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Introduction to Linear Algebra with Mathematica
Glossary
Preface
Chebyshev polynomials are usually used for either approximation of continuous functions or function expansion. They were invented by the Russian mathematician Pafnuty Chebyshev (1821--1894) in 1853–1854 (with early forms appearing in his work on hinge and linkage mechanisms, such as steam engines). 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 the recursive Clenshaw method.
This section presents some properties of Chebyshev polynomials needed for orthogonal expansions that are among the most remarkable and useful tools in numerical computations. These polynomials constitute a special case of Jacobi polynomials \( \displaystyle \quad P_n^{(\alpha , \beta )} (x) .\quad \) The latter depends on two parameters α, β > −1. The Jacobi polynomials are orthogonal on the interval [−1, 1], with respect to the weight function w(x) = (1 −x)α (1 + x)β, and satisfy the Jacobi ODE
\[
\left( 1 - x^2 \right) y'' + \left[ \beta - \alpha - \left( \alpha + \beta + 2 \right) x \right] y' + n \left( n + \alpha + \beta + 1 \right) y = 0 .
\]
In particular, we consider expansions with respect to Chebyshev polynomials
of the first kind \( \quad T_n (x) = \frac{P_n^{(-1/2,-1/2)}(x)}{P_n^{(-1/2,-1/2)}(1)} , \quad \) of the second kind \( \quad U_n (x) = \left( n+1 \right) \frac{P_n^{(1/2,1/2)}(x)}{P_n^{(1/2,1/2)}(1)} , \quad \) of the third kind \( \quad V_n (x) = \frac{P_n^{(-1/2,1/2)}(x)}{P_n^{(-1/2,1/2)}(1)} = \frac{4^n \left( n! \right)^2}{(2n)!}\ P_n^{(-1/2,1/2)}(x) , \quad \) and of the fourth kind \( \quad W_n (x) = \left( 2n+1 \right) \frac{P_n^{(1/2,-1/2)}(x)}{P_n^{(1/2,-1/2)}(1)} . \quad \)
Chebyshev Polynomials
The Chebyshev polynomials of the first kind \( T_n (x) = \cos \left( n\,\mbox{arccos} x \right) \) are polynomial solutions of the (singular) differential equation
\[
\left( 1- x^2 \right) y'' -x\,y' + n^2 y =0 , \quad y(1) = 1 .
\]
The Chebyshev polynomials of the second kind \( \displaystyle \quad U_n (x) = \frac{\sin \left[ (n+1)\,\arccos x \right]}{\sqrt{1 - x^2}} \quad \) are polynomial solutions of the (singular) differential equation
\[
\left( 1- x^2 \right) y'' -3x\,y' + n\left( n+2 \right) y =0 , \quad y(1) = n+1 .
\]
The Chebyshev polynomials of the third kind \( \displaystyle \quad V_{n} (x) = \sqrt{\frac{2}{1+x}}\,T_n \left( \sqrt{\frac{x+1}{2}} \right) \quad \) are polynomial solutions of the (singular) differential equation
\[
\left( 1- x^2 \right) y'' - \left( 2x -1 \right) y' + n\left( n+1 \right) y =0 , \quad y(1) = 1 .
\]
The Chebyshev polynomials of the fourth kind \( \displaystyle \quad W_n (x) = U_{2n} \left( \sqrt{\frac{x+1}{2}} \right) \quad \) are polynomial solutions of the (singular) differential equation
\[
\left( 1- x^2 \right) y'' - \left( 2x +1 \right) y' + n\left( n+1 \right) y =0 , \quad y(1) = 2n+1 .
\]
Chebyshev polynomials of the third and fourth kind are expressed via Chebyshev polynomials of the first two kinds:
\[
V_n (x) = T_n (x) - \left( 1 - x \right) U_{n-1} (x) , \quad W_n (x) = T_n (x) + \left( 1 + x \right) U_{n-1} (x) .
