Example 1: The ⁠Rademacher system, named after Hans Rademacher (1892--1969), is an orthogonal sequence of square-wave functions \[ r_n (x) = \mbox{sgn}\left( 2^n \pi x \right) , \qquad n \in \mathbb{N} = 0,1,2,\ldots, \] defined on the unit interval [0, 1]. Key attributes include being an incomplete orthonormal system, stochastically independent, and forming the foundation for the complete Walsh system.

Each Rademacher function alternates signs between −1 and +1 over intervals of length 2⁻⁽ⁿ⁺¹⁾. The n-th Rademacher function can also be defined as \[ r_n (t) = (-1)^{\left\lfloor 2^n t \right\rfloor} , \qquad t \in [0, 1] . \] In particular,

• r₀(𝑥) changes sign once at 𝑥 = ½,
• r₁(𝑥) changes sign at 𝑥 = ¼, ½, ¾,
• r₂(𝑥) changes sign every ⅛.

• The Rademacher functions are orthogonal in 𝔏²([0, 1]): \[ \langle r_m \mid r_n \rangle = \int_0^1 r_m (x)\,r_n (x)\,{\text d}x = \delta_{m,n} = \begin{cases} 1, \qquad& m=n , \\ 0, \qquad& m\ne n . \end{cases} \]
• They are normalized: ∥r∥₂ = 1.
• They are independent random variables when [0, 1] is equipped with Lebesgue measure.
• The Rademacher system is incomplete.

To prove that the Rademacher system is not a complete basis (not a full basis) for 𝔏²([0,1]), we must find a non-zero function 𝑓 ∈ 𝔏²([0,1]) that is orthogonal to every Rademacher function rₙ(𝑡) for all n = 0, 1, 2, …. If the system were complete, the only function orthogonal to every element would be the zero function.

The Explicit Counterexample Function: Define the function 𝑓(𝑡) on the interval [0, 1] as the product of the first two Rademacher functions: \[ f(t) = r_0 (t) \cdot r_1 (t) . \] Visualizing the function, we get

• r₀ is +1 on (0., 0.5) and −1 on (0.5, 1).
• r₁ is +1 on (0., 0.25) ∪ (0.5, 0.75) and −1 on (0.25, 0.5) ∪ (0.75, 1).
• Multiplying them yields 𝑓(𝑡), which takes the following values: \[ f(t) = \begin{cases} +1 , \qquad& t \in (0, 0.25), \\ -1 , \qquad& t \in (0.25, 0.5) , \\ -1 , \qquad& t \in (0.5, 0.75) , \\ +1 , \qquad& t \in (0.75, 1.) . \end{cases} \]

Proving Orthogonality of 𝑓(𝑡) to r₀(𝑡) and r₁(𝑡). We compute the inner product \( \displaystyle \quad \left\langle f \mid r_n \right\rangle = \int_0^1 f(t)\,r_n (t)\,{\text d}t \ \) for the first two cases:

• For n = 0: \[ \left\langle f , r_0 \right\rangle = \int_0^1 \,\left( r_0 (t) \cdot r_1 (t) \right) r_0 (t)\,{\text d}t = \int_0^1 \,\left( r_0 (t) \right)^2 \,r_1 (t) \,{\text d}t . \] Since \( \displaystyle \quad \left( r_0 (t) \right)^2 = 1 \ \) almost everywhere, we get \[ \left\langle f , r_0 \right\rangle = \int_0^1 \,r_1 (t)\,{\text d}t = 0. \]
• For n = 1: \[ \left\langle f , r_1 \right\rangle = \int_0^1 \,\left( r_0 (t) \cdot r_1 (t) \right) r_1 (t)\,{\text d}t = \int_0^1 \,\left( r_1 (t) \right)^2 \,r_0 (t) \,{\text d}t = \int_0^1 \,r_0 (t)\,{\text d}t = 0 . \]

