Guerilla street-spam fighter

On 17 April 2013, I requested from my local municipal councillor, Frances Nunziata, that some street-spam be removed from my residential neighbourhood. The councillor's administrative assistant, Ji Na Park, said they would refer my complaint to the Municipal Licensing and Standards department. "The signs are removed from the posts and the companies are contacted to be advised that this should not continue."

On 2 May 2013, I asked: "What I would like to know now is what sort of time-frame I should expect for your implied action to take place (if, that is, it will ever take place), since two weeks has proven insufficient for it. Perhaps you were mistaken in thinking that the Municipal Licensing and Standards department actually does this." Ji Na replied: "We apologize for the delay. As the signs are considered non-urgent, they are classified as non-priority and may take a number of weeks."

Two days later I removed 30 of the signs myself. I informed the councillor's office but added: "There were three signs that I have been unable to remove because they were too high up, and seriously bolted into the telephone poles (not just nailed)... May I have the city's co-operation in having these three signs removed?" Ji Na's 7 May 2013 reply: "We have reported in the 3 signs to Municipal Licensing and Standards for removal. Please let us know when you see improvements. As it is peak season, there will likely be more signs going up in the future. We will try and get them removed as soon as possible."

Today, after 55 weeks of waiting, I asked Catherine's brother (here for a few hours en route to England, leaving his truck in our yard) to help transport a ladder to the crime scenes. In the photo you see me, utility knife in hand, evaluating the job to be done:

Alternating sign

Simple continued-fraction representations of constants are a source of much mathematical amusement. The picture is a scan of the cover of Carl Douglas Olds' 1963 classic Continued Fractions which served as my introduction to the topic around 1967.

The sequence [3; 7, 15, 1, 292, 1, 1, ...] is of course the start of the continued-fraction expansion for π. I wish C.D. Olds had used (and insisted on) that semicolon (instead of a comma) after the initial 3. The book shows us how to calculate from these numbers convergents to π, numerators and denominators of the convergents being derived separately. For numerators we pretend things start with [0, 1, ...] and for denominators, [1, 0, ...]. The numerator- and denominator-sequences are then calculated by taking an entry from the continued-fraction sequence (starting with 3), multiplying it by the previous number, and adding the previous-to-previous number.

So, for numerators: [0, 1, 3*1+0 = 3; 7*3+1 = 22, 15*22+3 = 333, ...]
For denominators: [1, 0, 3*0+1 = 1; 7*1+0 = 7, 15*7+1 = 106, ...]

Dropping our pretend prepends and dividing: [3/1; 22/7, 333/106, 355/113, ...]. These are the well-known convergents to π. Note that the convergents alternate in their relation to π: [<π; >π, <π, >π, ...]. Subtracting π: [<0; >0, <0, >0, ...]. Or, if you will, their sign: [-1; 1, -1, 1, ...].

One can do the same thing for τ (pronounced tau), π's less well-known parent (more commonly referred to as 2π). The continued-fraction expansion for τ is [6; 3, 1, 1, 7, 2, 146, ...]. The convergents are: [6; 19/3, 25/4, 44/7, ...]. Like π, these convergents alternate in their relation to τ: [<τ; >τ, <τ, >τ, ...], their sign likewise.

What happens when one divides the convergents to τ by the convergents to π? The resulting sequence is [2; 133/66, 1325/666, 4972/2485, ...], numbers very close or equal to 2. To generate their sign, subtract 2: [0; 1/66, -7/666, 2/2485, ...], hence [0; 1, -1, 1, ...]. The twist is that the sign sequence is now littered with zeros (in seemingly random — but decidedly clustered — locations). How do those zeros affect the sign alternation? The surprising (to me) answer is: not at all. Remove all zeros from Convergents[τ]/Convergents[π]-2 and what remains will alternate sign, regardless of the number or placement of the previously intervening, single-or-consecutive zeros.

Zipp

A photo of my new Zipp speaker, unzipped.

A long time ago I had set up speaker wires to two ceiling corners of the kitchen to which were attached small, inexpensive (but effective) Radio Shack speakers. An alternate pair of wires had gone (via a different route) outside the house and ended up on the eave soffit by the back deck. I could detach the speakers from their brackets in the kitchen and attach them to the brackets outside.

The outside wires were removed I-can't-remember-when and the inside speakers came down last year when the kitchen was "refreshed". To listen to music on the back deck now meant turning up the volume in my adjacent computer-room and leaving the window wide open.

But that volume was now coming from my iMac speakers, because when it replaced my old under-the-desk Power Mac G5, I lost the optical-out connection to my five-speaker Logitech setup. Yea, I know: I needed a minijack-to-optical inbetweener, but I don't get out much. :P

Anyways, I finally ordered a Belkin Toslink cable that came with that inbetweener and ordered the Zipp speaker at the same time. So I can listen to music again on my Logitech speakers in my room, and on the Zipp outside — or anywhere else within my local-network WiFi reach.

The French connection

Catherine's father appears not to have been particularly interested in his parents' family connections, a point that was driven home to me last year when I discovered that his mother had (until 1940) a sister living in Winnipeg, of whom he himself (when I asked him about her) seemed unaware. I have since come into possession of a good number of handwritten letters from that sister to his mother, as well correspondence from France and newspaper articles that have (collectively) allowed me to flesh out some of the history.