\]
All Chebyshev polynomials are expressed via trigonometric functions upon substitution x = cos(θ):
\[
T_n (\cos\theta ) = \cos (n\,\theta ), \quad U_n (\cos\theta ) = \frac{\sin \left( (n+1)\,\theta\right)}{\sin\theta} , \quad V_n (\cos\theta ) = \frac{\cos \left( \frac{2n+1}{2}\,\theta \right)}{\cos \left( \frac{\theta}{2} \right)} , \quad W_n (\cos\theta ) = \frac{\sin \left( \frac{2n+1}{2}\,\theta \right)}{\sin \left( \frac{\theta}{2} \right)} .
\]
Orthogonality of Chebyshev Polynomials
Chebyshev polynomials form orthogonal sets in weighed vector space 𝔏²([−1, 1],w) 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} = \begin{cases}
0, &\quad \mbox{if }n \ne m , \\
\pi , &\quad \mbox{if } n = m = 0, \\
2\pi , &\quad \mbox{if } n = m \ne 0, \\
\end{cases}
\\
\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
\[
\delta_{n,m} = \begin{cases} 0, & \ \mbox{if } n \ne m ,
\\
1 , & \ \mbox{if } n = m ; \end{cases} \qquad\quad
{\cal E}_n = \begin{cases} 1 , & \ \mbox{if } n =0, \\
2, & \ \mbox{if } n \ge 1. \end{cases}
\]
For a function on the finite interval [-1, 1],
there are four kinds of Chebyshev expansions depending which kind of Chebyshev function is used.
| Family | Trigonometric definition (x = cos θ) |
Weight w(x) on [-1,1] | Endpoint values | Three-term recurrence | Generating function |
|---|---|---|---|---|---|
| First kind Tn(x) |
Tn(cos θ) = cos(nθ) | w(x) = (1 - x²)-1/2 |
Tn(1) = 1 Tn(-1) = (-1)n |
T0=1, T1=x Tn+1=2xTn-Tn-1 |
∑ Tn(x)tn = (1 - xt)/(1 - 2xt + t²) |
| Second kind Un(x) |
Un(cos θ) = sin((n+1)θ)/sin θ | w(x) = (1 - x²)1/2 |
Un(1) = n+1 Un(-1) = (-1)n(n+1) |
U0=1, U1=2x Un+1=2xUn-Un-1 |
∑ Uₙ(x)tn = 1/(1 - 2xt + t²) |
| Third kind Vn(x) |
Vₙ(cos θ) = cos((n+½)θ)/cos(θ/2) | w(x) = (1 - x)-1/2(1 + x)½ |
Vₙ(1) = 1 Vₙ(-1) = (-1)n(1+2n) |
V0=1, V1=2x-1 Vn+1=2xVn-Vn-1 |
∑ Vn(x)tn = (1 - t)/(1 - 2xt + t²) |
| Fourth kind Wn(x) |
Wₙ(cos θ) = sin((n+½)θ)/sin(θ/2) | w(x) = (1 - x)½(1 + x)-1/2 |
Wₙ(1) = 2n+1 Wₙ(-1) = (-1)n |
W₀=1, W₁=2x+1 Wn+1=2xWn-Wn-1 |
∑ Wₙ(x)tn = (1 + t)/(1 - 2xt + t²) |
Chebyshev Series
Chebyshev polynomials form a distinct class of polynomials particularly suited for approximating other functions. They are widely used in many areas of numerical analysis: uniform approximation, least-squares approximation, and the numerical solution of ordinary and partial differential equations (the so-called spectral or pseudospectral methods). In the appropriate Sobolev space, the set of Chebyshev polynomials forms an orthogonal basis, so that a function in that space can be expressed via the expansion on −1 ≤ x ≤ 1:
\[
f(x) = \frac{a_0}{2} + \sum_{k\ge 1} \,a_k \,T_k (x) = \sum_{k\ge 0} \,b_k \,U_k (x) =
\sum_{k\ge 0} \,c_k \,V_k (x) = \sum_{k\ge 0} \,d_k \,W_k (x) .
