PROJECT 02 · MATHEMATICAL MUSIC

Mathematical Coincidences
in Musical Structure

Development of the Myerthall–Owens Theorem

A General Theory Developed Using Flowers in the Rain

James MyerthallConcept Originator: Ryaed Owens

September 2, 2026

1

1 / INTRODUCTION

Can mathematical structure point to musical change?

The paper develops two related constructions for proposing salient locations in a piece: an exponential sequence based on the golden ratio, and trigonometric candidates built from a duration-adjusted oscillation. It then combines them, applies a separate circle-based index filter, and compares the retained candidates with score changes read by the author.

The interactive figures below make those steps inspectable. They show how the formulas generate positions and how sensitive the reported agreement is to the selected duration, filter, and error rule. Numerical resemblance motivates the model; the score annotations provide exploratory comparisons rather than proof of a general law.

2 / OWENS’ THEORY

Choose a scale, then divide it by ϕ.

The first move is to adjust a piece’s bar count by a factor built from the golden ratio and a nearby unit-circle angle. Repeated division by ϕ then proposes progressively earlier musical landmarks.

A / REWRITE THE RATIO

Two forms make the scale components visible.

Start with ϕ = (1 + √5) / 2. Rewriting ϕ as a/b gives a₁ = 1 + √5 and b₁ = 2. Rewriting it once more gives ϕ = (3 + √5) / (1 + √5), so a₂ = 3 + √5 and b₂ = 1 + √5.

GOLDEN RATIOϕ = (1 + √5) / 2 = a₁ / b₁
REWRITTEN FORMϕ = (3 + √5) / (1 + √5) = a₂ / b₂
UNIT-CIRCLE DIAMETERb₁ = 2
B / PICK A NEARBY ANGLEθc IN STEPS OF π/6

The construction compares a₁ and a₂ with angles spaced every 30°. Move the selector to see the distance from each candidate.

Comparison of golden-ratio construction values with nearby circle angles A dial marks a1, a2, and the selected angle in increments of pi over six.
|θc − a₁|2.000000
|θc − a₂|0.000080

At θc = 5π/3, a₂ differs by about 0.000080 radians. This is the paper’s closest match on its chosen π/6 grid.

2.1 / TRUE DURATION OF THE PIECE

TRUE DURATION RULEtO = L · 3(3 + √5) / 5π
59-BAR EXAMPLE59.000904 bars
SCALE CHANGE100.0015%

The adjustment is almost neutral in length; choosing the factor above 1 is a modelling convention, not a meaningful change to the score duration.

2.2 / EXPONENTIALLY OCCURRING SALIENT POINTS

C / EXPONENTIAL LANDMARKSτq = tO · ϕ⁻q
Owens’ exponential landmark sequence A falling exponential curve marks the first five golden-ratio divisions of the 59-bar piece.
Bar positions shrink geometrically; the spacing between neighbouring marks also shrinks.
τ0 / (L ÷ 5π/3)3 + √5 = a₂
τ1 / (L ÷ 5π/3)1 + √5 = a₁ = b₂
τ2 / (L ÷ 5π/3)2 = b₁

These values reappear because the same golden-ratio construction defines the sequence. The paper presents them as consistency checks, not independent evidence.

SCORE EVIDENCE / FLOWERS IN THE RAIN

The paper points to melody changes near bars 8, 14, and 36, and a major-to-minor shift around bar 23. Those are the author’s score readings used to interpret the predicted positions.

See all reported score readings and matches ↓

2.4 / MEASURED MOVEMENT-LENGTH CORRELATIONS

A close fit is still an approximation.

The paper links its scale choices to a golden triangle, a unit-circle projection, and the angles of a pentagon. Adjust the measurements to see where the numerical coincidences hold.

2.4.1 / GOLDEN-TRIANGLE RELATIONSHIPFLOWERS IN THE RAIN

The three movements are 59, 95, and 95 bars. In an ideal 36° / 72° / 72° triangle, the sine rule gives b/a = sin 72° / sin 36° = 2 cos 36° = ϕ.

