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Polynomial calculations

  • Thread starter TiberiusK
  • Start date
  • #1
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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?
 

Answers and Replies

  • #2
1,006
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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?
 
  • #3
24
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.
 
  • #4
24
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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)
 
  • #5
HallsofIvy
Science Advisor
Homework Helper
41,833
955


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?
 
  • #6
24
0


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?
 
  • #7
Dick
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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.
 
  • #8
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so f will be even
 
  • #9
Dick
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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?
 
  • #10
24
0


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

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