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In ancient times (i.e. 2nd year undergrad) I recall being very impressed with Tschebyscheff polynomials for designing lowpass filters. I’d used Tschebyscheff filters for the hardware we used for a speech recognition system our group built in the design lab. One of the benefits of these polynomials is that the oscillation in the \( \Abs{x} < 1 \) interval is strictly bounded. This same property, as well as the unbounded nature outside of the \( [-1,1] \) interval turns out to have applications to antenna array design.

The Tschebyscheff polynomials are defined by

\begin{equation}\label{eqn:tschebyscheff:40}

T_m(x) = \cos\lr{ m \cos^{-1} x }, \quad \Abs{x} < 1 \end{equation} \begin{equation}\label{eqn:tschebyscheff:60} T_m(x) = \cosh\lr{ m \cosh^{-1} x }, \quad \Abs{x} > 1.

\end{equation}

### Range restrictions and hyperbolic form.

Prof. Eleftheriades’s notes made a point to point out the definition in the \( \Abs{x} > 1 \) interval, but that can also be viewed as a consequence instead of a definition if the range restriction is removed. For example, suppose \( x = 7 \), and let

\begin{equation}\label{eqn:tschebyscheff:160}

\cos^{-1} 7 = \theta,

\end{equation}

so

\begin{equation}\label{eqn:tschebyscheff:180}

\begin{aligned}

7

&= \cos\theta \\

&= \frac{e^{i\theta} + e^{-i\theta}}{2} \\

&= \cosh(i\theta),

\end{aligned}

\end{equation}

or

\begin{equation}\label{eqn:tschebyscheff:200}

-i \cosh^{-1} 7 = \theta.

\end{equation}

\begin{equation}\label{eqn:tschebyscheff:220}

\begin{aligned}

T_m(7)

&= \cos( -m i \cosh^{-1} 7 ) \\

&= \cosh( m \cosh^{-1} 7 ).

\end{aligned}

\end{equation}

The same argument clearly applies to any other value outside of the \( \Abs{x} < 1 \) range, so without any restrictions, these polynomials can be defined as just

\begin{equation}\label{eqn:tschebyscheff:260}

\boxed{

T_m(x) = \cos\lr{ m \cos^{-1} x }.

}

\end{equation}

### Polynomial nature.

Eq. \ref{eqn:tschebyscheff:260} does not obviously look like a polynomial. Let’s proceed to verify the polynomial nature for the first couple values of \( m \).

- \( m = 0 \).\begin{equation}\label{eqn:tschebyscheff:280}

\begin{aligned}

T_0(x)

&= \cos( 0 \cos^{-1} x ) \\

&= \cos( 0 ) \\

&= 1.

\end{aligned}

\end{equation} - \( m = 1 \).\begin{equation}\label{eqn:tschebyscheff:300}

\begin{aligned}

T_1(x)

&= \cos( 1 \cos^{-1} x ) \\

&= x.

\end{aligned}

\end{equation} - \( m = 2 \).\begin{equation}\label{eqn:tschebyscheff:320}

\begin{aligned}

T_2(x)

&= \cos( 2 \cos^{-1} x ) \\

&= 2 \cos^2 \cos^{-1}(x) – 1 \\

&= 2 x^2 – 1.

\end{aligned}

\end{equation}

To examine the general case

\begin{equation}\label{eqn:tschebyscheff:340}

\begin{aligned}

T_m(x)

&= \cos( m \cos^{-1} x ) \\

&= \textrm{Re} e^{ j m \cos^{-1} x } \\

&= \textrm{Re} \lr{ e^{ j\cos^{-1} x } }^m \\

&= \textrm{Re} \lr{ \cos\cos^{-1} x + j \sin\cos^{-1} x }^m \\

&= \textrm{Re} \lr{ x + j \sqrt{1 – x^2} }^m \\

&=

\textrm{Re} \lr{

x^m

+ \binom{ m}{1} j x^{m-1} \lr{1 – x^2}^{1/2}

– \binom{ m}{2} x^{m-2} \lr{1 – x^2}^{2/2}

– \binom{ m}{3} j x^{m-3} \lr{1 – x^2}^{3/2}

+ \binom{ m}{4} x^{m-4} \lr{1 – x^2}^{4/2}

+ \cdots

} \\

&=

x^m

– \binom{ m}{2} x^{m-2} \lr{1 – x^2}

+ \binom{ m}{4} x^{m-4} \lr{1 – x^2}^2

– \cdots

\end{aligned}

\end{equation}

This expansion was a bit cavaliar with the signs of the \( \sin\cos^{-1} x = \sqrt{1 – x^2} \) terms, since the negative sign should be picked for the root when \( x \in [-1,0] \). However, that doesn’t matter in the end since the real part operation selects only powers of two of this root.

