Introduction to Linear Algebra with Mathematica

Preface

Chebyshev polynomials are usually used for either approximation of continuous functions or function expansion. For the case of functions that are solutions of linear ordinary differential equations with polynomial coefficients (a typical case for special functions), the problem of computing Chebyshev series is efficiently solved by means of Clenshaw’s method. This section presents some properties of the most remarkable and useful in numerical computations Chebyshev polynomials of first kind $$T_n (x)$$ and second kind $$U_n (x) .$$ Both Chebyshev polynomials are eigenfunctions of the corresponding singular Sturm--Liouville problems. Other two Chebyshev polynomials of the third kind and the fourth kind are not so popular in applications.

Chebyshev Polynomials

The Chebyshev polynomials of first kind $$T_n (x) = \cos \left( n\,\mbox{arccos} x \right)$$ are solutions of the differential equation
$\left( 1- x^2 \right) y'' -x\,y' + n^2 y =0 ,$

The Chebyshev polynomials of the second kind $$\displaystyle U_n (x) = \frac{\sin \left[ (n+1)\,\arccos x \right]}{\sqrt{1 - x^2}}$$ are solutions of the differential equation

$\left( 1- x^2 \right) y'' -3x\,y' + n\left( n+2 \right) y =0 ,$

The Chebyshev polynomials of the third kind $$\displaystyle V_{n} (x) = \frac{\cos \left( \frac{2n+1}{2}\,\theta \right)}{\cos \left( \frac{\theta}{2} \right)} ,$$ x = cos(θ), are solutions of the differential equation

$\left( 1- x^2 \right) y'' - \left( 2x -1 \right) y' + n\left( n+1 \right) y =0 ,$

The Chebyshev polynomials of the fourth kind $$\displaystyle W_n (x) = \frac{\sin \left( \frac{2n+1}{2}\,\theta \right)}{\sin \left( \frac{\theta}{2} \right)} , \qquad x= \cos\theta ,$$ are solutions of the differential equation $\left( 1- x^2 \right) y'' - \left( 2x +1 \right) y' + n\left( n+1 \right) y =0 ,$

Orthogonality of Chebyshev Polynomials

Chebyshev polynomials form orthogonal sets in weighed vector space 𝔏²[−1, 1] with respect to various weight functions as in the following inner products:
\begin{align*} \langle T_n (x), T_m (x) \rangle = \int_{-1}^1 \left( 1 - x^2 \right)^{-1/2} T_n (x) T_m (x) \,{\text d}x &= \frac{\pi}{{\cal E}_n} \, \delta_{n,m} , \\ \langle U_n (x), U_m (x) \rangle = \int_{-1}^1 \left( 1 - x^2 \right)^{1/2} U_n (x) U_m (x) \,{\text d}x &= \frac{\pi}{2} \, \delta_{n,m} , \\ \langle V_n (x), V_m (x) \rangle = \int_{-1}^1 \left( \frac{1+x}{1-x} \right)^{1/2} V_n (x) V_m (x) \,{\text d}x &= \pi \, \delta_{n,m} , \\ \langle W_n (x), W_m (x) \rangle = \int_{-1}^1 \left( \frac{1-x}{1+x} \right)^{1/2} W_n (x) W_m (x) \,{\text d}x &= \pi \, \delta_{n,m} , \end{align*}
where
$\delta_{n,m} = \begin{cases} 0, & \ \mbox{if } n \ne m , \\ 1 , & \ \mbox{if } n = m ; \end{cases} \qquad\quad {\cal E}_n = \begin{cases} 1 , & \ \mbox{if } n =0, \\ 2, & \ \mbox{if } n \ge 1. \end{cases}$

There are four kinds of Chebyshev expansions for a function on the finite interval [-1, 1] depending which kind of Chebyshev function is used.

and they are orthogonal on the interval [-1, 1] with weight function $$\rho (x) = \left( 1- x^2 \right)^{-1/2} :$$

$\int_{-1}^1 \frac{T_m (x)\, T_n (x)}{\sqrt{1-x^2}} \,{\text d} x = \begin{cases} \pi , & \quad \mbox{if n=m=0}, \\ \pi /2 , & \quad \mbox{if n=m\ne 0}, \\ 0 , & \quad \mbox{if n\ne m}. \end{cases}$
The Chebyshev polynomials of the second kind $$U_n (x)$$ are solutions of the differential equation
$\left( 1- x^2 \right) y'' -3x\,y' + n(n+2)\, y =0 ,$
and they are orthogonal on the interval [-1, 1] with weight function $$\rho (x) = \left( 1- x^2 \right)^{1/2} :$$
$\int_{-1}^1 U_m (x)\, U_n (x)\, \sqrt{1-x^2} \,{\text d} x = \begin{cases} \pi /2 , & \quad \mbox{if n=m}, \\ 0 , & \quad \mbox{if n\ne m}. \end{cases}$

Chebyshev Series

Chebyshev polynomials form a special class of polynomials especially suited for approximating other functions. They are widely used in many areas of numerical analysis: uniform approximation, least-squares approximation, numerical solution of ordinary and partial differential equations (the so-called spectral or pseudospectral methods), and so on. In the appropriate Sobolev space, the set of Chebyshev polynomials form an orthonormal basis, so that a function in the same space can, on −1 ≤ x ≤ 1 be expressed via the expansion
$f(x) = \frac{a_0}{2} + \sum_{k\ge 1} a_k T_k (x) = \sum_{k\ge 0} b_k U_k (x) = \sum_{k\ge 0} c_k V_k (x) = \sum_{k\ge 0} d_k W_k (x) .$
The Chebyshev polynomials (of any kind) form an orthogonal basis that (among other things) implies that the coefficients can be determined easily through the application of an inner product. This sum is called a Chebyshev series or a Chebyshev expansion. All of the theorems, identities, etc. that apply to Fourier series have a Chebyshev counterpart.

Arbitrary continuous function can be approximated by Chebyshev interpolation and Chebyshev series that converges pointwise. If f(x) is continuous in the interval [−1,1] apart from a finite number of step discontinuities in the interior, then its Chebyshev series expansion converges to f wherever f is continuous, and to the average of the left and right limiting values at each discontinuity. To obtain uniform convergence of the Chebyshev series, continuous function f should have bounded variation, similar to Fourier series (Dirichlet or Dini--Lipschitz condition). If a function is no more than barely continuous, then we can derive uniformly convergent approximations from its Chebyshev expansion by averaging out the partial sums, and thus forming ‘Cesàro sums’ of the Chebyshev series. For the case of solutions of linear ordinary differential equations with polynomial coefficients (a typical case for special functions), the problem of computing Chebyshev series is efficiently solved by means of Clenshaw’s method.

Series Expansions over Chebyshev Polynomials of the First Kind

If function f(x) is defined over closed interval {−1, 1], is absolutely integrable with the weight $$w (x) = \left( 1- x^2 \right)^{-1/2} ,$$ then the function can be expanded into Fourier--Chebyshev
$$\label{EqCheb1.1} \frac{f(x+0) + f(x-0)}{2} = \lim_{\varepsilon \downarrow 0} \frac{f(x+ \varepsilon ) + f(x- \varepsilon )}{2} = \frac{a_0}{2} + \sum_{k\ge 1} \,a_k\, T_{k} (x ) ,$$
where
$$\label{EqCheb1.2} a_k = \frac{2}{\pi}\,\int_{-1}^1 \,\frac{f(x)\, T_{k} (x )}{\sqrt{1-x^2}}\, {\text d} x = \frac{2}{\pi}\,\int_{0}^{\pi} \, f\left( \cos \,\theta \right) \cos \left( k\,\theta \right) {\text d} \theta , \qquad k=0,1,2,\ldots .$$

Since expansion \eqref{EqCheb1.1} resembles classical Fourier series expansion, we expect to see the Gibbs phenomenon at the points of discontinuity. It is not a surprise that the amounts of overshoot and undershoot at discontinuities are exactly the same as for Fourier series.

