Difference between revisions of "Chebyshev polynomials of the first kind"

Revision as of 21:20, 11 March 2022

The Chebyshev polynomials of the first kind are defined recursively by $\begin{align*} T_0(x) &= 1, \\ T_1(x) &= x, \\ T_{n+1}(x) &= 2xT_n(x) - T_{n-1}(x), \end{align*}$ or equivalently by $T_n(x) = \cos (n \arccos x).$

Proof of equivalence of the two definitions

In the proof below, $T_n(x)$ will refer to the recursive definition. Let $y = \arccos x$ , so that $x = \cos y$ .

For the $n = 0$ base case, $\cos 0 = 1 = T_0(x);$ for the $n = 1$ base case, $\cos y = x = T_1(x).$

Now for the inductive step, we assume that $\cos ((n-1)y) = T_{n-1}(x)$ and $\cos ny = T_n(x)$ , and we wish to prove that $\cos ((n+1)y) = T_{n+1}(x)$ .

From the cosine sum and difference identities we have $\cos ((n+1)y) = \cos (ny+y) = \cos ny \cos y - \sin ny \sin y$ and $\cos ((n-1)y) = \cos (ny-y) = \cos ny \cos y + \sin ny \sin y.$ The sum of these equations is $\cos ((n+1)y) + \cos ((n-1)y) = 2 \cos ny \cos y;$ rearranging, $\cos ((n+1)y) = 2 \cos y \cos ny - \cos ((n-1)y).$ Substituting our assumptions yields $\cos ((n+1)y) = 2xT_n(x) - T_{n-1}(x) = T_{n+1}(x),$ as desired.

Composition identity

For nonnegative integers $m$ and $n$ , the identity $T_{mn} = T_m(T_n(x))$ holds.

First proof

By the trigonometric definition, $T_m(T_n(x)) = \cos(m(\arccos(\cos(n(\arccos(x))))))$ .

As before, let $\arccos x = y$ . We have $\arccos(\cos(ny)) = 2k\pi \pm ny$ for some integer $k$ . Multiplying by $m$ and distributing gives $2mk\pi \pm mny$ ; taking the cosine gives $\cos (2mk\pi \pm mny) = \cos mny = \cos ( mn (\arccos x)) = T_{mn}(x)$ .

For now this proof only applies where the trigonometric definition is defined; that is, for $x \in [-1,1]$ . However, $T_{mn}(x)$ is a degree- $mn$ polynomial, and so is $T_m(T_n(x))$ , so the fact that $T_{mn}(x) = T_m(T_n(x))$ for some $mn + 1$ distinct $x \in [-1,1]$ is sufficient to guarantee that the two polynomials are equal over all real numbers.

Second proof (Induction)

First we prove a lemma: that $T_{k+n}(x) = 2T_n(x)T_k(x) - T_{k-n}(x)$ for all $n \leq k$ . To prove this lemma, we fix $k$ and induct on $n$ .

For all $k$ , we have $T_{k}(x) = 2T_{k}(x) - T_k(x) = 2T_0(x)T_{k}(x) - T_{k}(x),$ and for all $k \geq 1$ , $T_{k+1}(x) = 2xT_k(x) - T_{k-1}(x) = 2T_1(x)T_k(x) - T_{k-1}(x),$ proving the lemma for $n = 0$ and $n = 1$ respectively.

Suppose the lemma holds for $n - 1$ and $n$ ; that is, $T_{k+n}(x) = 2T_{n}(x)T_k(x) - T_{k-n}(x)$ and $T_{k+n-1}(x) = 2T_{n-1}(x)T_k(x) - T_{k-n+1}(x)$ . Then multiplying the first equation by $2x$ and subtracting the second equation gives $2xT_{k+n}(x) - T_{k+n-1}(x) = 2(2xT_{n}(x) - T_{n-1}(x))T_k(x) - (2xT_{k-n}(x) - T_{k-n+1}(x)),$ which simplifies to $T_{k+n+1}(x) = 2T_{n+1}(x)T_k(x) - T_{k-n-1}(x)$ using the original recursive definition, as long as $k-n-1 \geq 0$ . Thus, the lemma holds for $n + 1$ (as long as $n + 1 \leq k$ ), completing the inductive step.

To prove the claim, we now fix $n$ and induct on $m$ .