\]
These series become especially powerful when interval [-1,1] is transported to [0, π] by the substitution x = cosθ. This transformation is a smooth, strictly decreasing bijection between (0,π) and (-1,1).
This is the basic mechanism behind all the Chebyshev systems: they are trigonometric systems in disguise, transported from θ-space to x-space by x = cosθ.
Many orthogonal polynomial systems—including all four Chebyshev families—are shadows of the trigonometric system under this map, with weights arising from the Jacobian dx = −sinθ dθ.
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
\begin{equation} \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 ) ,
\end{equation}
where
\begin{equation} \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 .
\end{equation}
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 overshot 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 a 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 XIII 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:
\begin{align*}
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) ,
\\
&= \frac{1}{2^{n-1}}\,^{\prime}\, \sum_{j=0}^n \,\binom{n}{\frac{n-j}{2}}\,T_j (x) . \qquad n=1,2,\ldots ;
\end{align*}
where \( \binom{n}{k} = \frac{n!}{k! \left( n-k \right) !} \quad \) is the binomial coefficient and \( \lfloor \,\cdot \,\rfloor \) is the floor function.
As usual, the prime on the summation symbol indicates that the contribution of j = 0 needs to be halved if it appears.
\[
x^n = 2^{-n} \left[ U_n (x) + \left( n-1\right) U_{n-2} (x) + \cdots \right] , \qquad n=1,2,\ldots .
\]
■
End of Example 1
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.
■
End of Example 2
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}]
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];
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]
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]
\[
e^x \approx \sum_{k=0}^n b_k U_k (x) ,
\]
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}]
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}]
 |
 |
End of Example 3
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}]
Plot[{HeavisideTheta[t], heaviside[10, t]}, {t, -1.01, 1.01}, PlotStyle -> {Blue, Orange}]
\[
H(t) = \frac{1}{2} + \frac{4}{\pi} \, \sum_{k\ge 0} \frac{(-1)^k\, k}{(2k+1)(2k+3)}\, U_{2k+1} (t) .
\]
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}]
Plot[{HeavisideTheta[t], chebsum[10]}, {t, -1.01, 1.01}, PlotStyle -> {Blue, Orange}]
End of Example 4
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]
■
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]
End of Example 5
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}{a-x} &= \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) , \qquad a > 1 .
\end{align*}
Check with Mathematica:
For 𝑎 = 2 we have
s20[x_] =
1/Sqrt[3] + (2/Sqrt[3])*
Sum[(2 - Sqrt[3])^n * ChebyshevT[n, x], {n, 1, 20}];
Plot[{1/(2 - x), s20[x]}, {x, -1, 1}, PlotStyle -> Thick]
\begin{align*}
\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:
\[
\delta (x) = \frac{1}{\pi} + \frac{2}{\pi} \sum_{n\ge 1} (-1)^n T_{2n} (x) .
\]
■
End of Example 6
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:
\begin{equation} \label{EqCheb2.1}
f(x) = \sum_{k\ge 0} \,b_k\, U_{k} (x ) ,
\end{equation}
where
\begin{equation} \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 .
\end{equation}
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 \( \displaystyle f(\cos \theta )\,\sin\theta . \)
Parseval's identity for expansion \eqref{EqCheb2.1} is
\[
\int_{-1}^1 \,|f(x)|^2 \,\sqrt{1 - x^2} \,{\text d}x = \frac{\pi}{2}\,\sum_{n\ge 0} \, b_n^2 .
\]
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
\[
\sqrt{1-x} = - \frac{2}{\pi}\,\sum_{n\ge 0} \frac{16 \left( 1 + n \right)}{(4n^2 -1)(3 + 2n)(5 + 2n)} \, U_n (x) .