Formula \eqref{EqCheb1.2} shows that the Chebyshev expansion \eqref{EqCheb1.1} is actually cosine Fourier expansion of the even function $$f( \cos \theta )$$ over interval [0, π]. Therefore, all cosine Fourier expansions that were found previously can be automatically applied to the series expansions over Chebyshev polynomials of the first kind. Therefore, the conditions of pointwise convergence of Chebyshev series follow from similar results about Fourier series. In particular, we get Parseval’s identity:

$\frac{a_0^2}{2} + \sum_{n\ge 1} a_n^2 = \frac{2}{\pi}\, \| f \|^2 = \frac{2}{\pi}\, \int_{-1}^1 \left( 1 - x^2 \right)^{-1/2} \left\vert f (x) \right\vert^2 \,{\text d}x .$

For example, we found previously in section VIII that

$\frac{1}{2} - \frac{\pi}{4}\,|\theta | = \sum_{n\ge 1} \frac{1}{(2n-1)^2} \,\cos (2n-1) \theta , \qquad 0 < \theta < \pi .$
We claim that
$\frac{1}{2} - \frac{\pi}{4}\,|\arccos x| = \sum_{n\ge 1} \frac{1}{(2n-1)^2} \,T_{2n-1} (x) , \qquad -1 < x < 1 .$
Theorem 1: When a function f has m+1 continuous derivatives on [−1,1] and vanishes at endpoints, where m is a positive integer, then $$\left\vert f(x) - S_n (x) \right\vert = O\left( n^{-m} \right)$$ as n → ∞ , where
$S_n (x) = \frac{a_0}{2} + \sum_{k= 1}^n a_k T_k (x)$
is the partial Chebyshev sum.

We can interpolate arbitrary polynomial pn(x) of degree n:

$p_n (x) = \,\sum_{k=0}^{n} \,^{\prime} \, a_{k} \, T_{k} (x) = \frac{a_0}{2} + \sum_{k= 1}^n \,a_k\, T_{k} (x ) ,$
where the prime indicates that the term for k = 0 is to be halved and

$a_{k} = \frac{2}{n+1} \,\sum_{j=0}^n \,p(x_j )\, T_{k} (x_j ) , \qquad x_j = \cos \left( \frac{2j+1}{2n+2}\, \pi \right) .$

Example 1A: The power function xn can be expressed in terms of Chebyshev polynomials as follows:

$x^n = 2^{1-n} \,\sum_{k=0}^{\lfloor n/2 \rfloor} \,^{\prime} \, \binom{n}{k} \, T_{n-2k} (x) = T_n (x) + 2^{1-n} \,\sum_{k=1}^{\lfloor n/2 \rfloor} \, \binom{n}{k} \, T_{n-2k} (x) , \qquad n=1,2,\ldots ;$
where $$\binom{n}{k} = \frac{n!}{k! \left( n-k \right) !}$$ is the binomial coefficient and $$\lfloor \,\cdot \,\rfloor$$ is the floor function. As usual, the prime in the sum sign indicates that the term for k = 0 is to be halved. Similarly,

$x^n = 2^{-n} \left[ U_n (x) + \left( n-1\right) U_{n-2} (x) + \cdots \right] , \qquad n=1,2,\ldots .$

Example 2A: Consider a continuous function $$f(x) = 4\,x^3 -1 .$$ Since it is a polynomial of degree n = 3, we can use the corresponding formula to determine the coefficients in Chebyshev series:

$c_0 = \frac{2}{4} \,\sum_{j=0}^3 \,f\left( \cos \frac{2j+1}{8}\,\pi \right) = \frac{2}{\pi}\,\int_{0}^{\pi} \, f\left( \cos \,\theta \right) {\text d} \theta = -2,$
$c_1 = \frac{2}{4} \,\sum_{j=0}^3 \,f\left( \cos \frac{2j+1}{8}\,\pi \right) T_1 \left( \cos \frac{2j+1}{8}\,\pi \right) = \frac{2}{\pi}\,\int_{0}^{\pi} \, f\left( \cos \,\theta \right) \cos \theta \, {\text d} \theta = 3,$
$c_2 = \frac{2}{4} \,\sum_{j=0}^3 \,f\left( \cos \frac{2j+1}{8}\,\pi \right) T_2 \left( \cos \frac{2j+1}{8}\,\pi \right) = \frac{2}{\pi}\,\int_{0}^{\pi} \, f\left( \cos \,\theta \right) \cos 2\theta \, {\text d} \theta = 0,$
$c_3 = \frac{2}{4} \,\sum_{j=0}^3 \,f\left( \cos \frac{2j+1}{8}\,\pi \right) T_3 \left( \cos \frac{2j+1}{8}\,\pi \right) = \frac{2}{\pi}\,\int_{0}^{\pi} \, f\left( \cos \,\theta \right) \cos 3\theta \, {\text d} \theta = 1.$
Therefore, we get the Chebyshev expansion of the given function:
$4x^3 -1 = -1 + 3\, T_1 (x) + T_3 (x) = -1 +3x + \left( 4\,x^3 -3x \right) ,$
which is an exact equation. We do similar job with Chebyshev polynomials of the second kind:
$4\,x^3 -1 = b_0 \,U_0 + b_1 \,U_1 (x) + b_2 \, U_2 (x) + b_3 \, U_3 (x) = b_0 + b_1 \,2x + b_2 \left( 4x^2 -1 \right) + b_3 \left( 8x^3 -4x \right) ,$
where
\begin{align*} b_0 &= \frac{2}{\pi} \,\int_{-1}^1 \left( 4x^3 -1 \right) \sqrt{1-x^2} \,{\text d} x = \frac{2}{\pi} \,\int_{0}^{\pi} \left( 4\cos^3 \theta -1 \right) \sin^2 \theta \, {\text d} \theta = -1, \\ b_1 &= \frac{2}{\pi} \,\int_{-1}^1 \left( 4x^3 -1 \right) \sqrt{1-x^2} \,2x\, {\text d} x = \frac{2}{\pi} \,\int_{0}^{\pi} \left( 4\cos^3 \theta -1 \right) \sin \theta \,\sin 2\theta {\text d} \theta = 1, \\ b_2 &= \frac{2}{\pi} \,\int_{-1}^1 \left( 4x^3 -1 \right) \sqrt{1-x^2} \,2x\, {\text d} x = \frac{2}{\pi} \,\int_{0}^{\pi} \left( 4\cos^3 \theta -1 \right) \sin \theta \,\sin 3\theta {\text d} \theta = 0, \\ b_3 &= \frac{2}{\pi} \,\int_{-1}^1 \left( 4x^3 -1 \right) \sqrt{1-x^2} \,2x\, {\text d} x = \frac{2}{\pi} \,\int_{0}^{\pi} \left( 4\cos^3 \theta -1 \right) \sin \theta \,\sin 4\theta {\text d} \theta = \frac{1}{2} . \end{align*}
Therefore,
$4\,x^3 -1 = -U_0 + U_1 (x) + \frac{1}{2} \, U_3 (x) = -1 + 2x + \frac{1}{4} \left( 8x^3 -4x \right) ,$
which is the identity.    ■