Proving Orthogonality to Higher Rademacher Functions n ≥ 2. For any n ≥ 2, the function rₙ(𝑡) oscillates much faster than 𝑓(𝑡). We look at the inner product: \[ \left\langle f \mid r_n \right\rangle = \int_0^1 \,r_0 (t) \,r_1 (t) \,r_n (t)\,{\text d} t . \] Because Rademacher functions are stochastically independent random variables with an expected value (integral) of 0, the integral of their product equals the product of their integrals: \[ \int_0^1 \,r_0 (t) \,r_1 (t) \,r_n (t)\,{\text d} t = \left( \int_0^1 \,r_0 (t) \,{\text d} t \right) \cdot \left( \int_0^1 \,r_1 (t) \,{\text d} t \right) \cdot \left( \int_0^1 \,r_n (t) \,{\text d} t \right) = 0 \cdot 0 \cdot 0 = 0 . \] Alternatively, you can observe that on each of the four quadrants of length 0.25 where 𝑓(𝑡) is constant (+1 or −1), the function rₙ(𝑡) completes one or more full symmetric square-wave cycles. The integral of over each quadrant is exactly 0, causing the total integral to vanish.

This definitively proves that the Rademacher system lacks the completeness required to form a full basis for 𝔏².

Now we show that Rademacher functions are stochastically independent random variables with an expected value (integral) of 0.

Expected Value (Integral) is 0. For any Rademacher function , the function takes the value on exactly half of the interval [0, 1] and −1 on the other half. \[ \int_0^1 \,r_n (t)\,{\text d}t = \left( 1 \times \frac{1}{2} \right) + \left( -1 \times \frac{1}{2} \right) = 0 . \] In probability terms, the expected value E[rₙ] = 0.

Stochastic Independence via Dyadic Intervals: To prove that r₀, r₁, …, rₙ are stochastically independent, we must show that for any choice of signs ε ₖ ∈ {−1, +1}, the measure (length) of the set where they simultaneously hold those values matches the product of their individual probabilities: \[ \mu \left( \left\{ t \in [0, 1]\ : \ r_0 (t) = \varepsilon_0 \cdot r_1 (t) = \varepsilon_1 , \ldots , r_n (t) = \varepsilon_n \right\} \right) = \frac{1}{2^{n+1}} . \] because each successive Rademacher function splits the previous subintervals exactly in half, choosing a sequence of signs (ε₀, ε₁, … , εₙ) pinpoints exactly one specific dyadic interval of length 1/2ⁿ⁺¹. Since this matches \( \displaystyle \quad \prod_{k=0}^n\,\Pr [r_k = \varepsilon_k ] = (1/2)^{n+1} ,\quad \) the functions are mutually stochastically independent.

Rigorous Step for the Joint Integral of the inner product: For any measurable, independent random variables, the expectation of their product equals the product of their expectations (E[XYZ] = E[X] E[Y] E[Z]). Using the joint distribution (the previous step) to evaluate the integral for n ≥ 2: \begin{align*} \left\langle f(t) , r_n (t) \right\rangle &= \int_0^1 \,r_0 (t)\,r_1 (t)\,r_n (t) \,{\text d}t \\ &= \sum_{\varepsilon_0 , \varepsilon_1 , \varepsilon_n \in \{ 1, -1 \}} \ \left( \varepsilon_0 \cdot \varepsilon_1 \cdot \varepsilon_n \right) \cdot \mu \left( r_0 = \varepsilon_0 , \ r_1 = \varepsilon_1 , \ r_n = = \varepsilon_n \right) \\ &= \sum_{\varepsilon_0 , \varepsilon_1 , \varepsilon_n \in \{ 1, -1 \}} \ \left( \varepsilon_0 \cdot \varepsilon_1 \cdot \varepsilon_n \right) \cdot \left( \frac{1}{2}\cdot \frac{1}{2} \cdot \frac{1}{2} \right) \\ &= \left( \sum_{\varepsilon_0} \,\frac{\varepsilon_0}{2} \right) \cdot \left( \sum_{\varepsilon_1} \,\frac{\varepsilon_1}{2} \right) \cdot \left( \sum_{\varepsilon_n} \,\frac{\varepsilon_n}{2} \right) \\ &= \left( \int_0^1 \,r_0 (t)\,{\text d}t \right) \cdot \left( \int_0^1 \,r_1 (t)\,{\text d}t \right) \cdot \left( \int_0^1 \,r_n (t)\,{\text d}t \right) \\ &= 0\cdot 0 \cdot 0 = 0 . \end{align*} This clean separation completes the missing logical leap in the counterexample.

