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Show that \(g(x, t)=g(-x,-t)\) for both Hermite and Legendre polynomials. Now expand \(g(x, t)\) and \(g(-x,-t)\) and compare the coefficients of \(t^{n}\) to obtain the parity relations for these polynomials: $$H_{n}(-x)=(-1)^{n} H_{n}(x) \text { and } P_{n}(-x)=(-1)^{n} P_{n}(x) .$$

Short Answer

Expert verified
Yes, both Hermite and Legendre polynomials satisfy the parity relations. The Hermite polynomials \(H_n(-x) = (-1)^n H_n(x)\) and the Legendre polynomials \(P_n(-x) = (-1)^n P_n(x)\), which implies that \(g(x,t) = g(-x,-t)\) is indeed true for both Hermite and Legendre polynomials.

Step by step solution

01

Evaluation of Hermite Polynomials

The Hermite polynomials \(H_n(x)\) are defined by the Rodrigues formula: \(H_n(x) = (-1)^n e^{x^2} \frac{d^n}{dx^n} e^{-x^2}\). If one replaces \(x\) with \(-x\), it is clear that this replacement leads to a change of sign in the exponent of the exponential function in the derivative. This change will introduce a factor of \((-1)^n\). Thus \(H_n(-x) = (-1)^n H_n(x)\), fulfilling the required identity.
02

Evaluation of Legendre Polynomials

The Legendre polynomials \(P_n(x)\) are defined by the Rodrigues formula: \(P_n(x) = \frac{1}{2^n n!} \frac{d^n}{dx^n} (x^2 - 1)^n\). By replacing \(x\) with \(-x\), it is easily seen that this does not change the polynomial, since \(x^2\) becomes \((-x)^2\), which is still \(x^2\). Thus, \(P_n(-x) = P_n(x)\) and because of the \(-1\) factor in the definition of \(P_n(x)\), this satisfies the identity \(P_n(-x) = (-1)^n P_n(x)\).
03

Conclusion

Therefore, both Hermite and Legendre polynomials satisfy the proposed parity relation, \(H_n(-x) = (-1)^n H_n(x)\) and \(P_n(-x) = (-1)^n P_n(x)\), which shows that \(g(x,t) = g(-x,-t)\) for both types of polynomial.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Hermite polynomials
Hermite polynomials, denoted as \( H_n(x) \), are a sequence of orthogonal polynomials important in probability, combinatorics, and mathematical physics. One key property of these polynomials is their parity relation, which describes their symmetric behavior around the origin. This property can be expressed as \( H_n(-x) = (-1)^n H_n(x) \).

Hermite polynomials are defined using the Rodrigues formula:
\[ H_n(x) = (-1)^n e^{x^2} \frac{d^n}{dx^n} e^{-x^2} \]
  • Orthogonality: Hermite polynomials are orthogonal with respect to the weight function \( e^{-x^2} \) on the infinite interval \((-fty, \infty)\).
  • Application: These polynomials are used in solutions to quantum mechanics problems, like the quantum harmonic oscillator.
When considering symmetry, the parity relation arises from replacing \( x \) with \( -x \). The exponential part of the Rodrigues formula adapts symmetrically due to the even powers in the exponent. This naturally introduces a factor of \( (-1)^n \) due to the properties of derivatives acting on exponential functions, explaining the parity relation fully.
Legendre polynomials
Legendre polynomials \( P_n(x) \) play a significant role in physics and engineering, particularly in solving problems related to spherical symmetry like gravitational and electric fields. Like Hermite polynomials, Legendre polynomials also demonstrate unique symmetry properties through their parity relations, expressed as \( P_n(-x) = (-1)^n P_n(x) \).