Example 3A: It can be shown that Fourier series for exponential function is

$e^{a\,\cos\theta} = I_0 (a) + 2 \sum_{n\ge 1} I_n (a)\, \cos n\theta ,$
where In(𝑎) is the Bessel function of imaginary argument. From this equality follows
$e^{a\,x} = I_0 (a) + 2 \sum_{n\ge 1} I_n (a)\, T_n (x) , \qquad x \in [-1, 1] .$
In particular, when (𝑎 = 1, we have
$e^{x} = I_0 (1) + 2 \sum_{n\ge 1} I_n (1)\, T_n (x) , \qquad x \in [-1, 1] .$
Using pure imaginary (𝑎, we obtain
\begin{align*} \cos ax &= I_0 (a) + 2 \sum_{k\ge 1} (-1)^k J_{2k} (a)\, T_{2k} (x) , \\ \sin ax &= 2 \sum_{k\ge 0} (-1)^k J_{2k+1} (a)\, T_{2k+1} (x) . \end{align*}

Consider an exponential function $$f(x) = e^x .$$ Its third order Chebyshev approximation is

$e^x \approx \frac{c_0}{2} + \sum_{k=1}^3 c_k T_k (x) .$
We determine the coefficients by integration:
\begin{align*} c_0 &= \frac{2}{\pi} \, \int_0^{\pi} \, e^{\cos \theta} \, {\text d} \theta = 2\, I_0 (1) \approx 2.53213 , \\ c_1 &= \frac{2}{\pi} \, \int_0^{\pi} \, e^{\cos \theta} \,\cos \theta \, {\text d} \theta = 2\, I_1 (1) \approx 1.13032 , \\ c_2 &= \frac{2}{\pi} \, \int_0^{\pi} \, e^{\cos \theta} \,\cos 2\theta \, {\text d} \theta = 2\, I_2 (1) \approx 0.271495 , \\ c_3 &= \frac{2}{\pi} \, \int_0^{\pi} \, e^{\cos \theta} \,\cos 2\theta \, {\text d} \theta = 2\, I_3 (1) \approx 0.0443368 , \end{align*}
where $$I_n (1)$$ is the modified Bessel function of order n evaluated at x = 1. When we plot Chebyshev approximation along with the given function, we cannot determine any difference, which means that the error is less than 1%. Indeed, its mean square error is about 0.000029615.
c3 = (2/Pi)*Integrate[Exp[Cos[theta]]*Cos[3*theta], {theta , 0, Pi}]
approx = c0/2 + c1*ChebyshevT[1, x] + c2*ChebyshevT[2, x] + c3*ChebyshevT[3, x]
Plot[{Exp[x], approx}, {x, -1, 1}, PlotStyle -> {Blue, Orange}]
NIntegrate[(Exp[x] - approx)^2, {x, -1, 1}]
ChebyshevApprox[n_Integer?Positive, f_Function, x_] :=
Module[{c, xk}, xk = Pi (Range[n] - 1/2)/n;
c[j_] = 2*Total[Cos[j*xk]*(f /@ Cos[xk])]/n;
Total[Table[c[k]*ChebyshevT[k, x], {k, 0, n - 1}]] - c[0]/2];

f = Exp[#] &
ChebyshevApprox[3, f, x] // Simplify
Out[2]= 1/3 (3 - E^(-(Sqrt[3]/2)) (Sqrt[3] - 2 x) x - 4 x^2 + E^(Sqrt[3]/2) x (Sqrt[3] + 2 x))
GraphicsGrid[
Partition[
Table[Plot[{f[x], ChebyshevApprox[n, f, x]}, {x, -1, 1},
Frame -> True, Axes -> False, PlotStyle -> {Blue, Red},
PlotRange -> {-2, 10},
Epilog -> Text["n = " <> ToString[n], {0.25, 5}]], {n, 9}], 3],
ImageSize -> 500]
So we see that all approximations with $$n \ge 3$$ terms give very good approximations. Now we use Chebyshev polynomials of the second kind:
$e^x \approx \sum_{k=0}^n b_k U_k (x) ,$
where
$b_k = \int_{-1}^1 e^x \,U_k (x)\, \sqrt{1-x^2} \, {\text d} x = 2\left( k+1 \right) I_{k+1} (1), \quad k=0,1,2,\ldots .$
Therefore, on the interval [-1,1], we have
$e^x = 2\,\sum_{k\ge 0} \left( k+1 \right) I_{k+1} (1) \, U_k (x) .$
Upon plotting finite sum approximation, we see that approximation with three terms is enough. Its mean square error is about 0.0000268334.
exp[m_, x_] := 2*Sum[(k + 1)*BesselI[k + 1, 1]*ChebyshevU[k, x], {k, 0, m}]
Plot[{Exp[x], exp[2, x]}, {x, -1.01, 1.01}, PlotStyle -> {{Thick, Blue}, {Thick, Red}},Epilog -> Text["n = " <> ToString[2], {-0.75, 2}]
NIntegrate[(Exp[x] - exp[3, x])^2, {x, -1, 1.0}]
Exponential approximation with N = 2 and N = 3 terms. ■

Example 4A: Consider a problem for expanding the Heaviside function into Chebyshev's series:

$H(t) = \frac{c_0}{2} + \sum_{k\ge 1} c_k T_k (t) .$
The coefficients can be evaluated explicitly:
\begin{align*} c_0 &= \frac{2}{\pi} \, \int_0^1 \frac{{\text d}x}{\sqrt{1- x^2}} =1, \\ c_n &= \frac{2}{\pi} \, \int_0^1 \frac{T_n (x)\,{\text d}x}{\sqrt{1- x^2}} = \left\{ \begin{array}{ll} 0 , & \ \mbox{if $n=2k$ is even} \\ (-1)^k \,\frac{2}{n\pi} , & \ \mbox{if $n = 2k+1$ is odd. } \end{array} \right. \end{align*}
Therefore,
$H(t) = \frac{1}{2} + \frac{2}{\pi} \,\sum_{k\ge 0} \frac{(-1)^k}{2k+1}\, T_{2k+1} (t) .$
We plot its approximation with 11 terms:
heaviside[m_, t_] = 1/2 + (2/Pi)* Sum[(-1)^k *ChebyshevT[2*k + 1, t]/(2*k + 1) , {k, 0, m}];
Plot[{HeavisideTheta[t], heaviside[10, t]}, {t, -1.01, 1.01}, PlotStyle -> {Blue, Orange}]
We repeat calculations with Chebyshev polynomials of the second kind:
$H(t) = \frac{1}{2} + \frac{4}{\pi} \, \sum_{k\ge 0} \frac{(-1)^k\, k}{(2k+1)(2k+3)}\, U_{2k+1} (t) .$
The following graph presents 10 term approximation with Chebyshev polynomials of second kind.
chebsum[m_] := 1/2 - (4/Pi)* Sum[(-1)^k *k/(2*k - 1)/(2*k + 1) *ChebyshevU[2*k - 1, t], {k, 1, m}]
Plot[{HeavisideTheta[t], chebsum[10]}, {t, -1.01, 1.01}, PlotStyle -> {Blue, Orange}]
We observe Gibbs phenomenon near point of discontinuity t = 0 in both expansions with respect to Chebyshev polynomials of the first and second kind, as well as bad convergence at end points.    ■