Isosceles triangle made from three movement lengths A 59-bar base and two adjustable equal movement lengths form an isosceles triangle.
MEASURED RATIO1.610169
GOLDEN RATIO ϕ1.618034
RELATIVE GAP0.486%

Measured angles: 36.182° / 71.909° / 71.909°. The ideal triangle is 36° / 72° / 72°.

2.4.2 / UNIT-CIRCLE AND PENTAGONAL CONNECTIONPENTAGONAL ANGLES

A pentagon divides a turn into 72° sectors. Bisecting gives 36°, where twice the cosine equals ϕ; halving again gives the reciprocal identity.

Adjustable angle and its projection on the unit circle A radius is decomposed into horizontal cosine and vertical sine components.
2 cos θ compare to ϕ1.618034
2 sin(θ/2) compare to 1/ϕ0.618034
UNIT RADIUS1.000000

At exactly 36°, x = cos 36° ≈ 0.809 and y = sin 36° ≈ 0.588; x² + y² = 1. The diameter is 2, and 2 sin 18° = 1/ϕ ≈ 0.618034.

3 / MYERTHALL’S THEORY

Turn the duration into a wave.

Owens’ divisions identify only a few points. Myerthall starts from a second near-match and asks whether sine and cosine can supply more candidate positions between them.

A / A SECOND NEAR-MATCH

ϕ sits close to π/2, where sine is almost flat.

ϕ ≈ 1.618 radians and π/2 ≈ 1.571 radians, a gap of about 0.04724. Since sine reaches its maximum at π/2, their sine values differ by only about 0.00112.

ANGLE GAP|ϕ − π/2| ≈ 0.04724
SINE GAP|sin(ϕ) − 1| ≈ 0.001115

3.1 / TRUE DURATION

MYERTHALL DURATIONtM = L · ϕ / (π/2) = L(1 + √5)/π
59-BAR EXAMPLE60.774273 bars

Here the scale is about 103.007% of the notated length. As with tO, this is an explicit construction choice.

B / WHY THE SINE VALUES ARE CLOSEy = sin θ
Sine curve near pi over two and phi The sine function is near its maximum at both pi over two and phi; the markers differ slightly in angle and height.
Near its maximum, the sine curve changes slowly: the angles are visibly apart while their function values nearly coincide.
C / GENERATE MORE CANDIDATES

One cycle becomes four.

The first attempt uses one full cycle across the adjusted duration. Intersections, maxima, and minima become candidate landmarks.

fs(n) = sin(2πn/tM)    fc(n) = cos(2πn/tM)
Trigonometric candidate generation A plot will compare trigonometric functions and mark their candidate roots and extrema.
One period uses the full Myerthall duration tM.
What the first pass reveals: denser candidates can sit nearer a reported change, but they also create false candidates. For Flowers in the Rain, 10 of the 15 raw points lie within one bar under the paper’s earlier offset convention; the later filter keeps eight indices, seven of which the paper associates with boundaries within one bar.

3.2.2 / REFINEMENT WITH THE THREE-GAP THEOREM

Repeated steps. Three gap sizes.

Wrap an irrational step around the unit circle. The points keep arriving in a new order, while the spaces between neighbours settle into at most three lengths.

Rotation points and gaps on a unit circle Points created by repeatedly applying the selected circle rotation. Each arc represents the gap to the next point.
Golden rotation points m/ϕ, m = 1…15.

The model adds a separate filter. It retains candidate index m when {m/ϕ} ≤ ½. That semicircle cutoff is a modelling choice, not a consequence of the Three-Gap Theorem.

THE SAME ROTATION IN PITCH SPACE

Multiplying a frequency becomes adding a circle step.

Take C = 256 Hz and a just-intonation perfect fifth G = 384 Hz, so ρ = 3/2. Repeating the interval multiplies by ρ; taking log₂ turns that multiplication into a repeated step, then the fractional part removes complete octaves.

