# Understanding the Recurrence Relation of H_n(x) and Its Properties

• TiberiusK
In summary, the polynomials are defined as follows: first, set H_0(x) = 1 and H_1(x) = -x; then, for n >= 2, H_n is defined by the recurrence equation H_n(x) = -xH_(n-1)(x) - (n - 1)H_(n-2)(x). The student is having trouble proving that the functions is odd for n odd and even for n even, and also needs to calculate H_4(x) and H_5(x). Additionally, they need to show that H_n is an even function when n is even, and an odd function when n is odd
TiberiusK

## Homework Statement

The polynomials are defined as follows H_n(x) for n = 0; 1; : : : as follows: fi rst, set H_0(x) =1 and H_1(x) = -x; then, for n >= 2, H_n is defined by the recurrence
H_n(x) = -xH_(n-1)(x) - (n - 1)H_(n-2)(x): (1)
I have to use (1) to verify that H_2(x) = x^2 -1 and H_3(x) = 3x-x^3, and calculate H_4(x)and H_5(x).
I also have to show that that H_n is an even function when n is even, and that it is an odd function when n is odd (maybe using (1) and induction on n).
Also that also (maybe I should still use induction and (1)) that
H_2k(0) = (-1)^k(2k -1)(2k -3) ...1:
What is the value of H_n(0) when n is odd?

## Homework Equations

All above...I'm having trouble proving that the functions is odd for n odd and even for n even.Also I need to know if my calculations are correct

## The Attempt at a Solution

I proved that H_2(x) = x^2 -1 and H_3(x) = 3x-x^3 by substituting in the formula.
Also H_4(x)==6x^2 +x^4-3 and H_5(x)=-6x^3-x^5-3x
For n=odd=2k+1
Base case H_1(-x)=-x=>it is true.We pressume that n is true and prove for n+1.
H_(n+1)(-x) = -(-x)H_(n)(-x) - (n)H_(n-1)(-x)
=we know that H_n(-x) is odd so xH_(n)(-x)...and therefore (n)H_(n-1)(x) is also odd...here I need help?

For n=even=2k
ase case H_0(-x)=1=>it is true.We pressume that n is true and prove for n+1.
H_(n+1)(-x) = -(-x)H_(n)(-x) - (n)H_(n-1)(-x)...again same thing
And of course 'la piece de resistance'
What is the value of H_n(0) when n is odd?

Some things to figure out: is the product of two even functions even or odd? What about the product of two odd functions? What about the product of an even and an odd function? What about the sum of two even functions, or the sum of two odd functions?

An odd function satisfies f(-x) = -f(x). So what is the value of any odd function at x = 0?

I see,I'm just going to use these properties,The sum of two even functions is even,of an even and odd function is neither even nor odd, unless one of the functions is identically zero, of two odd functions is odd, and any constant multiple of an odd function is odd.The product of two even functions is an even function,of two odd functions is an even function,of an even function and an odd function is an odd function.

For example...when n=odd=2k+1...
-(-x)H_(n)(-x) is odd becase the product of 2 odd functions is odd
(n)H_(n-1)(-x)-a costant multiplied with an odd function is also odd
I can write this like -(-x)H_(n)(-x) - (n)H_(n-1)(-x)=>-(-x)H_(n)(-x)+( - (n)H_(n-1)(-x)) which is odd...is this ok?(if so the same will apply for n=even)

TiberiusK said:

## Homework Statement

The polynomials are defined as follows H_n(x) for n = 0; 1; : : : as follows: fi rst, set H_0(x) =1 and H_1(x) = -x; then, for n >= 2, H_n is defined by the recurrence
H_n(x) = -xH_(n-1)(x) - (n - 1)H_(n-2)(x): (1)
I have to use (1) to verify that H_2(x) = x^2 -1 and H_3(x) = 3x-x^3, and calculate H_4(x)and H_5(x).
I also have to show that that H_n is an even function when n is even, and that it is an odd function when n is odd (maybe using (1) and induction on n).
Also that also (maybe I should still use induction and (1)) that
H_2k(0) = (-1)^k(2k -1)(2k -3) ...1:
What is the value of H_n(0) when n is odd?

