.083333 |
| \(\mathrm{sech}\,x\) | 1.831931 | −0.083335 | −0.083334 | −0.083335 |
| \(1/(1+x^2)^2\) | 0.500000 | −0.083340 | −0.083335 | −0.083334 |
The core of §03
★★★ The divergent term’s coefficient \(A\) ranges from 0.443 to 1.832. The constant is \(-0.083333\) every time (largest departure from \(-1/12\): 6.7e−6).
★★ Episode 45’s footnote can now be closed.
★ And a phrasing worth correcting: \(-1/12\) is not “the value of” \(1+2+3+\cdots\). That series diverges. \(-1/12\) is what remains after the divergent term is subtracted. ★★★ The point is that the term you must subtract depends on the cutoff, and the remainder does not.
★ Two stumbles along the way, recorded here.
| stumble | symptom | cause |
|---|---|---|
| ① computed \(A\) numerically | \(1/(1{+}x^2)^2\) gave \(-0.25\) | a 4e−6 error in \(A\), times \(1/\delta^2=4\times10^4\) |
| ② switched to the exact \(A\) | still \(-0.0846\) | sum truncation and integral range did not match |
★★★ Both are one and the same thing: when you subtract a divergent term, the precision of what you subtract is amplified by \(1/\delta^2\). That is why Casimir calculations are delicate.
04★★ Where Episode 60’s running variance parts company
| Casimir (Ep. 45) | variance of \(1/f^\beta\) (Ep. 60) | |
|---|---|---|
| divergent term | \(A/\delta^2\) (depends on the cutoff) | \(N^{\beta-1}\) (depends on record length) |
| finite remainder | \(-1/12\) (cutoff-independent) | none |
| taking a difference | cancels (walls present vs absent) | does not cancel |
| result | a physical quantity is determined | you must state the record length |
The core of §04
Whether a running quantity can be given meaning
depends on whether a cutoff-independent remainder survives.
★★ This is what Episode 60’s Exercise 2 meant by “the same structure, but renormalisation is not available” ── now shown in numbers.
★ Episode 29’s renormalisation, Episode 45’s Casimir effect and Episode 60’s running variance all depend on where you cut ── but only the first two leave a remainder that does not.
05The handbook goes 7 → 10 → 14 steps
| step | from | content |
|---|---|---|
| ①–⑦ | Ep. 49 | measure it / plot semi-log / measure the slope, etc. |
| ⑧–⑩ | Ep. 55 | split the error / buy range / count the assumptions |
| ★⑪ | Ep. 59, 60 | subtract a degree 1–2 trend before the periodogram |
| ★⑫ | Ep. 60 | check the two conditions separately (\(\beta\le2q+3\) and \(\beta<2\alpha\)) |
| ★⑬ | Ep. 59 | measure the error’s shape before its size (coverage) |
| ★⑭ | Ep. 60 | when you state a variance, state the record length |
★ Steps ⑪–⑭ all come down to declaring where you cut. What Part XIX found is that the limits of a measurement are set almost entirely by that choice.
06Still undone
| item | status |
|---|---|
| ★★★ apply it to real data | Episodes 56–60 are all synthetic |
| (done) regulator-independence of \(-1/12\) | Episode 45’s footnote → closed in §03 here |
| an independent measurement of \(D(E)\) | Episode 56’s decisive test; literature not consulted |
| \(1/f\) measured against temperature | Episode 56’s proposal; unverified |
| pooling many short records | Episode 60 Exercise 3; independence doubtful |
| coverage for non-Gaussian processes | Episode 59 used Gaussian synthesis only |
| other window families | Episode 59 used \(\cos^p\) only |
★★★ The first is the big one ── this series has not once touched real measured data. Writing that down is what Episode 55’s step ⑩ asks for.
Exercises
- §02 has four routes landing on the same \(6\alpha\). Is that deep?
Show answer
No ── all four reduce to one Fourier transform.
★ Differentiation multiplies by \(i\omega\), fractional differentiation by \((i\omega)^\alpha\), and both a window’s sidelobes and a detrending gain are \(\omega^\alpha\) behaviour as \(\omega\to\infty\) or \(\omega\to0\). Every one is “multiply by a power in Fourier space.”
★★ Laying them side by side is still worth doing: only then can you see they are the same, and carry a result from one to another. ★ Episode 60’s \(\beta\le2q+3\) was obtained exactly that way ── by carrying Episode 58’s window argument over to detrending.
★★★ In Episode 55’s terms, an analogy is not a claim. “They all land on \(6\alpha\)” is a fact, but it is not new physics ── it is tidying the tools, and should be written as such. - §03 checks that \(-1/12\) is regulator-independent. Why is it?
