Chebyshev Polynomials - Definition and Properties

The Chebyshev polynomials are a sequence of orthogonal polynomials that are related to De Moivre's formula. They have numerous properties, which make them useful in areas like solving polynomials and approximating functions.

The \(n^\text{th}\) Chebyshev polynomial of the first kind, denoted by \(T_n(x)\), is defined as

\[ T_n (x) = \cos \left( n \cos^{-1} x \right),\]

or equivalently

\[T_n ( \cos \theta) = \cos n \theta.\ _\square \]

Since we know that \[\begin{align} \cos 0 \theta &= 1\\ \cos 1\theta & = \cos \theta\\ \cos 2 \theta &= 2 \cos^2 \theta - 1\\ \cos 3 \theta &= 4 \cos^3 \theta - 3 \cos \theta, \end{align}\] we can conclude that \[\begin{align} T_0(x) &= 1\\ T_1(x) &= x\\ T_2(x) &= 2x ^2 - 1\\ T_3 (x) &= 4 x^3 - 3x. \end{align}\]

Find \(T_4 (x)\) above.

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To find \(T_4(x)\), we can equivalently find a function of \(\cos 4\theta\) in terms of \(\theta\).

Using the cosine sum formula, we get

\[\cos 4\theta = \cos\theta\cos3\theta - \sin\theta\sin 3\theta.\]

Recall that \(\sin 3\theta = 3\sin\theta - 4\sin^3\theta\). Then

\[ \begin{align*}\cos\theta\cos3\theta - \sin\theta\sin 3\theta&=\cos\theta\left(4\cos^3\theta - 3\cos\theta\right) - 3\sin^2\theta - 4\sin^4\theta\\ &= 4\cos^4\theta - 3\cos^2\theta +3\left(1-\cos^2\theta\right) - 4\left(1-\cos^2\theta\right)^2\\ &= 8\cos^4\theta - 8\cos^2\theta + 1. \end{align*}\]

Thus, \[T_4(x)=8x^4-8x^2+1.\ _\square\]

Find \(T_5 (x)\) above.

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To find \(T_5(x)\), we can, just as in the previous example, find a function of \(\cos 5\theta\) in terms of \(\theta\).

Using the cosine sum formula again, we get \[\cos 5\theta = \cos\theta\cos4\theta - \sin\theta\sin 4\theta.\] But then we have to replace \(\cos4\theta\) with \(8\cos^4\theta-8\cos^2\theta+1,\) and then manually compute \(\sin 4\theta,\) and then...

Forget it. This is turning into a hopeless bash; we can't be doing this for \(T_6(x)\) or \(T_7(x)\), and we definitely can't easily generalize this to \(T_n(x)\). We could always use De Moivre's formula, but the calculation is also very extensive.

If only there were an easier way... \(_\square\)

How would we obtain a more general formula? In answering the previous question, most people tried to expand

\[ \cos (n+1) \theta = \cos n \theta \cos \theta - \sin n \theta \sin \theta. \]

We can easily convert the first 2 terms into the \( T_n \) form. However, the issue with this approach is that \( \sin n \theta \sin \theta \) is not easy to deal with, and will (currently) require much further expansion.

Instead, we will use the fact (from trigonometric sum and product formulas) that

\[\cos ( n+1) \theta + \cos (n-1) \theta = 2 \cos \theta \cos n \theta .\]

This gives us the recurrence relation:

\[ T_{n+1} (x) = 2x T_n (x) - T_{n-1} (x).\ _\square \]

The following is a table of initial values of \( T_{n} (x) \):

\[ \begin{align} T_0(x) &= 1 \\ T_1(x) &= x \\ T_2(x) &= 2x^2 - 1 \\ T_3(x) &= 4x^3 - 3x \\ T_4(x) &= 8x^4 - 8x^2 + 1 \\ T_5(x) &= 16x^5 - 20x^3 + 5x \\ T_6(x) &= 32x^6 - 48x^4 + 18x^2 - 1 \\ T_7(x) &= 64x^7 - 112x^5 + 56x^3 - 7x \\ T_8(x) &= 128x^8 - 256x^6 + 160x^4 - 32x^2 + 1 \\ T_9(x) &= 256x^9 - 576x^7 + 432x^5 - 120x^3 + 9x. \\ T_{10}(x) &= 512x^{10} - 1280x^8 + 1120x^6 - 400x^4 + 50x^2-1. \\ \end{align} \]

We can make the following conjectures about these coefficients:

