Sunday, March 2, 2014
I came across this picture I took while visiting the bell tower of the Old Post Office in Washington, D.C. a couple of years ago. It describes how the bells are occasionally rung in a “full peal,” consisting of all possible permutations of their tones. With 6 bells, you get 6! = 6*5*4*3*2*1 = 720 permutations, which could be rung in under an hour, but with 7 bells, you get 7! = 7*720 = 5040 patterns to ring!
I thought maybe this would involve listing the set of permutations in increasing numerical order:
But In the example they showed, it seems that the pattern is to first swap neighbors in 3 groups:
12 34 56
21 43 65
And then to keep the first and last positions in place but swap neighbors in the 2 groups inside:
2 14 36 5
2 41 63 5
I can’t tell if that’s all there is to the method or not, and I wonder if there’s a proof that the method produces all permutations. A quick test with 4 bells shows that you don’t get all 24 permutations before it repeats.
Saturday, February 22, 2014
Last weekend I had the opportunity to help my old math department finish up the digits of pi mural we began in 2009.
While I was there I convinced them to let me put Archimedes on the wall by the staff bathrooms (even though we don’t have bathtubs in them).
Friday, January 24, 2014
A lot of times the applications we come across for algebraic topics feel contrived; here’s one that occurred naturally in my very own home yesterday.
My wife works in a clinic with two other midwives. They all do the same amount of “on-call” time, but have different workloads when it comes to clinic hours. And sometimes they’re on call while they’re in the clinic. She knows their salaries, but didn’t know the hospital’s pay rate for clinic hours, on-call hours, or combo hours – she was interested in finding this out to see if it was worthwhile to pick up some extra shifts at the clinic.
Calling the number of on-call hours X, the number of clinic hours Y, and the number of combo hours Z, she set up an equation for each midwife and was able to figure out the pay rate for each type of work.
Math to the rescue!
Wednesday, January 1, 2014
Dr. Conway developed a sweet method (called the Doomsday Algorithm) to determine the day of the week for any day in history. It takes some practice, but dealing with the current year is pretty simple, provided you know the Doomsday.
Doomsday 2013 is Friday*.
- This means that the following dates are all Thursdays: 4/4, 6/6, 8/8, 10/10, 12/12.
- If you work nine to five at the local 7-11, you can remember four more Thursdays: 9/5, 7/11, 5/9, 11/7.
- Einstein’s birthday = Pi Day = 3/14 = Thursday.
- The last day of February (2/29 this year) is a Thursday.
- Lastly, 1/3 is Thursday (3 out of 4 years it’s the 3rd… but on leap years it’s the 4th).
Adding days of the week is easier when you treat them as numbers: mONEday, TWOsday, 3ednesday, FOURsday, 5day, S6turday (Treat Sunday as 0 - "nunday"). What's ten days past Wednesday? 10 + 3 = 13. Take out any multiples of 7 and you have 6 (Saturday).
*In case you’re interested, the link up top does a thorough job of teaching the method, but here’s the quick version. Start with the century day. For 2000 it is Tuesday (2). Then look at the 2-digit year, 14, and ask how many 12’s are in it. (1). What is the remainder? (2) How many 4’s are in this number? (0). Add up (2)+(1)+(2)+(0) = 5 = 5th day of the week = Friday.
Craziness? How about one more example from the future: 2055. Century day is still (2). How many 12’s in 55? (4). What is the remainder? (7). How many 4’s in this number? (1). Add ‘em up: (2)+(4)+(7)+(1) = 14. Take out multiples of 7 and you get 0. The 0th day of the week is Sunday. Now you can find the day of the week for any date in 2055.
Tuesday, December 3, 2013
“The World of Mathematics” by Mathigon is a free online eBook that explores a diverse array of math ideas in an accessible and visually engaging way. Taking advantage of the digital medium, Mathigon contains fun interactive components in each of its 30 sections, many of which are still in development.
Tuesday, November 26, 2013
It’s not particularly rewarding to play, but maybe you can improve on it – right now it does three things:
- draws a pig at a random distance
- asks for angle and “force”
- plots points until they go off the screen
//FIRST, THE SETUP
- Lbl B
(random number between 4 and 8 for the pig’s x-coordinate)
- Circle (P,0,1)
(this represents the pig. I made a more detailed pig face since I had some extra time, which I’ve included at the bottom).
(starting values for plotting the bird path)
(acceleration due to gravity… units are arbitrary, so I just tweaked this until it looked good)
- Pause: ClrHome
- Disp “ANGLE?”: Prompt T
- cos(T)→D: sin(T)→E
(D and E will tell us the x and y increments for the bird’s path)
- Disp “FORCE (1-10)?”: Prompt F
- F/10*D→D: F/10*E→E
(D and E are scaled by 1/10 of the “force” …also arbitrary)
- Lbl A
- X+D→X: Y+E→Y
(If D and E both remain constant, the bird will go in a line. The curve is produced by changing E, the increment for y.)
(You could think of E as velocity and G and acceleration)
- If Y>0 and X<10: Goto A
(Keep plotting points until you go off-screen to the right or hit the ground)
- Goto B
(This forms a hexagon for the pig’s head)
- Line (P+0.4,0.1,P+0.4,-0.7)
(This forms a rectangle nose)
- Pt-On(P+0.5,1.2,2): Pt-On(P-0.5,1.2,2)
(Ears are points with “style 2” so they’re tiny squares)
- Pt-On(P+0.5,0.5,3): Pt-On(P+-0.4,0.5,3)
(Eyes have “style 3” so they’re tiny crosses)
- Pt-On(P-0.15,-0.3): Pt-On(P+0.15,-0.3)