## Homework Equations

All above...I'm having trouble proving that the functions is odd for n odd and even for n even.Also I need to know if my calculations are correct

## The Attempt at a Solution

I proved that H_2(x) = x^2 -1 and H_3(x) = 3x-x^3 by substituting in the formula.
Also H_4(x)==6x^2 +x^4-3 and H_5(x)=-6x^3-x^5-3x
For n=odd=2k+1
Base case H_1(-x)=-x=>it is true.We pressume that n is true and prove for n+1.
H_(n+1)(-x) = -(-x)H_(n)(-x) - (n)H_(n-1)(-x)
=we know that H_n(-x) is odd so xH_(n)(-x)...and therefore (n)H_(n-1)(x) is also odd...here I need help?
Your mistake is trying to prove the even and 0dd cases separately.
Prove, instead, "If n is even H_n is even and if n is odd, H_n is odd" as a single induction:
For the "base case" prove both H_1 is odd and H_2 is even.

For the induction step, break into n even and n odd cases.

For n=even=2k
ase case H_0(-x)=1=>it is true.We pressume that n is true and prove for n+1.
H_(n+1)(-x) = -(-x)H_(n)(-x) - (n)H_(n-1)(-x)...again same thing
And of course 'la piece de resistance'
What is the value of H_n(0) when n is odd?

ok H_1 (x) is odd and H_2(x) is even.
So If n => even H_n is even and if n is odd =>H_n is odd
-(-x)H_(n)(-x) If n => even H_n is even the this is even if nis odd=>the product is odd
(n)H_(n-1)(-x) .If n => even H_(n-1) is odd and multiplied by an even n = even product(same goes for odd)...
-(-x)H_(n)(-x) - (n)H_(n-1)(-x)=>-(-x)H_(n)(-x)+( - (n)H_(n-1)(-x)) which is odd or even depending or n.
Also "An odd function satisfies f(-x) = -f(x). So what is the value of any odd function at x = 0(What is the value of H_n(0) when n is odd)"doesn't that depend on the function?

If f(x) is odd, then f(-x)=(-f(x)). Put x=0. No, f(0) doesn't depend on f when f is odd.

so f will be even

TiberiusK said:
so f will be even

I thought you were asking about the question what is H_n(0) when n is odd. Which makes H_n an odd function. Which makes H_n(0)=0. What are you asking?

You are right...I wasn't thinking straight...thanks for the reply

## What is a polynomial?

A polynomial is a mathematical expression made up of variables and coefficients, combined using the operations of addition, subtraction, and multiplication. It is characterized by having one or more terms that are raised to non-negative integer powers.

## What are the basic operations involved in polynomial calculations?

The basic operations involved in polynomial calculations are addition, subtraction, and multiplication. Division can also be performed, but it is considered more complex and is often done using other methods.

## How do you identify the degree of a polynomial?

The degree of a polynomial is determined by looking at the term with the highest power. For example, in the polynomial 4x3 + 2x2 + 5x + 1, the term with the highest power is 4x3, making the degree of the polynomial 3.

## What is the FOIL method and how is it used in polynomial calculations?

The FOIL method is a technique used to multiply two binomials. It stands for First, Outer, Inner, Last, and is used to help remember the order in which the terms should be multiplied. For example, to multiply (x + 2)(x + 3), we would multiply the First terms (x * x), then the Outer terms (x * 3), then the Inner terms (2 * x), and finally the Last terms (2 * 3). This results in the polynomial x2 + 3x + 2x + 6, which can then be simplified to x2 + 5x + 6.

## Can polynomials be graphed?

Yes, polynomials can be graphed. The graph of a polynomial is a smooth curve that can have various shapes and behaviors depending on the degree and coefficients of the polynomial. The degree of the polynomial determines the number of possible turns or bends in the graph, while the coefficients affect the steepness and direction of the curve.

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