Small problem 90points

Plz there is a small pb, i have to proove it, 90points or more if you like,,, plllllllllllllllz

thanks ...

Plz see the attached file

In case someone doesn't have a PDF reader, here's a GIF version:

First case (for (0,0)----

if you calculate the sum function you ll get N/N---> 1


all other cases if you calculate the sum and get the limit you ll get 0 / infinity goes ---->0

good luck

Thanks,
But i have to give the proof of this ... I simulated it and it's ok, BUT I HAVE TO GIVE THE PROOF ... :-(

Fourier/Laplace tranform the first equation, defining it only over the domain of definition, you'll getF F(αn)=1/n ∑ δ(h(ωi-ωj),h²(Φi-Φj))

Do the inverse transform:
αn=1/n ∑ F-1{δ(h(ωi-ωj),h²(Φi-Φj))}

Take the limit as N -> ∞

an=lim 1/n ∑ F-1{δ(h(ωi-ωj),h²(Φi-Φj))}


When ωi=ωj and Φi=Φj
From the graphical representation, you'll see the limit means the number of Dirac Delta's in the 2D transform domain will increase to N peaks all on top of each other; From the transform you'll find that the area under all these peaks; As the Dirac Delta's area is 1 then the total area enclosed = N*1

From this the limit as N-->∞ = Lim (1/N)*N*1=1


Else:
As h(ωi-ωj) and h²(Φi-Φj) increase, you'll find that this throws them out of the range of the integration, i.e. they won't be covered by the inverse transform area of integration, causing a finite area to lie under the integral, but as Lim N-->∞, this'll decrease the area to zero, as any finite constant divided by ∞ = 0.

May I know where is this problem??

Hope this helps\]

dear elmolla, lp

thanks for your help,

For the first case if wi=wj and Φj=Φi it's clear that the sum is equal to N and the limite to 1

But, i really did not understand your explanation ::: what do you mean by '' they won't be covered by the inverse transform area of integration, causing a finite area to lie under the integral, ''

thanks again ...

Nab ...

The inverse transform, let ωi-ωj=x and Φi-Φj=y
F(α(x,y))=lim[N->∞]((1/N)∑δ(Ω-xh,Φ-yh²))

if x,y=0,0
F-1(F(α(x,y)))=lim[N->∞]((1/N)∑∫∫δ(Ω,Φ)dΩdΦ)=lim[N->∞]((1/N)∑(1))=lim[N->∞](N/N)=1

if(x,y)≠0
F-1(α(x,y))=lim[N->∞]((1/N)∑(from h=0 to N-1)∫∫δ(Ω-xh,Φ-yh²)dΩdΦ)

Substituting for N-1 by ∞ in the summation
=lim[N->∞]((1/N)∑(from h=0 to ∞)∫∫δ(Ω-xh,Φ-yh²)dΩdΦ)
=lim[N->∞]((1/N)∑(from h=0 to ∞)∫δ(Φ-yh²)dΦ)
As h²>h, then ∞²>∞ [This can thought off graphically; Let J be the bound of infinity, the integral from -∞ to ∞ is like the integral from -J to J taking the limit as J ->∞
so the integral is like
=lim[N,J->∞]((1/N)∑(from h=0 to J)∫(from -J to J)δ(Φ-yJ²)dΦ)

As J²>2J as J->∞, then the Dirac delta δ(Φ-yJ²) will be out of the bound set by ±J so the integral will vanish as sum of zeros, hence it'll be zeros.

I know it looks a bit awkward, but the proper definition of infinite integrals is using the limit, so I think its better to think of it this way

Hope its clear

Dear elmolla,

I read your proof, i would like to say something:

1- Thank you for your answer, i really appreciated ...

2- Your proof is magic , but unfortunatly, i don't agree with you (or i have to say, there is some points that i did not understand )

i- You wrote F(α(x,y))=lim[N->∞]((1/N)∑δ(Ω-xh,Φ-yh²)) , i think that it's not true, you calcul the transform fourrier of which variable ( x ,y or h ), (please look at the definition of the inverse transform), what does (Φ,Ω) mean , what do you mean by δ(Ω-xh,Φ-yh²)? it's the same definition that i wrote before?

ii- When you wrote lim[N->∞]((1/N)∑(from h=0 to ∞)∫∫δ(Ω-xh,Φ-yh²)dΩdΦ)
=lim[N->∞]((1/N)∑(from h=0 to ∞)∫δ(Φ-yh²)dΦ) , i did not agree with you too, because you can not say that ∞²>∞ (bcz ∞²=∞).

iii-=lim[N,J->∞]((1/N)∑(from h=0 to J)∫(from -J to J)δ(Φ-yJ²)dΦ) it's not true also, because ∫ is from -∞ to +∞ and you can not say that J->∞ and then J²->∞² and than you said ∫(from -J to J)δ(Φ-yJ²)dΦ, it does not make sens.