\]
■
End of Example 1B
Example 2B: ■
End of Example 2B
Example 3B: We consider two hyperbolic functions on interval [−1, 1]:
\[
f(x) = \sinh (x) = \frac{1}{2} \left( e^x - e^{-x} \right) , \qquad g(x) = \cosh (x) = \frac{1}{2} \left( e^x + e^{-x} \right) .
\]
■
End of Example 3B
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}]
Plot[{HeavisideTheta[t], heaviside[10, t]}, {t, -1.01, 1.01}, PlotStyle -> {Blue, Orange}]
\[
H(t) = \frac{1}{2} + \frac{4}{\pi} \, \sum_{k\ge 0} \frac{(-1)^k\, k}{(2k+1)(2k+3)}\, U_{2k+1} (t) .
\]
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}]
Plot[{HeavisideTheta[t], chebsum[10]}, {t, -1.01, 1.01}, PlotStyle -> {Blue, Orange}]
End of Example 4B
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]
■
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]
End of Example 5B
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
\begin{equation} \label{EqCheb3.1}
f(x) = \sum_{k\ge 0} \,c_k\, V_{k} (x ) , \qquad -1 < x < 1 ,
\end{equation}
where
\begin{equation} \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 .
\end{equation}
Parseval's identity for Chebyshev polynomials of the third kind reads
\[
\sum_{n\ge 0}\, c_n^2 = \frac{1}{\pi} \,\int_{-1}^1 \,\left( \frac{1+x}{1-x} \right)^{1/2} \,|f(x) |^2 \,{\text d} x .
\]
In the integral \eqref{EqCheb3.2}, 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
\begin{equation} \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 .
\end{equation}
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
\[
\sqrt{1-x} = - \frac{2}{\pi}\,\sum_{n\ge 0} \frac{16 \left( 1 + n \right)}{(4n^2 -1)(3 + 2n)(5 + 2n)} \, U_n (x) .
\]
■
End of Example 1C
Example 2C: ■
End of Example 2C
Example 3C: We consider two hyperbolic functions on interval [−1, 1]:
\[
f(x) = \sinh (x) = \frac{1}{2} \left( e^x - e^{-x} \right) , \qquad g(x) = \cosh (x) = \frac{1}{2} \left( e^x + e^{-x} \right) .
\]
■
End of Example 3C
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}]
Plot[{HeavisideTheta[t], heaviside[10, t]}, {t, -1.01, 1.01}, PlotStyle -> {Blue, Orange}]
\[
H(t) = \frac{1}{2} + \frac{4}{\pi} \, \sum_{k\ge 0} \frac{(-1)^k\, k}{(2k+1)(2k+3)}\, U_{2k+1} (t) .
\]
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}]
Plot[{HeavisideTheta[t], chebsum[10]}, {t, -1.01, 1.01}, PlotStyle -> {Blue, Orange}]
End of Example 4C
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]
■
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]
End of Example 5C
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
\begin{equation} \label{EqCheb4.1}
f(x) = \sum_{k\ge 0} \,d_k\, W_{k} (x ) , \qquad -1 < x < 1 ,
\end{equation}
where
\begin{equation} \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 .
\end{equation}
Parseval's identity for Chebyshev-IV expansion is
\[
\sum_{n\ge 0}\, d_n^2 = \frac{1}{\pi} \,\int_{-1}^1 \,\left( \frac{1-x}{1+x} \right)^{1/2} \,|f(x) |^2 \,{\text d} x .
\]
Under the substitution x = cosθ, this weight becomes
\[
w(\cos \theta )\, |\sin \theta | = 1- \cos\theta = 2 \left( \sin\frac{\theta}{2} \right)^2 .
\]
We check with Mathematica:
FullSimplify[Sqrt[(1 - Cos[t])/(1 + Cos[t])]* Sin[t]]
Sin[t] Sqrt[Tan[t/2]^2]
FullSimplify[Sin[t]*Tan[t/2]]
1 - Cos[t]
Thus, the Hilbert space 𝔏²([-1,1], w(x) dx)
is unitarily equivalent to 𝔏²([0,π], (1 − cosθ) dθ).
