Question

Suppose that it is desired to construct a set of polynomials \(P_{0}(x), P_{1}(x), \ldots, P_{k}(x), \ldots,\) where \(P_{k}(x)\) is of degree \(k,\) that are orthogonal on the interval \(-1 \leq x \leq 1 ;\) see Problem 7. Suppose further that \(P_{k}(x)\) is normalized by the condition \(P_{k}(1)=1 .\) Find \(P_{0}(x), P_{1}(x),\) \(P_{2}(x),\) and \(P_{3}(x)\). Note that these are the first four Legendre polynomials (see Problem 24 of Section 5.3 ).

Short answer

The first four Legendre polynomials are: $$P_0(x) = 1$$ $$P_1(x) = x$$ $$P_2(x) = \frac{3}{2}x^2 - \frac{1}{2}$$ $$P_3(x) = \frac{5}{2}x^3 - \frac{3}{2}x$$

Step-by-step solution

Understand Orthogonality Condition

Orthogonal polynomials satisfy the condition \(\int_{-1}^{1} P_i(x)P_j(x)dx = L_k\delta_{ij}\) for \(i, j = 0, 1, 2, ...\) and \(L_k\) is the normalization constant. Here, \(\delta_{ij}\) is the Kronecker delta where \(\delta_{ij} = 1\) if \(i = j\) and \(\delta_{ij} = 0\) if \(i \neq j\).

Start with Polynomial of Degree 0

First, we need to find a polynomial of degree 0 that starts the sequence and is orthogonal to all others. The only possible polynomial of degree 0 is a constant. We can choose \(P_0(x) = 1\) since it meets the normalization condition \(P_0(1) = 1\).

Determine Polynomial of Degree 1

For the next polynomial, \(P_1(x)\), orthogonal to constant \(P_0(x)\), we can choose a linear function \(P_1(x) = ax +b\). Then, we have to satisfy the orthogonality condition: $$\int_{-1}^{1} (ax+b)dx = 0$$ This integral simplifies to: $$\int_{-1}^{1} ax dx + \int_{-1}^{1} b dx = 0$$ The first integral evaluates to 0, and the second evaluates to \(2b\) for \(x \in [-1, 1]\). Therefore, we have: $$2b = 0 \Rightarrow b = 0$$ So \(P_1(x) = ax\) for some constant \(a\). To satisfy the normalization condition \(P_1(1)=1\), we can simply choose \(a = 1\). So, \(P_1(x) = x\).

Determine Polynomial of Degree 2

For \(P_2(x)\), we assume a quadratic polynomial \(P_2(x) = ax^2 + bx + c\). To make the polynomial orthogonal to both \(P_0(x)\) and \(P_1(x)\), we use the orthogonality conditions: 1. \(\int_{-1}^{1} (ax^2 + bx + c)dx = 0\) 2. \(\int_{-1}^{1} x(ax^2 + bx + c)dx = 0\) After evaluating the integrals and solving the system of equations, we get \(c = 0\), and \(a = \frac{3}{2}\), i.e. \(P_2(x) =\frac{3}{2}x^2 - \frac{1}{2}\).

Determine Polynomial of Degree 3

For the polynomial \(P_3(x)\), we assume a cubic polynomial \(P_3(x) = ax^3 + bx^2 + cx + d\). We use the orthogonality conditions with respect to \(P_0(x)\), \(P_1(x)\), and \(P_2(x)\) to find values of \(a, b, c,\) and \(d\): 1. \(\int_{-1}^{1} (ax^3 + bx^2 + cx + d)dx = 0\) 2. \(\int_{-1}^{1} x(ax^3 + bx^2 + cx + d)dx = 0\) 3. \(\int_{-1}^{1}(ax^3 + bx^2 + cx + d)(\frac{3}{2}x^2 - \frac{1}{2})dx = 0\) After evaluating the integrals and solving the system of equations, we have: \(a=\frac{5}{2}\), \(b=0\), \(c=-\frac{3}{2}\), and \(d=0\), i.e., \(P_3(x) =\frac{5}{2}x^3 - \frac{3}{2}x\).

Conclusion

We have found the first four Legendre polynomials: $$P_0(x) = 1$$ $$P_1(x) = x$$ $$P_2(x) = \frac{3}{2}x^2 - \frac{1}{2}$$ $$P_3(x) = \frac{5}{2}x^3 - \frac{3}{2}x$$

Key concepts

Orthogonal Polynomials

Orthogonal polynomials are special types of polynomials that maintain a unique relationship with one another. They are "orthogonal" because they satisfy a specific integral condition. For a set of polynomials \( P_i(x) \) and \( P_j(x) \), if you integrate their product over a certain interval (such as \(-1 \leq x \leq 1\) for Legendre polynomials), they must satisfy the condition: \[ \int_{-1}^{1} P_i(x)P_j(x)dx = L_k \delta_{ij} \] Where \( \delta_{ij} \) is the Kronecker delta. If \( i eq j \), the integral equals zero, indicating the polynomials are orthogonal—a bit like seeing two vectors at a right angle to one another. If \( i = j \), it equals \( L_k \), a constant value derived from the nature of the polynomials. This parameter helps to distinguish their scale, similar to how the length of a vector shows its size.

Normalization Condition

The normalization condition for polynomials ensures that each polynomial is scaled in a specific way. When working with Legendre polynomials, the normalization condition states that \( P_k(1) = 1 \). This means, for each polynomial of degree \( k \), when you plug in \( x = 1 \) into the polynomial, the result should be 1.

• This helps to set a standard "size" for the polynomial, making calculations and comparisons more straightforward.
• It ensures that all polynomials in the set conform to a uniform rule when evaluated at \( x = 1 \).

This standardization is crucial when you want to measure or compare these polynomials or use them in further mathematical processes.

Kronecker Delta

Found within the orthogonality condition, the Kronecker delta \( \delta_{ij} \) symbol is very important. It acts kind of like a switch, having two values based on inputs.

• When \( i = j \), \( \delta_{ij} = 1 \).
• When \( i eq j \), \( \delta_{ij} = 0 \).

This binary behavior aids in defining the relations in sets of orthogonal functions or sequences by distinguishing when two elements are the same versus different. It's invaluable in streamlining the appearance and manageability of expressions when dealing with sums or integrals involving orthogonal functions.

Polynomial Degree

The degree of a polynomial is one of its defining features, denoting the highest power of the variable \( x \) it contains. In the Legendre polynomials context, each polynomial \( P_k(x) \) has a degree \( k \).

• For \( P_0(x) \), the degree is 0, resembling a constant.
• For \( P_1(x) \), the degree is 1, taking a linear form.
• Similarly, \( P_2(x) \) is quadratic and \( P_3(x) \) is cubic.

Each increase in degree adds complexity and offers more flexibility in how these polynomials model data or solve equations. The interplay between polynomial degrees is essential for calculations in series expansions, numerical methods, and solving differential equations. Understanding the degree helps in recognizing the polynomial's complexity and behavior.