Yes, it's "matemática". Also, for the record, I'm a native spanish speaker I usually know very well these things.
Yes, it's "matemática". Also, for the record, I'm a native spanish speaker I usually know very well these things.
Hey guys, here's one of the 50 best young mathematicians of the Ecuador and I'm sure I can answer that question...
...
Whoever told you it's equal to 1/2 is right for one possible answer. The "formula" if you'll call it that of an arithmetic series is determined by the following equation:
((a1+an)d)/2 where n represents the number of the last element in the series (the seventh and last, for example) and d represents the difference between one term and another.
Since the series +1, -1, +1, -1, +1...+-1 could very well be rounded up like +1, +1-1, +1, +1-1,...0 then replacing this in the equation would mean (1+0)+-1/2 which is equal to +-1/2. I'm not too sure of this since I've not received classes about cyclical progressions yet (and that is what this is), but I believe this is how it goes.
But if you count it like +1, -1, +1,...+1 the answer would be 1. If it's +1, -1, +1, ...-1 it'd be 0.
That's it hope it helps.
EDIT: Forgot to add, that all the formulas used are for the additions of all the terms in the series. There will never actually be a 1/2 in the series...
1-1+1-1+1.... does NOT go to 1/2 you msut have misunderstood. That series is divergent. Alternating series only converge if the numbers go closer and closer to 0 as the series goes on and on. You must have been lied to, or misunderstood something.
Tag(thx fkusumot) - "Yet again I completely fail to see your point..."
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dtewi said: It concerns the properties of infinity. The series of 1-1+1-1+1-1+1-1+1-1+1...... is equal to 1/2, or so I've been told. My question is this: Why? Why exactly does alternating between 0 and 1 for infinity cause it to end at 1/2? Why? |
Schrödinger's cat.
Oh shit, is that Stephen Hawking with a shotgun?
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