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&#039;(x)

where f&#039;(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&#039;(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| &gt; \epsilon

\tilde{f}(x) = K if |x| &lt; \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