Example 5A: Let us consider the signum function on the closed interval [-1,1], which we expand into Chebyshev series:

\begin{align*} \mbox{sign}(x) = \sum_{n\ge 1} a_n T_n (x) = \sum_{n\ge 1} b_n U_n (x) = \sum_{n\ge 0} c_n V_n (x) = \sum_{n\ge 0} d_n W_n (x) , \end{align*}
where
\begin{align*} a_n &= - \frac{2}{\pi} \int_{-1}^0 \frac{T_n (x)}{\sqrt{1-x^2}} \, {\text d}x + \frac{2}{\pi} \int_0^1 \frac{T_k (x)}{\sqrt{1-x^2}} \, {\text d}x = \frac{4}{\pi} \times \begin{cases} \frac{1}{n} , & \ \mbox{ if $n$ is odd}, \\ 0 , & \ \mbox{ if $n$ is even}; \end{cases} \\ b_n &= - \frac{2}{\pi} \int_{-1}^0 U_n (x) \,\sqrt{1-x^2} \, {\text d}x + \frac{2}{\pi} \int_0^1 U_n (x) \,\sqrt{1-x^2} \, {\text d}x = \frac{8}{\pi} \times \begin{cases} 0, & \ \mbox{if $n$ is even}, \\ \frac{(-1)^k k}{(2k-1)(2k+1)} , & \ \mbox{ if $n= 2k-1$ is odd}; \end{cases} \\ c_n &= \frac{1}{\pi} \int_0^1 \left( \frac{1+x}{1-x} \right)^{1/2} V_n (x) \, {\text d}x - \frac{1}{\pi} \int_{-1}^0 \left( \frac{1+x}{1-x} \right)^{1/2} V_n (x) \, {\text d}x = \\ d_n &= \frac{1}{\pi} \int_0^1 \left( \frac{1-x}{1+x} \right)^{1/2} W_n (x) \, {\text d}x - \frac{1}{\pi} \int_{-1}^0 \left( \frac{1+x}{1-x} \right)^{1/2} W_n (x) \, {\text d}x = \end{align*}
We plot partial sums using Mathematica:
SA[x_] = (4/Pi)*Sum[ChebyshevT[2*k - 1, x]/(2*k - 1), {k, 0, 10}] Plot[SA[x], {x, -1, 1}]
CT[n_, x_] := Piecewise[{{1, -1 < x < 0}, {0, 0}, {-1, 0 < x < 1}}];
CT[n, x] = Piecewise[{{1, -1 < x < 0}, {0, 0}, {-1, 0 < x < 1}}];
If[n == -1, 1];
Expand[2*x*ChebyshevT[n - 1, x] - ChebyshevT[n - 2, x]];
CCT0[x_] = unitstep[Mod[x, Pi, -Pi]];
ct[x_] = Piecewise[{{1, -1 <= x < 0}, {-1, 0 < x <= 1}}];
ct6ff0 = FourierSinSeries[CCT0[x], x, 6];
ctff0[x_, nmax_] := Sum[-Sin[(2 n + 1) .65 x]/(2 n + 1), {n, 0, nmax}] 1.275
stepplot[n_] := Plot[{ct[x], ChebyshevT[1, ctff0][x, n]}, {x, -1, 1}, Ticks -> {{-1, -.5, 0, .5, 1}}, PlotStyle -> {{Thick, Blue}, {Thick, Red}}, Exclusions -> None]
stepplot[50]
stepplot[500]
■

Example: Some other examples of Chebyshev series:

\begin{align*} \frac{1}{a^2 - x^2} &= \frac{1}{a \sqrt{a^2 -1}} \left[ 1 - 2\sum_{k\ge 1} \left( a - \sqrt{a^2 -1} \right)^{2k} T_{2k} (x) \right] , \qquad a > 1, \\ \frac{1}{x-a} &= \frac{1}{\sqrt{a^2 -1}} + \frac{2}{\sqrt{a^2 -1}} \sum_{n\ge 1} \left( a - \sqrt{a^2 -1} \right)^{n} T_{n} (x) \\ \mbox{sign}(x) &= -\frac{4}{\pi} \,\sum_{k \ge 1} \frac{(-1)^k}{2k-1} \,T_{2k-1} (x) , \\ |x| &= \frac{2}{\pi} - \frac{4}{\pi} \,\sum_{k\ge 1} \frac{(-1)^k}{4k^2 -1} \,T_{2k} (x) \\ &= \frac{1}{2} - \sum_{k\ge 1} (-1)^k \frac{(2k-3)!!}{(2k+2)!!} \, (4k+1)\,P_{2k} (x) , \\ \sin ax &= J_1 (a) + 2 \sum_{k\ge 1} (-1)^k J_{2k+1} (a)\, T_{2k+1} (x) , \\ \sinh ax &= I_1 (a) + 2 \sum_{k\ge 1} I_{2k+1} (a) \, T_{2k+1} (x) , \\ \cosh ax &= I_0 (a) + 2 \sum_{k\ge 1} I_{2k} (a) \, T_{2k} (x) , \\ \mbox{arccos}(x) &= \frac{\pi}{2} - \frac{4}{\pi} \,\sum_{k\ge 1} \frac{1}{(2k-1)^2}\, T_{2k-1} (x) , \\ \mbox{arcsin}(x) &= \frac{4}{\pi} \,\sum_{k\ge 1} \frac{1}{(2k-1)^2}\, T_{2k-1} (x) . \\ \arctan x &= 2 \sum_{k \ge 0} (-1)^k \frac{p^{2k+1}}{2k+1} \,T_{2k+1} (x) , \qquad p = \sqrt{2} -1 , \\ \sqrt{1-x^2} &= \frac{2}{\pi} - \frac{4}{\pi} \sum_{n\ge 1} \frac{T_{2n} (x)}{4 n^2 -1} , \\ \frac{1}{4}\,\ln \left( \frac{1+x}{1-x} \right) &= \sum_{n\ge 1} \frac{T_{2n -1} (x)}{2n-1} , \qquad -1 < x < 1 , \\ \frac{1-rx}{1 -2rx + r^2} &= \sum_{n\he 0} r^n T_n (x) , \qquad 0 \le r < 1 , \\ \ln \left( 1 -2rx + r^2 \right) &= -2 \sum_{n\he 1} \frac{r^n}{n}\, T_n (x) , \\ \ln \frac{1 + 2rx + r^2}{1 - 2rx + r^2} &= 4 \sum_{k\ge 0} \frac{r^{2k+1}}{2k+1}\,T_{2k+1} (x) . \end{align*}
Bessel functions:
\begin{align*} J_0 (ax) &= \sum_{k\ge 0} {\cal E}_k (-1)^k J_k^2 \left( a/2 \right) T_{2k} (x) , \\ J_1 (ax) &= 2 \sum_{k\ge 0} (-1)^k J_k \left( a/2 \right) T_{2k+1} (x) .1 \end{align*}
Delta function:
$\delta (x) = \frac{1}{\pi} + \frac{2}{\pi} \sum_{n\ge 1} (-1)^n T_{2n} (x) .$

Series Expansions over Chebyshev Polynomials of the Second Kind

We are tempted to expand a function into the series over Chebyshev polynomials of the second kind:
$$\label{EqCheb2.1} f(x) = \sum_{k\ge 0} \,b_k\, U_{k} (x ) ,$$
where
$$\label{EqCheb2.2} b_k = \frac{2}{\pi}\,\int_{-1}^1 \,f(x)\, U_{k} (x )\,\sqrt{1-x^2}\, {\text d} x = \frac{2}{\pi}\,\int_{0}^{\pi} \, f\left( \cos \,\theta \right) \sin \left( (k+1)\,\theta \right) \sin \theta \,{\text d} \theta , \qquad k=0,1,2,\ldots .$$

Formula \eqref{EqCheb2.2} show that the Chebyshev expansion with respect to Chebyshev polynomials of the second kind is actually the sine-Fourier series for the odd function $$\displaystyle f(\cos \theta )\,\sin\theta .$$

Since U2k+1(0) = 0, we get the expansion of the Dirac delta function:

