Maths 2.0

Page 1 of 1

 1089006.  Sun Aug 10, 2014 2:08 pm Starting with an 'M': A number is a Munchausen number if when you raise each one of its digits to its own power (ex. 4^4, 5^5 etc) and total up the results for each digit, you get back to the original number. 1 is a Munchausen number as 1^1=1. 321 is not as 3^3 +2^2+1^1 ≠321. You can go ahead and work your way up from 1 looking for these numbers, but you will have to look hard, as out of the whole cacophony of infinite numbers only two exist: 1 and 3435. The use for Munchausen numbers may be minimal, but what about a method for solving that everyday problem that we all face - estimating pi. You're in a room with a floor composed of strips of wood of the same width, all running parallel, and you are given a needle. How can you estimate pi? This is known as Buffon's needle problem. It can actually be calculated through dropping the needle repeatedly, and seeing if it touches a line separating the strips of wood! It can be shown through some not too tricky calculus that: π=2L/xp where L is the length of the needle, x the width of the strip and p the proportion of needles that touch a line. I think it's fantastic that the ratio of a circle's diameter to its circumference can be estimated just by repeatedly dropping needles! Continuing with odd ways to spend a weekend, instead of dropping needles suppose you're trying to comb a hairy ball so that all the hair it lies flat all over its surface. This is actually impossible to do - at some point there will always be a tuft due to a theorem from topology known (brilliantly) as the hairy ball theorem. We can actually model the wind system across the earth as a vector field, analogous to the hairs on the ball, and the HBT says that there must be at least one point where the vector is zero i.e. one point of the earth where the wind isn't blowing. From hairy balls, to something a bit more mature - is there an ideal voting method? Somehow we feel instinctively that there must be, but maths says otherwise! If by 'ideal' we mean a method that satisfies the three criteria: 1. No single voter can determine the whole groups choice. 2. If every voter prefers John to Jim, the outcome should rank John above Jim. 3.The relative ranking of John and Jim should not change if the voters change the rank of other candidates, but not the relative ranking of John and Jim. then Arrow's Impossibility theorem says that for elections with three of more candidates where voters can rank all candidates in order of preference, there is no way to produce a ranked list of all candidates that represents the 'will of the people'! (Note that this does NOT mean that there are no fair election methods). From voting to traffic, Braess's paradox states that adding a new road to a heavy traffic area may seem like a sensible idea, but doing so can actually increase your journey’s time! Travellers act selfishly and switch routes to try to improve their time, and so overall everybody loses, analogous to the Prisoner's Dilemma

1089007.  Sun Aug 10, 2014 2:17 pm

 tommyk wrote: From voting to traffic, Braess's paradox states that adding a new road to a heavy traffic area may seem like a sensible idea, but doing so can actually increase your journey’s time! Travellers act selfishly and switch routes to try to improve their time, and so overall everybody loses, analogous to the Prisoner's Dilemma

That effect has had an impact in Seoul, where they recently reversed their transport strategy and began replacing some roads with public walkways - the results from that experiment do seem to confirm that Braess's paradox works in the reverse direction too (i.e. reducing the capacity of the road network and replacing it with an alternative transport route actually reduced congestion).

 1089009.  Sun Aug 10, 2014 2:53 pm Yes I wanted to mention that but ran out of word space haha! Also, when the traffic is really high, Braess's Paradox actually disappears! This is counter intuitive as we would expect more traffic to equal more congestion. At high numbers, the wisdom of crowds comes into effect, and travellers avoid the new road. See http://phys.org/news203665202.html

1109642.  Wed Jan 07, 2015 3:33 am

 tommyk wrote: ..... only two exist: 1 and 3435....

Actually, if you take 0^0 = 0, then there would be four in total: 0, 1, 3435, and 438579088.

 1109795.  Wed Jan 07, 2015 12:07 pm But who does? That first Wiki article appears to say that if we insist on 0^0 having a meaning at all, then it equals 1 under just about every approach that it is possible to take. Yet the second Wiki article notes the "convention" that 0^0 is 0, and claims this as "standard usage". There seems to be an inconsistency here.

 1132131.  Mon May 04, 2015 7:19 pm Yes, I've never heard of the "convention" that 0^0=0; considering the rules of exponentiation I fail to see how it would make sense at all really. I'm tempted to think that this "convention" exists only for people who want there to be four Münchhausen's numbers ...

Page 1 of 1

All times are GMT - 5 Hours

Display posts from previous:

Forum tools
User tools