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Timeline for Asymptotic series [duplicate]

Current License: CC BY-SA 3.0

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May 27, 2015 at 2:39 history closed Jens
dr.blochwave
C. E.
bbgodfrey
m_goldberg
Duplicate of Multivariable Taylor expansion does not work as expected
May 26, 2015 at 17:10 review Close votes
May 27, 2015 at 2:42
Jan 20, 2014 at 13:36 answer added Emilio Pisanty timeline score: 2
Jan 20, 2014 at 10:54 comment added ogledala Plot3D[(Erf[Sqrt[2 x^2 + 5 y^2]]/Sqrt[2 x^2 + 5 y^2]), {x, -3, 3}, {y, -3, 3}] But has also to satisfy another condition, to have 1/x and 1/y as a major term in the given two series expansion: FullSimplify[Series[Erf[Sqrt[a x^2 + b y^2]]/Sqrt[a x^2 + b y^2], {y, Infinity,2}]] to get here 1/y as the main term and here FullSimplify[Series[Erf[Sqrt[a x^2 + b y^2]]/Sqrt[a x^2 + b y^2], {x, Infinity,2}]] 1/x is the main term.
Jan 20, 2014 at 10:20 comment added ogledala @episanty In order to capture the nature of a system I am describing, I needed a function of two variables (lets call it z[x,y] whose 3D plot "would look like a mountain", decaying differently from maximum in two directions: z[x,0] and [y,0]. The function: (Erf[Sqrt[a x^2 + b y^2]]/Sqrt[a x^2 + b y^2]), where x and y are variables, a and b parameters satisfies that that could be seen from this plot:
Jan 17, 2014 at 16:45 answer added Daniel Lichtblau timeline score: 2
Jan 17, 2014 at 16:42 comment added Daniel Lichtblau Is there some reason not to change variables e.g. w=a x^2 + b y^2?
Jan 17, 2014 at 15:59 comment added Emilio Pisanty It would be helpful if you stated precisely what kind of result you are expecting. You want a 'simplified' version of $\text{erf}(r)/r$; how should this simplified version relate to the original?
Jan 17, 2014 at 15:38 history edited Dr. belisarius CC BY-SA 3.0
added 13 characters in body
Jan 17, 2014 at 15:34 history asked ogledala CC BY-SA 3.0