$\delta (x) = \frac{2}{\pi}\,\sum_{k\ge 0} (-1)^k U_{2k} (x) .$

If function g(x) is continuous on the interval [−1, 1] and vanish at end points so that $$g(x)/\sqrt{1-x^2}$$ is quadratically/absolutely integrable on the interval, then we can expand it as

$g(x) = \sqrt{1-x^2} \,\sum_{k\ge 0} \,b_k\, U_{k} (x ) ,$
where
$b_k = \frac{2}{\pi}\,\int_{-1}^1 \,g(x)\, U_{k} (x )\, {\text d} x , \qquad k=0,1,2,\ldots .$

Example 1B: Let us consider the root function $$f(x) = \sqrt{1 - x^2} .$$ To expand it into Chebyshev series over second kind polynomials, we need to calculate Fourier-Chebyshev coefficients according to \eqref{EqCheb2.2}:

$b_n = \frac{2}{\pi}\,\int_{-1}^1 \sqrt{1-x^2} \,\sqrt{1- x^2} \,U_n (x) \,{\text d}x = \frac{2}{\pi}\,\int_{-1}^1 \left( 1 - x^2 \right) U_n (x) \,{\text d}x = \frac{2}{\pi}\cdot \frac{2}{3 + n - 3n^2 - n^3} \left[ 1 + (-1)^n \right] .$
Indeed, Mathematica is able to calculate this integral
Integrate[(1 - x^2)*ChebyshevU[n, x], {x, -1, 1}]
(2 + 2 Cos[n $Pi]])/(3 + n - 3 n^2 - n^3) Since \[ 1 + (-1)^n = \begin{cases} 2 , & \ \mbox{ if } \ n = 2k, \\ 0, & \ \mbox{ if } \ n = 2k+1 . \end{cases}$
So we get
$\sqrt{1- x^2} = \frac{2}{\pi}\, \sum_{k\ge 0} \frac{4}{3 + 2k -12 k^2 - 8 k^3} \, U_{2k} (x) .$

Now we consider another root function $$g(x) = \sqrt{1 - x} .$$ The corresponding Fourier--Chebyshev coefficients are evaluated according to Eq.\eqref{EqCheb2.2}:

$b_n = \frac{2}{\pi}\,\int_{0}^{\pi} \sin \frac{\theta}{2} \,\sin [(n+1) \theta ] \,\sin\theta \, {\text d}\theta$
because $$\sqrt{1 - \cos\theta} = \sin\frac{\theta}{2} .$$ We again ask Mathematica to rescure
Integrate[Sin[t/2]*Sin[(n + 1)*t]*Sin[t], {t, 0, Pi}]
(4 (-4 (1 + n) + (1 + 8 n + 4 n^2) Sin[n $Pi]]))/((-1 + 2 n) (1 + 2 n) (3 + 2 n) (5 + 2 n)) This yields \[ b_n = - \frac{2}{\pi}\,\frac{16 \left( 1 + n \right)}{(4n^2 -1)(3 + 2n)(5 + 2n)} , \qquad n=0,1,2,\ldots .$
So the required expansion becomes
$\sqrt{1-x} = - \frac{2}{\pi}\,\sum_{n\ge 0} \frac{16 \left( 1 + n \right)}{(4n^2 -1)(3 + 2n)(5 + 2n)} \, U_n (x) .$

Example 4B: Consider a problem for expanding the Heaviside function into Chebyshev's series:

$H(t) = \frac{c_0}{2} + \sum_{k\ge 1} c_k T_k (t) .$
The coefficients can be evaluated explicitly:
\begin{align*} c_0 &= \frac{2}{\pi} \, \int_0^1 \frac{{\text d}x}{\sqrt{1- x^2}} =1, \\ c_n &= \frac{2}{\pi} \, \int_0^1 \frac{T_n (x)\,{\text d}x}{\sqrt{1- x^2}} = \left\{ \begin{array}{ll} 0 , & \ \mbox{if $n=2k$ is even} \\ (-1)^k \,\frac{2}{n\pi} , & \ \mbox{if $n = 2k+1$ is odd. } \end{array} \right. \end{align*}
Therefore,
$H(t) = \frac{1}{2} + \frac{2}{\pi} \,\sum_{k\ge 0} \frac{(-1)^k}{2k+1}\, T_{2k+1} (t) .$
We plot its approximation with 11 terms:
heaviside[m_, t_] = 1/2 + (2/Pi)* Sum[(-1)^k *ChebyshevT[2*k + 1, t]/(2*k + 1) , {k, 0, m}];
Plot[{HeavisideTheta[t], heaviside[10, t]}, {t, -1.01, 1.01}, PlotStyle -> {Blue, Orange}]
We repeat calculations with Chebyshev polynomials of the second kind:
$H(t) = \frac{1}{2} + \frac{4}{\pi} \, \sum_{k\ge 0} \frac{(-1)^k\, k}{(2k+1)(2k+3)}\, U_{2k+1} (t) .$
The following graph presents 10 term approximation with Chebyshev polynomials of second kind.
chebsum[m_] := 1/2 - (4/Pi)* Sum[(-1)^k *k/(2*k - 1)/(2*k + 1) *ChebyshevU[2*k - 1, t], {k, 1, m}]
Plot[{HeavisideTheta[t], chebsum[10]}, {t, -1.01, 1.01}, PlotStyle -> {Blue, Orange}]
We observe Gibbs phenomenon near point of discontinuity t = 0 in both expansions with respect to Chebyshev polynomials of the first and second kind, as well as bad convergence at end points.    ■

Example 5B: Let us consider the signum function on the closed interval [-1,1], which we expand into Chebyshev series:

\begin{align*} \mbox{sign}(x) = \sum_{n\ge 1} a_n T_n (x) = \sum_{n\ge 1} b_n U_n (x) = \sum_{n\ge 0} c_n V_n (x) = \sum_{n\ge 0} d_n W_n (x) , \end{align*}
where
\begin{align*} a_n &= - \frac{2}{\pi} \int_{-1}^0 \frac{T_n (x)}{\sqrt{1-x^2}} \, {\text d}x + \frac{2}{\pi} \int_0^1 \frac{T_k (x)}{\sqrt{1-x^2}} \, {\text d}x = \frac{4}{\pi} \times \begin{cases} \frac{1}{n} , & \ \mbox{ if $n$ is odd}, \\ 0 , & \ \mbox{ if $n$ is even}; \end{cases} \\ b_n &= - \frac{2}{\pi} \int_{-1}^0 U_n (x) \,\sqrt{1-x^2} \, {\text d}x + \frac{2}{\pi} \int_0^1 U_n (x) \,\sqrt{1-x^2} \, {\text d}x = \frac{8}{\pi} \times \begin{cases} 0, & \ \mbox{if $n$ is even}, \\ \frac{(-1)^k k}{(2k-1)(2k+1)} , & \ \mbox{ if $n= 2k-1$ is odd}; \end{cases} \\ c_n &= \frac{1}{\pi} \int_0^1 \left( \frac{1+x}{1-x} \right)^{1/2} V_n (x) \, {\text d}x - \frac{1}{\pi} \int_{-1}^0 \left( \frac{1+x}{1-x} \right)^{1/2} V_n (x) \, {\text d}x = \\ d_n &= \frac{1}{\pi} \int_0^1 \left( \frac{1-x}{1+x} \right)^{1/2} W_n (x) \, {\text d}x - \frac{1}{\pi} \int_{-1}^0 \left( \frac{1+x}{1-x} \right)^{1/2} W_n (x) \, {\text d}x = \end{align*}
We plot partial sums using Mathematica:
SA[x_] = (4/Pi)*Sum[ChebyshevT[2*k - 1, x]/(2*k - 1), {k, 0, 10}] Plot[SA[x], {x, -1, 1}]
CT[n_, x_] := Piecewise[{{1, -1 < x < 0}, {0, 0}, {-1, 0 < x < 1}}];
CT[n, x] = Piecewise[{{1, -1 < x < 0}, {0, 0}, {-1, 0 < x < 1}}];
If[n == -1, 1];
Expand[2*x*ChebyshevT[n - 1, x] - ChebyshevT[n - 2, x]];
CCT0[x_] = unitstep[Mod[x, Pi, -Pi]];
ct[x_] = Piecewise[{{1, -1 <= x < 0}, {-1, 0 < x <= 1}}];
ct6ff0 = FourierSinSeries[CCT0[x], x, 6];
ctff0[x_, nmax_] := Sum[-Sin[(2 n + 1) .65 x]/(2 n + 1), {n, 0, nmax}] 1.275
stepplot[n_] := Plot[{ct[x], ChebyshevT[1, ctff0][x, n]}, {x, -1, 1}, Ticks -> {{-1, -.5, 0, .5, 1}}, PlotStyle -> {{Thick, Blue}, {Thick, Red}}, Exclusions -> None]
stepplot[50]
stepplot[500]
■

