You can show that the "jumps" in the numerator happen in such a way that you always get two terms of value 2^(-n) for n>1 and one oddball term of 2^0
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12-17-10 Mohamed Bouazizi NEVER FORGET
Stadtluft Macht Frei
Killing it is the new killing it
Ultima Ratio Regum
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Don't worry; I spent the better part of an hour doing this just now.12-17-10 Mohamed Bouazizi NEVER FORGET
Stadtluft Macht Frei
Killing it is the new killing it
Ultima Ratio Regum
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[QUOTE] Originally posted by Ramo
y'(x) = 21x^6 - 20x^3 - 6x
y''(x) = 126x^5 - 60x^2 - 6
||y''(5)/|| = too much arithmatic
Is it my turn yet?[/QUOTE
Unfortunately, looks like youy did not pass Calc III, that is not the way to kind kappa, the value of the curavture of a line.]Talk and chat in the Freebie and Webmaster Discussion Forums
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I got like a 35 (out of possible 120) when I took it, and I was 2nd highest at McGill.
Darn. You get at least my congratulations... I took the stupid thing Freshman year when I was a lot more arrogant then I am now, and unsurprisingly got the most popular score on the Putnam, the good old zero.
Anyway, I've got a few questions for everybody, except that my Geometry teacher probably wouldn't approve of me posting questions from the take-home test before I hand it in. Anyway, the problem that should have been the easiest is probably the most annoying right now... a stupid little proof in projective geometry about some points being collinear, and it's such a problem that you don't even have to be inventive with placing points you know the coordinates of and using variables. It's just straight calculation with numbers, and it ISN'T COMING OUT! In fact, the solution keeps getting worse and worse. It started out with of the 3 points, two of the lines between them were the same. I made an adjustment after discovering one mistake, and there were still 2 lines the same. And after finding yet another one (or possibly 2), I now have 3 completely different lines going between these points.
Anyway, here's an enjoyable little problem: Does there exist such a function that f(x) is rational on every irrational number and irrational on every rational number? If there is, give an example, if there isn't, explain why.All syllogisms have three parts.
Therefore this is not a syllogism.
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Originally posted by SnowFire
Anyway, here's an enjoyable little problem: Does there exist such a function that f(x) is rational on every irrational number and irrational on every rational number? If there is, give an example, if there isn't, explain why.<p style="font-size:1024px">HTML is disabled in signatures</p>
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Unfortunately, looks like youy did not pass Calc III, that is not the way to kind kappa, the value of the curavture of a line."Beware of the man who works hard to learn something, learns it, and finds himself no wiser than before. He is full of murderous resentment of people who are ignorant without having come by their ignorance the hard way. "
-Bokonon
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Originally posted by MasterBob The Elder
Unfortunately, looks like youy did not pass Calc III, that is not the way to kind kappa, the value of the curavture of a line.<p style="font-size:1024px">HTML is disabled in signatures</p>
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Solve this differential equation for r:
r''(t) = -C*r(t)/||r(t)||^3"Beware of the man who works hard to learn something, learns it, and finds himself no wiser than before. He is full of murderous resentment of people who are ignorant without having come by their ignorance the hard way. "
-Bokonon
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something like
r³=6*(-C*(t²/2))
(after 2 ruf integrations)
maybe you like this one:
create 24 by using +,-,x,:
and the numbers 1,3,4,6
You may only use the numbers 1 time each and you have to use them 1 time each.
have fun
Shadeex-president of Apolytonia former King of the Apolytonian Imperium
"I have not failed. I've just found 10,000 ways that won't work." --Thomas Alva Edison (1847-1931)
shameless plug to my site:home of Civ:Imperia(WIP)
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Originally posted by uncle_funk
Answer to mine:
n^2 - (n - 1)^3
I think the correct equation would be:
2*([n-1]^2 + n) - 1Rome rules
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Okay, this is vaguely irrelevant to the game (which I do not wish to participate in), but I was wondering about this the other day: What would be the ultimate shape of a bowl/glass? Imagine you only have a certain amount of material and for simplicity's sake the thickness of the walls of the container would be uniform, what shape would be able to hold the greatest amount of liquid considering a downward gravitational pull? I'm assuming it'd be circular from the top, so to simplify, what shape given a certain length of wire would give the greatest area to the shape contained by it, if you were allowed to keep the top of the shape open? And why?Världsstad - Dom lokala genrenas vän
Mick102, 102,3 Umeå, Måndagar 20-21
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A sphere has the most efficient volume/surface area ratio.
You can prove this locally (show it must have constant Gaussian curvature), but it requires a pretty decent knowledge of differential geometry. There might be an easier way to prove it, but this is the method I've seen.
Hmmm...
for the top of the item open, I'm not sure...
Let me think about it for a few seconds.12-17-10 Mohamed Bouazizi NEVER FORGET
Stadtluft Macht Frei
Killing it is the new killing it
Ultima Ratio Regum
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