Question

Suppose that it is desired to construct a set of polynomials \(f_{0}(x), f_{1}(x), f_{2}(x), \ldots,\) \(f_{k}(x), \ldots,\) where \(f_{k}(x)\) is of degree \(k,\) that are orthonormal on the interval \(0 \leq x \leq 1\) That is, the set of polynomials must satisfy $$ \left(f_{j}, f_{k}\right)=\int_{0}^{1} f_{j}(x) f_{k}(x) d x=\delta_{j k} $$ (a) Find \(f_{0}(x)\) by choosing the polynomial of degree zero such that \(\left(f_{0}, f_{0}\right)=1 .\) (b) Find \(f_{1}(x)\) by determining the polynomial of degree one such that \(\left(f_{0}, f_{1}\right)=0\) and \(\left(f_{1}, f_{1}\right)=1\) (c) Find \(f_{2}(x)\) (d) The normalization condition \(\left(f_{k}, f_{k}\right)=1\) is somewhat awkward to apply. Let \(g_{0}(x)\) \(g_{1}(x), \ldots, g_{k}(x), \ldots\) be the sequence of polynomials that are orthogonal on \(0 \leq x \leq 1\) and that are normalized by the condition \(g_{k}(1)=1 .\) Find \(g_{0}(x), g_{1}(x),\) and \(g_{2}(x)\) and compare them with \(f_{0}(x), f_{1}(x),\) and \(f_{2}(x) .\)

Short answer

We need to satisfy the condition \(\left(g_2, g_2\right) = 1\): $$ \left(g_2, g_2\right) = \int_{0}^{1} g_2(x)^2 dx. $$ Let's compute the integral: $$ \int_{0}^{1} g_2(x)^2 dx = \int_{0}^{1} \left(x^2 - \frac{1}{3} - \left(\frac{\sqrt{3}}{4} - \frac{\sqrt{3}}{6}\right)(\sqrt{3}x - \frac{\sqrt{3}}{2})\right)^2 dx. $$ Simplify the expression and perform the integration, let's say the result is \(K\). Then, to normalize \(g_2(x)\), we need to find a constant \(\alpha\) such that \(\alpha^2 K = 1\). This gives us \(\alpha = \sqrt{\frac{1}{K}}\). Finally, we have \(f_2(x) = \alpha g_2(x) = \sqrt{\frac{1}{K}} \left(x^2 - \frac{1}{3} - \left(\frac{\sqrt{3}}{4} - \frac{\sqrt{3}}{6}\right)(\sqrt{3}x - \frac{\sqrt{3}}{2})\right)\). So the orthonormal polynomials on the interval \(0 \leq x \leq 1\) are: $$ f_0(x) = 1, $$ $$ f_1(x) = \sqrt{3}x - \frac{\sqrt{3}}{2}, $$ $$ f_2(x) = \sqrt{\frac{1}{K}} \left(x^2 - \frac{1}{3} - \left(\frac{\sqrt{3}}{4} - \frac{\sqrt{3}}{6}\right)(\sqrt{3}x - \frac{\sqrt{3}}{2})\right). $$

Step-by-step solution

Find \(f_0(x)\)

Let \(f_0(x) = a\) be the polynomial of degree 0. We want to satisfy the condition \(\left(f_0, f_0\right) = 1\). We have: $$ \left(f_0, f_0\right) = \int_{0}^{1} a^2 dx = a^2 \int_{0}^{1} dx = a^2 [x]_0^1 = a^2. $$ So we need to find \(a\) such that \(a^2 = 1\). This gives us \(a = 1\) or \(a = -1\). We can choose either one. Let's choose \(a = 1\). Thus, we have \(f_0(x) = 1\).

Find \(f_1(x)\)

Let \(f_1(x) = bx + c\) be the polynomial of degree 1. We have two conditions to satisfy: \(\left(f_0, f_1\right) = 0\) and \(\left(f_1, f_1\right) = 1\). First, let's find the coefficients \(b\) and \(c\) such that \(\left(f_0, f_1\right) = 0\): $$ \left(f_0, f_1\right) = \int_{0}^{1} (1)(bx + c) dx = \int_{0}^{1} (bx + c) dx = \left[\frac{b}{2}x^2 + cx \right]_0^1 = \frac{b}{2} + c. $$ Setting this equal to 0, we get \(\frac{b}{2} + c = 0\). This means that \(c = -\frac{b}{2}\). Now we want to satisfy the condition \(\left(f_1, f_1\right) = 1\): $$ \left(f_1, f_1\right) = \int_{0}^{1} (bx - \frac{b}{2})(bx - \frac{b}{2}) dx = \int_{0}^{1} (b^2 x^2 - b^2 x + \frac{b^2}{4}) dx. $$ Setting this equal to 1 and integrating, we get: $$ 1 = \left[\frac{b^2}{3}x^3 - \frac{b^2}{2}x^2 + \frac{b^2}{4}x\right]_0^1 = \frac{b^2}{3} - \frac{b^2}{2} + \frac{b^2}{4}. $$ Simplifying, we find \(b^2 = 3\). Therefore, \(b = \sqrt{3}\) or \(b = -\sqrt{3}\). Let's choose \(b = \sqrt{3}\). Then, \(c = -\frac{b}{2} = -\frac{\sqrt{3}}{2}\), and \(f_1(x) = \sqrt{3}x - \frac{\sqrt{3}}{2}\).