Series Expansions over Chebyshev Polynomials of the Third Kind

A function f(x) that satisfies the Dirichlet conditions on the interval [−1, 1] can be expanded into Chebyshev series
$$\label{EqCheb3.1} f(x) = \sum_{k\ge 0} \,c_k\, V_{k} (x ) , \qquad -1 < x < 1 ,$$
where
$$\label{EqCheb3.2} c_k = \frac{1}{\pi}\,\int_{-1}^1 \,f(x)\, V_{k} (x )\left( \frac{1+x}{1-x} \right)^{1/2} {\text d} x , \qquad k=0,1,2,\ldots .$$
In the integral, we make substitution $$x = \cos \theta \quad \Longrightarrow \quad {\text d}x = -\sin\theta\,{\text d}\theta .$$ The radical becomes
$\left( \frac{1+x}{1-x} \right)^{1/2} = \frac{1+x}{\sqrt{1 - x^2}} = \frac{1 + \cos\theta}{\sin\theta} = \frac{2\,\cos^2 \frac{\theta}{2}}{\sin\theta} .$
The Chebyshev polynomial of the third kind in new variable θ is the ratio:
$V_n (\cos \theta ) = \frac{\cos \left( \frac{2n+1}{2}\,\theta \right)}{\cos \left( \frac{\theta}{2} \right)} .$
Therefore, the kernel of the Fourier--Chebyshev coefficient is the product of these expressions:
$C_n = \frac{1 + \cos\theta}{\sin\theta} \cdot \frac{\cos \left( \frac{2n+1}{2}\,\theta \right)}{\cos \left( \frac{\theta}{2} \right)} \cdot \sin\theta = 2\,\cos \left( \frac{\theta}{2} \right) \cdot \cos \left( \frac{2n+1}{2}\,\theta \right)$
because $$1 + \cos \theta = 2\,\cos^2 \frac{\theta}{2} .$$ We replace the product of two cosine functions with the sum:
$2\,\cos\alpha\,\cos\beta = \cos \left( \alpha + \beta \right) + \cos \left( \alpha - \beta \right) .$
This yields
$$\label{EqCheb3.3} c_n = \frac{1}{\pi}\,\int_{0}^{\pi} \,f(\cos\theta )\left[ \cos n\theta + \cos (n+1)\theta \right] {\text d} \theta , \qquad n=0,1,2,\ldots .$$

This formula shows that the coefficient cn is the average of two ajacent coefficients of cosine Fourier series for the even function f(cosθ).

Example 1C: Let us consider the root function $$f(x) = \sqrt{1 - x^2} .$$ To expand it into Chebyshev series over third kind polynomials, we need to calculate Fourier-Chebyshev coefficients according to \eqref{EqCheb3.3}:

$c_n = \frac{1}{\pi}\,\int_{0}^{\pi} \sqrt{1-\cos^2 \theta} \left[ \cos n\theta + \cos (n+1)\theta \right] {\text d} \theta = \frac{(-1)^n -1}{n(2+n)} + \frac{1 + (-1)^n}{1 - n^2} .$
Indeed, Mathematica is able to calculate this integral
Integrate[ Sin[t]*(Cos[n*t] + Cos[(n + 1)*t]), {t, 0, Pi}
(-1 + Cos[n $Pi]])/(n (2 + n)) + (1 + Cos[n \[Pi]])/(1 - n^2) Since \[ 1 + (-1)^n = \begin{cases} 2 , & \ \mbox{ if } \ n = 2k, \\ 0, & \ \mbox{ if } \ n = 2k+1 ; \end{cases} \qquad \mbox{and} \qquad (-1)^n -1 = \begin{cases} \phantom{-}0 , & \ \mbox{ if } \ n = 2k, \\ -2, & \ \mbox{ if } \ n = 2k+1 . \end{cases}$
So we get
$\sqrt{1- x^2} = \frac{2}{\pi}\, \sum_{k\ge 0} \frac{1}{1 - 4 k^2} \, V_{2k} (x) - \frac{2}{\pi}\, \sum_{k\ge 0} \frac{1}{(2k+1)(3+2k)}\, V_{2k+1} (x) .$

Now we consider another root function $$g(x) = \sqrt{1 - x} .$$ The corresponding Fourier--Chebyshev coefficients are evaluated according to Eq.\eqref{EqCheb2.2}:

$b_n = \frac{2}{\pi}\,\int_{0}^{\pi} \sin \frac{\theta}{2} \,\sin [(n+1) \theta ] \,\sin\theta \, {\text d}\theta$
because $$\sqrt{1 - \cos\theta} = \sin\frac{\theta}{2} .$$ We again ask Mathematica to rescue
Integrate[Sin[t/2]*Sin[(n + 1)*t]*Sin[t], {t, 0, Pi}]
(4 (-4 (1 + n) + (1 + 8 n + 4 n^2) Sin[n $Pi]]))/((-1 + 2 n) (1 + 2 n) (3 + 2 n) (5 + 2 n)) This yields \[ b_n = - \frac{2}{\pi}\,\frac{16 \left( 1 + n \right)}{(4n^2 -1)(3 + 2n)(5 + 2n)} , \qquad n=0,1,2,\ldots .$
So the required expansion becomes
$\sqrt{1-x} = - \frac{2}{\pi}\,\sum_{n\ge 0} \frac{16 \left( 1 + n \right)}{(4n^2 -1)(3 + 2n)(5 + 2n)} \, U_n (x) .$

Example 4C: Consider a problem for expanding the Heaviside function into Chebyshev's series:

