5.1. Chebyshev polynomials¶
5.1.1. Cosines¶
We know that
\[\cos(\alpha + \beta) = \cos\alpha \cos\beta - \sin \alpha \sin\beta\]Put \(\alpha = n\theta\) and \(\beta = \theta\)
\[\cos((n+1)\theta) = \cos n\theta \cos\theta - \sin n\theta \sin \theta\]Put \(\alpha = n\theta\) and \(\beta = -\theta\)
\[\cos((n-1)\theta) = \cos n\theta \cos\theta + \sin n\theta \sin \theta\]Adding
\[ \begin{align}\begin{aligned}\cos((n-1)\theta) + \cos((n+1)\theta) = 2\cos n\theta \cos\theta\\\cos((n+1)\theta) = 2\cos n\theta \cos\theta - \cos((n-1)\theta)\end{aligned}\end{align} \]We claim that for each non-negative integer \(n\), there exist integers \(c_i\) s.t.
\[\cos n\theta = \sum_{i=0}^{n} c_i \cos^i (\theta)\]
5.1.2. Chebyshev polynomials¶
Essentially \(\cos n\theta\) is a polynomial in \(\cos \theta\).
For a fixed \(n\) we define \(n^{th}\) Chebyshev polynomial as:
\[cos(n\theta) = T_n(cos\theta))\]Putting \(x = \cos\theta\), \(T_n(x)\) is a polynomial.
\[T_n(x) = \cos(n \arccos(x)) \quad \forall x \in [-1,1]\]Basic Chebyshev polynomials:
\[\begin{split}&T_0(x) = \cos 0 \theta = 1 \\ &T_1(x) = \cos 1 \theta = x\end{split}\]Recurrence relation:
\[\begin{split}&T_{n-1}(x) + T_{n+1}(x) = 2 T_1(x) T_n(x) = 2x T_n(x)\\ &T_{n+1}(x) = 2x T_n(x) - T_{n-1}(x)\end{split}\]or
\[x T_n(x) = \frac{T_{n-1}(x) + T_{n+1}(x)}{2}\]This has resemblance to space shift operator in algebraic signal processing.
Recurrence examples:
\[\begin{split}&T_2(x) = 2x T_1(x) - T_0(x) = 2x^2 - 1\\ &T_3(x) = 2 T_2(x) - T_1(x) = 2x(2x^2 -1) - x = 4x^3 - 3x\end{split}\]Chebyshev polynomials upto 7th order
\[\begin{split}&T_0(x) = 1 \\ &T_1(x) = x \\ &T_2(x) = 2x^2 - 1\\ &T_3(x) = 4x^3 - 3x\\ &T_4(x) = 8x^4 - 8x^2 + 1\\ &T_5(x) = 16x^5 - 20x^3 + 5x\\ &T_6(x) = 32x^6 - 48x^4 + 18x^2 - 1\\ &T_7(x) = 64x^7 - 112x^5 + 56x^3 -7x\end{split}\]
5.1.3. Properties¶
Maximum value
Composition
Proof:
Zeros
Zeros of \(T_n(x)\) are given by:
All the roots of \(T_n(x)\) are real and lie in the interval \([-1,1]\).
Recurrence
Thus:
Symmetry
Lets compute \(T_{-1}(x)\).
\[\begin{split}&T_1(x) = 2 xT_0(x) - T_{-1}(x) \\ \implies &T_{-1}(x) = 2 xT_0(x) - T_1(x)\\ &= 2x - x = x = T_1(x)\end{split}\]It can be shown that:
\[T_n(x) = T_{-n}(x)\]The sequence of Chebyshev polynomials is symmetric.
5.1.4. Power form¶
Euler formula:
Define \(u = e^{j\theta}\)
\[x = \cos \theta = \frac{u^{1} + u^{-1}}{2}\]Thus
\[T_n(x) = \cos n\theta = \frac{u^{n} + u^{-n}}{2}\]
5.1.5. Differential equation¶
Even/odd polynomials
Even degree Chebyshev polynomials are even functions.
Odd degree Chevyshev polynomials are odd functions.
Differential equation
It can be shown that:
\[(1-x^2)T''_n(x) - xT'_n(x) + n^2T_n(x) = 0\]Assume that \(y = \sum_{k=0}^{n}t_k x^k\) is a solution to above differential equation. Then:
\[(n^2 - k^2)t_k + (k+1)(k+2)t_{k+2} = 0\]We already know that \(t_n = 2^{n-1}\)
Solving further we get:
\[t_{n-2m} = (-1)^m 2^{n - 2m -1} \frac{n}{n - m} {n -m \choose m}\]
5.1.6. Orthogonality¶
The integral
is zero unless \(m=n\).
If \(m=n\)
If \(m=n=0\), integral is \(\pi\).
Otherwise, integral is \(\pi/2\)
Putting \(x = \cos\theta\)
Chebyshev polynomials are orthogonal over \([-1,1]\) w.r.t. weight \((1-x^2)^{-1/2}\)
The sequence \(\frac{1}{\pi}T_0, \frac{2}{\pi}T_1, \frac{2}{\pi}T_3,\dots\) is an orthonormal system.
Discrete orthogonality condition
Let \(x_j\) be roots of \(T_N\)
The sum:
\[\sum_{k=1}^{N} T(mx_k) T(nx_k)\]Is zero if \(m \neq n\)
\(N\) if \(m=n=0\)
\(N/2\) otherwise
5.1.7. Chebyshev polynomials of different kinds¶
So far we have looked at Chebyshev polynomials of first kind.
Chebyshev polynomials are generated by the recurrence relation:
\[C_{n+1}(x) = 2xC_n(x) - C_{n-1}(x)\]\(C_0 = 1, C_1 = x\) generates the Chebyshev polynomials of first kind denoted by \(T_n(x)\).
Chebyshev polynomials of second, third and forth kind are described below.
5.1.8. References¶
Change log
- Last Modified
$Id: chebyshev.rst 249 2012-08-05 06:17:57Z shailesh $