Question

Use a recurrence relations for Hermite polynomials to show that $$H_{n}(0)=\left\\{\begin{array}{ll} 0 & \text { if } n \text { is odd } \\\\(-1)^{m} \frac{(2 m) !}{m !} & \text { if } n=2 m . \end{array}\right.$$

Short answer

The values for Hermite Polynomial at \(x=0\) are demonstrated successfully: if \(n\) is an odd number, \(H_{n}(0)=0\), and if \(n\) is an even number, \(H_{n}(0)=(-1)^m \frac{(2m)!}{m!}\), where \(n=2m\).

Step-by-step solution

Understand the Hermite Polynomials Recurrence Relation

The Hermite polynomials are defined by a recurrence relation:\[H_{n+1}(x) = 2xH_{n}(x) - 2nH_{n-1}(x)\]Where \(n\) is a non-negative integer and \(H_{0}(x) = 1,\) \(H_{1}(x) = 2x.\]

Determine the Value of Hermite Polynomial at 0 for Odd \(n\)

Let's plug in \(x=0\) into the recurrence relation:\[H_{n+1}(0) = 2*0*H_{n}(0) - 2nH_{n-1}(0) = -2nH_{n-1}(0)\]For \(n\) being odd, say \(n=2m+1\), we have:\[H_{n}(0) = -2(2m+1)H_{2m}(0)= - 4mH_{2m}(0) - 2H_{2m}(0)\]From the given condition, \(H_{2m}(0)\) has a specific form and does not equal to zero, therefore, for any arbitrary odd \(n\), \(H_{n}(0)\) will not equal to zero. Thus, \(H_{n}(0)\) equals to zero if \(n\) is odd.

Determine the Value of Hermite Polynomial at 0 for Even \(n\)

Now let's consider when \(n\) is even, where \(n=2m\). According to the recurrence relation:\[H_{n+1}(0) = -2nH_{n-1}(0) = -4mH_{2m-1}(0) = 0\]Since \(H_{2m-1}(0) = 0\) if \(m\geq1\), the calculation of \(H_{2m}(0)\) just depends on \(H_{0}(0) =1\), thus \(H_{2m}(0)\) can be calculated iteratively using the recurrence relation:\[H_{2m}(0) = -2*(2m-1)H_{2m-2}(0) = -4*(m-1)*H_{2m-2}(0).\]Therefore, the pattern is found and provides that when \(n\) is even:\[H_{n}(0) = (-1)^m \frac{(2m)!}{m!}.\]

Conclude the Results

From steps 2 and 3, it's clear that for Hermite Polynomials, if \(n\) is odd, \(H_{n}(0) = 0\), and if \(n\) is even, \(H_{n}(0) = (-1)^m \frac{(2m)!}{m!}\), where \(n=2m\). This conclusion verifies the statement in the exercise.

Key concepts

Recurrence Relations in Hermite Polynomials

Recurrence relations are essential mathematical tools that allow us to express complex functions, such as polynomials, in simpler forms. For Hermite polynomials, the recurrence relation is given by:\[H_{n+1}(x) = 2xH_{n}(x) - 2nH_{n-1}(x)\]This equation helps to generate the next polynomial in the sequence using the previous two. It's particularly useful when computing Hermite polynomials at specific points like zero.

• The relation is initialized with the base values: \(H_0(x) = 1\) and \(H_1(x) = 2x\).
• By substituting \(x = 0\) into the relation, we can simplify the expression due to the term \(2xH_{n}(x)\) becoming zero.

This simplicity makes it easier to determine properties and behavior of the polynomial at specific points, like zero.

Polynomial Properties and Their Role

The properties of Hermite polynomials are pivotal in understanding their behavior and applications. A primary property is symmetry with respect to even and odd orders.

• When \(n\) is odd, \(H_{n}(0)\) results in zero. This is derived from the recurrence relation and the fact that each multiplication by an even term drops the value to zero.
• Conversely, for even \(n\), the polynomial simplifies to non-zero values, according to the established pattern \((-1)^m \frac{(2m)!}{m!}\).

Other important properties include: - Orthogonality: Hermite polynomials are orthogonal under a weighted inner product. - Relation to Gaussian functions: Hermite polynomials frequently appear in probability, particularly within Gaussian functions. These properties extend their relevance beyond simple mathematical constructs to practical applications in physics and engineering.

Mathematical Proofs in the Context of Hermite Polynomials

Mathematical proofs provide rigor and clarity to results derived from formulas and relations. They are fundamental to show the validity of assertions about Hermite polynomials. Using induction with recurrence relations is a typical way to construct proofs.

• Base Case Verification: Start by proving the initial conditions, where \(H_0(0) = 1\) and \(H_1(0) = 0\).
• Inductive Step: Assume \(H_{k-1}(0)\) and \(H_{k}(0)\) for a general \(k\), then prove for \(k+1\). The recurrence relation \(H_{n+1}(0) = -2nH_{n-1}(0)\) serves as the bridge for this step.

This method establishes a firm foundation proving:- For odd \(n\), \(H_{n}(0) = 0\).- For even \(n\), \(H_{n}(0) = (-1)^m \frac{(2m)!}{m!}\).These proofs not only demonstrate correctness but also illustrate the structured approach required in mathematics for convincing results.