ObsessiveMathsFreak
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But there is in the limiting case as the number of particles goes to infinity.Tomaz Kristan said:There is NO leftmost ball at all.
But there is in the limiting case as the number of particles goes to infinity.Tomaz Kristan said:There is NO leftmost ball at all.
In the limit as the number of balls goes towards infinity. As we logically increase the number of balls, in each of our cases, there is a leftmost ball.Tomaz Kristan said:Limit for what? For the "leftmost" ball speed after a second? For the force between the two "leftmost" balls?
Nothing like that exists.
ObsessiveMathsFreak said:In the limit as the number of balls goes towards infinity. As we logically increase the number of balls, in each of our cases, there is a leftmost ball.
There is, in the nonstanard model. It contains H balls, where H is a transfinite (hyper)integer. The H-th ball is the leftmost.There is NO leftmost ball at all.
That is an accurate statement.Gelsamel Epsilon said:So basically transfinite numbers are numbers which are larger then any finite number but smaller then infinity?
Hurkyl said:That is an accurate statement.
(resisting urge to go into what is probably unnecessary detail)
It does go that way. That's what an infinite sum means.Tomaz Kristan said:No, it doesn't go that way and you know that.
ObsessiveMathsFreak said:It does go that way. That's what an infinite sum means.
When we write \sum_{n=0}^{\infty} a_n, what we mean is;
\lim_{k \to \infty } \sum_{n=0}^{k} a_n
It's clear as crystal. In each limiting case, there is a leftmost ball and the center of mass remains fixed. In the limit as k \to \infty, the accelleration of the center of mass is zero.
Tomaz Kristan said:You argument is false, OMF.
If it hadn't been, you could just as well proved, that the biggest natural number exists. Bigger than any other.
Well, it's a basic mistake on your side, trust me!
Alkatran said:Therefore, to calculate the value of your system, you HAVE to take the limit as n approaches infinity, which means you always consider the left most (nth) ball, your forces cancel, and the problem disappears.
Except people haven't even been working with an axiomatic system; they've only been working with an informally defined system.That's what is called an inconsistent axiomatic system.
Alkatran said:Important point that's probably already been made: the sum of an infinite sequence is equal to the limit of the sum of its first n terms as n goes to infinity. We need this definition to avoid problems such as:
This sequence converges
<br /> \sum_n=1^\infty \frac{(-1)^n}{n}<br />
However, because 1/n diverges as we sum to infinity, we can just rearrange the terms so that we get a series of the form
1/2 + 1/4 + 1/6 + 1/8 ... (add until we reach at least 3) - 1/1 + ... (add until we reach at least 6) - 1/3 + ...
As you can see, this series diverges to infinity, even though it as the exact same elements as our alternating converging series. The order we sum the elements in makes a HUGE difference!
Therefore, to calculate the value of your system, you HAVE to take the limit as n approaches infinity, which means you always consider the left most (nth) ball, your forces cancel, and the problem disappears.
Hurkyl said:Of course, the usual procedure when that happens is to figure out what the conflicting assumptions are, so you can fix things, rather than broadcast to the world "Hey, look at me, I've made conflicting assumptions".
Tomaz Kristan said:OK. So nothing is wrong with this paradox, after all. It stands.
Where do you suggest me, to publish it?![]()
HallsofIvy said:After having been told precisely what is wrong with it, you say this? Lovely.
Tomaz Kristan said:OK. So nothing is wrong with this paradox, after all. It stands.
Where do you suggest me, to publish it?![]()
Alkatran said:I hate it when people read the responses but don't pay attention to them. This has very little to do with Newtonian mechanics. It has everything to do with simple first year calculus.