$H(t) = \frac{c_0}{2} + \sum_{k\ge 1} c_k T_k (t) .$
The coefficients can be evaluated explicitly:
\begin{align*} c_0 &= \frac{2}{\pi} \, \int_0^1 \frac{{\text d}x}{\sqrt{1- x^2}} =1, \\ c_n &= \frac{2}{\pi} \, \int_0^1 \frac{T_n (x)\,{\text d}x}{\sqrt{1- x^2}} = \left\{ \begin{array}{ll} 0 , & \ \mbox{if $n=2k$ is even} \\ (-1)^k \,\frac{2}{n\pi} , & \ \mbox{if $n = 2k+1$ is odd. } \end{array} \right. \end{align*}
Therefore,
$H(t) = \frac{1}{2} + \frac{2}{\pi} \,\sum_{k\ge 0} \frac{(-1)^k}{2k+1}\, T_{2k+1} (t) .$
We plot its approximation with 11 terms:
heaviside[m_, t_] = 1/2 + (2/Pi)* Sum[(-1)^k *ChebyshevT[2*k + 1, t]/(2*k + 1) , {k, 0, m}];
Plot[{HeavisideTheta[t], heaviside[10, t]}, {t, -1.01, 1.01}, PlotStyle -> {Blue, Orange}]
We repeat calculations with Chebyshev polynomials of the second kind:
$H(t) = \frac{1}{2} + \frac{4}{\pi} \, \sum_{k\ge 0} \frac{(-1)^k\, k}{(2k+1)(2k+3)}\, U_{2k+1} (t) .$
The following graph presents 10 term approximation with Chebyshev polynomials of second kind.
chebsum[m_] := 1/2 - (4/Pi)* Sum[(-1)^k *k/(2*k - 1)/(2*k + 1) *ChebyshevU[2*k - 1, t], {k, 1, m}]
Plot[{HeavisideTheta[t], chebsum[10]}, {t, -1.01, 1.01}, PlotStyle -> {Blue, Orange}]
We observe Gibbs phenomenon near point of discontinuity t = 0 in both expansions with respect to Chebyshev polynomials of the first and second kind, as well as bad convergence at end points.    ■

Example 5C: Let us consider the signum function on the closed interval [-1,1], which we expand into Chebyshev series:

\begin{align*} \mbox{sign}(x) = \sum_{n\ge 1} a_n T_n (x) = \sum_{n\ge 1} b_n U_n (x) = \sum_{n\ge 0} c_n V_n (x) = \sum_{n\ge 0} d_n W_n (x) , \end{align*}
where
\begin{align*} a_n &= - \frac{2}{\pi} \int_{-1}^0 \frac{T_n (x)}{\sqrt{1-x^2}} \, {\text d}x + \frac{2}{\pi} \int_0^1 \frac{T_k (x)}{\sqrt{1-x^2}} \, {\text d}x = \frac{4}{\pi} \times \begin{cases} \frac{1}{n} , & \ \mbox{ if $n$ is odd}, \\ 0 , & \ \mbox{ if $n$ is even}; \end{cases} \\ b_n &= - \frac{2}{\pi} \int_{-1}^0 U_n (x) \,\sqrt{1-x^2} \, {\text d}x + \frac{2}{\pi} \int_0^1 U_n (x) \,\sqrt{1-x^2} \, {\text d}x = \frac{8}{\pi} \times \begin{cases} 0, & \ \mbox{if $n$ is even}, \\ \frac{(-1)^k k}{(2k-1)(2k+1)} , & \ \mbox{ if $n= 2k-1$ is odd}; \end{cases} \\ c_n &= \frac{1}{\pi} \int_0^1 \left( \frac{1+x}{1-x} \right)^{1/2} V_n (x) \, {\text d}x - \frac{1}{\pi} \int_{-1}^0 \left( \frac{1+x}{1-x} \right)^{1/2} V_n (x) \, {\text d}x = \\ d_n &= \frac{1}{\pi} \int_0^1 \left( \frac{1-x}{1+x} \right)^{1/2} W_n (x) \, {\text d}x - \frac{1}{\pi} \int_{-1}^0 \left( \frac{1+x}{1-x} \right)^{1/2} W_n (x) \, {\text d}x = \end{align*}
We plot partial sums using Mathematica:
SA[x_] = (4/Pi)*Sum[ChebyshevT[2*k - 1, x]/(2*k - 1), {k, 0, 10}] Plot[SA[x], {x, -1, 1}]
CT[n_, x_] := Piecewise[{{1, -1 < x < 0}, {0, 0}, {-1, 0 < x < 1}}];
CT[n, x] = Piecewise[{{1, -1 < x < 0}, {0, 0}, {-1, 0 < x < 1}}];
If[n == -1, 1];
Expand[2*x*ChebyshevT[n - 1, x] - ChebyshevT[n - 2, x]];
CCT0[x_] = unitstep[Mod[x, Pi, -Pi]];
ct[x_] = Piecewise[{{1, -1 <= x < 0}, {-1, 0 < x <= 1}}];
ct6ff0 = FourierSinSeries[CCT0[x], x, 6];
ctff0[x_, nmax_] := Sum[-Sin[(2 n + 1) .65 x]/(2 n + 1), {n, 0, nmax}] 1.275
stepplot[n_] := Plot[{ct[x], ChebyshevT[1, ctff0][x, n]}, {x, -1, 1}, Ticks -> {{-1, -.5, 0, .5, 1}}, PlotStyle -> {{Thick, Blue}, {Thick, Red}}, Exclusions -> None]
stepplot[50]
stepplot[500]
■

Series Expansions over Chebyshev Polynomials of the Fourth Kind

A function f(x) that satisfies the Dirichlet conditions on the interval [−1, 1] can be expanded into Chebyshev series
$$\label{EqCheb4.1} f(x) = \sum_{k\ge 0} \,d_k\, W_{k} (x ) , \qquad -1 < x < 1 ,$$
where
$$\label{EqCheb4.2} d_k = \frac{1}{\pi}\,\int_{-1}^1 \,f(x)\, W_{k} (x )\left( \frac{1-x}{1+x} \right)^{1/2} {\text d} x , \qquad k=0,1,2,\ldots .$$
Using expression
$W_n (\cos \theta ) = \frac{\sin \left( \frac{2n+1}{2}\,\theta \right)}{\sin \left( \frac{\theta}{2} \right)}$
we rewrite the kernel of coefficient \eqref{EqCheb4.2} as
$D_n = \frac{\sin\theta}{1 + \cos\theta} \cdot \frac{\sin \left( \frac{2n+1}{2}\,\theta \right)}{\sin \left( \frac{\theta}{2} \right)} \cdot \sin\theta = \frac{\sin\theta}{\cos^2 \frac{\theta}{2}} \cdot \frac{\sin \left( \frac{2n+1}{2}\,\theta \right)}{\sin \left( \frac{\theta}{2} \right)} \cdot \sin\theta = 2\,\sin \left( \frac{\theta}{2} \right) \sin \left( \frac{2n+1}{2}\,\theta \right) .$
This allows us to rewrite the coefficient dn as
$$\label{EqCheb4.3} d_k = \frac{1}{\pi}\,\int_{-1}^1 \,f(\cos\theta )\left[ \cos n\theta - \cos [(n+1) \theta ] \right] {\text d}\theta , \qquad n=0,1,2,\ldots .$$

Example 1D: Let us consider the root function $$f(x) = \sqrt{1 - x^2} .$$ To expand it into Chebyshev series over second kind polynomials, we need to calculate Fourier-Chebyshev coefficients according to \eqref{EqCheb2.2}:

$b_n = \frac{2}{\pi}\,\int_{-1}^1 \sqrt{1-x^2} \,\sqrt{1- x^2} \,U_n (x) \,{\text d}x = \frac{2}{\pi}\,\int_{-1}^1 \left( 1 - x^2 \right) U_n (x) \,{\text d}x = \frac{2}{\pi}\cdot \frac{2}{3 + n - 3n^2 - n^3} \left[ 1 + (-1)^n \right] .$
Indeed, Mathematica is able to calculate this integral
Integrate[(1 - x^2)*ChebyshevU[n, x], {x, -1, 1}]
(2 + 2 Cos[n $Pi]])/(3 + n - 3 n^2 - n^3) Since \[ 1 + (-1)^n = \begin{cases} 2 , & \ \mbox{ if } \ n = 2k, \\ 0, & \ \mbox{ if } \ n = 2k+1 . \end{cases}$
So we get
$\sqrt{1- x^2} = \frac{2}{\pi}\, \sum_{k\ge 0} \frac{4}{3 + 2k -12 k^2 - 8 k^3} \, U_{2k} (x) .$

