
#1
Jun1108, 08:01 AM

P: 193

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 (xE_{n}) [/tex] 



#2
Jun1108, 08:05 AM

Sci Advisor
HW Helper
P: 9,398

Clearly it isn't (and I assume that you mean to have an x in the exponent on the LHS).




#3
Jun1108, 09:15 AM

Sci Advisor
HW Helper
P: 9,398

Perhaps you meant to integrate the RHS over the real line?




#4
Jun1108, 10:37 AM

HW Helper
P: 2,618

series identity 



#5
Jun1108, 01:10 PM

Sci Advisor
HW Helper
P: 9,398

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
Jun1108, 03:11 PM

P: 2,265

This is true
[tex] \sum_{n=\infty}^{+\infty} e^{i 2 \pi n x} = \sum_{k=\infty}^{+\infty} \delta(xk) [/tex] but to generalize it with arbitrary coefficients (that are placed on both sides) is not a true equality. 



#7
Jun1208, 02:49 AM

P: 193

rbj and matt were right only this
[tex] \sum_{n=\infty}^{+\infty} e^{i 2 \pi n x} = \sum_{k=\infty}^{+\infty} \delta(xk) [/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(xk) [/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
Jun1208, 05:13 AM

P: 193

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] 


Register to reply 
Related Discussions  
Fourier series: Parseval's identity... HELP!  Calculus & Beyond Homework  11  
New recursive series identity  Linear & Abstract Algebra  14  
[Identity relations] Need help at some odd identity relation problem  Calculus & Beyond Homework  1  
Using an identity to find the sum to n terms of a series  Calculus & Beyond Homework  7 