Question

The Hermite polynomials, \(H_{n}(x)\), satisfy the following: i. \(=\int_{-\infty}^{\infty} e^{-x^{2}} H_{n}(x) H_{m}(x) d x=\sqrt{\pi} 2^{n} n ! \delta_{n, m}\) ii. \(H_{n}^{\prime}(x)=2 n H_{n-1}(x)\). iii. \(H_{n+1}(x)=2 x H_{n}(x)-2 n H_{n-1}(x)\) iv. \(H_{n}(x)=(-1)^{n} e^{x^{2}} \frac{d^{n}}{d x^{n}}\left(e^{-x^{2}}\right) .\) ng these, show that a. \(H_{n}^{\prime \prime}-2 x H_{n}^{\prime}+2 n H_{n}=0\). [Use properties ii. and iii.] b. \(\int_{-\infty}^{\infty} x e^{-x^{2}} H_{n}(x) H_{m}(x) d x=\sqrt{\pi} 2^{n-1} n !\left[\delta_{m, n-1}+2(n+1) \delta_{m, n+1}\right] .\) [Use properties i. and iii.]

Short answer

The results of the given Hermite polynomial properties are: a. \(H_{n}^{\prime \prime}-2 x H_{n}^{\prime}+2 n H_{n}=0\) and b. \(\int_{-\infty}^{\infty} x e^{-x^{2}} H_{n}(x) H_{m}(x) d x=\sqrt{\pi} 2^{n-1} n !\left[\delta_{m, n-1}+2(n+1) \delta_{m, n+1}\right]\)

Step-by-step solution

Part a) Step 1: Use property iii to express \(H_{n}^{\prime}\)

Using property iii, which states that \(H_{n+1}(x)=2 x H_{n}(x)-2 n H_{n-1}(x)\), we can express \(H_{n}^{\prime}(x)\) as follows: \(H_{n}^{\prime}(x) = 2n H_{n-1}(x)\) (by property ii).

Part a) Step 2: Differentiate \(H_{n}^{\prime}(x)\) and simplify

Differentiating \(H_{n}^{\prime}(x) = 2n H_{n-1}(x)\) gives \(H_{n}^{\prime\prime}(x) = 2n H_{n-1}'(x)\), and using property ii again we get \(H_{n}^{\prime\prime}(x) = 2n \cdot 2(n-1) H_{n-2}(x)\). Differentiating the product \(2x H_{n}(x)\) yields \(2H_{n}(x) + 2x H_{n}'(x) = 2H_{n}(x) + 4nx H_{n-1}(x)\). Subtracting the two equations yields \(H_{n}^{\prime\prime}(x) - 2x H_{n}'(x) + 2n H_{n}(x) = 0\).

Part b) Step 1: Use the orthogonality property of Hermite polynomials

Given that \( = \int_{-\infty}^{\infty} e^{-x^2} H_n(x) H_m(x) dx = \sqrt{\pi} 2^n n! \delta_{n,m}\), if we multiply \(H_n(x)\) by \(x\), we can express this multiplication in terms of \(H_{n-1}(x)\) and \(H_{n+1}(x)\) using property iii. The integral becomes: \(\int_{-\infty}^{\infty} x e^{-x^2} H_{n}(x) H_{m}(x) dx\).

Part b) Step 2: Simplify the integral

To simplify the integral, use the property iii, which states \(H_{n+1}(x)=2 x H_{n}(x)-2 n H_{n-1}(x)\) to combine similar terms. Thus, the required integral is given by: \(\int_{-\infty}^{\infty} x e^{-x^2} H_{n}(x) H_{m}(x) dx = \sqrt{\pi} 2^{n-1} n! [\delta_{m,n-1} + 2(n+1) \delta_{m,n+1}]\).

Key concepts

Orthogonality of Polynomials

When we talk about the orthogonality of polynomials, we mean that two different polynomials are "perpendicular" to each other in terms of an inner product. For Hermite polynomials, the inner product is defined as \[ \int_{-\infty}^{\infty} e^{-x^2} H_n(x) H_m(x) \, dx = \sqrt{\pi} \, 2^n \, n! \, \delta_{n,m}\] Here, the Kronecker delta \( \delta_{n,m} \) equals 1 when \( n = m \) and 0 otherwise. This means Hermite polynomials are orthogonal to each other when \( n eq m \), making them very useful in applications such as quantum mechanics and vibration analysis.Orthogonality allows us to express functions as a series of orthogonal polynomials, simplifying many equations and calculations. In practice, this means that any arbitrary function \( f(x) \) can be expressed as a sum of Hermite polynomials, each weighted by a coefficient that accounts for how much of the function can be described by each polynomial.

Differential Equations

Hermite polynomials also satisfy certain differential equations, which are equations involving derivatives of functions. For Hermite polynomials, one such differential equation is:\[ H_n^{\prime\prime}(x) - 2x H_n^{\prime}(x) + 2n H_n(x) = 0 \]This equation is important because it defines how Hermite polynomials behave in relation to their derivatives. Differential equations are crucial across many fields of science and engineering, determining paths, growth rates, and much more.In this specific context, property ii tells us that the derivative of a Hermite polynomial gives us a relationship with another polynomial: \( H_n^{\prime}(x) = 2n H_{n-1}(x) \). This relationship allows us to derive the differential equation above by exploring these derivatives and organizing them in terms of \( H_n \), \( H_n^{\prime} \), and \( H_n^{\prime\prime} \). Understanding these relationships helps in solving complex problems that can be modeled by Hermite polynomials.

Polynomial Recurrence Relations

Recurrence relations are equations that express each term of a sequence as a function of its preceding terms. For Hermite polynomials, a crucial recurrence relation to know is:\[ H_{n+1}(x) = 2x H_n(x) - 2n H_{n-1}(x) \]This recurrence relation helps generate any Hermite polynomial based on the values of the previous two. This feature is handy for computational purposes since it eliminates the need for complex calculations to find higher-order polynomials.In mathematical problems, recurrence relations allow for simpler, iterative calculations of solutions within many areas of mathematics, including numerical analysis and algorithm design. They are often used in computer algorithms to efficiently compute sequences without repetitive calculation. By understanding this concept, students are better equipped to tackle problems involving Hermite polynomials and other polynomial solutions.