The final result of the expansion above can be written

\begin{equation}\label{eqn:tschebyscheff:360}

\boxed{

T_m(x) = \sum_{k = 0}^{\lfloor m/2 \rfloor} \binom{m}{2 k} (-1)^k x^{m – 2 k} \lr{1 – x^2}^k.

}

\end{equation}

This clearly shows the polynomial nature of these functions, and is also perfectly well defined for any value of \( x \). The even and odd alternation with \( m \) is also clear in this explicit expansion.

### Plots

### Properties

In [1] a few properties can be found for these polynomials

\begin{equation}\label{eqn:tschebyscheff:100}

T_m(x) = 2 x T_{m-1} – T_{m-2}

\end{equation}

\begin{equation}\label{eqn:tschebyscheff:420}

0 = \lr{ 1 – x^2 } \frac{d T_m(x)}{dx} + m x T_m(x) – m T_{m-1}(x)

\end{equation}

\begin{equation}\label{eqn:tschebyscheff:400}

0 = \lr{ 1 – x^2 } \frac{d^2 T_m(x)}{dx^2} – x \frac{dT_m(x)}{dx} + m^2 T_{m}(x)

\end{equation}

\begin{equation}\label{eqn:tschebyscheff:440}

\int_{-1}^1 \inv{ \sqrt{1 – x^2} } T_m(x) T_n(x) dx =

\left\{

\begin{array}{l l}

0 & \quad \mbox{if \( m \ne n \) } \\

\pi & \quad \mbox{if \( m = n = 0 \) } \\

\pi/2 & \quad \mbox{if \( m = n, m \ne 0 \) }

\end{array}

\right.

\end{equation}

### Recurrance relation.

Prove \ref{eqn:tschebyscheff:100}.

### Answer.

To show this, let

\begin{equation}\label{eqn:tschebyscheff:460}

x = \cos\theta.

\end{equation}

\begin{equation}\label{eqn:tschebyscheff:580}

2 x T_{m-1} – T_{m-2}

=

2 \cos\theta \cos((m-1) \theta) – \cos((m-2)\theta).

\end{equation}

Recall the cosine addition formulas

\begin{equation}\label{eqn:tschebyscheff:540}

\begin{aligned}

\cos( a + b )

&=

\textrm{Re} e^{j(a + b)} \\

&=

\textrm{Re} e^{ja} e^{jb} \\

&=

\textrm{Re}

\lr{ \cos a + j \sin a }

\lr{ \cos b + j \sin b } \\

&=

\cos a \cos b – \sin a \sin b.

\end{aligned}

\end{equation}

Applying this gives

\begin{equation}\label{eqn:tschebyscheff:600}

\begin{aligned}

2 x T_{m-1} – T_{m-2}

&=

2 \cos\theta \Biglr{ \cos(m\theta)\cos\theta +\sin(m\theta) \sin\theta }

– \Biglr{

\cos(m\theta)\cos(2\theta) + \sin(m\theta) \sin(2\theta)

} \\

&=

2 \cos\theta \Biglr{ \cos(m\theta)\cos\theta +\sin(m\theta)\sin\theta) }

– \Biglr{

\cos(m\theta)(\cos^2 \theta – \sin^2 \theta) + 2 \sin(m\theta) \sin\theta \cos\theta

} \\

&=

\cos(m\theta) \lr{ \cos^2\theta + \sin^2\theta } \\

&= T_m(x).

\end{aligned}

\end{equation}

### First order LDE relation.

Prove \ref{eqn:tschebyscheff:420}.

### Answer.

To show this, again, let

\begin{equation}\label{eqn:tschebyscheff:470}

x = \cos\theta.

\end{equation}

Observe that

\begin{equation}\label{eqn:tschebyscheff:480}

1 = -\sin\theta \frac{d\theta}{dx},

\end{equation}

so

\begin{equation}\label{eqn:tschebyscheff:500}

\begin{aligned}

\frac{d}{dx}

&= \frac{d\theta}{dx} \frac{d}{d\theta} \\

&= -\frac{1}{\sin\theta} \frac{d}{d\theta}.

\end{aligned}

\end{equation}

Plugging this in gives

\begin{equation}\label{eqn:tschebyscheff:520}

\begin{aligned}

\lr{ 1 – x^2} &\frac{d}{dx} T_m(x) + m x T_m(x) – m T_{m-1}(x) \\

&=

\sin^2\theta

\lr{

-\frac{1}{\sin\theta} \frac{d}{d\theta}}

\cos( m \theta ) + m \cos\theta \cos( m \theta ) – m \cos( (m-1)\theta ) \\

&=

-\sin\theta (-m \sin(m \theta)) + m \cos\theta \cos( m \theta ) – m \cos( (m-1)\theta ).

