Wasp with Cicada

I just witnessed something unusual, so I might as well record it. I was bending close to the ground and a huge wasp landed with a buzz at my feet. Before I could barely register that it was carrying an entire cicada underneath it, it rushed into a nearby hole and disappeared underground. The tailings from the hole had looked like an anthill.

Number of segments derived from the glance function

For glance function H, with last jump at diameter, we have 

H(diameter)=number of segments

Several ways to see this is true without the (lost) argument by induction based on the (false) analysis of adding a new (segment+gap) at one end of the collection of (one fewer) segments. One is to simply examine the form of the glance function for these segments (N=3, diam=a+b+c+d+e):

With segments of lengths a, c, and e; separated by gaps of lengths b and d, then the glance function has rows with +/- jumps at these places (sorry about the shitty notation, inserting the "h" is awkward):

+a - (a+b) + (a+b+c) - (a+b+c+d) + (a+b+c+d)

       +b      -   (b+c)    + (b+c+d) - (b+c+d+e)

                       +c      -   (c+d)   + (c+d+e)

                                    +d     -    (d+e)

                                              + e

For each segment row: the row ends with + and all other +/- cancel. For a gap: all the +/- cancel. Hence the height of the glance function at (diam) is the same as the number of rows that start with a segment length - one per segment.

Deriving the total length of segments from the glance function

Theorem:

For glance function

H = ∑ c_a*h_a

where the c_a are constants and the h_a are Heaviside functions, with impulse at a.

Then we have

(*)          ∑ c_a*a = sum of lengths of the segments

The proof is a generalization of brute force calculation, which is always of the same form. So for example, if we have:

With segments of lengths a, c, and e; separated by gaps of lengths b and d, then the left hand side of (*) is:

a - (a+b) + (a+b+c) - (a+b+c+d) + (a+b+c+d)

       b      -   (b+c)    + (b+c+d) - (b+c+d+e)

                       c      -   (c+d)   + (c+d+e)

                                       d      -    d+ e

                                                     e

Here, everything in the second row cancels most of what is above it, leaving alternating +/- a's. Being odd in number, the final sum is +a.

Now the second row was used up completely, in those cancelations. So the third row starts with c and because  of cancelations from the fourth row, result in a final value of +c. The fourth row was used up canceling things in the third row. Now the fifth row simply adds +e. In total we get a+c+e. This is the sum we wanted.

It follows that if two domains in the 2D plane have the same glance functions over all lines, then they have the same Radon transforms. So the domains are the same, up to the sorts of details usually assumed away.

Note: (sum of the gaps) = D - (sum of length of segments)

Evolution of a picture of Hollyhocks

A simple theorem of glance functions

THIS IS NAIVELY WRONG BUT MAY CONTAIN AN IDEA

A little Theorem about glance functions [described in my Hypothesis Testing... article].

Let h_a be the glance function of a simple interval of length a.

If H is the glance function for a collections of intervals separated by gaps, and a new interval of length a is added after a gap of length b, then the glance function of the new collection of intervals is

(*)                                      H -> H – H^b  + H^a + h_b + h_a  - h_a+b

Where the superscript on a function H indicates a term-wise shift of the independent variable by adding that amount, so  H^a(x)= H(a+x).

Corollary: the value of the glance function after the last step equals the number of intervals being glanced.

Proof: It is true for one interval, since h_a equals 1 at it's last step. If true for H, with N final steps, then we note the terms in the above (*) has final step values: +N , -N, +N,  +1, +1, -1. This totals to N+1. Hence the corollary is true by induction.

Update: The correct formula is

(**)                     H à H – H^b  + H^a+b + (h_a + h_b – h_a+b)

The induction argument is the same. We can name the first part the "shifted H" part. The rest is the "self-contained" info about the added segment.

Note: this encourages thinking about re-constructing the original segments, starting from their glance function. Because we know how many segments are involved and the maximum length from first to last end points of the segments.

Compare me to a paremecium

I was reading Feynman about his observing paramecium behavior to be quite diverse and unpredictable versus what the textbook said about paramecium moving and bouncing off of obstructions.

As I sit here, doing what I always did on a happy Saturday morning [except it is Thursday] - sitting with a piece of paper and thinking about abstractions, I realize that sitting at my desk is what I spend almost the entirety of my time doing. I move around a bit. I eat and sleep. But from the pov of an alien observer, they would conclude that my sitting with a piece of paper is what I do. Everything else in my life would be reduced to something not mentioned in the observer's textbook.

June Hollyhocks