Proving A_n(r) with Integer Coeff Polynomials: Q&A

[physicsRookie]

Define a sequence
$$A_n(r) = \int_{-1}^1(1-x^2)^n \cos(rx)\, dx, \qquad n \in \mathbb{N}, r \in \mathbb{R}.$$

Prove that
$$A_n(r) = \frac{n!}{r^{2n+1}}[P_n(r)\sin r - Q_n(r)\cos r]$$

where $$P_n$$ and $$Q_n$$ are two polynomials with integer coefficients. What is the degree of $$P_n$$ and of $$Q_n$$?


Can anyone help me? Thanks.

 

[matt grime]

Induction. And an educated guess for the degrees, followed by induction.

 

[tacman]

Sure, I can help you with this question. Let's break it down step by step.

First, let's recall the definition of the sequence A_n(r) given in the question. It is defined as the integral of the function (1-x^2)^n multiplied by the cosine of rx, over the interval [-1,1]. This sequence depends on two variables, n and r, and we can see that n must be a positive integer and r can be any real number.

Next, we want to prove that A_n(r) can be expressed as a polynomial with integer coefficients, multiplied by a sine and cosine term. In other words, we want to show that A_n(r) has the form:

A_n(r) = P(r)sin(r) - Q(r)cos(r)

where P(r) and Q(r) are polynomials with integer coefficients. To do this, we will use mathematical induction.

First, let's consider the base case where n = 1. In this case, we have:

A_1(r) = \int_{-1}^1(1-x^2)\cos(rx)\, dx

To evaluate this integral, we can use integration by parts with u = (1-x^2) and dv = cos(rx)dx. This gives us:

A_1(r) = (1-x^2)\frac{\sin(rx)}{r}\bigg|_{-1}^1 - \int_{-1}^1 \frac{-2x}{r}\sin(rx)\, dx

Simplifying this expression, we get:

A_1(r) = \frac{2\sin(r)}{r}

We can see that this is in the form we wanted, with P(r) = 0 and Q(r) = 2/r. Therefore, the base case is true.

Next, let's assume that the statement is true for some positive integer k. In other words, we assume that:

A_k(r) = P_k(r)\sin(r) - Q_k(r)\cos(r)

Now, we want to prove that this statement is also true for k+1. We have:

A_{k+1}(r) = \int_{-1}^1 (1-x^2)^{k+1}\cos(rx)\, dx

Using integration by parts again, we get:

A_{k+1}(r) = (1