In [2]:
using Plots, ComplexPhasePortrait, ApproxFun, SingularIntegralEquations, DifferentialEquations
gr();
Dr. Sheehan Olver
s.olver@imperial.ac.uk
Website: https://github.com/dlfivefifty/M3M6LectureNotes
This lecture we do the following:
Given a family of orthogonal polynomials $p_k(x)$ with respect to the weight $w(x)$ on $(a,b)$, we always know it satisfies a three-term recurrence: \begin{align*} x p_0(x) &= a_0 p_0(x) + b_0 p_1(x) \\ x p_k(x) &= c_k p_{k-1}(x) + a_k p_k(x) + b_k p_{k+1}(x) \end{align*}
Consider now the Cauchy transform of the weighted orthogonal polynomial: $$ Q_k(z) := {\cal C}_{[a,b]}[p_k w](z) = {1 \over 2 \pi \I} \int_a^b {p_k(x) w(x) \over x -z} \dx $$
Theorem (Three-term recurrence Cauchy transform of weighted OPs) $C_k(z)$ satisfies the same recurrence relationship as $p_k(x)$ for $k=1,2,\ldots$: \begin{align*} z C_0(z) &= a_0 C_0(z) + b_0 C_1(z) - {1 \over 2 \pi \I} \int_a^b w(x) \dx \\ z C_k(z) &= c_k C_{k-1}(z) + a_k C_k(z) + b_k C_{k+1}(z) \end{align*}
Proof
\begin{align*}
z C_k(z) &= {1 \over 2 \pi \I} \int_a^b {z p_k(x) w(x) \over x -z} \dx = {1 \over 2 \pi \I} \int_a^b {(z -x) p_k(x) w(x) \over x -z} \dx + \int_a^b {x p_k(x) w(x) \over x -z} \dx \\
&= -{1 \over 2 \pi \I} \int_a^b p_k(x) w(x) \dx + \int_a^b {(c_k p_{k-1}(x) + a_k p_k(x) + b_k p_{k+1}(x) w(x) \over x -z} \dx \\
&= -{1 \over 2 \pi \I} \int_a^b p_k(x) w(x) \dx + c_k C_{k-1}(z) + a_k C_k(z) + b_k C_{k+1}(z)
\end{align*}
when $k > 0$, the integral term disappears.
⬛️
This gives a convenient way to calculate the Cauchy transforms: if we know $C_0(z) ={\cal C}w(z)$ and $\int_a^b w(x) \dx$, solve the lower triangular system: $$ \begin{pmatrix} 1 \\ a_0-z & b_0 \\ c_1 & a_1-z & b_1 \\ &c_2 & a_2-z & b_2 \\ && c_3 & a_3-z &\ddots\\ &&&\ddots & \ddots \end{pmatrix}\begin{pmatrix}C_0(z) \\C_1(z) \\C_2(z) \\C_3(z) \\ \vdots \end{pmatrix} = \begin{pmatrix}C_0(z) \\{1 \over 2 \pi \I} \int_a^b w(x) \dx \\0 \\0 \\ \vdots \end{pmatrix} $$
Example (Chebyshev Cauchy transform)
Consider the Chebyshev case $w(x) = {1 \over \sqrt{1-x^2}}$, which satisfies $\int_{-1}^1 w(x) \dx = {\pi}$. Recall that $$ C_0(z) ={\cal C}w(z) = { \I \over 2\sqrt{z-1}\sqrt{z+1}} $$ Further, we have \begin{align*} x T_0(x) = T_1(x) \\ x T_k(x) = {T_{k-1}(x) \over 2} + { T_{k+1}(x) \over 2} \end{align*} hence \begin{align*} z C_0(z) = C_1(z) - {1 \over 2 \I} \\ z C_k(z) = {C_{k-1}(z) \over 2} +{C_{k+1}(z) \over 2} . \end{align*} In other words, we want to solve $$ \begin{pmatrix} 1 \\ -z & 1 \\ 1/2 & -z & 1/2 \\ &1/2 & -z & 1/2 \\ && 1/2 & -z &\ddots\\ &&&\ddots & \ddots \end{pmatrix}\begin{pmatrix}C_0(z) \\C_1(z) \\C_2(z) \\C_3(z) \\ \vdots \end{pmatrix} = \begin{pmatrix} { \I \over 2\sqrt{z-1}\sqrt{z+1}} \\{1 \over 2 \I} \\0 \\0 \\ \vdots \end{pmatrix} $$ with forward substitution.
