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Show that $sum_{N-1}^{k=0} left | F_{k} right |^{2} = frac{2N^{2} + 3N + 1}{6}$ for DFT coefficients $F_{k}$

0

$begingroup$

Let $F_{k}$ be the DFT of the sequence $1, 2, ..., N$. Show that

$$sum_{N-1}^{k=0} left | F_{k} right |^{2} = frac{2N^{2} + 3N + 1}{6}$$

Any advice is appreciated.

fourier-transform

share|cite|improve this question

asked Mar 21 at 5:18

hegerberhegerber

11

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  One way you could do it would be to note a relationship between the norm of the DFT and of the original sequence (do you know the relationship?), and use the fact that $1^2 + 2^2 + 3^2 +cdots + N^2 = frac{N(N+1)(2N+1)}{6}$.
  $endgroup$
  – Minus One-Twelfth
  Mar 21 at 5:27
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MinusOne-Twelfth thanks for the sum of squares tip, that was very helpful! I noticed that $f_{n}$ would be equal to $[1:N]$. However, I have trouble isolating $e^{-ifrac{2pi}{N}nk}$ out. Do you have any suggestions for that?
  $endgroup$
  – hegerber
  Mar 21 at 6:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You probably don't need to worry about that, you can just use the result here: math.stackexchange.com/questions/3156356/….
  $endgroup$
  – Minus One-Twelfth
  Mar 21 at 6:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MinusOne-Twelfth I wish I could upvote you. Thanks so much!
  $endgroup$
  – hegerber
  Mar 21 at 23:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're welcome! $ddot{smile}$
  $endgroup$
  – Minus One-Twelfth
  Mar 21 at 23:52
  

  
  
  
  

add a comment | 

0

$begingroup$

Let $F_{k}$ be the DFT of the sequence $1, 2, ..., N$. Show that

$$sum_{N-1}^{k=0} left | F_{k} right |^{2} = frac{2N^{2} + 3N + 1}{6}$$

Any advice is appreciated.

fourier-transform

share|cite|improve this question

asked Mar 21 at 5:18

hegerberhegerber

11

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  One way you could do it would be to note a relationship between the norm of the DFT and of the original sequence (do you know the relationship?), and use the fact that $1^2 + 2^2 + 3^2 +cdots + N^2 = frac{N(N+1)(2N+1)}{6}$.
  $endgroup$
  – Minus One-Twelfth
  Mar 21 at 5:27
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MinusOne-Twelfth thanks for the sum of squares tip, that was very helpful! I noticed that $f_{n}$ would be equal to $[1:N]$. However, I have trouble isolating $e^{-ifrac{2pi}{N}nk}$ out. Do you have any suggestions for that?
  $endgroup$
  – hegerber
  Mar 21 at 6:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You probably don't need to worry about that, you can just use the result here: math.stackexchange.com/questions/3156356/….
  $endgroup$
  – Minus One-Twelfth
  Mar 21 at 6:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MinusOne-Twelfth I wish I could upvote you. Thanks so much!
  $endgroup$
  – hegerber
  Mar 21 at 23:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're welcome! $ddot{smile}$
  $endgroup$
  – Minus One-Twelfth
  Mar 21 at 23:52
  

  
  
  
  

add a comment | 

$begingroup$

Let $F_{k}$ be the DFT of the sequence $1, 2, ..., N$. Show that

$$sum_{N-1}^{k=0} left | F_{k} right |^{2} = frac{2N^{2} + 3N + 1}{6}$$

Any advice is appreciated.

fourier-transform

share|cite|improve this question

asked Mar 21 at 5:18

hegerberhegerber

11

$endgroup$

share|cite|improve this question

asked Mar 21 at 5:18

hegerberhegerber

11

share|cite|improve this question

asked Mar 21 at 5:18

hegerberhegerber

11

asked Mar 21 at 5:18

hegerberhegerber

11

asked Mar 21 at 5:18

hegerberhegerber

11