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Is Math A Sport?
theodp writes "The close of the International Mathematical Olympiad prompts Slate to question if math is a sport, wondering if mathletes might someday compete in the Olympics alongside track stars and basketball players."
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Math:
The study of the measurement, properties, and relationships of quantities and sets, using numbers and symbols.
Sport:
An activity involving physical exertion and skill that is governed by a set of rules or customs and often undertaken competitively.
who | grep -i blond | date cd ~; unzip; touch; strip; finger; mount; gasp; yes; uptime; umount; sleep
Math is interesting, math is fun, math is usefull, but math is not a sport.
From WordNet (r) 2.0:
sport
n 1: an active diversion requiring physical exertion and
competition [syn: athletics]
Many people cite it as a "non-sport," but synchronized swimming is incredibly difficult, both athletically and otherwise. Here's a way for you to find out:
swim 60 meters underwater.
stay underwater 3 out of 5 minutes.
train in a pool 7 days a week in addition to a periodized weight regimen and plyometrics.
Those things are just auxiliary. As a prerequisite, you must to have incredible overall swimming skills, cardiovascular and muscular endurance, great strength, agility, balance, discipline and superbly-honed technique.
As a British Junior Invitational Mathematical Olympiad (Yes, really.) I must say, unequivocally, no.
Just a guess here...
The problem isn't that x = 0. If it continued as written, every group of 3 terms ( (1,2,3) (4,5,6) , etc...) would sum to something strictly positive.
The problem is the exact definition of 'x'. Try to expand out the series for x just a little bit further and still get the identity x = x/2 to work out after adding the brackets. I think that you're going to have some trouble.
Your deduction assumes that if
...
x = 1 - 1/2 - 1/4 + 1/3 - 1/6 - 1/8 + 1/5 - 1/10
then
x = 1 - 1/2 + 1/3 - 1/4 + 1/5 -
as that's what you do in your substitution step. But those two infinite sums are totally different. Rearranging numbers in an infinite sum is not allowed without very careful consideration. It be like saying:
1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 - 1/2 + 1/8 + 1/9 + 1/10 + 1/11 + 1/12 + 1/13 + 1/14 - 1/3 + 1/15...
was the same as 1 + 1/2 - 1/2 + 1/3 - 1/3 + 1/4 - 1/4....
The idea is the same - your are promoting the subtractions well before they would ever happen. Before you ever reached the -1/4, you would have already added up between 1/1 to 1/28 and only subtracted out 1/1 to 1/4, leaving a pretty significant sum. In fact, the sum should become roughly sum 1 + sum(1/(k+1) to 1/n) s.t. k = n/7 using integer division (rough meaning it's 10PM here and I'd rather go to bed than be precise). In fact, I think this sum goes to infinity, while the other goes to 1!
In short, alcohol and calc do not mix: do not drink and derive.
Matt Fahrenbacher
James Tiberius Kirk: "Spock, the women on your planet are logical. No other planet in the galaxy can make that claim."
This isn't using divisions by zero. This particular "proof" relies on the fact that rearranging the terms of a nonabsolutely convergent series does not necessarily give the same sum. In fact, such a series can be transformed into a series with any given sum simply by rearranging the terms.
This isn't using divisions by zero. This particular "proof" relies on the fact that rearranging the terms of a nonabsolutely convergent series does not necessarily give the same sum. In fact, such a series can be transformed into a series with any given sum simply by rearranging the terms.
1) The series diverges, and the algebraic manipulations of its "value" were meaningless.
2) The series converges to 0, and there was division by 0.
3) This particular series doesn't follow that pattern past the first few terms, and the equation 2x=x doesn't follow, even if it does converge.
In any case, Riemann's rearrangement result is unnecessary.