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seen Nov 11 '13 at 15:56

Jul
2
awarded  Curious
May
8
comment Showing that the riemanian metric $\frac{g}{\sqrt{x^2+y^2+z^2}}$ is complete
Thank you very much for this solution as well. The calculation is very informative!
May
8
comment Showing that the riemanian metric $\frac{g}{\sqrt{x^2+y^2+z^2}}$ is complete
Perfect - thank you!
May
2
accepted Showing that the riemanian metric $\frac{g}{\sqrt{x^2+y^2+z^2}}$ is complete
Apr
22
comment Showing that the riemanian metric $\frac{g}{\sqrt{x^2+y^2+z^2}}$ is complete
Please forgive the long delay. This strikes me as a wonderful solution. Any chance you have a reference for that criterion?
Apr
9
comment Showing that the riemanian metric $\frac{g}{\sqrt{x^2+y^2+z^2}}$ is complete
What additional hypotheses could one place on $g$ beyond stating that it is the standard Euclidean metric?
Apr
9
asked Showing that the riemanian metric $\frac{g}{\sqrt{x^2+y^2+z^2}}$ is complete
Feb
28
comment Complexifying a group action of SL(n, R) to a group action of SL(n, C)
that makes sense in the context of the rest of the article (the goal is to conjugate the action to its linear part in a neighborhood of the origin, i.e. locally in $\mathbb{R}^m$, but for all of $SL(n, \mathbb{R}^m)$). Thank you for your help! Are there any common methods one might employ to establish convergence for g far from I? (Not in detail, just the general idea)
Feb
28
accepted Complexifying a group action of SL(n, R) to a group action of SL(n, C)
Feb
28
awarded  Commentator
Feb
28
comment Complexifying a group action of SL(n, R) to a group action of SL(n, C)
Right, the action is not assumed to be linear but only to fix the origin
Feb
27
comment Complexifying a group action of SL(n, R) to a group action of SL(n, C)
An analytic action of $SL(n, \mathbb{R})$ means that the map induced by an element $g$ that sends $x\in\mathbb{R}^m$ to $gx$ is given by a convergent power series in the coordinates of $x\in\mathbb{R}^m$, but the series would be different for each element of $SL(n, \mathbb{R})$, right? When you say that we can have one convergent series that describes the action of all elements, that is made possible because we are "working only locally in both the group and the space on which it acts"? Does this correspond to the author's suggestion that we obtain a "local holomorphic" action?
Feb
27
asked Complexifying a group action of SL(n, R) to a group action of SL(n, C)
Jan
18
comment show $\ln \frac{1-x}{1+x}$ is $L^2$ but not $L^1$ using its taylor expansion
Thank you robjohn. I'm afraid I needed more hand holding than I had originally thought and I couldn't figure out the solution from just your hints, although now it all makes sense. I appreciate your help.
Jan
18
accepted show $\ln \frac{1-x}{1+x}$ is $L^2$ but not $L^1$ using its taylor expansion
Jan
18
revised show $\ln \frac{1-x}{1+x}$ is $L^2$ but not $L^1$ using its taylor expansion
Added expansion for |x|>1 based on robjohn's suggestion + corrected spelling mistake
Jan
18
comment show $\ln \frac{1-x}{1+x}$ is $L^2$ but not $L^1$ using its taylor expansion
Thank you, that is a cool trick, I will remember that! How may I use these expansions to show that ln(1-x)(1+x) is not L^1 but is L^2?
Jan
18
asked show $\ln \frac{1-x}{1+x}$ is $L^2$ but not $L^1$ using its taylor expansion
Jan
9
comment Evaluate $\int_0^1{\frac{y}{\sqrt{y(1-y)}}dy}$
A little more work but thank you for this suggestion as well, I was just as rusty with partial fractions
Jan
9
revised Evaluate $\int_0^1{\frac{y}{\sqrt{y(1-y)}}dy}$
Added solution according to Mr Strochyk's suggestion