On divergence


by metroplex021
Tags: divergance, divergence
metroplex021
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#1
Jan22-14, 02:06 PM
P: 115
Hi folks -- could anyone think of a justification of the idea that if a function's arguments diverge (i.e. are taken to infinity), there's a high probability that the function too will diverge?

This would be really helpful for thinking about fundamental theories in particle physics, so any help much appreciated!
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economicsnerd
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#2
Jan22-14, 02:38 PM
P: 206
Not about "high probability", but if your function is nice enough, complex analysis could be of use. Liouville's theorem at least tells you that non-constant differentiable functions [itex]\mathbb C \to \mathbb C[/itex] are unbounded. I'd guess there's some related machinery that would help more.
HallsofIvy
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Jan22-14, 02:45 PM
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Quote Quote by metroplex021 View Post
Hi folks -- could anyone think of a justification of the idea that if a function's arguments diverge (i.e. are taken to infinity), there's a high probability that the function too will diverge?

This would be really helpful for thinking about fundamental theories in particle physics, so any help much appreciated!
That statement will make sense if you have some way of "measuring" sets of functions so that you can talk about "probability" in relation to sets of functions.

mfb
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#4
Jan22-14, 02:48 PM
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On divergence


Quote Quote by economicsnerd View Post
Not about "high probability", but if your function is nice enough, complex analysis could be of use. Liouville's theorem at least tells you that non-constant differentiable functions [itex]\mathbb C \to \mathbb C[/itex] are unbounded. I'd guess there's some related machinery that would help more.
Unbounded, but they don't have to go to infinity. sin(z) for z on the real axis is an example of a function that stays bounded for an argument that goes to infinity.


I agree with HallsofIvy, without some way to define a probability this does not work.
This would be really helpful for thinking about fundamental theories in particle physics
Why?


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