# Chebyshev Polynomials: $$T_n(x)$$

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Chebyshev polynomials are very useful for interpolating functions. Formally, the Chebyshev polynomial of degree $$n$$ is defined as

\begin{equation*}
T_n(x) = \cos(n\cos^{-1}x), \qquad \text{for } x\in [-1,\;1]
\end{equation*}
At first look, this expression does not resemble a polynomial at all!
In this note we will follow two different approaches to show that $$T_n(x)$$is indeed a polynomial. We start with the easy one, which only requires some basic trigonometric identities in real numbers, and where we will find a recursive definition of the Chebyshev polynomials. The not-so-easy approach requires working with complex numbers, but it will give us a closed formula for computing the polynomials.

### The easy path

Using the definition of the Chebyshev polynomials and a couple of trigonometric results we can find a convenient recursion:
\begin{align*}
T_{n+1}(x) + T_{n-1}(x) &= \cos[(n+1)\cos^{-1}x] + \cos[(n-1)\cos^{-1}x]\\
&= \cos[n\cos^{-1}x + \cos^{-1}x] + \cos[n\cos^{-1}x – \cos^{-1}x]\\ \\
\text{apply the identity } & \cos(u+v)+\cos(u-v) = 2\cos u\cos v \text{ to get}\\ \\
&= 2\cos(n\cos^{-1}x)\cos(\cos^{-1}x)\\
&= 2xT_n(x)
\end{align*}
That is, given two consecutive Chebyshev polynomials, the next one can be defined by the recursion
\begin{equation*}
T_{n+1}(x) = 2xT_{n}(x) – T_{n-1}(x)
\end{equation*}
The first two elements of the sequence can be computed easily:
\begin{align*}
T_0(x) &= \cos(0\cos^{-1}x) = 1\\
T_1(x) &= \cos(1\cos^{-1}x) = x
\end{align*}
Therefore, an alternative definition, where the polynomials are more evident, is given by the recursion:
\begin{align*}
T_0(x) &= 1 \\
T_1(x) &= x \\
T_{n+1}(x) &= 2xT_n(x) – T_{n-1}(x),\qquad\text{for } n=1,2,\dots
\end{align*}

### The challenging path

In this path, we will need the binomial and de Moivre’s formulas.

The binomial formula for complex numbers is a straightforward generalization of the corresponding formula for real numbers:

#### Theorem: Binomial formula for complex numbers

For $$z_1$$and $$z_2$$complex numbers
\begin{equation*}
(z_1 + z_2)^n = \sum_{k=0}^{n} {n \choose k} z_1^{n-k}\;z_2^k
\end{equation*}

De Moivre’s formula is easily derived from Euler’s formula $$e^{ix} = \cos(nx) + i\sin(nx)$$, where $$i$$is the imaginary unit.

#### Theorem: de Moivre’s formula

\begin{equation*}
(\cos\theta + i\sin\theta)^n = \cos n\theta + \sin n\theta
\end{equation*}

We are now ready to show that $$T_n(x)$$is a polynomial.

#### Theorem: Chebyshev polynomials

For $$-1\leq x \leq 1$$the expression
\begin{equation*}
T_n(x) = \cos(n\cos^{-1}x)
\end{equation*}
is a polynomial of degree $$n$$in the variable $$x$$.

#### Proof

Let’s define the integer $$m$$as:
\begin{equation*}
m = \begin{cases}
n/2 & \text{if $$n$$is even},\\
(n-1)/2 & \text{if $$n$$is odd}.
\end{cases}
\end{equation*}

\begin{align*}
\cos n\theta + i\sin n\theta &= (\cos\theta + i\sin\theta)^n\\
&= \sum_{k=0}^{n} {n \choose k} \cos^{n-k}\theta \;(i\sin\theta)^k\\
&= \sum_{k=0,\text{even}}^{n} {n \choose k} \cos^{n-k}\theta \;i^k\sin^k\theta + \sum_{k=0,\text{odd}}^{n} {n \choose k} \cos^{n-k}\theta \;i^k\sin^k\theta\\
&= \sum_{k=0}^{m} {n \choose 2k} \cos^{n-2k}\theta \;i^{2k}\sin^{2k}\theta + \sum_{k=0}^{m} {n \choose 2k+1} \cos^{n-2k-1}\theta \;i^{2k+1}\sin^{2k+1}\theta\\
&= \sum_{k=0}^{m} {n \choose 2k} (-1)^k \cos^{n-2k}\theta \;\sin^{2k}\theta + \sum_{k=0}^{m} {n \choose 2k+1} (-1)^k i\cos^{n-2k-1}\theta\;\sin^{2k+1}\theta\\
\end{align*}

Equating the real parts:
\begin{align*}
\cos n\theta &= \sum_{k=0}^{m} {n \choose 2k} (-1)^k \cos^{n-2k}\theta \;\sin^{2k}\theta \\
&= \sum_{k=0}^{m} {n \choose 2k} (-1)^k \cos^{n-2k}\theta \;(1-\cos^2\theta)^{k} \\
&= \sum_{k=0}^{m} {n \choose 2k} \cos^{n-2k}\theta \;(\cos^2\theta – 1)^{k} \\
\end{align*}
Write $$x = \cos\theta$$and suppose that $$0 \leq \theta \leq \pi$$, in which case $$-1\leq x \leq 1$$and $$\theta = \cos^{-1}x$$. Substitute
\begin{align*}
\cos (n\cos^{-1}x) &= \sum_{k=0}^{m} {n \choose 2k} x^{n-2k} \;(x^2 – 1)^{k} \\ \\
\text{using the} & \text{binomial formula once more} \\ \\
&= \sum_{k=0}^{m} \left[{n \choose 2k} x^{n-2k} \; \sum_{h=0}^{k}{k \choose h}(-1)^h (x^2)^{k-h} \right]\\
&= \sum_{k=0}^{m} \sum_{h=0}^{k}{n \choose 2k} \;{k \choose h}(-1)^h x^{n-2h} \\
\end{align*}
From this formula it is clear that the highest degree monomials are obtained when $$h=0$$. To complete the proof, all we need to show is that the $$x^n$$monomials do not cancel each other: setting $$h=0$$we obtain the leading coefficient $$\alpha_n$$:
\begin{align*}
\alpha_n &= \sum_{k=0}^{m} {n \choose 2k} \;{k \choose 0}(-1)^0 \\
&= \sum_{k=0}^{m} {n \choose 2k} \\
&= 2^{n-1}
\end{align*}

### Useful properties

#### Roots and extrema

Since $$T_n(x) = \cos(n\cos^{-1}x)$$, it is clear that $$-1 \leq T_n(x) \leq 1$$(remember that the polynomial is only defined for $$x \in [-1,1]$$). Extrema occurs when $$n\cos^{-1}x = k\pi$$for some integer $$k$$:
\begin{equation*}
\end{equation*}

On the other hand, the zeros of the polynomial occur when $$n\cos^{-1}x = k\pi + \frac{\pi}{2}$$for some integer $$k$$:
\begin{equation*}
Brown and Churchill (1996) Complex Variables and Applications, $$6^{th}$$edition. McGraw-Hill.