Summary Rademacher functions are independent because any intersection of their sign states yields a single dyadic interval of length 2⁻⁽ⁿ⁺¹⁾. Because they are independent and centered (E[rₖ] = 0), the integral of their product strictly separates into the product of their individual integrals, forcing ⟨ r₀r₁ ∣ rₙ ⟩ = 0.    ■

Example 2: The Rademacher system is closely related to the Walsh system. In fact, the Walsh functions are obtained as finite products of Rademacher functions. If \[ n = \sum_{k\ge 0} \varepsilon_k\,2^k , \qquad \varepsilon_k \in \{ 0, 1 \} , \] is the binary expansion of the integer n, then the n-th Walsh function is \[ w_n (x) = \prod_{k\ge 0} \,r_k (x)^{\varepsilon_k} , \] where only finitely many factors differ from 1.

⁠Khintchine's inequality states that for any sequence of real coefficients (𝑎₁, 𝑎₂, … , 𝑎ₙ) ∈ ℝⁿ and any 0 < p ≤ ∞, the norm of a linear combination of Rademacher functions is strictly equivalent to its 𝔏² norm. Spesifically, there exist positive constants A_p and B_p depending only on p such that \[ A_p \left( \sum_{i=1}^n \ a_i^2 \right)^{1/2} \le \left( \int_0^1 \left\vert \sum_{i=1}^n a_i r_i (t) \right\vert^p \,{\text d}t\right)^{1/p} \le B_p \left( \sum_{i=1}^n \ a_i^2 \right)^{1/2} . \] This inequality shows that for Rademacher series, all topologies coincide. This is highly unusual for general functional series, where higher moments typically grow much faster. Therefore, the Rademacher functions mimic independent fair coin flips, this inequality bounds the absolute moments of random walks and serves as a pillar in functional analysis and Banach space theory.

The Rademacher system is incomplete because it cannot represent functions that lack symmetry across dyadic intervals. To fix this, Raymond Paley (1907--1933) introduced a method to complete the system by taking all possible finite products of Rademacher functions, forming the complete orthonormal Walsh system.

The Paley Ordering Rule

To construct the m-th Walsh function in Paley order, look at the binary expansion of the integer m: \[ m = \sum_{k=0}^j \ b_k 2^k , \qquad b_k \in \{ 0, 1 \} . \] The function w_m(t) is then defined as \[ w_m (t) = \prod_{k=0}^j \ \left( r_k (t) \right)^{b_k} . \] Step-by-Step Generation Example

• w₀(𝑡): Binary representation is 0. The empty product equals 1.
• w₁(𝑡): Binary is 1 = 1·2⁰. Thus, w₁(𝑡) = r₀(𝑡).
• w₂(𝑡): Binary is 2 = 1·2¹ + 0·2⁰. Thus, w₂(𝑡) = r₁(𝑡).
• w₃(𝑡): Binary is 3 = 1·2¹ + 1·2⁰. Thus, w₃(𝑡) = r₁(𝑡) · r₀(𝑡).

Example 3:

Let us consider a sequence of Zernike polynomials that are orthogonal over the area of a two-dimensional unit disk. Named after Dutch optical physicist Frits Zernike (1888–1966), laureate of the 1953 Nobel Prize in Physics and inventor of phase-contrast microscopy, these polynomials are indispensable in advanced beam optics, imaging, and ophthalmology for modeling wavefront aberrations.

