Relating Chebyshev Equation/Polynomials with Cosines
The Chebyshev Equation is another type of second-order linear ODE that has the form of
\begin{align}
(1-x^2) y^{\prime\prime} -xy’ +n^2y = 0 \tag{1}
\end{align}
where \( n \) is a given integer. Like the Legendre equation, it has three regular singular points \( x = -1,1,\infty \). We can, of course, use the series solution method to obtain the two linearly independent solutions at the ordinary point \( x=0 \) just as before (see Exercise). But it will be more instructive to let \( x = \cos \theta \), so \( dy/dx = (d\theta/dx) (dy/d\theta) = (-1/\sin\theta) dy/d\theta\), and
\begin{align}
\frac{d^2y}{dx^2} &= \frac{-1}{\sin\theta} \frac{d}{d\theta} (\frac{-1}{\sin\theta} \frac{dy}{d\theta}) \\
&= \frac{1}{\sin^2\theta} \frac{d^2y}{d\theta^2} -\frac{\cos \theta}{\sin^3\theta}\frac{dy}{d\theta} \tag{2}
\end{align}
So (1) is transformed into
\begin{align}
(1-\cos\theta ^2) ( \frac{1}{\sin^2\theta} \frac{d^2y}{d\theta^2} -\frac{\cos \theta}{\sin^3\theta}\frac{dy}{d\theta}) -\cos\theta (\frac{-1}{\sin\theta} \frac{dy}{d\theta}) +n^2y &= 0 \\
\frac{d^2y}{d\theta^2} -\frac{\cos \theta}{\sin\theta}\frac{dy}{d\theta} + \frac{\cos \theta}{\sin\theta} \frac{dy}{d\theta} +n^2y &= 0 \\
\frac{d^2y}{d\theta^2} +n^2y &= 0 \tag{3}
\end{align}
So the Chebyshev equation is just a simple harmonic equation in disguise when the independent variable is now in terms of the cosine. From previous knowledge, the solutions of (3) will simply be \( \cos (n\theta) \) and \( \sin (n\theta) \), and hence in the context of (1), they are
\begin{align}
T_n(x) = \cos(n \cos^{-1}x), V_n(x) = \sin(n \cos^{-1}x) \tag{4}
\end{align}
We will then be able to compute their series form by considering De Moivre’s formula:
\begin{align}
e^{i n\theta} &= \cos(n\theta) + i\sin(n\theta) = T_n(x) + iV_n(x) \\
e^{i n\theta} &= (\cos\theta + i\sin\theta)^n = (x + i\sqrt{1-x^2})^n \tag{5}
\end{align}
By expanding (5) with the Binomial Theorem and equating the real/imaginary parts, we have
\begin{align}
T_n(x) + iV_n(x) &= x^n + iC^{n}_{1}x^{n-1}\sqrt{1-x^2} \\
&\quad -C^{n}_{2}x^{n-2}(1-x^2) -iC^{n}_{3}x^{n-3}(1-x^2)^{\frac{3}{2}} \\
&\quad + C^{n}_{4}x^{n-4}(1-x^2)^2 + \cdots \tag{6}
\end{align}
i.e.
\begin{align}
T_n(x) &= x^n -C^{n}_{2}x^{n-2}(1-x^2) + C^{n}_{4}x^{n-4}(1-x^2)^2 – \cdots \\
&= \sum_{m=0}^{\lfloor \frac{n}{2} \rfloor} (-1)^m C^{n}_{2m}x^{n-2m}(1-x^2)^m \\
&= \sum_{m=0}^{\lfloor \frac{n}{2} \rfloor} (-1)^m C^{n}_{2m}x^{n-2m} \sum_{l=0}^{m} C^{m}_l (-1)^l x^{2l} \\
&= \sum_{m=0}^{\lfloor \frac{n}{2} \rfloor}\sum_{l=0}^{m} C^{n}_{2m}C^{m}_l (-1)^{m+l} x^{n-2m+2l} \\
&= \sum_{l=0}^{\lfloor \frac{n}{2} \rfloor}\sum_{m=0}^{\lfloor \frac{n}{2} \rfloor-l} (-1)^{m+2l} C^{n}_{2(m+l)}C^{m+l}_l x^{n-2(m+l)+2l} \\
&= \sum_{m=0}^{\lfloor \frac{n}{2} \rfloor}\sum_{l=0}^{\lfloor \frac{n}{2} \rfloor-m} (-1)^m C^{n}_{2m+2l}C^{m+l}_l x^{n-2m} \tag{7}
\end{align}
This represents a set of finite polynomials which shall be known as the Chebyshev polynomials of the first kind. The first few of them are
\begin{align}
\begin{aligned}
T_0(x) &= 1 & T_1(x) &= x \\
T_2(x) &= 2x^2 -1 & T_3(x) &= 4x^3 -3x \\
T_4(x) &= 8x^4 -8x^2 +1 & T_5(x) &= 16x^5 -20x^3 +5x \\
\end{aligned} \tag{8}
\end{align}
Similar to the Legendre polynomials, they obey the normalization requirement of \( T_n(1) = 1 \). They are also either odd or even following \( n \), and \( T_n(-1) = (-1)^n \). Moreover, the leading term of \( T_n(x) \) has a coefficient of \( 2^{n-1} \) for \( n \geq 1 \). Meanwhile, the other solution
\begin{align}
V_n(x) &= \sqrt{1-x^2}(C^n_1x^{n-1} -C^n_3x^{n-3}(1-x^2) \\
& \quad +C^n_5x^{n-5}(1-x^2)^2 – \cdots) \tag{9}
\end{align}
is not a polynomial due to the \( \sqrt{1-x^2} \) factor. To circumvent this, we can define
\begin{align}
U_n(x) &= (1-x^2)^{-\frac{1}{2}} V_n(x) \tag{10}
\end{align}
which are subsequently known as the Chebyshev polynomials of the second kind. However, we will not go into details for them.
