Problem 69 - Calculus Single Variable

Question

Involve the factorial numbers \(n !\), which were introduced in Section \(1.4 .\) They can be defined by \(0 !=1\) and, for a positive integer \(n, n !=n(n-1) \ldots 3 \ldots 2 \cdot 1\). The Legendre polynomials \(P_{n},\) introduced in 1782 by A.M. Legendre, can be defined by the following formula, which was discovered by Benjamin Olinde Rodrigues $$\begin{array}{l}(1794-1850 \text { or } 1851): \\\P_{n}(x)=\frac{1}{2^{n} n !} \frac{d^{n}}{d x^{n}}\left(x^{2}-1\right)^{n} \quad n=0,1,2,3, \ldots\end{array}$$ Calculate \(P_{0}(x), P_{1}(x), P_{2}(x), P_{3}(x),\) and \(P_{4}(x)\)

Step-by-Step Solution

Answer

\(P_0(x) = 1\), \(P_1(x) = x\), \(P_2(x) = \frac{3x^2 - 1}{2}\), \(P_3(x) = 5x^3 - 3x\), \(P_4(x) = \frac{35x^4 - 30x^2 + 3}{8}\).

Step 1: Calculate \(P_0(x)\)

Start with \(n = 0\). Substituting in Rodrigues' formula: \(P_0(x) = \frac{1}{2^0 \cdot 0!} \frac{d^0}{dx^0}((x^2-1)^0)\). Note that \(d^0/dx^0\) of any function is the function itself, and \((x^2 - 1)^0 = 1\). Therefore, \(P_0(x) = 1\).

Step 2: Calculate \(P_1(x)\)

For \(n = 1\), substitute into the formula: \(P_1(x) = \frac{1}{2^1 \cdot 1!} \frac{d^1}{dx^1}((x^2-1)^1)\). Calculate \((x^2-1) = x^2 - 1\). The derivative \(\frac{d}{dx}(x^2-1) = 2x\). Thus, \(P_1(x) = \frac{2x}{2} = x\).

Step 3: Calculate \(P_2(x)\)

For \(n = 2\), use the formula: \(P_2(x) = \frac{1}{2^2 \cdot 2!} \frac{d^2}{dx^2}((x^2-1)^2)\). First expand \((x^2-1)^2 = x^4 - 2x^2 + 1\). The first derivative is \(\frac{d}{dx}(x^4 - 2x^2 + 1) = 4x^3 - 4x\) and the second derivative is \(\frac{d^2}{dx^2}(4x^3 - 4x) = 12x^2 - 4\). Then, \(P_2(x) = \frac{12x^2 - 4}{8} = \frac{3x^2 - 1}{2}\).

Step 4: Calculate \(P_3(x)\)

For \(n = 3\), use: \(P_3(x) = \frac{1}{2^3 \cdot 3!} \frac{d^3}{dx^3}((x^2-1)^3)\). Begin with \((x^2-1)^3 = x^6 - 3x^4 + 3x^2 - 1\). Calculate the derivatives: \(\frac{d}{dx} = 6x^5 - 12x^3 + 6x\), \(\frac{d^2}{dx^2} = 30x^4 - 36x^2 + 6\), and \(\frac{d^3}{dx^3} = 120x^3 - 72x\). Therefore, \(P_3(x) = \frac{120x^3 - 72x}{48} = 5x^3 - 3x\).

Step 5: Calculate \(P_4(x)\)

For \(n = 4\), apply: \(P_4(x) = \frac{1}{2^4 \cdot 4!} \frac{d^4}{dx^4}((x^2-1)^4)\). Start with \((x^2-1)^4 = x^8 - 4x^6 + 6x^4 - 4x^2 + 1\). The derivatives are: \(\frac{d}{dx} = 8x^7 - 24x^5 + 24x^3 - 8x\), \(\frac{d^2}{dx^2} = 56x^6 - 120x^4 + 72x^2 - 8\), \(\frac{d^3}{dx^3} = 336x^5 - 480x^3 + 144x\), and \(\frac{d^4}{dx^4} = 1680x^4 - 1440x^2 + 144\). Substituting: \(P_4(x) = \frac{1680x^4 - 1440x^2 + 144}{384} = \frac{35x^4 - 30x^2 + 3}{8}\).

Key Concepts

Factorial

Factorials are fundamental in mathematics and play a crucial role in many calculations, including those involving Legendre polynomials. A factorial, denoted by the symbol "!", is the product of an integer and all the integers below it. For example:

• The factorial of 3, written as 3!, is calculated as: \(3! = 3 \times 2 \times 1 = 6\).
• The factorial of 5, or 5!, is: \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\).

Factorials grow very rapidly and are used in a variety of fields, from combinatorics to calculus. They help us determine permutations and combinations and are essential in defining sequences and series. In the context of Legendre polynomials, factorials are used as part of Rodrigues’ formula to calculate polynomial terms efficiently.

Rodrigues' Formula

Rodrigues' Formula provides an elegant method for generating Legendre polynomials, which are solutions to Legendre's differential equation. The formula, introduced by Benjamin Olinde Rodrigues, is given by: \[ P_n(x) = \frac{1}{2^n n!} \frac{d^n}{dx^n}((x^2 - 1)^n) \]Here’s what each part means:

• \(2^n n!\): This is a normalization factor ensuring that each polynomial maintains certain standard properties.
• \(\frac{d^n}{dx^n}\): This denotes the n-th derivative, which involves differentiating the function \((x^2 - 1)^n\) n times.

Rodrigues' formula simplifies the complex task of deriving Legendre polynomials by breaking it down into manageable steps. It uses basic calculus operations combined with factorial scaling, making it a powerful tool for mathematicians.

Derivatives

Derivatives are a cornerstone of calculus, providing insights into rates of change and the slope of curves. In the context of Legendre polynomials, derivatives help in transforming the function \((x^2 - 1)^n\) through Rodriguez's formula. Derivatives are like instructions to "take apart" functions to understand their behavior:

• The first derivative \(\frac{d}{dx}\) gives the rate of change of the function. For instance, if \(f(x) = x^2\), then \(\frac{df}{dx} = 2x\).
• The second derivative \(\frac{d^2}{dx^2}\) measures how the rate of change itself changes.
• Higher-order derivatives, like \(\frac{d^n}{dx^n}\), continue to track changes of progressively higher order.

In Rodrigues’ formula, taking derivatives ensures the polynomial solutions reflect an appropriate degree and symmetry characteristic of Legendre polynomials. Each ordered derivative creates more complex expressions that, when normalized, form polynomial solutions.

Polynomial Solutions

Polynomial solutions are functions that express polynomials in terms of their roots and coefficients. Legendre polynomials are a specific class of orthogonal polynomials. These are especially significant in physics and engineering as they often emerge in solving problems with spherical symmetries:

• Legendre polynomials are denoted \(P_n(x)\) and have important properties such as orthogonality over the interval \([-1, 1]\).
• They provide solutions to the Legendre differential equation, a critical method in physics for potential function problems.

The challenge in generating polynomial solutions often lies in ensuring they adhere to specific properties. Using Rodrigues' formula, the polynomials develop systematically from the basic equation \((x^2 - 1)^n\), simplified and normalized into usable solutions for scientific problems.