INTERVAL RATIOρ = νN / νCG/C = 384/256 = 3/2
LOGARITHMIC STEPα = log₂ ρone fifth = 0.584963 octaves
WRAP INTO ONE OCTAVExm = {m log₂ ρ}discard the integer octaves
AFTER TWO FIFTHS(3/2)² = 9/4 = 2 × (9/8)

The factor 2 is one complete octave. Removing it leaves the just-intonation whole tone 9/8, at circle position {log₂(9/8)} ≈ 0.16993.

m 0C0.00000m 1G0.58496m 2D0.16993m 3A0.75489m 4E0.33985m 5B0.92481m 6F♯0.50978m 7C♯0.09474m 8G♯0.67970

An octave has log₂(2) = 1 and returns to the same pitch-class point; an equal-tempered semitone has step 1/12 and repeats after 12 steps. The exact 3/2 fifth has an irrational log₂ step, so its circle rotation never closes exactly.

4 / UNIFIED THEORY

The Myerthall–Owens Theorem.

The unified function multiplies Owens’ exponential envelope by Myerthall’s trigonometric oscillation. Its zeros and stationary points are ordered first, then filtered by the same circle-index rule.

FROM FUNCTION TO CANDIDATE LISTL = 59 bars
THE FUNCTIONf(n) = (tO/2) e−λn sin(ωn)λ = ln ϕ   ·   ω = 8π/tM
DIFFERENTIATEf′(n) = (tO/2)e−λn[ω cos(ωn) − λ sin(ωn)]Set the bracket to zero for stationary points.
ROOT FAMILIESuk = kπ/ω    vk = (atan(ω/λ)+kπ)/ωZeros interleave with maxima and minima.
λ0.481212ω0.413606 rad/barfirst stationary phase1.71666 bars
ORDERED cₘ / INSIDE THIS SCORE

Each cₘ is the next positive zero or stationary point after sorting both families together. Only cₘ ≤ L belongs to the score. Since tM > L, c₁₆ = tM is already outside the domain; the paper’s retained set through c₁₅ is {2, 4, 5, 7, 10, 12, 13, 15}. The damping uses absolute bar position, so the candidates are recalculated for each piece length rather than copied as fixed percentages. The circle theorem does not imply the separate filter {m/ϕ} ≤ ½.

LANDMARK RULE

UNIFIED FUNCTIONExponential decay scales a sine wave. Its positive zeros and stationary points form the ordered candidate sequence cₘ.

f(n) = tO/2 · ϕ⁻ⁿ · sin(8πn/tM)
59-BAR SCORE / UNIFIED CANDIDATESCONTINUOUS CANDIDATE ↔ REPORTED SCORE BAR
Mathematical landmarks compared with score boundaries An interactive plot will show the selected model over the piece’s bar coordinate and list reported boundaries. Interactive model plot
The curve is normalized vertically to show its decay; bar locations retain their paper values.HOVER OR FOCUS A MARK FOR ITS VALUE
REPORTED BOUNDARYKEPT BY FILTERFILTERED OUTOWENS POINT
OWENS TRUE DURATION59.001bars · adjusted scale
MYERTHALL TRUE DURATION60.774bars · one full period
RETAINED CANDIDATES8 / 15retained points inside duration
DIRECT MATCHES4 / 8at ±1.00 bar

PAPER VALUES / SELECTED RULE

Reported candidate alignment

Signed offset = chosen integer candidate − reported score boundary. The paper reports the exact offsets below.

INDEX mMODEL POSITION cₘSCORE BOUNDARY rTABLE OFFSET dCONTINUOUS e = cₘ − rWITHIN ±1?
Loading the paper’s reported values…

Two residuals, two readings. The earlier tables choose either floor(cₘ) or ceil(cₘ) according to which integer is closer to r, then report d = chosen integer − r. Section 5.2 deliberately drops that integer step and evaluates e = cₘ − r directly. Use the selector above to see how the totals change.

FINAL CROSS-PIECE EVALUATION / DIRECT CONTINUOUS ERROR

The paper’s stated headline uses the continuous residual at a one-bar tolerance.

CANDIDATE TRUE POSITIVES18 / 32
PRECISION56.3%
95% WILSON INTERVAL39.3–71.8%
EARLIER TABLE OFFSETS29 / 3290.6% · more permissive rule

The outcomes come from only four pieces, and candidates within a piece may not be independent. The paper treats this interval as descriptive, not a precise estimate of performance on music generally.

5 / THE MYERTHALL–OWENS THEOREM IN PRACTICE

Applications across four scores.