Using expression for the Chebyshev polynomial of the fourth kind
\[
W_n (\cos\theta ) = \frac{\sin \left( \frac{2n+1}{2}\,\theta \right)}{\sin \left( \frac{\theta}{2} \right)} ,
\]
we rewrite the kernel of coefficient \eqref{EqCheb4.2} as
\[
D_n = \frac{\sin\theta}{1 + \cos\theta} \cdot \frac{\sin \left( \frac{2n+1}{2}\,\theta \right)}{\sin \left( \frac{\theta}{2} \right)} \cdot \sin\theta = \frac{\sin\theta}{\cos^2 \frac{\theta}{2}} \cdot \frac{\sin \left( \frac{2n+1}{2}\,\theta \right)}{\sin \left( \frac{\theta}{2} \right)} \cdot \sin\theta = 2\,\sin \left( \frac{\theta}{2} \right) \sin \left( \frac{2n+1}{2}\,\theta \right) .
\]
Since the sequence of functions Wₙ(cosθ) correspond to the shifted sine system sin((n+1)θ), and the sine system is complete in its natural weighted space, the span of { Wₙ } is dense in the Chebyshev–IV space.
This allows us to rewrite the coefficient dn as
\begin{equation} \label{EqCheb4.3}
d_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 .
\end{equation}
We check again with Mathematica:
FullSimplify[
2*Sin[t/2]*Sin[(2*n + 1)*t/2] == Cos[n*t] - Cos[(n + 1)*t]]
True
Using formula \eqref{EqCheb1.2} for Chebyshev polynomials of the first kind, we derive
\begin{equation} \label{EqCheb4.4}
d_n = \frac{1}{\pi}\,\int_{-1}^1 \,f(x)\, W_{n} (x )\left( \frac{1-x}{1+x} \right)^{1/2} {\text d} x = \frac{1}{2} \left[ a_n - a_{n+1} \right] ,
\end{equation}
where 𝑎ₙ is the Fourier--Chebyshev coefficient of the first kind. When n = 0, the coefficient becomes
\[
d_0 = \frac{1}{\pi}\,\int_{-1}^1 \,f(x)\,\sqrt{\frac{1-x}{1+x}} \ {\text d}x = \frac{1}{\pi}\,\int_{0}^{\pi} \,f(\cos\theta )\,\left( 1 - \cos\theta \right) {\text d}\theta .
\]
Example 1D: Let us consider the root function \( f(x) = \sqrt{1 - x^2} . \) To expand it into Chebyshev series over fourth kind polynomials, we need to calculate Fourier-Chebyshev coefficients according to \eqref{EqCheb2.2} or \eqref{EqCheb4.3}:
\begin{align*}
d_n &= \frac{1}{\pi}\,\int_{-1}^1 \sqrt{1-x^2} \,\frac{\sqrt{1- x}}{\sqrt{1+x}} \,W_n (x) \,{\text d}x = \frac{1}{\pi}\,\int_{-1}^1 \left( 1 - x \right)\,W_n (x) \,{\text d}x
\\
&= \frac{1}{\pi}\,\int_{0}^{\pi} \,\sin\theta\,2\,\sin \left( \frac{\theta}{2} \right) \sin \left( \frac{2n+1}{2}\,\theta \right) {\text d}\theta .
\end{align*}
Using identity Wₙ(x) = Tₙ(x) + (1 + x) Un-1(x), we evaluate the integral
\[
d_n = \frac{1}{\pi}\,\int_{-1}^1 \left( 1 - x \right)\,\left[ T_n (x) + \left( 1 + x \right) U_{n-1} (x) \right] {\text d}x , \qquad n= 1,2, \ldots .
