We start the year 2015 with a nice topic in polynomials, in this case let’s talk about Chebyshev Polynomial, only we take the first kind, we left the Second kind for another opportunity. So, which is it?

## Definition

Chebyshev polynomials are polynomials with the largest possible leading coefficient, with the condition that their absolute value in the interval $[-1, 1]$ is bounded by $1$. The chebyshev polynomial of the first kind are defined by the recurrence relation.

$T_{0}(x) = 1$
$T_{1}(x) = x$
$T_{n+1}(x) = 2xT_{n}(x) - T_{n-1}(x)$

These are characterized by the property:

We can raise the following question

Prove that for each $n \in \mathbb{N}$ there is a polynomial $T_{n}$ with integer coefficients and the leading $2^{n-1}$ such that $T_{n}(\cos{x}) = \cos{nx}$ for all $x$

Solution

We know that $T_{0}(x)=1$ and $T_{1}(x)=x$. For $n>1$ we use induction on $n$.

Since $\cos{(n+1)}x = 2\cos{x}\cos{(nx)} - \cos{(n-1)}x$, we can define $T_{n+1} = 2T_{1}T_{n} - T_{n-1}$ Since $T_{1}T_{n}$ and $T_{n-1}$ are of degrees $n+1$ and $n-1$ respectively, $T_{n+1}$ is of degree $n+1$ and has the leading coefficient $2\times2^{n} = 2^{n+1}$. It also follows from the construction that all its coefficients are integers.

Actually, these sequence of orthogonal polynomials are related to De Moivre’s formula

## De Moivre’s Theorem

For any complex number $x$ and any integer $n$.

Proof:

We prove this formula by induction on $n$ and by applying the trigonometric sum and product formulas. We first consider the non-negative integers. The base case $n=0$ is clearly true. For the induction step, observe that:

Here we have a problem about chebyshev polynomial:

Prove that the maximum in absolute value of any monic real polynomial of $n$-th degree on $[-1, 1]$ is not less than $\frac{1}{2^{n-1}}$

Solution

Note that equality holds for a multiple of the n-th Chebyshev polynomial $T_{n}(X)$ The leading coefficient of $T_{n}$ equals $2^{n-1}$, so $C_{n}(X) = \frac{1}{2^{n-1}}T_{n}(X)$ is a monic polynomial and

Moreover the values of $T_{n}$ at points $1, \cos{\frac{\pi}{n}}, \cos{\frac{2\pi}{n}}, \dots , \cos{\frac{(n-1)\pi}{n}}, -1$ are alternately $\frac{1}{2^{n-1}}$ and $-\frac{1}{2^{n-1}}$

Now suppose that $P \neq T_{n}$ is a monic polynomial such that $% $ Then $P(X)-C_{n}(X)$ at points $1, \cos{\frac{\pi}{n}}, \dots , \cos{\frac{(n-1)\pi}{n}}, -1$ alternately takes positive and negative values. Therefore the polynomial $P-C_{n}$ has at least $n$ zeros, namely, at least one is very interval between two adjacent points. However, $P-C_{n}$ is a polynomial of degree $n-1$ as the monomial $x^n$ is canceled, so we have arrived at a contradiction.

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Now, we describe a theorem generated by the Chebyshev polynomials.

## Chebyshev’s theorem

For fixed $n\geq1$, the polynomial $2^{-n+1}T_{n}(x)$ is the unique monic $n$th-degree polynomial satisfying

for any other monic $n$th-degree polynomial $P(x)$. Let’s see an example.

Let $A_{1}, A_{2}, \dots, A_{n}$ be points in the plane. Prove that on any segment of length $l$ there is a point $M$ such that

Solution

We can assume that $l=2$. Associate complex coordinates to points in such a way that the segment coincides with the interval $[-1, 1]$. Then

where $P(z)$ is a monic polynomial with complex coefficients, and $z \in [-1, 1]$.

We can write $P(z) = R(z) + iQ(z)$, where $R(z)$ is the real part and $Q(z)$ is the imaginary part of the polynomial. Since $z$ is real, we have $|P(z)| \geq |R(z)|$. The polynomial $R(z)$ is monic, so on the interval [-1, 1] it varies away from zero at least as much as the Chebyshev polynomial. Thus we can find $z$ in this interval such that $|R(z)| \geq \frac{1}{2^{n-1}}$. This implies $|P(z)| \geq 2\frac{1}{2^n}$, and rescaling back we deduce the existence in the general case of a point $M$ satisfying the inequality from the statement.

Stepping aside from the classical picture, let us also consider the families of polynomials $T_{n}(x)$ defined by $T_{0}(x) = 2, T_{1}(x) = x, T_{n+1}(x) = xT_{n}(x) - T_{n-1}(x)$. It’s determined by the equality:

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Let $a_{i} \in \mathbb{R} (i=1, 2, \dots, n)$ and $f(x) = \sum_{i=0}^{n}a_{i}x^{i}$ such that if $|x| \leq 1$, then $|f(x)| \leq 1$ on MathStackExchange