Now we consider another root function $$g(x) = \sqrt{1 - x} .$$ The corresponding Fourier--Chebyshev coefficients are evaluated according to Eq.\eqref{EqCheb2.2}:

$b_n = \frac{2}{\pi}\,\int_{0}^{\pi} \sin \frac{\theta}{2} \,\sin [(n+1) \theta ] \,\sin\theta \, {\text d}\theta$
because $$\sqrt{1 - \cos\theta} = \sin\frac{\theta}{2} .$$ We again ask Mathematica to rescure
Integrate[Sin[t/2]*Sin[(n + 1)*t]*Sin[t], {t, 0, Pi}]
(4 (-4 (1 + n) + (1 + 8 n + 4 n^2) Sin[n $Pi]]))/((-1 + 2 n) (1 + 2 n) (3 + 2 n) (5 + 2 n)) This yields \[ b_n = - \frac{2}{\pi}\,\frac{16 \left( 1 + n \right)}{(4n^2 -1)(3 + 2n)(5 + 2n)} , \qquad n=0,1,2,\ldots .$
So the required expansion becomes
$\sqrt{1-x} = - \frac{2}{\pi}\,\sum_{n\ge 0} \frac{16 \left( 1 + n \right)}{(4n^2 -1)(3 + 2n)(5 + 2n)} \, U_n (x) .$

Example 4D: Consider a problem for expanding the Heaviside function into Chebyshev's series:

$H(t) = \frac{c_0}{2} + \sum_{k\ge 1} c_k T_k (t) .$
The coefficients can be evaluated explicitly:
\begin{align*} c_0 &= \frac{2}{\pi} \, \int_0^1 \frac{{\text d}x}{\sqrt{1- x^2}} =1, \\ c_n &= \frac{2}{\pi} \, \int_0^1 \frac{T_n (x)\,{\text d}x}{\sqrt{1- x^2}} = \left\{ \begin{array}{ll} 0 , & \ \mbox{if $n=2k$ is even} \\ (-1)^k \,\frac{2}{n\pi} , & \ \mbox{if $n = 2k+1$ is odd. } \end{array} \right. \end{align*}
Therefore,
$H(t) = \frac{1}{2} + \frac{2}{\pi} \,\sum_{k\ge 0} \frac{(-1)^k}{2k+1}\, T_{2k+1} (t) .$
We plot its approximation with 11 terms:
heaviside[m_, t_] = 1/2 + (2/Pi)* Sum[(-1)^k *ChebyshevT[2*k + 1, t]/(2*k + 1) , {k, 0, m}];
Plot[{HeavisideTheta[t], heaviside[10, t]}, {t, -1.01, 1.01}, PlotStyle -> {Blue, Orange}]
We repeat calculations with Chebyshev polynomials of the second kind:
$H(t) = \frac{1}{2} + \frac{4}{\pi} \, \sum_{k\ge 0} \frac{(-1)^k\, k}{(2k+1)(2k+3)}\, U_{2k+1} (t) .$
The following graph presents 10 term approximation with Chebyshev polynomials of second kind.
chebsum[m_] := 1/2 - (4/Pi)* Sum[(-1)^k *k/(2*k - 1)/(2*k + 1) *ChebyshevU[2*k - 1, t], {k, 1, m}]
Plot[{HeavisideTheta[t], chebsum[10]}, {t, -1.01, 1.01}, PlotStyle -> {Blue, Orange}]
We observe Gibbs phenomenon near point of discontinuity t = 0 in both expansions with respect to Chebyshev polynomials of the first and second kind, as well as bad convergence at end points.    ■

Example 5D: Let us consider the signum function on the closed interval [-1,1], which we expand into Chebyshev series:

\begin{align*} \mbox{sign}(x) = \sum_{n\ge 1} a_n T_n (x) = \sum_{n\ge 1} b_n U_n (x) = \sum_{n\ge 0} c_n V_n (x) = \sum_{n\ge 0} d_n W_n (x) , \end{align*}
where
\begin{align*} a_n &= - \frac{2}{\pi} \int_{-1}^0 \frac{T_n (x)}{\sqrt{1-x^2}} \, {\text d}x + \frac{2}{\pi} \int_0^1 \frac{T_k (x)}{\sqrt{1-x^2}} \, {\text d}x = \frac{4}{\pi} \times \begin{cases} \frac{1}{n} , & \ \mbox{ if $n$ is odd}, \\ 0 , & \ \mbox{ if $n$ is even}; \end{cases} \\ b_n &= - \frac{2}{\pi} \int_{-1}^0 U_n (x) \,\sqrt{1-x^2} \, {\text d}x + \frac{2}{\pi} \int_0^1 U_n (x) \,\sqrt{1-x^2} \, {\text d}x = \frac{8}{\pi} \times \begin{cases} 0, & \ \mbox{if $n$ is even}, \\ \frac{(-1)^k k}{(2k-1)(2k+1)} , & \ \mbox{ if $n= 2k-1$ is odd}; \end{cases} \\ c_n &= \frac{1}{\pi} \int_0^1 \left( \frac{1+x}{1-x} \right)^{1/2} V_n (x) \, {\text d}x - \frac{1}{\pi} \int_{-1}^0 \left( \frac{1+x}{1-x} \right)^{1/2} V_n (x) \, {\text d}x = \\ d_n &= \frac{1}{\pi} \int_0^1 \left( \frac{1-x}{1+x} \right)^{1/2} W_n (x) \, {\text d}x - \frac{1}{\pi} \int_{-1}^0 \left( \frac{1+x}{1-x} \right)^{1/2} W_n (x) \, {\text d}x = \end{align*}
We plot partial sums using Mathematica:
SA[x_] = (4/Pi)*Sum[ChebyshevT[2*k - 1, x]/(2*k - 1), {k, 0, 10}] Plot[SA[x], {x, -1, 1}]
CT[n_, x_] := Piecewise[{{1, -1 < x < 0}, {0, 0}, {-1, 0 < x < 1}}];
CT[n, x] = Piecewise[{{1, -1 < x < 0}, {0, 0}, {-1, 0 < x < 1}}];
If[n == -1, 1];
Expand[2*x*ChebyshevT[n - 1, x] - ChebyshevT[n - 2, x]];
CCT0[x_] = unitstep[Mod[x, Pi, -Pi]];
ct[x_] = Piecewise[{{1, -1 <= x < 0}, {-1, 0 < x <= 1}}];
ct6ff0 = FourierSinSeries[CCT0[x], x, 6];
ctff0[x_, nmax_] := Sum[-Sin[(2 n + 1) .65 x]/(2 n + 1), {n, 0, nmax}] 1.275
stepplot[n_] := Plot[{ct[x], ChebyshevT[1, ctff0][x, n]}, {x, -1, 1}, Ticks -> {{-1, -.5, 0, .5, 1}}, PlotStyle -> {{Thick, Blue}, {Thick, Red}}, Exclusions -> None]
stepplot[50]
stepplot[500]
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Clenshaw's method for the evaluation of a Chebyshev sum

1. Clenshaw, C.W., Norton, H.J.: The solution of nonlinear ordinary differential equations in chebyshev series. The Computer Journal, 1963, {\bf 6}, Issue 1, 88–92; https://doi.org/10.1093
2. Schweizer, W., Special Functions in Physics with MATLAB, 2021, Springer. https://link.springer.com/chapter/10.1007%2F978-3-030-64232-7_18