\]
\begin{align*}
d_0 &= \frac{1}{\pi}\,\int_{-1}^1 \left( 1 - x \right)\,{\text d}x = \frac{2}{\pi} ,
\\
d_1 &= \frac{1}{\pi}\,\int_{-1}^1 \left( 1 - x \right)\left( 2x + 1 \right) {\text d}x = \frac{2}{3\pi} ,
\\
d_n &= \frac{1}{\pi}\,\int_{-1}^1 \left( 1 - x \right)\left[ T_n (x) + \left( 1+ x \right) U_{n-1} (x) \right]\,{\text d}x = -\frac{\left(2n +4\right) (-1)^n/}{\left( n+2 \right) \left( n^2 -1 \right) \pi} .
\end{align*}
Integrate[1 - x, {x, -1, 1}]/Pi
2/\[Pi]
Integrate[(1 - x)* (2*x + 1), {x, -1, 1}]/Pi
2/(3 \[Pi])
Integrate[(1 -
x)*(ChebyshevT[n, x] + (x + 1)*ChebyshevU[n - 1, x]), {x, -1,
1}]/Pi
(-1 - 2 n + (1 - 2 n (1 + n)) Cos[n \[Pi]])/(n (2 + n) (-1 +
n^2) \[Pi])
We check this formula using trigonometric form of Chebyshev coefficient.
Indeed, Mathematica is able to calculate this integral much faster than the previous one.
Integrate[Sin[t]*2*Sin[t/2]*Sin[(2*n + 1)*t/2], {t, 0, Pi}]
(-1 - 2 n + (1 - 2 n (1 + n)) Cos[n \[Pi]])/((-1 + n) n (1 + n) (2 +
n))
With all coefficients at hand, we determine the Chebyshev series for the target function:
\[
\sqrt{1-x^2} = \frac{2}{\pi} + \frac{2}{3\pi} \left( 2x + 1 \right) - \frac{2}{\pi}\,\sum_{n\ge 2} \,\frac{(-1)^n/}{\left( n^2 -1 \right)} \left[ T_n (x) + \left( 1 + x \right) U_{n-1} (x) \right] .
\]
Finally, we plot the partial sum with 50 terms:
s50[x_] =
2/Pi + 2/3/Pi -
2*Sum[(-1)^
Plot[{Sqrt[1 - x^2], s50[x]}, {x, -1, 1}, PlotStyle -> Thick] n /(n^2 - 1) *(ChebyshevT[n, x] + (1 + x)*
ChebyshevU[n - 1, x]), {n, 2, 50}]/Pi;
Example B: Now we consider another root function \( \quad g(x) = \sqrt{1 - x} . \) The corresponding Fourier--Chebyshev coefficients are evaluated according to Eq.\eqref{EqCheb4.2} or \eqref{EqCheb4.3}:
\[
d_n = \frac{1}{\pi}\,\int_{0}^{\pi} \sin \frac{\theta}{2} \,2\,\sin \left( \frac{\theta}{2} \right) \sin \left( \frac{2n+1}{2}\,\theta \right)
\, {\text d}\theta
\]
because \( \sqrt{1 - \cos\theta} = \sin\frac{\theta}{2} . \) We again ask Mathematica to rescue
????
Integrate[Sin[t/2] * 2 *Sin[t/2] * Sin[(2*n + 1)*t/2], {t, 0, Pi}]/Pi
(2 (-4 + (-2 + 8 n + 8 n^2) Sin[n \[Pi]]))/((-1 + 2 n) (1 + 2 n) (3 +
2 n) \[Pi])
Then \( \displaystyle \quad
g(\cos \theta )=\sqrt{2}\, \sin (\theta /2),\quad \) and the coefficients become
\[
d_n = \frac{2}{\pi }\int _0^{\pi }\sqrt{2}\, \sin (\theta /2)\, \frac{\sin ((n+1)\theta )}{\sin (\theta /2)}\,{\text d}\theta =\frac{2\sqrt{2}}{\pi }\int _0^{\pi }\sin ((n+1)\theta )\,{\text d}\theta .
