SOTM#170 - Rayman Legends

[Ambidextroid]

I vote Serza and Hoodcom!

[PluMGMK]

I vote Harpic, Master and Serza5 w/ Hoodcom.

[Serza5]

7 sigs means 3 votes maximum and no minumum.

Code: Select all

Harpic fraîcheur :  ||
Ambidextroid :      |
Master :            |
CHRdutch :          0
PluMGMK :           |
Serza5 w/ Hoodcom : ||
Hoodcom w/ Serza5 : |

Because the votes started slightly later i'll add one day so the last day of votes will be the 28th and the results will be on the 29th.

I vote Serza and Hoodcom!

I'm assuming this vote means for both sigs and not just one.

[Reese Riverson]

I will vote for Ambi's and Master's please.

[Dart]

I'm going with Serza and Hoodcom

[Master]

Very nicely done ones here. Tough decision indeed...
Serza5 w/ Hoodcom, Ambi and CHRdutch.

[Hunchman801]

Harpic, Ambi and PluM.

I'm curious about the math behind PluM's though.

[Tomazo]

I vote for CHRdutch and Ambidextroid.

[Lysol]

I'll vote for Harpic.

[PluMGMK]

Harpic, Ambi and PluM.

I'm curious about the math behind PluM's though.

Well according to Wikipedia if we define the integral the "normal" way, as a sum of two integrals where the lower limit of one matches the upper of the other, at some arbitrary finite value, and the other two limits tend to plus and minus infinity, then it is undefined. But we can cheat and instead have it as an integral from a to -a, and let a tend to infinity, and then it has the value you'd intuitively expect, which is the centre of the peak. This is how I solved my problem, before realizing I'd been overcomplicating it.
As for that thing about the Uncertainty Principle, I'm a bit rusty about why that's the case, but I believe it's related to Fourier transforms.

[Serza5]

7 sigs means 3 votes maximum and no minumum.

Code: Select all

Harpic fraîcheur :  ||||
Ambidextroid :      |||||
Master :            ||
CHRdutch :          ||
PluMGMK :           ||
Serza5 w/ Hoodcom : ||||
Hoodcom w/ Serza5 : ||

Because the votes started slightly later i'll add one day so the last day of votes will be the 28th and the results will be on the 29th.

[Eparcyl]

I vote for Harpic Fraîcheur

[Hunchman801]

Well according to Wikipedia if we define the integral the "normal" way, as a sum of two integrals where the lower limit of one matches the upper of the other, at some arbitrary finite value, and the other two limits tend to plus and minus infinity, then it is undefined. But we can cheat and instead have it as an integral from a to -a, and let a tend to infinity, and then it has the value you'd intuitively expect, which is the centre of the peak. This is how I solved my problem, before realizing I'd been overcomplicating it.
As for that thing about the Uncertainty Principle, I'm a bit rusty about why that's the case, but I believe it's related to Fourier transforms.

Well yeah, you pretty much created an indeterminate form from one that wasn't.

I'm not too sure how this applies to the function in your signature though, because it's neither even nor odd and its integral from 0 to infinity does not converge.

[PluMGMK]

Took me a few days to realize it wasn't obvious that I had done a change of variables! I actually integrated using λ' = λ - λ0. This allowed me to split it into two, one of which was just λ0 times the area under the Lorentz curve, which is unity, and the other of which was the integral of an odd function. So actually my limits in the original variable were ±a + λ0, my bad!

[Hunchman801]

Took me a few days to realize it wasn't obvious that I had done a change of variables! I actually integrated using λ' = λ - λ0. This allowed me to split it into two, one of which was just λ0 times the area under the Lorentz curve, which is unity, and the other of which was the integral of an odd function. So actually my limits in the original variable were ±a + λ0, my bad!

Well you're right, I tried it and you end with λ0 plus an improper integral of an odd function. What's interesting is that the actual meaning of the improper integral is ambiguous, so we use the Cauchy principal value, which is indeed 0.

I don't think it's a case of analytic continuation (used everywhere in physics) though, just an ambiguous notation.

[Serza5]

7 sigs means 3 votes maximum and no minumum.

Code: Select all

Harpic fraîcheur :  |||||
Ambidextroid :      |||||
Master :            ||
CHRdutch :          ||
PluMGMK :           ||
Serza5 w/ Hoodcom : ||||
Hoodcom w/ Serza5 : ||

Because the votes started slightly later i'll add one day so the last day of votes will be the 28th and the results will be on the 29th.

[PluMGMK]

I don't think it's a case of analytic continuation (used everywhere in physics) though, just an ambiguous notation.

Yeah. Actually this conversation inspired me to do some more research on this and apparently there are good physical reasons for this distribution not to have a mean. So just a case of my not really knowing what I was doing! The funny thing is though, I wasn't actually looking for the mean, it just fell out when I manipulated another integral that I had used to produce some information that was actually useful.

[Hunchman801]

Interesting! My knowledge of wave propagation is way too rusted to make sense of the physical interpretation (it's been a good ten years at least) but I'm sure it all makes sense.

[Rulez]

Vote for Harpicques!

[Indy]

1 vote for Harpic!