These are defined by the Rodrigues formula:
\[ P_n(x) = \frac{1}{2^n n!} \frac{d^n}{dx^n} (x^2 - 1)^n \]
  • Orthogonality: Legendre polynomials are orthogonal with respect to the weight function \( 1 \) over the interval \([-1, 1]\).
  • Application: They are used extensively in solving Legendre's differential equation and in expansion series like the Legendre series.
The parity property stems from the nature of the even powers involved in \( (x^2 - 1)^n \). When substituting \( -x \), the expression \( (-x)^2 \) remains \( x^2 \), maintaining symmetry, and thus integrating the factor \( (-1)^n \) through the derivative operations, resulting in their parity relationship.
Rodrigues formula
The Rodrigues formula is a powerful tool for generating specific types of orthogonal polynomials, such as Hermite and Legendre polynomials. What makes Rodrigues formula so elegant is its ability to define these polynomial sequences in a compact form. Each polynomial type can be systematically derived by performing derivatives under particular conditions.

For Hermite polynomials, the Rodrigues formula is:
\[ H_n(x) = (-1)^n e^{x^2} \frac{d^n}{dx^n} e^{-x^2} \]
For Legendre polynomials, the formula is:
\[ P_n(x) = \frac{1}{2^n n!} \frac{d^n}{dx^n} (x^2 - 1)^n \]
  • General Approach: The idea is to apply the \( n \)-th derivative to a base function, weighted appropriately.
  • Simplicity: Rodrigues formula turns otherwise complex polynomials into simple, manageable forms using fundamental calculus operations.
The utility of the Rodrigues formula lies in its capability to construct polynomials showing symmetry and orthogonality properties crucial in various mathematical and physical contexts. This systematic approach simplifies understanding and deriving the subtle characteristics of the polynomial sequences.

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Most popular questions from this chapter

Using integration by parts several times, show that $$\int_{-1}^{1}\left(1-x^{2}\right)^{n} d x=\frac{2^{m} n(n-1) \cdots(n-m+1)}{3 \cdot 5 \cdot 7 \cdots(2 m-1)} \int_{-1}^{1} x^{2 m}\left(1-x^{2}\right)^{n-m} d x .$$ Now show that $$\int_{-1}^{1}\left(1-x^{2}\right)^{n} d x=\frac{2 \Gamma\left(\frac{1}{2}\right) n !}{(2 n+1) \Gamma\left(n+\frac{1}{2}\right)}$$

Let \(L_{n} \equiv L_{n}^{0}\). Now differentiate both sides of $$g(x, t)=\frac{e^{-x t /(1-t)}}{1-t}=\sum_{0}^{\infty} t^{n} L_{n}(x)$$ with respect to \(x\) and compare powers of \(t\) to obtain \(L_{n}^{\prime}(0)=-n\) and \(L_{n}^{\prime \prime}(0)=\frac{1}{2} n(n-1)\). Hint: Differentiate \(1 /(1-t)=\sum_{n=0}^{\infty} t^{n}\) to get an ex- pression for \((1-t)^{-2}\).

Use the expansion of the generating function for Hermite polynomials to obtain $$\sum_{m, n=0}^{\infty} e^{-x^{2}} H_{m}(x) H_{n}(x) \frac{s^{m} t^{n}}{m ! n !}=e^{-x^{2}+2 x(s+t)-\left(s^{2}+t^{2}\right)}$$ Then integrate both sides over \(x\) and use the orthogonality of the Hermite polynomials to get $$\sum_{n=0}^{\infty} \frac{(s t)^{n}}{(n !)^{2}} \int_{-\infty}^{\infty} e^{-x^{2}} H_{n}^{2}(x) d x=\sqrt{\pi} e^{2 s t}$$ Deduce from this the normalization constant \(h_{n}\) of \(H_{n}(x)\).

Apply the general formalism of the recurrence relations given in the book to find the following two relations for Laguerre polynomials: (a) \(n L_{n}^{v}-(n+v) L_{n-1}^{v}-x \frac{d L_{n}^{v}}{d x}=0 .\) (b) \((n+1) L_{n+1}^{v}-(2 n+v+1-x) L_{n}^{v}+(n+v) L_{n-1}^{v}=0 .\)

Show that for Legendre polynomials, \(k_{n}^{(n)}=2^{n} \Gamma\left(n+\frac{1}{2}\right) /\left[n ! \Gamma\left(\frac{1}{2}\right)\right]\). Hint: Multiply and divide the expression given in the book by \(n !\); take a factor of 2 out of all terms in the numerator; the even terms yield a factor of \(n !\), and the odd terms give a gamma function.

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