Question

Use the generalized Rodriguez formula to show that \(P_{0}(1)=1\) and \(P_{1}(1)=1\). Now use a recurrence relation to show that \(P_{n}(1)=1\) for all \(n .\) To be rigorous, you need to use mathematical induction.

Short answer

By using the generalized Rodriguez's formula, we can show that \(P_{0}(1)=1\) and \(P_{1}(1)=1\). Subsequently, by applying mathematical induction with a recurrence relation, we can show that \(P_{n}(1)=1\) for all \(n\).

Step-by-step solution

Generalized Rodriguez's formula

The generalized Rodriguez's formula for Legendre polynomials is given by \[P_{n}(x) = \frac{1}{2^n n!} \frac{d^n}{dx^n} [ (x^2 - 1)^n ]\] Run the formula for \(n=0\) and \(n=1\) to show that \(P_{0}(1)=1\) and \(P_{1}(1)=1\) respectively.

Applying Rodriguez's formula for \(n=0\)

For \(n=0\), the formula simplifies to \(P_{0}(x) = 1\). Hence, \(P_{0}(1) = 1\).

Applying Rodriguez's formula for \(n=1\)

For \(n=1\), the formula simplifies to \(P_{1}(x) = x\). Hence, \(P_{1}(1) = 1\).

Mathematical Induction: Base Case

The principle of mathematical induction involves two steps: confirming the base case and the inductive step. We've already confirmed the base case: \(P_{0}(1) = 1\) and \(P_{1}(1) = 1\).

Mathematical Induction: Inductive Hypothesis

For the inductive step, we'll assume that \(P_{k}(1) = 1\) for some integer \(k \geq 1\). This is known as our inductive hypothesis.

Mathematical Induction: Prove for \(n=k+1\)

Our goal now is to show that if the statement is true for \(k\), then it is also true for \(k+1\). The recurrence relation for Legendre polynomials is given by \((n+1)P_{n+1}(x) = (2n+1)xP_{n}(x)-nP_{n-1}(x)\) Use the recurrence relation to prove for \(n=k+1\).

Applying Recurrence Relation

Substitute \(n=k\) and \(x=1\) into the recurrence relation, it simplifies to \((k+1)P_{k+1}(1) = (2k+1)P_{k}(1)-kP_{k-1}(1)\). The hypothesis gives that \(P_{k}(1) = 1\) and \(P_{k-1}(1) = 1\), so the equation simplifies to \(P_{k+1}(1) = 1\). Therefore, if the statement is true for \(k\), then it is also true for \(k+1\).

Conclusion

We've shown that the statement is true for \(n=0\) and \(n=1\) (base cases), and that if it's true for some \(k\), then it's true for \(k+1\) (inductive step). Consequently, by mathematical induction, \(P_{n}(1) = 1\) for all \(n\).

Key concepts

Rodriguez Formula

The Rodriguez formula is a powerful tool in the world of mathematical analysis, playing a critical role in the study of Legendre polynomials. These polynomials arise in several areas of physics and engineering, particularly in solving problems dealing with spherical symmetry. The generalized formula is expressed as:
\[\begin{equation}P_{n}(x) = \frac{1}{2^n n!} \frac{d^n}{dx^n}[(x^2 - 1)^n],\end{equation}\]
where \(P_{n}(x)\) represents the Legendre polynomial of degree \(n\), and \(\frac{d^n}{dx^n}\) is the nth derivative with respect to \(x\).

• For \(n=0\), the formula simplifies, resulting in the constant polynomial \(P_{0}(x) = 1\), confirming that \(P_{0}(1)=1\).
• For \(n=1\), the formula becomes linear, or \(P_{1}(x) = x\), so at \(x=1\), \(P_{1}(1)=1\) holds true as well.

This result is crucial because it sets the stage for proving that Legendre polynomials evaluated at \(x=1\) yield 1 for any non-negative integer value of \(n\) using mathematical induction as explained in further sections.

Mathematical Induction

Mathematical induction is a fundamental proof technique that's especially helpful in establishing the truth of an infinite series of statements. With this method, one proves that if a statement holds for one particular natural number, and if assuming its truth for some number implies its truth for the next number, then the statement holds for all natural numbers.
The process involves two main steps:

• Base Case: Verify that the statement is true for the initial number, which often, but not always, is 0 or 1. For the Legendre polynomials case, the base cases have been shown to be true, as \(P_{0}(1) = 1\) and \(P_{1}(1) = 1\).
• Inductive Step: Assume that the statement holds for some arbitrary number \(k\), known as the induction hypothesis (e.g., \(P_{k}(1) = 1\)). Then show that based on this assumption, the statement must also hold for the next number, \(k+1\) (e.g., \(P_{k+1}(1) = 1\)).

In essence, mathematical induction bridges the step from one case to the next, creating a 'domino effect.' Once the base case and the inductive step are confirmed, the statement's validity cascades through all natural numbers.

Recurrence Relation

Recurrence relations are equations that express each element of a sequence as a function of the preceding elements. They are an integral part of combinatorics, numerical analysis, and discrete mathematics.
In the context of Legendre polynomials, the recurrence relation provides a way to express higher-degree polynomials based on lower-degree ones. The specific recurrence relation for Legendre polynomials is given by:
\[\begin{equation}(n+1)P_{n+1}(x) = (2n+1)xP_{n}(x) - nP_{n-1}(x),\end{equation}\]
Sub-headline: Proving \(P_{n}(1)=1\) Using the Recurrence Relation

• Start with the assumption that \(P_{k}(1) = 1\) and \(P_{k-1}(1) = 1\) for an arbitrary natural number \(k\).
• By substituting \(n = k\) and \(x = 1\) into the recurrence relation, we get a simplified equation which demonstrates that \(P_{k+1}(1) = 1\) also holds true.

In the context of this problem, the recurrence relation provides a recursive method to show that each subsequent Legendre polynomial, when evaluated at 1, is equal to 1, thus proving the original statement for all natural numbers by mathematical induction.