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visits member for 3 years, 11 months
seen 8 hours ago

Jan
30
awarded  Enthusiast
Jan
23
revised Alternative to Axler's “Linear Algebra Done Right”
edited tags
Jan
22
awarded  Nice Answer
Jan
22
comment Why is $y = \sqrt{x-4}$ a function? and $y = \sqrt{4 - x^2}$ should be a circle
@vincentbelkin, I edited my answer.
Jan
22
comment Why is $y = \sqrt{x-4}$ a function? and $y = \sqrt{4 - x^2}$ should be a circle
@AsafKaragila, thanks!
Jan
22
revised Why is $y = \sqrt{x-4}$ a function? and $y = \sqrt{4 - x^2}$ should be a circle
edit in response to follow up question
Jan
22
revised Derivative for log
improved formatting, capitalization
Jan
22
suggested approved edit on Derivative for log
Jan
22
awarded  Editor
Jan
22
revised Why is $y = \sqrt{x-4}$ a function? and $y = \sqrt{4 - x^2}$ should be a circle
added more info relevant to the question
Jan
22
revised Why is $y = \sqrt{x-4}$ a function? and $y = \sqrt{4 - x^2}$ should be a circle
improved formatting
Jan
22
answered Why is $y = \sqrt{x-4}$ a function? and $y = \sqrt{4 - x^2}$ should be a circle
Jan
22
suggested approved edit on Why is $y = \sqrt{x-4}$ a function? and $y = \sqrt{4 - x^2}$ should be a circle
Jan
16
awarded  Yearling
Jan
10
accepted Defining the determinant of linear transformations as multilinear alternating form
Jan
3
accepted Is there a geometric argument that the Legendre transform of a convex function is convex?
Oct
7
asked Is there a geometric argument that the Legendre transform of a convex function is convex?
Sep
1
comment Is there a simple way to bound this contour integral?
Thanks for your comment. I realized that I was having a hard time phrasing my question. Now that I think about it, I wanted to be able to get a sharper estimation - I wanted to be able to get a feel for how $\int e^{-R^2 \cos2\theta} R \, d\theta$ behaves for large values of $R$ (even though to evaluate the original integral, we just needed to show this was $o(1)$). But like you said, this estimation isn't too bad. How did you get that estimate of yours?
Sep
1
comment Bounding a function by its second derivative using Fourier series
I am sorry, but I am confused. I am asking if there is a way to bound a function by its second derivative with Fourier series, not when a function has a Fourier series.
Sep
1
asked Is there a simple way to bound this contour integral?