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Apr
14
awarded  Notable Question
Apr
12
awarded  Nice Question
Apr
11
awarded  Nice Question
Dec
28
comment If $f,g$ integrable then $f(x-y)g(y)$ integrable for almost every $x$
This is called the convolution. It is very important. Try to derive some of the other properties found at en.wikipedia.org/wiki/Convolution
Dec
10
comment Supremum of supremum…
Why did you delete your other question?
Dec
7
comment Help understanding Weyl's proof of Heisenberg's Inequality
@SamTugendhaft Or don't. That's cool too.
Dec
5
comment Help understanding Weyl's proof of Heisenberg's Inequality
@SamTugendhaft You're in the Schwarz class, so there shouldn't be boundary terms. Make sure to upvote and select my answer.
Dec
5
comment Help understanding Weyl's proof of Heisenberg's Inequality
Integration by parts?
Dec
4
answered Help understanding Weyl's proof of Heisenberg's Inequality
Nov
24
comment Andrew runs 6 miles everyday of the week except Sunday.
Andrew's knees are probably broken.
Nov
22
comment How many integers between $10000$ and $99999$, inclusive, are divisible by $3$ or $5$ or $7?$
@0.5772156649... Do you know about Project Euler?
Nov
17
comment Computing the double integral
@RobertIsrael Do you think this user knows about Lebesgue?
Nov
17
comment Computing the double integral
What have you done?
Nov
17
comment Factor $x^2+x+1+i$ in $\mathbb{C}[x]$.
If those are the roots... then $(x+i)(x-(-1+i))$?
Nov
17
comment Initial Boundary Value Problem for Wave Equation
What have you tried?
Nov
16
awarded  Popular Question
Nov
16
comment Find a matrix X∈V such that U∩W=span{X}.
Probably not how you're supposed to do it... but if you think of these as vectors in $\mathbb{R}^3$, then the span of them is a plane. So, the problem is equivalent to finding the intersection of two planes, which would be a line (or a vector times a parameter). Get this by taking the cross product for each set of two vectors to get the normal, and then take the cross product of THAT to get the vector you're looking for.
Oct
29
comment Evans 2ed Chapter 6 Problem 9
Since $u=0$ on $\partial \Omega$, there is no tangential part to the derivative.
Oct
26
comment Under what conditions is $|Du|=|\frac{\partial u}{\partial \nu}|$ at a point?
$u=0$ means there is no tangential derivative.
Oct
25
revised Obtaining Positive Solutions by the Method of Characteristics for a First Order Linear PDE
deleted 17 characters in body