Calculating DFT for specific polynomial

In summary, to find DFT2n(P(x^2)), we first need to find DFT2n(P(x)) using the given values of v_1, ..., v_n and then use the DFT2n formula to find DFT2n(P(x^2)).
  • #1
aka mor
1
0
Hello everyone,

This seems like a simple problem but I get the impression that I'm missing something.

1. Homework Statement

Given the values ## v_1,v_2,...,v_n ## such that DFTn ## (P(x)) = (v_1, v_2, \ldots, v_n) ## and ##deg(P(x)) < n##, find DFT2n## P(x^2)##

Homework Equations

The Attempt at a Solution



My attempt:
Let [itex]P(x) = \sum\limits_{i=0}^{k} a_ix^i[/itex] .
First, expand ##P(x^2)## :
[tex]P(x^2) = a_0 + a_1x^2 + a_2x^4 + \ldots + a_kx^{2k}[/tex]
Let's denote by ##W^j_n## the j'th root of unity of order n.
We have:
[tex]W^j_{2n} = e^{\frac{2\pi i j}{2n}} = e^{\frac{\pi i j}{n }}[/tex]
Using this information, let's expand a term p of ##P(x^2)## on ##W^j_{2n}##:
[tex]a_p(x^2) = a_p(e^\frac{\pi ij}{n})^{2p} = a_p(e^{2\pi i}) ^\frac{jp}{n} = a_p[/tex] (since ##e^{2\pi i} = 1## ).
So, when evaluating any term of the compound polynomial on some unity root of order 2n we get ##\sum\limits_{i=0} ^k ak## and therefore the final answer is:
[tex](\sum\limits_{i=0} ^k ak, \sum\limits_{i=0} ^k ak,\ldots, \sum\limits_{i=0} ^k ak)[/tex].

This doesn't seem right, I didn't even use the information about ## (v_1, \ldots, v_n) ## (answer should be in terms of those?).

Any help would be appreciated.
 
Last edited by a moderator:
Physics news on Phys.org
  • #2


Hello,

Your approach seems to be on the right track, but I believe you are missing a key step in your solution. In order to find DFT2n(P(x^2)), we need to first find the DFT2n of P(x). This can be done by using the DFT2n formula, which is:

DFT2n(P(x)) = (v_1, v_2, ..., v_n, v_{n+1}, ..., v_{2n})

Where v_{n+1}, ..., v_{2n} are the conjugates of v_1, ..., v_n respectively. So, in order to find DFT2n(P(x^2)), we need to first find DFT2n(P(x)) and then use the same formula to find DFT2n(P(x^2)). Your approach is correct in finding the DFT2n of P(x^2), but we need to first find DFT2n(P(x)) using the given values of v_1, ..., v_n. I hope this helps. Let me know if you have any further questions.
 

1. What is DFT in relation to polynomials?

DFT stands for Discrete Fourier Transform and is a mathematical tool used to convert a discrete signal or function into its equivalent representation in the frequency domain. In the context of polynomials, DFT can be used to calculate the coefficients or roots of a specific polynomial function.

2. How is DFT calculated for a specific polynomial?

To calculate DFT for a specific polynomial, we first need to represent the polynomial function as a discrete sequence of values. This sequence is then multiplied by a complex exponential function and summed up. The result of this summation gives us the coefficients or roots of the polynomial function in the frequency domain.

3. What is the significance of calculating DFT for a polynomial?

Calculating DFT for a polynomial can help us understand the behavior and properties of the polynomial function in the frequency domain. It can also be used to solve problems related to signal processing, data compression, and image processing.

4. Can DFT be applied to all types of polynomials?

Yes, DFT can be applied to all types of polynomials, including real and complex polynomials. However, the polynomial must be represented as a discrete sequence of values for DFT to be applied.

5. Are there any limitations or challenges in calculating DFT for specific polynomials?

One limitation of calculating DFT for specific polynomials is that it can be computationally intensive, especially for large polynomials. This can make the process time-consuming and may require powerful computing resources. Additionally, DFT may not give accurate results for polynomials with very high or very low frequency components.

Similar threads

  • Engineering and Comp Sci Homework Help
Replies
3
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
255
  • Calculus and Beyond Homework Help
Replies
1
Views
212
  • Calculus and Beyond Homework Help
Replies
4
Views
305
  • MATLAB, Maple, Mathematica, LaTeX
Replies
1
Views
1K
  • Linear and Abstract Algebra
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
15
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
1
Views
950
  • Advanced Physics Homework Help
Replies
1
Views
685
Back
Top