# Chebyshev polynomial (First kind)

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 is bounded by . The chebyshev polynomial of the first kind are defined by the recurrence relation.

These are characterized by the property:

We can raise the following question

**Prove that for each there is a polynomial with integer coefficients and the leading such that for all **

**Solution**

We know that and . For we use induction on .

Since , we can define Since and are of degrees and respectively, is of degree and has the leading coefficient . 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 and any integer .

**Proof:**

We prove this formula by induction on and by applying the trigonometric sum and product formulas. We first consider the non-negative integers. The base case 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 -th degree on is not less than **

**Solution**

Note that equality holds for a multiple of the n-th Chebyshev polynomial The leading coefficient of equals , so is a monic polynomial and

Moreover the values of at points are alternately and

Now suppose that is a monic polynomial such that Then at points alternately takes positive and negative values. Therefore the polynomial has at least zeros, namely, at least one is very interval between two adjacent points. However, is a polynomial of degree as the monomial is canceled, so we have arrived at a contradiction.

By the way, I would like to know other solutions for the last one, so I asked on MathStackExchange. If you want share something with me, please send me a message.

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

**Chebyshev’s theorem**

For fixed , the polynomial is the unique monic th-degree polynomial satisfying

for any other monic th-degree polynomial . Let’s see an example.

**Let be points in the plane. Prove that on any segment of length
there is a point such that**

**Solution**

We can assume that . Associate complex coordinates to points in such a way that the segment coincides with the interval . Then

where is a monic polynomial with complex coefficients, and .

We can write , where is the real part and is the imaginary part of the polynomial. Since is real, we have . The polynomial 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 in this interval such that . This implies , and rescaling back we deduce the existence in the general case of a point satisfying the inequality from the statement.

Stepping aside from the classical picture, let us also consider the families of polynomials defined by . It’s determined by the equality:

Also, I suggest visit the amazing answer for:

**Let and such that if , then **
on MathStackExchange