Properties of Chebyshev Polynomials
Orthogonality
The Chebyshev equation can be cast into Sturm-Liouville Form (refer to this tutorial):
\begin{align}
\sqrt{1-x^2}y^{\prime\prime} -\frac{x}{\sqrt{1-x^2}}y’ + \frac{n^2}{\sqrt{1-x^2}}y &= 0 \\
\frac{d}{dx}(\sqrt{1-x^2}y’) + \frac{n^2}{\sqrt{1-x^2}}y &= 0 \tag{11}
\end{align}
with \( p(x) = \sqrt{1-x^2} \) and \( \rho(x) = 1/\sqrt{1-x^2} \). Just like the Legendre equation, this means that the results for Sturm-Liouville operators are applicable. The natural interval is again \( [-1,1] \) and the Chebyshev polynomials are the bounded eigenfunctions in this situation.
Therefore, orthogonality between the Chebyshev polynomials is expected (Theorem 2 of this tutorial) with the new weighting:
\begin{align}
\int_{-1}^{1} \frac{T_m(x) T_n(x)}{\sqrt{1-x^2}} dx &= 0 & \text{if \( m \neq n \)} \tag{12}
\end{align}
The direct method to show this is very similar to the previous tutorial about Legendre polynomials:
\begin{align}
T_m(x)\frac{d}{dx}(\sqrt{1-x^2}T_n(x)’) + \frac{n^2}{\sqrt{1-x^2}}T_m(x)T_n(x) &= 0 \\
\int_{-1}^{1} T_m(x)\frac{d}{dx}(\sqrt{1-x^2}T_n(x)’) dx + \int_{-1}^{1} \frac{n^2}{\sqrt{1-x^2}}T_m(x)T_n(x) dx &= 0 \\
-\int_{-1}^{1} T_m'(x)\sqrt{1-x^2}T_n'(x) dx + \int_{-1}^{1} \frac{n^2}{\sqrt{1-x^2}}T_m(x)T_n(x) dx &= 0 \tag{13}
\end{align}
Exchanging the roles of \( l \) and \( m \) and subtraction gives the desired statement:
\begin{align}
\begin{aligned}
(-\int_{-1}^{1} T_m'(x)\sqrt{1-x^2}T_n'(x) dx + \int_{-1}^{1} \frac{n^2}{\sqrt{1-x^2}}T_m(x)T_n(x) dx ) \\
– (-\int_{-1}^{1} T_n'(x)\sqrt{1-x^2}T_m'(x) dx + \int_{-1}^{1} \frac{m^2}{\sqrt{1-x^2}}T_m(x)T_n(x) dx ) \end{aligned} &= 0 \\
\int_{-1}^{1} \frac{n^2}{\sqrt{1-x^2}}T_m(x)T_n(x) dx -\int_{-1}^{1} \frac{m^2}{\sqrt{1-x^2}}T_m(x)T_n(x) dx &= 0 \\
(n^2-m^2) \int_{-1}^{1} \frac{T_m(x)T_n(x)}{\sqrt{1-x^2}} dx &= 0 \\
\Rightarrow \int_{-1}^{1} \frac{T_m(x)T_n(x)}{\sqrt{1-x^2}} dx &= 0 \tag{14}
\end{align}
Normalization Factor
This time, we can derive the normalization factor without Rodrigues’ Formula (which will be derived later). The main observation is that under \( x = \cos \theta \), \( T_n(x) = \cos(n\theta) \), so we can use trigonometry to compute the norm. For \( n = 0 \), it is not hard to see that
\begin{align}
\int_{-1}^{1} \frac{T_0(x) T_0(x)}{\sqrt{1-x^2}} dx &= \int_{-1}^{1} \frac{1}{\sqrt{1-x^2}} dx \\
&= \int_{-1}^{1} \frac{1}{\sqrt{1-\cos^2\theta}} d(\cos\theta) \\
&= \int_{\pi}^{0} -\frac{\sin\theta}{\sin\theta} d\theta = -\int_{\pi}^{0} d\theta = \pi \tag{15}
\end{align}
and when \( n \geq 1 \),
\begin{align}
