What Does Autocorrelation Reveal About Highly Peaked Functions?

[stevendaryl]

If $f(x)$ is a square-integrable real-valued function on the reals, then we can define an auto-correlation function $C(\theta)$ via:

$C(\theta) = \int dx f(x) f(x+\theta)$

I'm trying to get some insight on what
$C(\theta)$ is like, for small values of $\theta$ in the case in which $f(x)$ is highly peaked at $x=0$; that is, $f(x)$ has a graph that looks roughly like this:

I don't actually have an analytic form for $f(x)$, so I can't explicitly do the integral, but I'm trying to get a qualitative feel for what the correlation function is like for such a function. I'm assuming that $C(\theta)$ can be approximated by a power series:
$C(\theta) = A_0 + A_1 \theta + A_2 \theta^2 + \ldots$
The question is: is $A_1$ nonzero (and if so, what is its sign?)

I have two different approaches to get a qualitative answer that both seem reasonable, but they give different answers, and I'm wondering why.

First approach: Assume that you can take the derivative through the integral:

$A_1 = \frac{d}{d \theta} C(\theta) |_{\theta = 0}<br /> = \int dx f(x) f'(x)$

where $f'(x) = \frac{d}{d \theta} f(x+\theta) |_{\theta = 0}$

It's immediately obvious that $A_1 = 0$, because $f(x)$ is an even function (let's assume that it is, anyway), so $f'(x)$ is an odd function, so the product is an odd function, and the integral of an odd function gives 0. So the conclusion is that $C(\theta)$ has no linear term near $\theta = 0$

Second approach: Approximate $f(x)$ by a step-function. That is, we approximate $f(x)$ by the function $\tilde{f}(x)$ defined by:

$\tilde{f}(x) = 0$ if $|x| > \epsilon$

$\tilde{f}(x) = K$ if $|x| < \epsilon$

The use of this function gives the following auto-correlation:

$\int dx \tilde{f}(x) \tilde{f}(x+\theta) = K^2 (2 \epsilon - \theta)$

With this approximation for $f(x)$, we get $A_1 = -K^2$, which is nonzero.

So if I assume that $f(x)$ is analytic, I get $A_1 = 0$, and if I assume that $f(x)$ is a step-function, I get $A_1$ is negative. Those two statements are not really contradictions, because a step function is not analytic. However, it seems to me that you can approximate a step function by an analytic function arbitrarily closely, so it seems that using such an approximation, I should get an approximation for $C(\theta)$ that is similar to the exact result for a step function.

So which argument is correct?

 

[Office_Shredder]

Your second approach assumed that $\theta$ was a positive number when doing the calculation for $C(\theta)$, the correct formula should be
$$K^2(2\epsilon - |\theta|)$$

and we see that C is no longer analytic, but does have a maximum at theta=0 so the results are fairly similar

 

[stevendaryl]

Your second approach assumed that $\theta$ was a positive number when doing the calculation for $C(\theta)$, the correct formula should be
$$K^2(2\epsilon - |\theta|)$$

and we see that C is no longer analytic, but does have a maximum at theta=0 so the results are fairly similar

Thanks! That was what I was missing. If you approximate $|\theta|$ by an analytic function, it would have to have derivative 0 at $\theta=0$