In [5]:
x = Fun()
w = 1/sqrt(1-x^2)
z = 0.1+0.1im
n = 10
L = zeros(ComplexF64,n,n)
L[1,1] = 1
L[2,1] = -z
L[2,2] = 1
for k=3:n
L[k,k-1] = -z
L[k,k-2] = L[k,k] = 1/2
end
C = L \ [ im/(2sqrt(z-1)sqrt(z+1)); 1/(2im); zeros(n-2)]
Out[5]:
In [6]:
T₅ = Fun(Chebyshev(), [zeros(5);1])
cauchy(T₅*w,z) - C[6]
Out[6]:
Warning This fails for large $n$ or large $z$:
In [11]:
x = Fun()
w = 1/sqrt(1-x^2)
z = 5+6im
n = 100
L = zeros(ComplexF64,n,n)
L[1,1] = 1
L[2,1] = -z
L[2,2] = 1
for k=3:n
L[k,k-1] = -z
L[k,k-2] = L[k,k] = 1/2
end
C = L \ [ im/(2sqrt(z-1)sqrt(z+1)); 1/(2im); zeros(n-2)]
T₂₀ = Fun(Chebyshev(), [zeros(20);1])
C[21], cauchy(T₂₀*w, z)
Out[11]:
Get around it by dropping the first row:
In [12]:
L[2:end,1:end-1]
Out[12]:
In [13]:
C = L[2:end,1:end-1]\ [1/(2im); zeros(n-2)]
C[6]- cauchy(T₅*w, z)
Out[13]:
Now consider the Hilbert transform of weighted orthogonal polynomials: $$ H_k(x) = \HH_{[a,b]}[p_k w](x) = {1 \over \pi} \int_a^b {p_k(t) w(t) \over t-x} \dt $$
Just like Cauchy transforms, the Hilbert transforms have
Corollary (Hilbert transform recurrence) \begin{align*} x H_0(x) &= a_0 H_0(x) + b_0 H_1(x) - {1 \over \pi} \int_a^b w(x) \dx\\ x H_k(x) &= c_k H_{k-1}(x) + a_k H_k(x) + b_k H_{k+1}(x) \end{align*}
Proof Recall $$ \CC^+ f(x) + \CC^- f(x) = -\I \HH f(x) $$ Therefore, we have $$ C_k^+(x) + C_k^-(x) = -\I \HH[w p_k](x) $$ hence we have \begin{align*} x H_0(x) &= \I x (C_0^+(x) + C_0^-(x)) = \I \left[a_0 (C_0^+(x) + C_0^-(x)) + b_0 (C_1^+(x) + C_1^-(x)) -{1 \over \pi \I} \int_a^b w(x) \dx \right]\\ &= a_0 H_0(x) + b_0 H_1(x) - {1 \over \pi} \int_a^b w(x) \dx \end{align*} Other $k$ follows by a similar argument.
⬛️
For $$ H_k(x) = \int_{-1}^1 {T_k(t) \over (t-x)\sqrt{1-t^2}} \dt $$ The recurrence gives us \begin{align*} x H_0(x) &= H_1(x) -1 \\ x H_k(x) &= {H_{k-1}(x) \over 2} + {H_k(x) \over 2} \\ \end{align*} In this case, we have $H_0(x) = \HH[w](x) = 0 $. Therefore, we can rewrite this recurrence as \begin{align*} H_1(x)&= 1 \\ x H_1(x) &= {H_2(x) \over 2} \\ x H_k(x) &= {H_{k-1}(x) \over 2} + {H_k(x) \over 2} \\ \end{align*} This is precisely the three-term recurrence satisfied by $U_{k-1}$! We therefore have $$ H_k(x) = U_{k-1}(x) $$
In [14]:
x = 0.1
T = Fun(Chebyshev(),[zeros(n);1])
hilbert(w*T,x) - Fun(Ultraspherical(1), [zeros(n-1);1])(x)
Out[14]:
This gives a very easy way to compute Hilbert transforms: if $$ f(x) = \sum_{k=0}^\infty f_k T_k(x) $$ then $$ \HH\left[{f \over \sqrt{1-\diamond^2}}\right](x) = \sum_{k=0}^\infty f_{k+1} U_k(x) $$