Euler....IS_ALIVE
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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