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