Is |H_n(x)| Always Less Than or Equal to |H_n(ix)| for Hermite Polynomials?

[Suvadip]

How to prove that\(\displaystyle |H_n(x)|<=|H_n(ix)| \)where \(\displaystyle H_n(x)\) is the Hermite polynomial?

 

[jvicens]

To prove that |H_n(x)|<=|H_n(ix)| where H_n(x) is the Hermite polynomial, we can use the properties of the Hermite polynomials and the triangle inequality.

First, let's recall the definition of the Hermite polynomials. The Hermite polynomials are a family of orthogonal polynomials that are solutions to the Hermite differential equation. They are defined as:

H_n(x) = (-1)^n e^(x^2/2) (d^n/dx^n) e^(-x^2/2)

Now, let's consider the absolute value of H_n(x):

|H_n(x)| = |-1|^n |e^(x^2/2)| |(d^n/dx^n) e^(-x^2/2)|

Since n is a positive integer, |-1|^n = 1. Also, the exponential functions do not change the magnitude of a number. Therefore, we can rewrite the absolute value of H_n(x) as:

|H_n(x)| = |(d^n/dx^n) e^(-x^2/2)|

Next, let's consider the absolute value of H_n(ix):

|H_n(ix)| = |(d^n/dx^n) e^(-x^2/2)| |e^(i^2x^2/2)|

Again, since n is a positive integer, |i^2|^n = 1. Also, the exponential functions do not change the magnitude of a number. Therefore, we can rewrite the absolute value of H_n(ix) as:

|H_n(ix)| = |(d^n/dx^n) e^(-x^2/2)|

Since both |H_n(x)| and |H_n(ix)| have the same magnitude, we can conclude that |H_n(x)|<=|H_n(ix)|.

Furthermore, we can use the triangle inequality to strengthen our proof. The triangle inequality states that for any two complex numbers, z and w, |z+w|<=|z|+|w|. In our case, we can consider z = H_n(x) and w = H_n(ix). Therefore, we can write:

|H_n(x)+H_n(ix)|<=|H_n(x)|+|H_n(ix)|

But, we know that H_n(x)+H_n(ix) = H_n(x+ix). Therefore, we