Perhapes i'm wrong, but even if your proof is wrong, i was impressed by your proof (really it's magic)

Anyway, thanks again, and even if I think that the proof isn't right, i can give you as much point as you want for you help,

thanks again

Hi Nab,

May be I'm not that good at maths . I believe there are a few misconceptions here.

1. I calculated the transform from the variable x to variable ? and from variable y to variable ?. i.e. the transform kernel is exp[-j(?x+?y)], not h.

This means that ? and ? are the new set of dimensions, or domain, after the transform.

2. I defined ?(t) as the Dirac Delta function, not the Kronecker Delta. You've defined it as Kronecker Delta, which is 1 when the argument of the function is zero, while I'm using it for the Dirac Delta which is defined as an impulse of infinite height and unit area at the position where the argument of the function is zero. I believe this is my mistake .

3. The Fourier transform of exp(jwot) is ?(w-wo), where ?(w) is the Dirac Delta as defined above.

4. Now with the integral from -? to +?; This is an improper integral. The proper definition of \[\int_{-\infty}^\infty \! f(x) \, dx\ = \lim_{T \to \infty} \int_{-T}^T \! f(x) \, dx \].


I'm not feeling easy with (?2 = ?) although I think its kinda right, this is actually an ill-defined quantity, we can't write it this way. I should have elaborated at this point better than what I've done.

Lets consider the limit as T increases intuitively, of the number of Delta's, whether Kronecker or Dirac, in the region [0,T] of the function ?(t-nT2) where n is a constant.
When T?1, T = T2 and this function is non zero in the region defined above, but as T goes to 10, we'll see the this function goes out of the boundary for all n>1 way more fast than the integral boundaries increase. With similar reasoning we can think of that
\[ \lim_{T \to \infty} \ \int_0^T \! \delta(t-nT^2) = 0\] for all n>1

I hope I've clarified things right now . Please let me know if I have any flaws or you see anything as not clear. I'm always open to criticism and discussion. No body is perfect .

Best regards

Dear elmolla,

thanks for your post and for your PM (i did not see you post untill today)

first ... do not worry, i think you are really good at maths

YEP, my mistake, i mix up the kroneker and the delta function:|, thanks for your clarification :idea:

BUT I have, again , some questions :

1. You said before that you calculated the transform from the variable x to variable Ω and from variable y to variable Φ. i.e. the transform kernel is exp[-j(Ωx+Φy)], not h.
Thus, this means that Ω and Φ are the new set of dimensions, or domain, after the transform. So why you wrote '' F(α(x,y))=lim[N->∞]((1/N)∑δ(Ω-xh,Φ-yh²)) ''? You have to take off the ''x'' and ''y''.
But it's not a problem, we will continue and suppose that F(α(x,y))(Ω,Φ)=lim[N->∞]((1/N)∑δ(Ω-h,Φ-h²))

2. When you calculated the invers transform you have to but exp[j(Ωx+Φy)], i mean instead
lim[N->∞]((1/N)∑(from h=0 to ∞)∫∫δ(Ω-xh,Φ-yh²)dΩdΦ)
you have to write
lim[N->∞]((1/N)∑(from h=0 to ∞)∫∫δ(Ω-h,Φ-h²) exp[j(Ωx+Φy)] dΩdΦ)

3. And my final question (or remark) :
You are right when you said that ∞²>∞ because ∞²/∞=∞

But, i think that (perheps i'm wrong) you are not alloweded to write this :
lim[N->∞]((1/N)∑(from h=0 to ∞)∫∫δ(Ω-xh,Φ-yh²)dΩdΦ)
=lim[N->∞]((1/N)∑(from h=0 to ∞)∫δ(Φ-yh²)dΦ)
Because even if
lim[N->∞] N²/N=∞ (it means that ∞²>∞)
we can not state (I think) ... that
lim[N->∞, J->∞] N²/J=∞ because J and N is not the same variable, for exemple you can take N=d and J=exp(d) ... then :

lim(N->∞) <=> lim(d->∞)
and
lim(J->∞) <=> lim(d->∞)

Thus,

lim[N->∞, J->∞] N²/J <=> lim[d->∞] N²/J

BUT
lim[d->∞] N²/J=0 Then even if [N->∞, J->∞] we can have lim[N->∞, J->∞] N²/J=0

I mean that you can say that ∞²>∞ if you have the same variable, otherwise it's not true ...

Thus, i think that you can not write :
lim[N->∞]((1/N)∑(from h=0 to ∞)∫∫δ(Ω-xh,Φ-yh²)dΩdΦ)
=lim[N->∞]((1/N)∑(from h=0 to ∞)∫δ(Φ-yh²)dΦ)
because we do not know how Φ, Ω and N will vary



Once again, I not really good on maths, but i was really surpise but your development (YOU ARE SMART...)...

Thanks I really appreciate your help...


Nab ...