Zernike polynomials map circular regions natively by separating coordinates into a radial component \(r\) and an azimuthal angle \(\theta\). They are split into angularly even and odd functions. For non-negative integers satisfying \(n \ge m \ge 0\) where the difference \(n-m\) is an even number, the odd Zernike polynomials are defined as:

and the corresponding even Zernike polynomials are defined as:

Here, the variables are bounded within polar parameters \(0 \le r \le 1\) and \(0 \le \theta < 2\pi\). The term \(R_n^m(r)\) represents the highly oscillatory radial polynomial mapping, defined explicitly by the combinatorial sum: \[ R_n^m (r) = \sum_{k=0}^{\frac{n-m}{2}} \frac{(-1)^k \, (n-k)!}{k! \left(\frac{n+m}{2}-k\right)! \left(\frac{n-m}{2}-k\right)!} \, r^{n-2k} . \]

To align with our least squares framework, the inner product over the two-dimensional circular domain contains an explicit area element \({\text d}A = r\,{\text d}r\,{\text d}\theta\), indicating that the radial variable \(r\) itself acts as the internal weight density function. The double integral orthogonality relationship over the entire unit disk evaluates as follows:

where δ is the Kronecker delta, and \(\epsilon_m\) is the Neumann factor (evaluating to \(2\) if \(m=0\), and \(1\) otherwise). This strict 2D orthogonality means any distorted optical phase wavefront can be reconstructed in the least-squares sense with absolutely zero informational overlap between its expansion coefficients.

Zernike polynomials are orthogonal over a 2D unit disk and are widely used in optics to model wavefront aberrations. They separate coordinates into a radial component \(r\) and an azimuthal angle \(\theta\), categorized into even and odd functions defined via radial polynomials \(R_n^m(r)\).

Explicit Low-Order Zernike Polynomial Formulas

The following table summarizes the explicit mathematical formulas for lower-order unnormalized Zernike polynomials \(Z_n^{\pm m}(r, \theta)\) and their corresponding optical aberration names:

The inner product and orthogonality properties over the circular domain allow wavefronts to be reconstructed accurately using least-squares expansion coefficients.

Example 4:

In introductory textbooks, least squares problems typically involve smooth, symmetric functions with clean, closed-form analytic solutions. However, in physical applications—such as localized atmospheric thermal turbulence or complex structural lens defects—the target function is often highly non-linear, off-center, and analytically non-integrable. In these cases, numerical computer calculations are strictly essential to evaluate the generalized Fourier coefficients.

Let us consider an asymmetrical, off-center thermal bloom pocket modeled as a localized Gaussian anomaly superimposed on a radial cubic deformation over the unit circular aperture: \[ f(r, \theta) = 3.5 \exp\left\{ -8 \left[ (r\cos\theta - 0.35)^2 + (r\sin\theta + 0.25)^2 \right] \right\} - 1.2r^3 . \]

Because of the shifted coordinates inside the exponential term, computing the projection inner product integrals \(\langle f, Z_n^{\pm m} \rangle\) analytically is impossible. We must deploy 2D numerical quadrature (numerical integration) to extract the best approximation coefficients in the mean-square sense.

Mathematica 14 Simulation & Optimization Script

You can execute the following complete, self-contained Mathematica 14 script to numerically compute the generalized Fourier-Zernike coefficients up to radial degree \(N=4\) via adaptive 2D quadrature and visualize the structural minimization of the mean square deviation.

Pedagogical Elements:

• The Role of the Jacobian Coordinate Weight: Inside the NIntegrate[] function loop, the expression explicitly contains * r matching the polar element metric rdr&thicksp;dθ. This demonstrates how polar space inherently alters the internal geometric weight density.
• Mean Convergence vs. Local Over-smoothing: Students will observe in the generated errorPlot that while global least-squares error metrics approach zero, localized spike errors persist near the coordinate point \((0.35, -0.25)\). Truncating at low radial frequencies (\(N=4\)) naturally smooths out steep structural gradients, highlighting why higher modal bounds are required to capture point-wise spatial variations.