Question

The Legendre polynomials satisfy $$ \int_{-1}^{1} \phi_{j}(x) \phi_{k}(x) d x=\left\\{\begin{array}{ll} 0, & j \neq k \\ \frac{2}{2 j+1}, & j=k \end{array}\right. $$ Suppose that the best fit problem is given on the interval \([a, b]\). Show that with the transformation \(t=\frac{1}{2}[(b-a) x+(a+b)]\) and a slight change of notation, we have $$ \int_{a}^{b} \phi_{j}(t) \phi_{k}(t) d t=\left\\{\begin{array}{ll} 0, & j \neq k, \\ \frac{b-a}{2 j+1}, & j=k \end{array}\right. $$

Short answer

Question: Show that with the given transformation, the integral properties of Legendre polynomials hold for an arbitrary interval [a, b] rather than [-1, 1]. Short Answer: By performing the substitution \(t=\frac{1}{2}[(b-a) x+(a+b)]\), we can rewrite and simplify the integral using the properties of Legendre polynomials. We found that when \(j=k\), the integral properties hold with \(\frac{b-a}{2j+1}\), and when \(j \neq k\), they hold with 0.

Step-by-step solution

Substitution

Perform the substitution \(t=\frac{1}{2}[(b-a) x+(a+b)]\). Let $$ t = \frac{1}{2}((b-a)x + (a + b)). $$ Then we have $$ x = \frac{2t - (a + b)}{b - a}. $$ It follows that, $$ dt = \frac{1}{2}(b-a)dx. $$

Rewrite the Integral

Rewrite the integral with the substitution and rewrite the Legendre polynomials using the transformed variable t. We are given, $$ \int_{-1}^{1} \phi_{j}(x) \phi_{k}(x) dx = \left\lbrace\begin{array}{ll} 0, & j \neq k, \\ \frac{2}{2 j+1}, & j=k. \end{array}\right. $$ Using substitution, the integral becomes $$ \int_{a}^{b} \phi_{j}\left(\frac{2t-(a+b)}{b-a}\right) \phi_{k}\left(\frac{2t-(a+b)}{b-a}\right) \frac{1}{2}(b-a) dt. $$

Simplify the Integral

Simplify the integral using the properties of Legendre polynomials. Notice that the Legendre polynomials only depend on the transformed variable t. So, we can write $$ \int_{a}^{b} \phi_{j}(t) \phi_{k}(t) \frac{1}{2}(b - a) dt, $$ with \(\phi_j(t) = \phi_j \left(\frac{2t - (a + b)}{b - a}\right)\), and similarly for \(\phi_k(t)\).

Analyze the Result

Identify whether the integral properties hold for the transformed variable t. Now let's analyze the integral properties. If \(j = k\), then $$ \int_{a}^{b} \phi_{j}(t) \phi_{k}(t) \frac{1}{2}(b - a) dt = \frac{1}{2}(b - a) \int_{-1}^{1} \phi_{j}(x) \phi_{k}(x) dx= \frac{b-a}{2j+1}. $$ If \(j \neq k\), then $$ \int_{a}^{b} \phi_{j}(t) \phi_{k}(t) \frac{1}{2}(b - a) dt = \frac{1}{2}(b - a) \int_{-1}^{1} \phi_{j}(x) \phi_{k}(x) dx = 0. $$ Hence, we have shown that $$ \int_{a}^{b} \phi_{j}(t) \phi_{k}(t) dt = \left\lbrace\begin{array}{ll} 0, & j \neq k, \\ \frac{b-a}{2j+1}, & j=k. \end{array}\right. $$

Key concepts

Legendre Polynomials

Legendre polynomials are a set of orthogonal polynomials. They play a crucial role in mathematical physics and engineering, especially in solving differential equations. These polynomials are defined on the interval [-1, 1]. Because of their orthogonality property, they satisfy a unique integral norm when integrated over this interval. Orthogonality means that the integral of the product of two different Legendre polynomials over [-1, 1] is zero, which is what makes them particularly useful in approximating functions.

• The formula for Legendre polynomials is given as: \[ \int_{-1}^{1} \phi_{j}(x) \phi_{k}(x) \, dx = \begin{cases} 0, & j eq k \ \frac{2}{2j+1}, & j = k \end{cases} \]
• This property is very useful in applications such as solving partial differential equations and in models like quantum mechanics.

Understanding these polynomials means recognizing their shape, symmetry, and how they simplify the resolution of complex mathematical problems.

Integration by Substitution

Integration by substitution is a technique used to solve integrals, especially when they involve a function of another function. This method can simplify complicated integrals by changing variables, making them easier to evaluate. In the context of our exercise, we substitute the variable \( x \) with a new variable \( t \), defined as \[ t = \frac{1}{2}[(b-a)x + (a+b)]. \]
By changing the variable, we can convert an unfamiliar problem into a familiar one, allowing us to utilize properties of orthogonal polynomials easily.

• First, determine the relationship between the variables. Here we have \( x = \frac{2t - (a + b)}{b - a} \).
• Calculate the differential of the new variable. For the substitution above, \( dt = \frac{1}{2}(b-a) \, dx.\)

Applying integration by substitution in this context involves recognizing the transformation makes complex intervals of varying lengths amenable to using properties derived from standard Legendre polynomials.

Best Fit Problem

The best fit problem involves finding a function that most closely represents a given set of data. It often utilizes polynomials, like Legendre polynomials, due to their favorable properties.
Generally, the best fit minimizes the error between the function and the data, often using a least squares method.

• When working on the interval \( [a, b] \), transformations, like the one in our exercise \( t=\frac{1}{2}[(b-a)x+(a+b)]\), allow the application of defined orthogonality conditions.
• This transformation ensures your data corresponds naturally to the potential approximating functions like Legendre polynomials, improving the accuracy and simplicity of solving the best fit problem.

By re-defining the interval, we align the problem neatly into the framework where the polynomials' orthogonal properties hold. Thus, the tedious process of finding the best fit becomes a manageable and efficient task.