Jean Baptiste Homard (born 1852), a "well-known contractor in northern France", married Marie Claire Vassel (born 1857) and they had (in 1879) a daughter, Marie Isabel [Jeanne] (anglicized to Mary Elizabeth [Jane], that unknown sister). Jean Baptiste and Marie Claire were on the steamer Labrador, arriving in New York (from Havre) on 28 Jul 1886. A daughter, Emélie Matilde (anglicized to Emily May, Catherine's grandmother), was born 1 Apr 1887 in Fort William, Ontario. Another daughter, Gabrielle, was born in 1889 — supposedly back in France. Marie Claire's mother, Catherine Bouquet Vassel (born in 1826) ended up back in Canada with them in 1891 (according to the 1911 census). Gabrielle died of scarlet fever (in 1903) in St. Norbert, Manitoba.

Jane had married Albert George Cowley in 1898 in Winnipeg and they ended up with three children: two boys with mostly-known, living descendants and a girl without offspring. Here they are (sans daughter, plus pet dog) circa 1910:

By the time Emily married Frank Powers (3 Jul 1912) in Calgary, Alberta, Jean Baptiste and Marie Claire Homard, and Catherine Vassel, had returned to France — to Thiaucourt. At the start of World War I, they became prisoners-of-war, ending up in Frankfurt-am-Main, Germany. Catherine died a year or so into her incarceration. Marie Claire and Jean Baptiste survived "well, but hungry" and were released at the war's end. Marie Claire died 12 Dec 1923 and Jean Baptiste, 5 Apr 1924. An 8 Apr 1924 letter to Emily from a cousin, Jeanne Vautrin in Thiaucourt, gives details of Jeanne's family, including a photograph circa 1920:

The Vautrin children are Jean (born ~1906), Michel (born ~1913), and Marie Blanche (born ~1915). The masculine-faced, dark complexioned adult in the photo must be Jeanne because her husband, Alfred François Vautrin (1 May 1880, Beney-en-Woëvre - 20 Aug 1914, Bréhain), would have been dead. Jeanne also mentioned a sister, Marie. Chances are good that there are living descendants in that part of France.

Canoeing in Slocan Lake

In July 2004 we ended up in British Columbia for a week. The photo shows Catherine just after noon on the 21st as we leave New Denver to paddle across Slocan Lake. Visible on the other side are some of the peaks of northern Valhalla Provincial Park, including the small glacier of Mount Denver. We pitched a tent at Wee Sandy Creek beach, spent the night, and on the following day paddled down to Nemo Creek beach, where we set up and spent another two nights before heading back to New Denver. A particular highlight was this waterfall on Nemo Creek:

I mention this, now almost ten years later, because of a just-happened, tragic New Denver canoeing mishap.

The mystery of the Bell network interface

Yesterday, Catherine discovered a "record of work completed" stuck behind the front screen-door of her father's recently-sold house. It suggested that a Bell technician had come that day and "completed all work" at no charge. The name given for the address was D. Bonnell and the telephone/order# was 248-4913. The back of the form suggested that a network interface device had been installed and, sure enough, Catherine noticed that such a thing was indeed at the side of the house.

Alas, we do not know this person that initiated the installation. We checked with the realtor to make sure that it had nothing to do with the purchaser of the house who will not take possession until the end of the month. Unfortunately this came on the heels of another incident where an unknown-to-us "Don Criss" was trying to use the house as a drop-off for a package delivery, so we were somewhat paranoid about fraudulent activity involving an obviously empty, just-sold residence.

So my son called Bell to have the device removed. As it was still there today, Catherine became very upset. She phoned Bell to inquire about the removal. Not only did they have no record of someone having requested the removal of the device the previous day, it took her about forty minutes to have a new request on record. She was told that the person claiming her father's residence for the installation was a "Donna" Bonnell. Catherine pointed out that the phone number leads also (using a different reverse-411 service) to a W. Abouelkheir but of course that may have been a previous owner of that particular number.

Update: One day later a Bell technician removed it. There's obviously some element of fraud here since Ms. Bonnell's address was never what it claimed to be!

Simple arithmetic

I rarely go for a walk without my camera. Once, someone stopped me to ask if I was a photographer. "No," I said, "I just like to take pictures." It's a little like that with numbers. I like to play with them but that doesn't make me a mathematician.

Some time ago, Peter Luschny asked me to help him with the English in a mathematical treatise he is writing. I agreed. More recently he wanted me to calculate additional terms of a sequence therein and (after he explained the program he had written for it) I managed to do that. There had been an obvious and not unexpected size-bias in that sequence's terms at even indices versus the terms at odd indices. In particular, Peter noted that the terms at positions 2, 6, 10, 14, etc. were zero. In what was for me an unusual stance, I decided to see if I could prove it.

Consider a list of n -1s and n +1s: {-1, -1, -1, ..., +1, +1, +1}, 2n terms. What Peter's program does is accumulate those terms, then accumulate again the terms just derived, and — for all possible permutations of our list — count how often the final number in that second accumulation is zero. I better explain what I mean by accumulate. In Mathematica, Accumulate[{a, b, c, d}] yields {a, a+b, a+b+c, a+b+c+d}. We don't actually need to know what the exact permutation of our 2n terms is, only that they are all odd:

{odd, odd, odd, odd, odd, odd, odd, odd, ..., odd, odd}.

The first accumulation yields:

{odd, even, odd, even, odd, even, odd, even, ..., odd, even} (2n terms)

because of course an odd plus an odd is an even and an even plus an odd is an odd. The second accumulation (knowing that an even plus an even is another even) yields:

{odd, odd, even, even, odd, odd, even, even, ..., even, even} for 2n divisible by 4, or
{odd, odd, even, even, odd, odd, even, even, ..., odd, odd} for 2n not divisible by 4.

The important thing here is that in the latter case the final term is odd, thus never zero. Since we are counting zeros, the total number for 2n = {2, 6, 10, 14, ...} will always be zero. The count in the other case, 2n = {0, 4, 8, 12, ...}, appears to be A063074.