\int_{-1}^{1} \frac{T_n(x) T_n(x)}{\sqrt{1-x^2}} dx &= \int_{-1}^{1} \frac{\cos^2(n\theta)}{\sqrt{1-\cos^2\theta}} d(\cos\theta) \\
&= \int_{\pi}^{0} -\frac{\sin\theta}{\sin\theta}\frac{1}{2}(1+\cos(2n\theta)) d\theta \\
&= -\frac{1}{2}\int_{\pi}^{0} d\theta -\frac{1}{2} \int_{\pi}^{0} \cos(2n\theta) d\theta \\
&= \frac{\pi}{2} + (0) = \frac{\pi}{2} \tag{16}
\end{align}
Rodrigues’ Formula
The corresponding Rodrigues’ Formula for Chebyshev polynomials is
\begin{align}
T_n(x) = (-1)^n\frac{2^nn!}{(2n)!} \sqrt{1-x^2} \frac{d^n}{dx^n}(1-x^2)^{n -\frac{1}{2}} \tag{17}
\end{align}
Just like before, let \( u = (1-x^2)^{n -\frac{1}{2}} \), differentiating once gives
\begin{align}
u’ &= -2(n-\frac{1}{2})x (1-x^2)^{n -\frac{3}{2}} \\
(1-x^2)u’ &= -2(n-\frac{1}{2})x (1-x^2)^{n -\frac{1}{2}} = -2(n-\frac{1}{2})xu \\
(1-x^2)u’ +2(n-\frac{1}{2})xu &= 0 \tag{18}
\end{align}
and using Leibniz Product Rule to differentiate \( n+1 \) more times:
\begin{align}
\begin{aligned} ((1-x^2)u^{(n+2)} -2x(n+1)u^{(n+1)} -n(n+1) u^{(n)}) \\
+ (2x(n-\frac{1}{2})u^{(n+1)} + 2(n-\frac{1}{2})(n+1)u^{(n)}) \end{aligned} &= 0 \\
(1-x^2)u^{(n+2)} -3xu^{(n+1)} + (n-1)(n+1)u^{(n)} &= 0 \tag{19}
\end{align}
Then, let \( w = \sqrt{1-x^2}u^{(n)} \), and hence
\begin{align}
\begin{aligned}
u^{(n+1)} &= \frac{w’}{\sqrt{1-x^2}} + \frac{xw}{(1-x^2)^{\frac{3}{2}}} \\
u^{(n+2)} &= \frac{w^{\prime\prime}}{\sqrt{1-x^2}} + \frac{2xw’}{(1-x^2)^{\frac{3}{2}}} + \frac{w(2x^2+1)}{(1-x^2)^{\frac{5}{2}}} \\
\end{aligned} \tag{20}
\end{align}
Substituting these back into (19) leads to
\begin{align}
\begin{aligned} (\sqrt{1-x^2}w^{\prime\prime} + \frac{2xw’}{\sqrt{1-x^2}} + \frac{w(2x^2+1)}{(1-x^2)^{\frac{3}{2}}}) \\
-3x(\frac{w’}{\sqrt{1-x^2}} + \frac{xw}{(1-x^2)^{\frac{3}{2}}} ) + (n-1)(n+1)\frac{w}{\sqrt{1-x^2}} \end{aligned} &= 0 \\
(1-x^2)w^{\prime\prime} -xw’ + w + (n-1)(n+1)w &= 0 \\
(1-x^2)w^{\prime\prime} -xw’ + n^2w &= 0 \tag{21}
\end{align}
This has the same form as (1), so \( w = \sqrt{1-x^2}u^{(n)} = \sqrt{1-x^2} \frac{d^n}{dx^n}(1-x^2)^{n -\frac{1}{2}} \) will be a solution to the Chebyshev equation. It is not hard to argue that it is a finite polynomial, and hence it must be equal to \( T_n(x) \) within a multiplicative factor. Also, it is not hard to see that the only term in \(w\) that does not contain any factor of \( (1-x^2) \) after expanding and will not vanish when \(x = 1\), will be \( (-2x)^{n} \sqrt{1-x^2} (n-\frac{1}{2})(n-\frac{3}{2})\cdots(\frac{1}{2})(1-x^2)^{-\frac{1}{2}} = (-2x)^{n} (2n)!/(2^{2n}n!) \). So \( w(1) = (-1)^n(2n)!/(2^nn!) \) and this matches the reciprocal of the factor in front of (17).
