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

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

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

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 $T_{n} (x)$:

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

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 $$\frac{ 1 - tx } { 1 - 2tx + t^2 } = \sum_{n=0} ^ \infty T_n (x) t^n.$$