Representation for Dirac delta?

[Arnau Lapera]

Hello,
I'm very interested in knowing whether

n^{3/2} e^{-i x sqrt{n} }

could be -in some sense- a series representation, as n->\infinity, for
Dirac's \delta(x) ? If useful at all, I'm happy to add a 'small'
imaginary part to x (ie, x -> x + i \epsilon).

[For example, sin(n x)/(x \pi) -> \delta(x). Proof: Let f(x) be a test
function, then
\lim_{n\to \infty} \int_{-\infty}^{+\infty} dx f(x) sin(n x)/(x \pi) =
(change of variable y=n x)
\lim_{n\to \infty} \int_{-\infty}^{+\infty} dy sin(y)/(y \pi) f(y/n)=
f(0) \int_{-\infty}^{+\infty} dy sin(y)/(y \pi) = f(0)
I've tried to apply similarly to my case, without much success (I
can't disprove it either though!)]

Many thanks in advance!

[Kallikanzarid]

I'm noob in math, but... :)

Let's substitute sqrt{n} = m.
Then your function becomes m^3 e^{-i x m}.
My naive attempt to apply this by taking an integral of m^3 f(x) e^{-i
x m} fails, because the result is obviously F[f](m), which IIRC not
necessarily converges to f(0) ;) A a matter of fact, in many cases it
converges to zero: http://en.wikipedia.org/wiki/Riemann%E2%80%93Lebesgue_lemma

[Kallikanzarid]

Sorry, it's F[m^3 f](m) - doesn't change much though.

[Arnau Lapera]

hi Kallikanzarid, thanks for your reply. It's precisely the Riemann-
Lebesgue theorem that gives me some hope that my series tends to
Dirac's delta. That is, \int_{-\infty}^{\infty} dx f(x) e^{-i x m}
indeed goes to zero as m->infty, and if it does so like f(0)/m^3 (the
contribution to the integral is the highest near x=0), that would
cancel out the m^3 in front to yield f(0), as I hope. Naively, though,
the integral goes like 1/m instead, but I've no rigorous proof for
this....?

[W^3]

In article
<164cb9ca-f5c0-4585...@s20g2000yqd.googlegroups.com>,
Kallikanzarid <lex....@gmail.com> wrote:

> On 1 дек, 21:22, Arnau Lapera <ar.lap...@gmail.com> wrote:
> > Hello,
> > I'm very interested in knowing whether
> >
> > n^{3/2} e^{-i x sqrt{n} }
> >
> > could be -in some sense- a series representation, as n->\infinity, for
> > Dirac's \delta(x) ?  If useful at all, I'm happy to add a 'small'
> > imaginary part to x (ie, x -> x + i \epsilon).
> >
> > [For example, sin(n x)/(x \pi) -> \delta(x). Proof: Let f(x) be a test
> > function, then
> > \lim_{n\to \infty} \int_{-\infty}^{+\infty} dx f(x) sin(n x)/(x \pi) =
> > (change of variable y=n x)
> > \lim_{n\to \infty} \int_{-\infty}^{+\infty} dy sin(y)/(y \pi) f(y/n)=
> > f(0) \int_{-\infty}^{+\infty} dy sin(y)/(y \pi) = f(0)
> > I've tried to apply similarly to my case, without much success (I
> > can't disprove it either though!)]
> >
> > Many thanks in advance!
>
> I'm noob in math, but... :)
>
> Let's substitute sqrt{n} = m.
> Then your function becomes m^3 e^{-i x m}.
> My naive attempt to apply this by taking an integral of m^3 f(x) e^{-i
> x m} fails, because the result is obviously F[f](m),

Acually m^3*F[f](m)

> which IIRC not
> necessarily converges to f(0) ;) A a matter of fact, in many cases it
> converges to zero: http://en.wikipedia.org/wiki/Riemann%E2%80%93Lebesgue_lemma

In fact n^{3/2}int_R e^{-ixsqrt{n}}f(x)dx -> 0 as n -> oo for every
smooth test function. That's because of what you noticed, namely the
above = n^{3/2}*F(f)(sqrt(n)). Now recall that if f is in the Schwarz
class, then so is F(f), which implies y^p*F(f)(y) -> 0 for any
positive p.

[Jon Slaughter]

Arnau Lapera wrote:
> Hello,
> I'm very interested in knowing whether
>
> n^{3/2} e^{-i x sqrt{n} }
>

I assume you mean w.r.t. x.

Of course. This is just the fourier kernel e^(-iwx) with w = f(n).

g(w)*dirac(w) is still a dirac delta function. Amplifying such a function
will not effect the results.

One way to think about it is

e^(-iwx) is zero for all w != 0 since there is an "equal" amount of negative
area as positive. (think of integrating over the periods of a sinusoid)

but when w = 0 we get e^(0) = 1.

Hence

int(e^(-iwx)) = 0 for all w except w = 0 which gives 1.