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seen Apr 12 at 18:24

Dec
24
awarded  Excavator
Dec
24
revised Plotting in the Complex Plane
removed thanks
Dec
24
suggested suggested edit on Plotting in the Complex Plane
Dec
14
awarded  Teacher
Dec
10
revised Is there an analytic approximation to the minimum function?
added 45 characters in body
Dec
10
comment Is there an analytic approximation to the minimum function?
If you take $\log(\exp(kx) + \exp(ky)$, multiply by $k$ and subtract the derivative with respect to $k$ divided by k, you end up with my proposed answer and the troubling $\log 2$ goes away.
Dec
10
answered Is there an analytic approximation to the minimum function?
Dec
7
comment Real approximation to the maximum using Laplace's method integral
Yes, I was able to show in many examples that (either the last two versions) converge to the minimum of the function. (Yes, $x_0$ must be in the interval). In the cases I am interested the function is defined everywhere and the interval is infinite. I am puzzled of what to do with the imaginary part of the result. After all I am looking for an elegant "soft absolute minimum" of arbitrary function. (where the softness is controlled by a single parameter)
Dec
2
revised Laplace integral - Asymptotic expansion
added tex code
Dec
2
suggested suggested edit on Laplace integral - Asymptotic expansion
Dec
2
revised Real approximation to the maximum using Laplace's method integral
added 44 characters in body; edited title
Nov
30
revised Real approximation to the maximum using Laplace's method integral
added 27 characters in body
Nov
30
revised Real approximation to the maximum using Laplace's method integral
deleted 45 characters in body
Nov
30
asked Real approximation to the maximum using Laplace's method integral
Sep
25
awarded  Critic
Sep
25
comment Calculating $\int\sqrt[3]{\tan(x)}dx$ and $\int\sqrt[4]{\tan(x)}dx$
$ \int \sqrt{\tanh (x)} \, dx = \tanh ^{-1}\left(\sqrt{\tanh (x)}\right)-\tan ^{-1}\left(\sqrt{\tanh (x)}\right)$
Sep
24
awarded  Scholar
Sep
24
accepted Any idea how to attack this integral $\int_0^{\infty} p^2 e^{-\beta\sqrt{p^2+1}} \, dp$
Sep
24
comment Any idea how to attack this integral $\int_0^{\infty} p^2 e^{-\beta\sqrt{p^2+1}} \, dp$
@DepeHb, none of the integrals of the terms in the series will converge (I tried it).
Sep
24
revised Any idea how to attack this integral $\int_0^{\infty} p^2 e^{-\beta\sqrt{p^2+1}} \, dp$
added 4 characters in body