1. Coefficients are integers.
2. Constant term is \( (-1) ^k \) for \( n = 2k \) and 0 for \( n = 2k+1 \).
3. Leading term is \( 2^{n-1} \).
4. Linear term of \( T_{2n+1} (x) \) is \( (-1)^{n} ( 2n+1) \).
5. The non-zero terms have exponents with the same parity as \(n\).
6. Sum of coefficients is 1.
7. Non-zero coefficients alternate in sign.
8. The coefficient of the \( x^2 \) term in \( T_{2k} (x) \) is \( (-1)^{k+1} 2k^2 \).
9. The sum of the absolute value of the coefficients is \( \frac{1}{2} \left( 1 + \sqrt{2} \right) ^n + \frac{1}{2} \left( 1 - \sqrt{2}\right) ^n \).
10. The roots of \( T_n \) are \( \cos \left(\dfrac{(2k+1)\pi}{2n}\right ), \) where \(k\in\mathbb{Z}\).

We will prove some of these conjectures. The rest are left as exercises for the reader.

Prove conjecture 2.

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We will prove this using induction. First, in the two base cases, we see that \(T_0(x)=1\) and \(T_1(x)=x\), which satisfies that the constant terms are \((-1)^0\) and \(0\), respectively. Now we must prove that given \(T_n(x)\) has a coefficient of \(0\) if \(n=2k+1\) and \((-1)^k\) if \(n=2k\), then \(T_{n+1}(x)\) also satisfies this.

Note that the constant term can be evaluated by plugging in \(x=0\). Doing so in the recurrence relation of \(T_n(x)\) gives \[T_{n+1}(0)=2\times 0\times T_n(0)-T_{n-1}(0)=-T_{n-1}(0).\]

This means that if \(T_{2k}(0)=(-1)^k\), then \(T_{2k+2}(0)=(-1)^{k+1}\); also, if \(T_{2k+1}(0)=0\), then \(T_{2k+3}(0)=0\), completing the induction. \(_\square\)

Prove conjecture 6.

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The sum of the coefficients of \(T_n(x)\) is just \(T_n(1)\). Recalling that \(T_n(\cos\theta)=\cos n\theta\), we see that we want to evaluate \(\cos n\theta\) when \(\cos\theta=1\). But this means \(\theta=0\), so \(\cos n\theta=\cos 0 = 1\).

Therefore, \(T_n(1)=1\) and we are done. \(_\square\)

Prove conjecture 10.

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We desire to find the roots \(x\) of \(T_n(x)=0\).

Substituting \(x=\cos\theta\), we want to instead find the roots of \[T_n(\cos\theta)=\cos n\theta =0.\]

This happens at \(n\theta=\dfrac{\pi}{2}+k\pi\) for \(k\in\mathbb{Z}\).

Thus, \[\theta=\dfrac{\pi}{2n}+\dfrac{k\pi}{n}=\dfrac{(2k+1)\pi}{2n}.\]

But this means \[x=\cos\theta =\cos\left(\dfrac{(2k+1)\pi}{2n}\right),\] so we are done. \(_\square\)

Show that if \(r\) is a rational number such that \( \cos ( r \pi ) \) is rational, then \( \cos (r \pi ) \in \{ 0, \pm \frac{1}{2}, \pm 1 \} \).

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Because we have a function relating \(\cos \theta\) to \(\cos n\theta\), it makes sense to suspect that there is such a function for \(\sin\theta\) and \(\sin n\theta\) too. Indeed, the Chebyshev polynomials of the second kind are exactly this:

The \(n^\text{th}\) Chebyshev polynomial of the second kind, denoted by \(U_n(x)\), is defined by

\[U_n(\cos \theta)= \frac{ \sin \left( ( n+1) \theta \right) } { \sin \theta} _\square\]

1. \( T_n ( -x ) = (-1) ^n T_n (x)\)
2. \(T_{2n} (0) = (-1) ^ n\)
3. \(T_n (1) = 1\)
4. \(T_{2n+1} (0) = 0\)
5. \( T_n (-1) = (-1)^n \)
6. Prove that \[ T_n(x) = \frac{ \left( x - \sqrt{ x^2 - 1} \right)^n + \left( x + \sqrt{ x^2 - 1 } \right) ^ n } { 2} .\]
7. Prove that \[ T_n (x) = \frac{ (-2)^n n! } { (2n)!} \sqrt{ \left(1-x^2 \right) } \frac{ d^n } { dx^n } \left( 1 - x^2 \right) ^ { \frac{n-1}{2} }. \]
8. Show that the generating function for \( T_n(x) \) is given by \[ \frac{ 1 - tx } { 1 - 2tx + t^2 } = \sum_{n=0} ^ \infty T_n (x) t^n. \]