\]
The integral evaluates to
\[
\int _0^{\pi }\cos (k\theta )\, d\theta =\left\{ \, \begin{array}{ll}\textstyle \pi ,&\textstyle k=0,\\ \textstyle [4pt]0,&\textstyle k\geq 1,\end{array}\right.
\]
?????????????????????/
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 C: Finally, we consider another root function f(x) = √{1+x}, x ∈ [-1,1].
Under x = cosθ, this function becomes
\[
f(\cos \theta )=\sqrt{1+\cos \theta }=\sqrt{2}\, \cos (\theta /2).
\]
The Chebyshev–IV coefficients are
\[
d_n = \frac{2}{\pi }\int _0^{\pi }f(\cos \theta )\, W_n(\cos \theta )\, \frac{d\theta }{\sin (\theta /2)}.
\]
Using the trigonometric form of W_n,
\[
W_n (\cos \theta ) =\frac{\sin ((n+1)\theta )}{\sin (\theta /2)},
\]
we obtain
\[
d_n = \frac{2}{\pi }\int _0^{\pi }\sqrt{2}\, \cos (\theta /2)\, \frac{\sin ((n+1)\theta )}{\sin (\theta /2)}\, d\theta =\frac{2\sqrt{2}}{\pi }\int _0^{\pi }\cot (\theta /2)\, \sin ((n+1)\theta )\, d\theta .
\]
Using the identity
\( \displaystyle \quad
\cot (\theta /2)\sin ((n+1)\theta ) =\cos (n\theta )+\cos ((n+1)\theta ), \quad \)
the integral becomes elementary:
\[
d_n = \frac{2\sqrt{2}}{\pi }\left( \int _0^{\pi }\cos (n\theta )\, d\theta +\int _0^{\pi }\cos ((n+1)\theta )\, {\text d}\theta \right) .
\]
Since
\[
\int _0^{\pi }\, \cos (k\theta )\, {\text d}\theta =\left\{ \, \begin{array}{ll}\textstyle \pi ,&\textstyle k=0,\\ \textstyle [4pt]0,&\textstyle k\geq 1,\end{array}\right.
\]
we find
\[
d_0=\frac{2\sqrt{2}}{\pi }(\pi +0) = 2\sqrt{2},\qquad d_n = 0\quad (n\geq 1).
\]
Thus, the Chebyshev–IV expansion of f(x) = √{1+x} collapses to a single term:
\[
\sqrt{1+x}=2\sqrt{2}\, W_0(x),
\]
and since W₀(x) = 1, this is consistent with the identity
\[
\sqrt{1+\cos \theta } = \sqrt{2}\cos (\theta /2).
\]
■
End of Example 1D
Example 2D: ■
End of Example 2D
Example 3D: We consider two hyperbolic functions on interval [−1, 1]:
\[
f(x) = \sinh (x) = \frac{1}{2} \left( e^x - e^{-x} \right) , \qquad g(x) = \cosh (x) = \frac{1}{2} \left( e^x + e^{-x} \right) .
\]
■
End of Example 3D
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}]
Plot[{HeavisideTheta[t], heaviside[10, t]}, {t, -1.01, 1.01}, PlotStyle -> {Blue, Orange}]
\[
H(t) = \frac{1}{2} + \frac{4}{\pi} \, \sum_{k\ge 0} \frac{(-1)^k\, k}{(2k+1)(2k+3)}\, U_{2k+1} (t) .
\]
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}]
Plot[{HeavisideTheta[t], chebsum[10]}, {t, -1.01, 1.01}, PlotStyle -> {Blue, Orange}]
End of Example 4D
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]
■
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]
End of Example 5D
Clenshaw's method for the evaluation of a Chebyshev sum
- 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
- Schweizer, W., Special Functions in Physics with MATLAB, 2021, Apringer. https://link.springer.com/chapter/10.1007%2F978-3-030-64232-7_18
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Return to the Part 4 Numerical Methods
Return to the Part 5 Fourier Series
Return to the Part 6 Partial Differential Equations
Return to the Part 7 Special Functions