For all $n$ , we have $T_0(T_n(x)) = 1 = T_0(x)$ and $T_1(T_n(x)) = T_n(x),$ proving the claim for $m = 0$ and $m = 1$ respectively.

Suppose the claim holds for $m - 1$ and $m$ ; that is, $T_{m-1}(T_n(x)) = T_{(m-1)n}(x)$ and $T_m(T_n(x)) = T_{mn}(x)$ . We may also assume $m \geq 2$ , since the smaller cases have already been proven, in order to ensure that $n \leq mn$ . Then by the lemma (with $k = mn$ ) and the original recursive definition, $\begin{align*} T_{(m+1)n}(x) &= T_{mn+n}(x) \\ &= 2T_n(x)T_{mn}(x) - T_{mn - n}(x) \\ &= 2T_n(x)T_{mn}(x) - T_{(m-1)n}(x) \\ &= 2T_n(x)T_m(T_n(x)) - T_{m-1}(T_n(x)) \\ &= T_{m+1}(T_n(x)), \\ \end{align*}$ completing the inductive step.

Roots

Since $T_n(\cos y) = \cos ny$ , and the values of $ny$ for which $\cos ny = 0$ are $\frac{2k+1}{2}\pi$ for integers $k$ , the roots of $T_n(x)$ are of the form $\cos \left( \frac{2k+1}{2n}\pi \right).$

These roots are also called Chebyshev nodes.

Connection to roots of unity

Because cosine is $1$ only at integer multiples of $2\pi$ , the roots of the polynomial $T_n(x) - 1$ follow the simpler formula $\cos \frac{2k\pi}{n}$ . The $n$ th roots of unity have arguments of $\frac{2k\pi}{n}$ and magnitude $1$ , so the roots of $T_n(x) - 1$ are the real parts of the $n$ th roots of unity. This lends intuition to several patterns.

All roots of $T_n(x) - 1$ are also roots of $T_{mn}(x) - 1$ , since all $n$ th roots of unity are also $mn$ th roots of unity. This can also be shown algebraically as follows: Suppose $T_n(x) - 1 = 0$ . Then $T_{mn}(x) - 1 = T_m(T_n(x)) - 1 = T_m(1) - 1 = 1 - 1 = 0,$ using the composition identity and the fact that $T_m(1) = \cos(m \arccos 1) = \cos 0 = 1$ for all $m$ .

Particular cases include that $1$ , being a root of $T_1(x) - 1 = x - 1$ , is a root of $T_n(x) - 1$ for all positive $n$ , and $-1$ , being a root of $T_2(x) - 1 = 2x^2 - 2$ , is a root of $T_n(x) - 1$ for all positive even $n$ . All other roots of $T_n(x) - 1$ correspond to roots of unity which fall into conjugate pairs with the same real part. One might therefore suspect that all remaining roots of $T_n(x) - 1$ are double roots. In fact, $T_{2n+1}(x) - 1= (x - 1)(W_n(x))^2$ and $T_{2n+2}(x) - 1 = 2(x - 1)(x + 1)(U_n(x))^2$ for polynomials $W_n(x)$ and $U_n(x)$ satisfying the same recurrence relation as $T_n(x)$ , but with the different base cases $W_1(x) = 2x + 1$ and $U_1(x) = 2x$ . The polynomials $U_n(x)$ are known as Chebyshev polynomials of the second kind.

Rational roots

The rational roots of $T_n(x) - 1$ for any $n$ must be elements of the set $S = \{ -1, -\frac{1}{2}, 0, \frac{1}{2}, 1 \}$ .

Any root other than $1$ of $T_{2n+1}(x) - 1$ must be a root of $W_n(x)$ . The polynomials $w_n(x) = W_n \left( \frac{x}{2} \right)$ satisfy the recursive formula $\begin{align*} w_0(x) &= 1 \\ w_1(x) &= x + 1 \\ w_{n+1}(x) &= xw_n(x) - w_{n-1}(x), \end{align*}$ ensuring that they have integer coefficients and are monic for all $n$ . Furthermore, their constant terms satisfy $w_{n+1}(0) = -w_{n-1}(0),$ and so are always either $-1$ or $1$ . Thus, by the Rational Root Theorem, any rational roots of $w_n(x)$ must be $-1$ or $1$ , so the only possible rational roots of $W_n(x)$ are $-\frac{1}{2}$ and $\frac{1}{2}$ . Thus, the only possible rational roots of $T_{2n+1}(x) - 1$ are $-\frac{1}{2}$ , $\frac{1}{2}$ , and $1$ , each an element of $S$ .