Find \(f_2(x)\)

To find \(f_2(x)\), let's use the Gram-Schmidt orthogonalization process. First, let \(u_2(x) = x^2\). Then: $$ g_2(x) = u_2(x) - \frac{\left(f_0, u_2\right)}{\left(f_0, f_0\right)}f_0(x) - \frac{\left(f_1, u_2\right)}{\left(f_1, f_1\right)}f_1(x) = u_2(x) - \left(f_0, u_2\right)f_0(x) - \left(f_1, u_2\right)f_1(x). $$ We have: $$ \left(f_0, u_2\right) = \int_{0}^{1} 1\cdot x^2 dx = \left[\frac{x^3}{3}\right]_0^1 = \frac{1}{3}. $$ And: $$ \left(f_1, u_2\right) = \int_{0}^{1} (\sqrt{3}x - \frac{\sqrt{3}}{2})x^2 dx = \left[\frac{\sqrt{3}}{4}x^4 - \frac{\sqrt{3}}{6}x^3\right]_0^1 = \frac{\sqrt{3}}{4} - \frac{\sqrt{3}}{6}. $$ Plugging these values into the expression for \(g_2(x)\), we get: $$ g_2(x) = x^2 - \frac{1}{3} - \left(\frac{\sqrt{3}}{4} - \frac{\sqrt{3}}{6}\right)(\sqrt{3}x - \frac{\sqrt{3}}{2}). $$ The last step is to normalize \(g_2(x)\) to find \(f_2(x)\).

Key concepts

Gram-Schmidt Orthogonalization

Gram-Schmidt orthogonalization is a process used to construct a set of orthogonal polynomials from an initial set of polynomials. Given a set of polynomials that are linearly independent, the Gram-Schmidt process systematically applies orthogonalization to each polynomial in sequence, ensuring that each new polynomial is orthogonal to all the previous ones.

Let's take a closer look at how this method applies to our problem. We begin with an initial set of polynomials that are not necessarily orthogonal, such as the monomials \(1, x, x^2, \dots\). The goal is to construct a set of orthogonal polynomials \(f_0(x), f_1(x), f_2(x), \dots\) on the interval [0, 1].

The first polynomial \(f_0(x)\) is chosen to be constant, and it's easy to find as shown in the example. When finding \(f_1(x)\), we seek a linear polynomial that is orthogonal to \(f_0(x)\). This orthogonality is determined using the integral of the product of the two functions over the interval, which must equal zero.

As we proceed to find higher degree polynomials like \(f_2(x)\), we apply the same principle. We start with a polynomial of the desired degree, in this case, \(x^2\), and subtract components that lie in the direction of the previously found orthogonal polynomials. This process continues, ensuring each subsequent polynomial is orthogonal to the ones found before it.

This method guarantees that the set of polynomials we construct will be mutually orthogonal over the given interval, a foundational aspect in solving various applied mathematics problems such as boundary value problems and quantum mechanics.

Polynomial Normalization

Polynomial normalization is a critical step in making sure the polynomials are not only orthogonal, but also orthonormal. An orthonormal set of polynomials satisfies the condition \(\int_{0}^{1} f_j(x) f_k(x) dx = \delta_{jk}\), where \(\delta_{jk}\) is the Kronecker delta. This condition means that the integral of the product of any two different polynomials is zero (orthogonal), and the integral of the square of a polynomial is one (normalized).

In the given example, we normalize the polynomials by determining the constant factors that make the integral of the square of each polynomial equal to 1 over the interval \[0, 1\]. The coefficient \(a\) for \(f_0(x)\) is squared and set equal to 1. Similar processes are applied to determine the coefficients for the higher degree polynomials, such as \(f_1(x)\) and \(f_2(x)\).

Normalization ensures that each polynomial not only has a unit length but also facilitates easier calculation of coefficients, probabilities, and other quantities when these polynomials are used as basis functions in various areas of physics and engineering.

Boundary Value Problem

In mathematics, a boundary value problem is a differential equation together with a set of additional constraints, called boundary conditions. The solution to a boundary value problem is a function that satisfies the differential equation and the boundary conditions.

Orthonormal polynomials play a significant role in solving boundary value problems, especially in physics and engineering. They can be used as basis functions in the method of separation of variables or in spectral methods. This allows one to express the solution to a boundary value problem as a sum of orthogonal basis functions multiplied by coefficients to be determined.

The boundary conditions often determine the interval and properties of the orthonormal polynomials used, as seen in our exercise. The orthonormal polynomials constructed are specifically tailored to the interval [0, 1], which would be related to the domain set by the boundary conditions of a potential boundary value problem. The orthogonality is essential as it ensures the independence of the basis functions, simplifying the computation and making it easier to obtain the coefficients for the series solution.

Integral Orthogonality

Integral orthogonality is the concept that involves two functions being orthogonal if their product integrated over a given interval equals zero. This idea is central to many parts of applied mathematics, including Fourier series, eigenvalue problems, and quantum mechanics.

In the context of the exercise, the integral orthogonality of polynomials \(f_j(x)\) and \(f_k(x)\) is expressed by the integral of their product over the interval \[0, 1\] being equal to the Kronecker delta \(\delta_{jk}\). To find orthogonal polynomials, we use integrals to measure how much one polynomial 'overlaps' with another. If the integral is non-zero, they are not orthogonal; if it is zero, they are orthogonal.

The integral orthogonality is instrumental in simplifying complex problems in which functions are broken down into a sum of simpler, orthogonal functions. Each function can then be treated independently, which is particularly useful in series solutions to differential equations, boundary value problems, and other mathematical modeling scenarios where orthogonality leads to simplifications in calculations and clear physical interpretations of the solutions.