Series identity

  • Thread starter mhill
  • Start date
  • #1
188
1

Main Question or Discussion Point

for every sequence of numbers a_n E_n is this identity correct ?

[tex] \sum_{n= -\infty}^{\infty}a_n e^{2\pi i E_{n}}= \sum_{n= -\infty}^{\infty}a_n \delta (x-E_{n}) [/tex]
 

Answers and Replies

  • #2
matt grime
Science Advisor
Homework Helper
9,395
3
Clearly it isn't (and I assume that you mean to have an x in the exponent on the LHS).
 
  • #3
matt grime
Science Advisor
Homework Helper
9,395
3
Perhaps you meant to integrate the RHS over the real line?
 
  • #4
Defennder
Homework Helper
2,591
5
Last edited by a moderator:
  • #5
matt grime
Science Advisor
Homework Helper
9,395
3
I suspect I know who this poster is, and the same advice I've given repeatedly still applies: have you tried it for any examples? Eg a_i=0 for i=/=0 and a_0=1, E_0=0, then the LHS is 1 and the RHS is d(x) (d for delta)...
 
  • #6
rbj
2,226
7
This is true

[tex] \sum_{n=-\infty}^{+\infty} e^{i 2 \pi n x} = \sum_{k=-\infty}^{+\infty} \delta(x-k) [/tex]

but to generalize it with arbitrary coefficients (that are placed on both sides) is not a true equality.
 
  • #7
188
1
rbj and matt were right only this

[tex]
\sum_{n=-\infty}^{+\infty} e^{i 2 \pi n x} = \sum_{k=-\infty}^{+\infty} \delta(x-k)
[/tex] (1)

is correct , however my question is if using Fourier analysis we could generalized to an identity

[tex]
\sum_{n=-\infty}^{+\infty}a_{n} e^{i 2 \pi n x} = \sum_{k=-\infty}^{+\infty} b_{n}\delta(x-k)
[/tex]

where the a_n and b_n are related by some way , this is interesting regarding an article of Functional equation for Dirichlet series, using (1) the author was able to proof the functional equation for Riemann Zeta, my idea was to develop a functional equation for almost every dirichlet series to see where they have the 'poles'
 
  • #8
188
1
If we have in the general case

[tex] \sum_{n=-\infty}^{+\infty}b_{n} e^{i 2 \pi n x} = D A(x) [/tex]

Where A(x) is the partial sum of a_n and D is the derivative operator , in case A(x)=[x] we recover usual delta identity , then i believe we can calculate b_n by the Fourier integral

[tex] b_n = \int_{0}^{1} dx DA(x) e^{-2i\pi x} [/tex]
 

Related Threads for: Series identity

  • Last Post
Replies
2
Views
980
  • Last Post
Replies
7
Views
3K
  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
5
Views
552
  • Last Post
Replies
20
Views
3K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
2
Views
10K
Top