By the composition identity, $T_{2n}(x) - 1 = T_{n}(T_{2}(x)) - 1$ . If $x$ is rational, then $T_2(x) = 2x^2 - 1$ is rational also. Therefore, if all rational roots of $T_n(x) - 1$ lie in $S$ , then any rational root of $T_{2n}(x) - 1$ must be a solution of $2x^2 - 1 = s$ for some $s \in S$ . The following five cases show that the rational roots of $T_{2n}(x) - 1$ must lie in $S$ as well: $\begin{align*} 2x^2 = 0 &\Longrightarrow x = 0, \\ 2x^2 - \frac{1}{2} = 0 &\Longrightarrow x = \pm \frac{1}{2}, \\ 2x^2 - 1 = 0 &\Longrightarrow x = \pm \frac{\sqrt 2}{2}, \\ 2x^2 - \frac{3}{2} = 0 &\Longrightarrow x = \pm \frac{\sqrt 3}{2}, \\ 2x^2 - 2 = 0 &\Longrightarrow x = \pm 1. \end{align*}$

Because any value $\cos \frac{a\pi}{b}$ for integers $a$ and $b$ is a root of $T_{2b}(x) - 1$ , the five elements of $S$ are the only rational values of the cosine of a rational multiple of $\pi$ . This result is known as Niven's Theorem.

Constructible roots

The $n$ th roots of unity form a regular polygon with $n$ sides and circumradius $1$ in the complex plane. Therefore, such a polygon can be constructed if and only if the position of some primitive $n$ th root of unity can be constructed. The root of unity in turn can be constructed if and only if its real part is a constructible number.

Just as primitive roots of unity are roots of $x^n - 1$ but not of $x^d - 1$ for any $d < n$ , the roots of $T_n(x) - 1$ that are the real parts of primitive roots of unity are roots of $T_n(x) - 1$ but not of $T_d(x) - 1$ for any $d < n$ . Therefore, a regular $n$ -gon can be constructed if and only if $T_n(x) - 1$ has a real root that is a constructible number and not a root of $T_d(x) - 1$ for any $d < n$ .

@@ Line 4: / Line 4: @@
 In the proof below, <math>T_n(x)</math> will refer to the recursive definition. Let <math>y = \arccos x</math>, so that <math>x = \cos y</math>.
-For the <math>n = 0</math> base case, <cmath>\cos(0y) = \cos 0 = 1 = T_0(x);</cmath> for the <math>n = 1</math> base case, <cmath>\cos(1y) = \cos y = x = T_1(x).</cmath>
+For the <math>n = 0</math> base case, <cmath>\cos 0 = 1 = T_0(x);</cmath> for the <math>n = 1</math> base case, <cmath>\cos y = x = T_1(x).</cmath>
-Now for the inductive step, We then assume that <math>\cos ((n-1)y) = T_{n-1}(x)</math> and <math>\cos ny = T_n(x)</math>, and we wish to prove that <math>\cos ((n+1)y) = T_{n+1}(x)</math>.
+Now for the inductive step, we assume that <math>\cos ((n-1)y) = T_{n-1}(x)</math> and <math>\cos ny = T_n(x)</math>, and we wish to prove that <math>\cos ((n+1)y) = T_{n+1}(x)</math>.
 From the [[Trigonometric identities#Angle addition identities|cosine sum and difference identities]] we have <cmath> \cos ((n+1)y) = \cos (ny+y) = \cos ny \cos y - \sin ny \sin y</cmath> and <cmath> \cos ((n-1)y) = \cos (ny-y) = \cos ny \cos y + \sin ny \sin y.</cmath> The sum of these equations is <cmath> \cos ((n+1)y) + \cos ((n-1)y) = 2 \cos ny \cos y;</cmath> rearranging, <cmath> \cos ((n+1)y) = 2 \cos y \cos ny  - \cos ((n-1)y).</cmath> Substituting our assumptions yields <cmath> \cos ((n+1)y) = 2xT_n(x) - T_{n-1}(x) = T_{n+1}(x),</cmath> as desired.

Page

Toolbox

Search