3 ms·

It's clearer from the Wikipedia article (https://en.wikipedia.org/w/index.php?title=Goat_problem&oldid=993368252 https://en.wikipedia.org/w/index.php?title=Goat

by svat 6y ago

It's clearer from the Wikipedia article (https://en.wikipedia.org/w/index.php?title=Goat_problem&oldid=993368252 https://en.wikipedia.org/w/index.php?title=Goat_problem&oldi...): earlier we only had expressions for r like https://wikimedia.org/api/rest_v1/media/math/render/svg/9758becd77e7b4102ccdf50ff0845a749d236dc2 https://wikimedia.org/api/rest_v1/media/math/render/svg/9758... and https://wikimedia.org/api/rest_v1/media/math/render/svg/a63430b98b5e138a3d73831a2b7ecb6825be7a5b https://wikimedia.org/api/rest_v1/media/math/render/svg/a634... (from which of course one could numerically compute the answer r=1.1587…), but now thanks to this paper we have the expression https://wikimedia.org/api/rest_v1/media/math/render/svg/5b7451ea92f107e56651328ad15bfcc3276f87f2 https://wikimedia.org/api/rest_v1/media/math/render/svg/5b74... that does not involve r on the right-hand side.

Sniffnoy 6y ago
Hm, if it's based on this ratio of contour integrals, shouldn't it be possible to do better than this? Like why would it be so hard to find the residues for these poles? Shouldn't that be just a bit of formal Laurent series manipulation? What am I missing here?
- Someone 6y ago
  If the series don’t cancel out nicely (likely, I would guess), wouldn’t you end up with some infinite sum? If so, as this example shows (contour integrals in a closed form?) “closed form” is loosely defined, but I think most would say something with an infinite sum wouldn’t be one (but then, https://en.wikipedia.org/wiki/Closed-form_expression#Analytic_expression https://en.wikipedia.org/wiki/Closed-form_expression#Analyti... says: “In particular, special functions such as the Bessel functions and the gamma function are usually allowed, and often so are infinite series and continued fractions. On the other hand, limits in general, and integrals in particular, are typically excluded.[citation needed]”
  - Sniffnoy 6y ago
    No, you shouldn't end up with an infinite sum; if you only want to know the series finitely far out, you only need to know finitely many terms. However, it seems the problem is hard for other reasons. I had assumed, without checking, that they'd put the center of the circle at 3pi/8 because it's a pole. Nope! As best I can tell from some graphing tools, there is indeed precisely one pole in the circle, and it's on the real line, but it's close to the right endpoint; I don't know that it has any nice form. So I imagine that getting any sort of exact series expansion around there -- or even just getting exactly the first few terms, i.e. the first few derivatives there, which would be all you'd need -- would be difficult for that reason. (Although, the higher the order of the zero, the more initial terms you'd need...) Edit: Actually, I guess it looks like a zero of order 1? Except that doesn't make sense, because then the top contour integral would be zero...
IshKebab 6y ago
Ha if only the article about mathematical equations had actually included any mathematical equations. This is like those articles that are about a picture but don't include the picture.

by svat 6y ago

Sniffnoy 6y ago

Hm, if it's based on this ratio of contour integrals, shouldn't it be possible to do better than this? Like why would it be so hard to find the residues for these poles? Shouldn't that be just a bit of formal Laurent series manipulation? What am I missing here?

Someone 6y ago
If the series don’t cancel out nicely (likely, I would guess), wouldn’t you end up with some infinite sum? If so, as this example shows (contour integrals in a closed form?) “closed form” is loosely defined, but I think most would say something with an infinite sum wouldn’t be one (but then, https://en.wikipedia.org/wiki/Closed-form_expression#Analytic_expression https://en.wikipedia.org/wiki/Closed-form_expression#Analyti... says: “In particular, special functions such as the Bessel functions and the gamma function are usually allowed, and often so are infinite series and continued fractions. On the other hand, limits in general, and integrals in particular, are typically excluded.[citation needed]”
- Sniffnoy 6y ago
  No, you shouldn't end up with an infinite sum; if you only want to know the series finitely far out, you only need to know finitely many terms. However, it seems the problem is hard for other reasons. I had assumed, without checking, that they'd put the center of the circle at 3pi/8 because it's a pole. Nope! As best I can tell from some graphing tools, there is indeed precisely one pole in the circle, and it's on the real line, but it's close to the right endpoint; I don't know that it has any nice form. So I imagine that getting any sort of exact series expansion around there -- or even just getting exactly the first few terms, i.e. the first few derivatives there, which would be all you'd need -- would be difficult for that reason. (Although, the higher the order of the zero, the more initial terms you'd need...) Edit: Actually, I guess it looks like a zero of order 1? Except that doesn't make sense, because then the top contour integral would be zero...

IshKebab 6y ago

Ha if only the article about mathematical equations had actually included any mathematical equations. This is like those articles that are about a picture but don't include the picture.