The coefficient lattice
Click a point━ p + q = n┄ p − q = k◎ Selected term
CONVOLUTION ATLAS THESIS WALKTHROUGH · FIRST SUBSECTION
Joukowski–Legendre Laurent expansion
One point on an ellipse. A finite collection of rotating powers.
Follow the geometry and the coefficient matching in thesis Lemma 1.17.
\(P_n(\mathit{Jouk}(z))=\displaystyle\sum_{j=0}^{n}\alpha_{n,j}z^{n-2j}\)\(\mathit{Jouk}(z)=\tfrac12(z+z^{-1}),\quad\alpha_{n,j}=\alpha_j\alpha_{n-j}\)
Connection to the thesis
This lesson follows Subsection 1.2.1 — Joukowski–Legendre Laurent expansion, in Section 1.2, Legendre coefficient bounds and projection tails on Bernstein ellipses.
Lemma 1.17 — Joukowski–Legendre Laurent expansion. For \(z\in\mathbb C\setminus\{0\}\) and \(n\in\mathbb N_0\),
\[P_n(\mathit{Jouk}(z))=\sum_{j=0}^{n}\alpha_{n,j}z^{n-2j},\qquad \alpha_{n,j}=\alpha_j\alpha_{n-j}.\]
The thesis attributes this identity to Bandtlow’s EDMD–Legendre Lemma 1.1. That is the cited source’s numbering; Lemma 1.17 is its number in the thesis. The manuscript remains the mathematical source.
The walkthrough uses the thesis notation: generating variable \(s\), circle radius \(\varrho\), and frequency \(m=n-2j\). The existing exploration views use the equivalent names \(u=s\), \(r=\varrho\), and \(k=m\). We study the unscaled mode \(P_n\) here.
Degree 3 · selected j = 1 · (p, q) = (2, 1) · frequency m = +1
One selection. Both views.
Click a lattice point or a frequency bar. Gold follows that same term into the geometry; the dashed green ring marks its reciprocal partner.
As t changes, the lattice and coefficient magnitudes stay fixed. The phase rotates each contribution, changing the pair sum and complete mode.
Selected coefficient Cⱼ
×Rotating phase eⁱᵏᵗ
=Contribution now · gold ring
Selected summand, all terms & the underlying map
Select a degree; read off its frequencies.
The violet diagonal selects the terms multiplying the same power of u. Each crossing with a dashed frequency line gives one Laurent term.
One point → one summand
Degree p + q
Frequency p − q
Structural weight
Radial factor rᵏ
The structural weight belongs to the mode shape. It is not the Legendre coefficient A.
All terms on this diagonal
| j | (p, q) | k | αⱼ αₙ₋ⱼ | Select |
|---|
Fourier coefficients of this one Legendre mode
Bar height = |A αⱼ αₙ₋ⱼ rᵏ|. The phase factor rotates the contribution; it does not change its magnitude.
● Positive k · anticlockwise● Negative k · clockwise● k = 0 · constant
Two counter-rotating terms. One ellipse.
Move t or press Play. Dashed circles are the individual component paths. Vectors placed head-to-tail show their sum, with equal horizontal and vertical scales.
The selected reciprocal pair
● Positive component● Negative component● Pair sum
The complete mode
All n + 1 terms● A Pₙ(Γᵣ(t))━ Selected summand
The underlying Joukowski map
\(z=re^{it}\ \longmapsto\ \Gamma_r(t)=\frac12(z+z^{-1})\)
The circle in the z-plane maps to the ellipse in the ω-plane. These are coordinates where the polynomial is evaluated; the plots above show the values of its components.
Why this sum is a convolution.
The auxiliary variable u keeps track of degree. Multiplying the two series combines degrees p and q; extracting uⁿ selects exactly p + q = n.
Factor the quadratic
\(1-2\omega u+u^2=(1-zu)(1-z^{-1}u)\)
Substitute 2ω = z + z⁻¹, then group:
(1 − zu) − z⁻¹u(1 − zu).
Expand each factor
\((1-zu)^{-1/2}=\sum_{p\ge0}\alpha_p z^p u^p\)
\((1-z^{-1}u)^{-1/2}=\sum_{q\ge0}\alpha_q z^{-q}u^q\)
Match the degree
\([u^n]\sum_{p,q\ge0}\alpha_p\alpha_q z^{p-q}u^{p+q}\)
Set q = j, p = n − j. The sum p + q fixes n; the difference p − q becomes n − 2j.
Watch the generating series converge
This compares the unscaled mode shapes Pₘ. The separate amplitude A is not part of the generating function.
Increase N to add more terms to the partial sum.
The plot fits the partial sums and exact value, with equal horizontal and vertical scales.
\(G_N(u)=\sum_{m=0}^{N}P_m(\Gamma_r(t))u^m\)
\(G(u)=(1-zu)^{-1/2}(1-z^{-1}u)^{-1/2}\)
- Partial sum (through degree N)
- Exact generating function G
- Absolute truncation error
- Geometric tail bound
The tail bound concerns truncation, not floating-point rounding. Both analytic branches equal 1 at u = 0. Here u is real, 0 ≤ u < 1/r, so both binomial expansions converge absolutely.
Where do the coefficients αⱼ come from?
Taylor-expand (1 − v)⁻¹ᐟ² at zero. Repeated differentiation gives
\(\frac{d^j}{dv^j}(1-v)^{-1/2}=\frac{1\cdot3\cdots(2j-1)}{2^j}(1-v)^{-j-1/2}.\)
Dividing the derivative at zero by j! gives the Taylor coefficient (the empty product is 1):
\(\alpha_j=\frac{1\cdot3\cdots(2j-1)}{2^j j!}=\frac{(2j)!}{4^j(j!)^2}=\frac{\binom{2j}{j}}{4^j}.\)
\(\alpha_0=1,\quad\alpha_{j+1}=\frac{2j+1}{2j+2}\alpha_j,\quad(1-v)^{-1/2}=1+\frac12v+\frac38v^2+\frac5{16}v^3+\cdots\)
The ratio of successive terms tends to |v|, so the series converges absolutely for |v| < 1. Analyticity of the branch at zero identifies its Taylor sum with the function.
Keep the amplitude separate from the shape.
\(\underbrace{a_n(f)}_{\text{one mode's amplitude}}\,\underbrace{P_n(\Gamma_r(t))}_{\text{mode shape}}=\sum_{j=0}^n\underbrace{a_n(f)\alpha_j\alpha_{n-j}r^{n-2j}}_{\text{complex Fourier coefficient}}\,\underbrace{e^{i(n-2j)t}}_{\text{rotating phase}}\)
The pull-back changes the representation of Pₙ, while its Legendre coefficient stays the same. Several Legendre modes can share a Fourier frequency; their contributions add in the full function.
How aₙ(f) is obtained
\(a_n(f)=\frac{2n+1}{2}\int_{-1}^1 f(x)P_n(x)\,dx\)
For f ∈ L²(−1,1), these are the Legendre coefficients on the interval. Evaluating the full expansion on an ellipse additionally requires an analytic extension and convergence there. This page explores a single polynomial mode, whose finite identity is valid for every z ≠ 0.