\end{aligned}

\end{equation}

Applying the cosine addition formula \ref{eqn:tschebyscheff:540} gives

\begin{equation}\label{eqn:tschebyscheff:560}

m \lr{ \sin\theta \sin(m \theta) + \cos\theta \cos( m \theta ) } – m

\lr{

\cos( m \theta) \cos\theta + \sin( m \theta ) \sin\theta

}

=0.

\end{equation}

} % answer

### Second order LDE relation.

Prove \ref{eqn:tschebyscheff:400}.

### Answer.

This follows the same way. The first derivative was

\begin{equation}\label{eqn:tschebyscheff:640}

\begin{aligned}

\frac{d T_m(x)}{dx}

&=

-\inv{\sin\theta}

\frac{d}{d\theta} \cos(m\theta) \\

&=

-\inv{\sin\theta} (-m) \sin(m\theta) \\

&=

m \inv{\sin\theta} \sin(m\theta),

\end{aligned}

\end{equation}

so the second derivative is

\begin{equation}\label{eqn:tschebyscheff:620}

\begin{aligned}

\frac{d^2 T_m(x)}{dx^2}

&=

-m \inv{\sin\theta} \frac{d}{d\theta} \inv{\sin\theta} \sin(m\theta) \\

&=

-m \inv{\sin\theta}

\lr{

-\frac{\cos\theta}{\sin^2\theta} \sin(m\theta) + \inv{\sin\theta} m \cos(m\theta)

}.

\end{aligned}

\end{equation}

Putting all the pieces together gives

\begin{equation}\label{eqn:tschebyscheff:660}

\begin{aligned}

\lr{ 1 – x^2 } &\frac{d^2 T_m(x)}{dx^2} – x \frac{dT_m(x)}{dx} + m^2 T_{m}(x) \\

&=

m

\lr{

\frac{\cos\theta}{\sin\theta} \sin(m\theta) – m \cos(m\theta)

}

– \cos\theta m \inv{\sin\theta} \sin(m\theta)

+ m^2 \cos(m \theta) \\

&=

0.

\end{aligned}

\end{equation}

### Orthogonality relation

Prove \ref{eqn:tschebyscheff:440}.

### Answer.

First consider the 0,0 inner product, making an \( x = \cos\theta \), so that \( dx = -\sin\theta d\theta \)

\begin{equation}\label{eqn:tschebyscheff:680}

\begin{aligned}

\innerprod{T_0}{T_0}

&=

\int_{-1}^1 \inv{\lr{1-x^2}^{1/2}} dx \\

&=

\int_{-\pi}^0 \lr{-\inv{\sin\theta}} -\sin\theta d\theta \\

&=

0 – (-\pi) \\

&= \pi.

\end{aligned}

\end{equation}

Note that since the \( [-\pi, 0] \) interval was chosen, the negative root of \( \sin^2\theta = 1 – x^2 \) was chosen, since \( \sin\theta \) is negative in that interval.

The m,m inner product with \( m \ne 0 \) is

\begin{equation}\label{eqn:tschebyscheff:700}

\begin{aligned}

\innerprod{T_m}{T_m}

&=

\int_{-1}^1 \inv{\lr{1-x^2}^{1/2}} \lr{ T_m(x)}^2 dx \\

&=

\int_{-\pi}^0 \lr{-\inv{\sin\theta}} \cos^2(m\theta) -\sin\theta d\theta \\

&=

\int_{-\pi}^0 \cos^2(m\theta) d\theta \\

&=

\inv{2} \int_{-\pi}^0 \lr{ \cos(2 m\theta) + 1 } d\theta \\

&= \frac{\pi}{2}.

\end{aligned}

\end{equation}

So far so good. For \( m \ne n \) the inner product is

\begin{equation}\label{eqn:tschebyscheff:720}

\begin{aligned}

\innerprod{T_m}{T_m}

&=

\int_{-\pi}^0 \cos(m\theta) \cos(n\theta) d\theta \\

&=

\inv{4} \int_{-\pi}^0

\lr{

e^{j m \theta}

+ e^{-j m \theta}

}

\lr{

e^{j n \theta}

+ e^{-j n \theta}

}

d\theta \\

&=

\inv{4} \int_{-\pi}^0

\lr{

e^{j (m + n) \theta}

+e^{-j (m + n) \theta}

+e^{j (m – n) \theta}

+e^{j (-m + n) \theta}

}

d\theta \\

&=

\inv{2} \int_{-\pi}^0

\lr{

\cos( (m + n)\theta )

+\cos( (m – n)\theta )

}

d\theta \\

&=

\inv{2}

\evalrange{

\lr{

\frac{\sin( (m + n)\theta )}{ m + n }

+\frac{\sin( (m – n)\theta )}{ m – n}

}

}{-\pi}{0} \\

&= 0.

\end{aligned}

\end{equation}

# References

[1] M. Abramowitz and I.A. Stegun. Handbook of mathematical functions with formulas, graphs, and mathematical tables, volume 55. Dover publications, 1964.