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

In summary: H_n(x+ix)|<=|H_n(x)|+|H_n(ix)|And since |H_n(x+ix)| = |H_n(ix)|, we can rewrite the inequality as:|H_n(ix)|<=|H_n(x)|+|H_n(ix)|This further proves that |H_n(x)|<=|H_n(ix)|.In summary, by using the properties of the Hermite polynomials and the triangle inequality, we can prove that |H_n(x)|<=|H_n(ix)| where H_n(x) is the Hermite polynomial.
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Suvadip
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How to prove that\(\displaystyle |H_n(x)|<=|H_n(ix)| \)where \(\displaystyle H_n(x)\) is the Hermite polynomial?
 
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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
 

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

What is the Modulus of Hermite polynomial?

The Modulus of Hermite polynomial is a mathematical concept used in the field of statistics and probability, specifically in the study of random variables. It is a measure of the spread or variability of a Hermite polynomial, which is a type of mathematical function used to model the probability distribution of a random variable.

How is the Modulus of Hermite polynomial calculated?

The Modulus of Hermite polynomial is calculated by taking the square root of the second moment of the Hermite polynomial. This can be expressed mathematically as √(∫ (x-μ)^2 f(x) dx), where μ is the mean of the random variable and f(x) is the probability density function.

What is the significance of the Modulus of Hermite polynomial?

The Modulus of Hermite polynomial is an important measure in statistics because it allows us to compare the variability of different Hermite polynomials. It is particularly useful in determining the shape of a probability distribution and making inferences about the likelihood of certain outcomes.

How does the Modulus of Hermite polynomial relate to other measures of variability?

The Modulus of Hermite polynomial is related to other measures of variability such as the variance and standard deviation. In fact, the Modulus of Hermite polynomial is essentially the square root of the variance, making it a more interpretable measure of spread and variability in the data.

Can the Modulus of Hermite polynomial be negative?

No, the Modulus of Hermite polynomial cannot be negative. It is always a positive value, as it represents the square root of the second moment of the Hermite polynomial which cannot be negative. Additionally, the Modulus of Hermite polynomial is always equal to or greater than zero, with a value of zero indicating no variability in the data.

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