Generating Function
The generating function of Chebyshev polynomials can be easily found by manipulating trigonometry and complex geometric series, which is
\begin{align}
\frac{1-xt}{1-2xt+t^2} = \sum_{n=0}^{\infty} T_n(x) t^n \tag{22}
\end{align}
To obtain this, recall that \( T_n(x) = \cos(n\theta) \) when \(x = \cos \theta \) and consider
\begin{align}
\sum_{n=0}^{\infty} \exp(in\theta) t^n &= \frac{1}{1-e^{i\theta} t} \\
&= \frac{1-e^{-i\theta} t}{(1-e^{-i\theta} t)(1-e^{i\theta} t)} \\
&= \frac{1-e^{-i\theta} t}{1 -2t\cos\theta + t^2} \tag{23}
\end{align}
Taking the real parts of both sides then yields (22).
Recurrence Relation
The main recurrence relation is also obvious due to trigonometry. Recall the trigonometric identity of
\begin{align}
\cos((n\pm 1)\theta) = \cos(n\theta) \cos\theta \mp \sin(n\theta)\sin\theta \tag{24}
\end{align}
Adding the two plus/minus cases of the above together then gives
\begin{align}
\cos((n+1)\theta) + \cos((n-1)\theta) = 2\cos(n\theta) \cos\theta \tag{25}
\end{align}
Again, note that \( T_n(x) = \cos(n\theta) \) with \(x = \cos \theta \) and rearrange to get
\begin{align}
T_{n+1}(x) = 2xT_n(x) -T_{n-1}(x) \tag{26}
\end{align}
which can be used to iteratively generate the Chebyshev polynomials from the initial values of \( T_0(x) = 1 \) and \( T_1(x) = x \).
Gram-Schmidt Orthogonalization
Alternatively, to produce the Chebyshev polynomials using Gram-Schmidt Orthogonalization, we follow the same procedure similar to that for the Legendre polynomials, but now rely on (12) so that we need a weighting of \( \rho = 1/\sqrt{1-x^2} \) when evaluating the inner products. Still, the steps for \( n=0,1 \) are trivial and we shall start from \( n=2 \), but it is also better to first derive the reduction formula for the following integral.
\begin{align}
I_n = \int_{-1}^1 \frac{x^n}{\sqrt{1-x^2}} dx \tag{27}
\end{align}
By symmetry, \( I_{2m+1} = 0 \) is zero for odd indices. Meanwhile, using integration by parts,
\begin{align}
I_{2m} &= \int_{-1}^1 \frac{x^{2m}}{\sqrt{1-x^2}} dx \\
&= -\int_{-1}^1 x^{2m-1} d(\sqrt{1-x^2}) \\
&= -[x^{2m-1} \sqrt{1-x^2}]_{-1}^1 + (2m-1)\int_{-1}^1 x^{2m-2}\sqrt{1-x^2} dx \\
&= (0) + (2m-1)\int_{-1}^1 \frac{x^{2m-2}(1-x^2)}{\sqrt{1-x^2}} dx \\
&= (2m-1)I_{2m-2} -(2m-1)I_{2m} \tag{28}
\end{align}
Therefore,
\begin{align}
I_{2m} = \frac{2m-1}{2m} I_{2m-2} \tag{29}
\end{align}
From (15), \( I_0 = \pi \), and hence \( I_2 = \pi/2 \), \( I_4 = 3\pi/8 \). Now we can proceed to the orthogonalization:
\begin{align}
T_2(x) &= x^2 -\frac{\langle x^2, T_0 \rangle_{\rho}}{\lVert T_0 \rVert_{\rho}^2} T_0 -\frac{\langle x^2, T_1 \rangle_{\rho}}{\lVert T_1 \rVert_{\rho}^2} T_1 \\
