August 10th is the 222nd day of the year on a non-leap year. Numbers like 222, or 33333, or 77 are called repdigits which is just a portmanteau (which is also an interesting word) shorthand for repeated digit numbers. The occurrence reminded me of a problem I was chasing that related to a special form of repdigit called a repunit, where all the repeating digits are ones, 11, or 1111, etc; but before I do, I should point out that August tenth is the date on which the cornerstone for the Royal Observatory in Greenwich, England (“Where time begins.”) was laid in 1675. It was also on this date in 1949 that the US Dept of Defense was created(Gotta’ pay homage to the folks who write my checks) by President Harry Truman, who strangely, was from the state of Missouri that was admitted to the Union on this same date as the 24th state (1821).
It was also a big day for repeated crime, in 1969 the Manson cult followed up the killing of Sharon Tate (on the ninth) with the twin murders of Leno and Rosemary La Bianca. On this day in 1977 they finally arrested the serial killer known as “Son of Sam” and “the 44 caliber killer”, after his weapon of choice. David Berkowitz after killing at least six women, and injuring several more without leaving significant clues, was arrested when he left his car too near a fire hydrant while carrying out one murder, and the police tracked him from the ticket.
Say “Happy Birthday”, (1602) to Gilles De Roberval, the French mathematician. He did some early work in the development of integral calculus. It is also the birthday of William Willett, an Englishman who developed the idea of daylight savings time.
Ok, now to the repunit idea...
I first met repunits as a recreational idea, and occasionally show my students this interesting, but limited bit of trivia:
12 = 1 and the sum of the digits is 1
112 = 121 and the sum of the digits is 4
1112= 12321 and the sum of the digits is 9
11112 = 1234321 and the sum of the digits is 16
There are also some interesting ideas about which repunits are prime, but other than that they didin't interest me too much, but then I began thinking about a problem about polynomials that had all ones as coefficients, x+1, x2+x+1, x3+x2+x+1, etc.. I called them repunit polynomials and used the expression rn to express the one with n+1 terms (so the highest exponent is n, R2 would be x2+x+1, for example)
All of them obviously go through the point (0,1), and all the ones with an odd degree (the highest exponent is odd) also pass through the point (-1,0). It was the others, the ones with even exponents, that made me wonder. They have no real valued solutions, and I was playing around with the minimum value when I produced the image above. The question of course, is what happens to the minimum value as the power goes to infinity. The graphs shown are the first five even n values r2, r4, …r10 and then r20. The window shows the domain and range values from -1<x<0 and 0<y<1.
In each case I found the minimum value and plotted it on the curve too. The progression of minima seems to march inexorably toward some limit…. But where? I tried to enter the values to umpteen decimal places and used regression, but while both the linear and quadratic had very high correlation coefficients (.98+), the residuals made me think that neither fully explanied the pattern of the minima.
Calculus offered a different type of sequence to play with. The derivative of r4 for example just gives 4x3+3x2+2x+1, and a similar thing would happen with all the others… so the question of where is the x-coordinate of the minimum for rn could also be asked as the root (there is only one) of the sequential polynomial 1+2x + 3x2+ …+ nxn-1 . (which I have taken the liberty of naming a natural polynomial after the natural numbers that are their coefficients. Unfortunately this seems as difficult to tame as the original.
I convinced myself that the limit of the x-coordinate of the minimum had to be > -1. (WHY??) Well, rn for an even n has an odd number of terms. All the terms with even exponents will turn into +1 when evaluated at x=-1, and all the terms with odd exponents will turn into -1. That means there is one more +1 than there is -1; so if n is even, the value of rn(-1) = 1 and the value of rn(0)= 1; with the minimum somewhere in between. (The derivative at -1 will be negative,in fact it is equal to -n/2, and at 0 it will be positive one). I think that also means that the limit of the x coordinate of the minimum as n goest to infinity will have to approach x=-1 also. As n increases, the slope at (-1,1) grows more and more negative and approaches infinity so I can't imagine how the limit could be any value > -1. I still have a problem proving this, and I'm not at all sure what to do to get a handle on the limit for the y coordinate, so if you have any clues, pass them on.