The paper applies the retained-candidate rule to four pieces ranging from 59 to 264 bars. The three additional works were chosen for recognition and varied duration. Its final precision calculation compares each continuous point directly with its matched score boundary.

5.1 / APPLICATIONS

DIRECT CONTINUOUS RESIDUAL e EARLIER TABLE OFFSET d

At ±1 bar, the change in comparison rule shifts the apparent precision. Section 5.2 explicitly discards the earlier integer choice for its aggregate calculation.

A / FOUR COMPOSITIONS

Different lengths, same eight retained indices.

The model carries m = {2, 4, 5, 7, 10, 12, 13, 15} across every score. Continuous residuals produce these one-bar results:

Flowers in the Rain · 59 bars4 / 8 · 50%
Arabesque No. 1 · 107 bars4 / 8 · 50%
Rondo alla Turca · 137 bars5 / 8 · 62.5%
Ballade No. 1 · 264 bars5 / 8 · 62.5%
5.2 / OVERALL PREDICTION PRECISION B / UNCERTAINTY

The range is wide.

For 18 true positives in 32 candidates, the paper retains a 95% Wilson score interval of 39.3% to 71.8%. The considered Wald interval was 39.1% to 73.4%; the paper uses Wilson for the small sample.

The paper cautions that the 32 candidate results come from only four pieces and may not be independent within a piece. It treats the interval as descriptive, not a general performance estimate.

C / SCORE-READING NOTESTHE OBSERVED SIDE IS HUMAN-ANNOTATED

Flowers in the Rain

  • 8Melody changes across bars 5–8.
  • 14Melody changes around bars 13–17.
  • 20Key signature and melody change across bars 18–22.
  • 25Tempo and dynamics change around bars 23–25.
  • 36Time signature and melody change around bars 36–39.
  • 46Key signature and melody change around bars 44–48.
  • 55Dynamics and tempo change entering bar 55.

Arabesque No. 1

  • 13Mood and dynamics shift.
  • 26–28Melody and tempo change; a new phrase begins.
  • 43A new phrase begins.
  • 67A quiet, slowing close builds suspense before the main melody returns.
  • 80A new phrase begins with a gradual crescendo.
  • 85A phrase shift accompanies a change in register.
  • 99The final main-melody repetition grows softer toward the end.

The paper includes detailed score excerpts for these two works. For Rondo alla Turca and Ballade No. 1 it reports candidate tables but omits corresponding score-excerpt verification; the page keeps those cases visible in the interactive table without inventing musical annotations.

D / THE ARGUMENT’S DECISION TRAIL
  1. Noticeϕ, 5π/3, π/2, and 36° are numerically close or geometrically linked.
  2. ConstructUse those choices to define two adjusted durations and candidate functions.
  3. RefineMerge trigonometric zeros and extrema; apply a separately proposed semicircle cutoff.
  4. InterpretCompare continuous positions with score changes identified by reading the score.
  5. RecalculateUse direct continuous error for the final 18/32 precision result.
  6. LimitAsk for more pieces and do not claim a universal law from four examples.

6 / CONCLUSION

A model worth testing, not a law established.

The equations construct predictions. The score annotations are exploratory observations. The gap between those things is where future testing belongs.

A

What is predicted

Continuous bar positions from a defined function. The retained Myerthall–Owens candidates are zeros and stationary points whose index passes the semicircle filter.

B

What is observed

Score changes identified by the author: phrase shifts, changes in melody, key, tempo, dynamics, or mood. Boundary placement is interpretive, and the matching procedure can assign nearby predictions to the same boundary.

C

What the final metric shows

With continuous residuals and a one-bar tolerance, 18 of 32 retained candidates match: 56.3%. The earlier rounded-offset tables produce a different, more permissive count.

D

What remains open

Four examples cannot establish a universal musical law. The paper calls for broader, more varied testing and treats mismatches as evidence to investigate.

SOURCE

James Myerthall, Mathematical Coincidences in Musical Structure: Development of the Myerthall–Owens Theorem, 2 September 2026. Concept originator: Ryaed Owens. The paper’s appendices include Owens’ whiteboard demonstration and score material for the four examples.

RETURN TO MODEL ↑Appendices A–E, original paper pages 34–47.