&= x^2 -(\frac{1}{\pi}\int_{-1}^{1} \frac{x^2}{\sqrt{1-x^2}} dx) (1) -(\frac{2}{\pi}\int_{-1}^{1} \frac{(x^2)(x)}{\sqrt{1-x^2}} dx) (x) \\
&= x^2 -\frac{1}{\pi}\frac{\pi}{2} -(0) \\
&= x^2 -\frac{1}{2} \tag{30}
\end{align}
Remember to normalize it with the condition \( T_n(1) = 1 \). For \( n = 3 \), we have
\begin{align}
T_3(x) &= x^3 -\frac{\langle x^3, T_0 \rangle_{\rho}}{\lVert T_0 \rVert_{\rho}^2} T_0 -\frac{\langle x^3, T_1 \rangle_{\rho}}{\lVert T_1 \rVert_{\rho}^2} T_1 -\frac{\langle x^3, T_2 \rangle_{\rho}}{\lVert T_2 \rVert_{\rho}^2} T_2 \\
&= x^3 -(\frac{1}{\pi}\int_{-1}^{1} \frac{x^3}{\sqrt{1-x^2}} dx) (1) -(\frac{2}{\pi}\int_{-1}^{1} \frac{(x^3)(x)}{\sqrt{1-x^2}} dx) (x) \\
&\quad -(\frac{2}{\pi}\int_{-1}^{1} \frac{(x^3)(2x^2-1)}{\sqrt{1-x^2}} dx) (2x^2-1) \\
&= x^3 -(0) -\frac{2}{\pi}\frac{3\pi}{8}x -(0) \\
&= x^3 -\frac{3}{4}x \tag{31}
\end{align}
and it can be continued beyond.
Exercise
Derive the Chebyshev polynomials by the direct method of series solution and compare that to (7).
Answer
Substitute (1)-(3) of this tutorial (with \(m\) in place of \(n\)) into (1) gives
\begin{align}
(1-x^2) \sum_{m=2}^{\infty} m(m-1) a_mx^{m-2} -x \sum_{m=1}^{\infty} m a_mx^{m-1} +n^2\sum_{m=0}^{\infty} a_m x^m &= 0 \\
\begin{aligned}\sum_{m=2}^{\infty} m(m-1) a_mx^{m-2} -\sum_{m=2}^{\infty} m(m-1) a_mx^{m} \\
– \sum_{m=1}^{\infty} m a_mx^{m} +n^2\sum_{m=0}^{\infty} a_m x^m \end{aligned} &= 0 \\
\begin{aligned}\sum_{m=0}^{\infty} (m+2)(m+1) a_{m+2}x^{m} -\sum_{m=2}^{\infty} m(m-1) a_mx^{m} \\
– \sum_{m=1}^{\infty} m a_mx^{m} +n^2\sum_{m=0}^{\infty} a_m x^m \end{aligned} &= 0
\end{align}
Comparing the coefficients, the recurrence relation is
\begin{align}
(m+2)(m+1) a_{m+2} -m(m-1) a_m -m a_m + n^2 a_m = 0 \\
a_{m+2} = -\frac{n^2-m^2}{(m+2)(m+1)}a_m = -\frac{(n-m)(n+m)}{(m+2)(m+1)}a_m
\end{align}
The series will be truncated when \( m = n \) and become the Chebyshev polynomial of degree \(n\) we want, which is then
\begin{align}
T_n(x) &= x^n -\frac{n(n-1)}{2(2n-2)} x^{n-2} \\
&\quad +\frac{n(n-1)(n-2)(n-3)}{(2)(4)(2n-2)(2n-4)} x^{n-4} -\cdots \\
&= \sum_{l=0}^{\lfloor \frac{n}{2} \rfloor} (-1)^l\frac{n!}{(n-2l)!}\frac{(n-l-1)!}{(2^l (n-1)!)(2^l l!)}x^{n-2l} \\
&= \sum_{l=0}^{\lfloor \frac{n}{2} \rfloor} (-1)^l\frac{nC^{n-l}_{l}}{(n-l)2^{2l}}x^{n-2l} \\
\end{align}
To account for the normalization factor (so that the leading coefficient is \( 2^{n-1} \)), it would become
\begin{align}
T_n(x) &= \frac{1}{2} \sum_{l=0}^{\lfloor \frac{n}{2} \rfloor} (-1)^l\frac{nC^{n-l}_{l}}{n-l}(2x)^{n-2l}
\end{align}
Hence by comparing this to (7), we obtain the following combinatoric identity:
\begin{align}
\sum_{l=0}^{\lfloor \frac{n}{2} \rfloor-m} C^{n}_{2m+2l}C^{m+l}_l = 2^{n-2m-1} \frac{nC^{n-m}_{m}}{n-m}
\end{align}
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