Show answer
Under a Mellin transform, \(-1/12\) emerges as a residue of \(\zeta(-1)\).
★ Writing \(\sum n f(\delta n)=\frac{1}{2\pi i}\int\delta^{-s}\zeta(s-1)F(s)\,ds\) (with \(F\) the Mellin transform of \(f\)), the pole at \(s=2\) gives \(A/\delta^2\) and the pole at \(s=0\) gives the constant.
★★ The residue at \(s=0\) is \(\zeta(-1)f(0)\) ── as long as \(f(0)=1\), the shape of \(f\) is irrelevant. ★ What does depend on \(f\) is \(A=F(2)\), which is the 0.443–1.832 in the table.
★★★ What this episode did numerically is confirm that analysis ── five cutoffs were tried, which is not the same as showing it for all of them. What was shown is “the five we tried agreed.” - §01 wrote too many digits again, despite the handbook. Why did the handbook not prevent it?
Show answer
Because step ⑩ says “count the assumptions,” not “vary the fitting range and see.”
★ Episode 56’s intercept is not conditional on a model ── it is purely computed. So it passes “count the assumptions” untouched.
★★ What caught it was the procedure built in Episode 59 ── vary a condition and see whether the number moves. Here we varied the range of \(T\) and found ±0.04.
★★★ So what should be added to ⑩ is “vary the conditions and measure how far it moves.” Episode 59 built that procedure to measure \(\beta\), but it works on any number.
★ §01 says writing it in the handbook does not stop you doing it. Writing is not enough; you only find out by varying ── that is this episode’s practical lesson.
Summary — closing Part XIX
★ The nine numbers of Episodes 56–60 were recomputed and all reproduce. But Episode 56’s intercept −2.1708 had too many digits: it moves by ±0.04 with the range of \(T\), so “about −2.2” is the right statement. One episode after Episode 55 issued that very warning, we did it again.
★★★ \(\text{dB/oct}=6\alpha\) returned the same value along four independent routes ──
differentiation, fractional differentiation, a window’s edge, and subtracting a trend.
★ And \(\alpha\) need not be an integer (order 0.5 gives 3.010 against a predicted 3.010). ★★ But all four share one Fourier transform at the root, so this is not new physics ── it is tidying the tools (Exercise 1).
★★★ And Episode 45’s open item is closed ── \(-1/12\) comes out the same for five cutoffs (largest departure 6.7e−6), while the divergent coefficient \(A\) ranges over 0.443–1.832. ★ \(-1/12\) is not “the value of” \(1+2+3+\cdots\); it is what remains once the divergent term is subtracted.
★★ That is where Episode 60’s running variance parts company ── the Casimir case leaves a cutoff-independent remainder and the \(1/f^\beta\) variance does not, so one determines a physical quantity and the other only forces you to state the record length. Whether a running quantity can be given meaning depends on whether such a remainder survives.
★ The handbook is now 14 steps. ⑪ subtract a trend, ⑫ check the two conditions separately, ⑬ measure the error’s shape first, ⑭ give a variance its record length ── all of them declarations of where you cut.
★★★ The biggest item is unchanged: this series has still never touched real measured data.
★ What that return showed is that the limits of a measurement are set almost entirely by where you cut ── the edge (Ep. 57), the window (Ep. 58), the record length (Ep. 60), and the cutoff on a divergence (Ep. 45 and 61).
★★ Counting from Episode 1’s one line, \(6\alpha\) has now been confirmed along four routes. ★ If there is a next step, it is the first item on the list ── real data.
This document is Episode 61 of the “Thinking in Waves” series, closing Part XIX. The computations are in kenshou/shime.js (own FFT, zero dependencies). On the numbers in this episode: the regression test of Episodes 56–60, the four routes to \(6\alpha\), and the five-cutoff check of \(-1/12\) are all computed here. Zeta-function regularisation and \(\zeta(-1)=-1/12\), its independence of the regulator, and the Fourier representation of fractional differentiation are all well-known results and are not discoveries of this episode ── what this episode did was verify numerically, with five cutoffs, the point Episode 45 marked as unchecked; find that Episode 56’s intercept carried too many digits; and gather the four operations that land on \(6\alpha\) into one table. §03 tried five cutoffs, which does not establish the result for all of them (the analytic reason is sketched in Exercise 2, and that argument is standard, not ours). The larger gaps in route ③ of §02 come from the coarse sampling of the sidelobe envelope. This episode, and every episode of Part XIX, uses synthetic data only and has not been applied to real measurements. ── To print, use your browser’s Print → Save as PDF.