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4h
answered Proof of the integral operator in $L^2(\mathbb{R})$ being self-adjoint “by hand”
21h
revised Can we prove that matrix multiplication by its inverse is commutative?
edited body
21h
answered Can we prove that matrix multiplication by its inverse is commutative?
1d
comment Wanted: Simple integration theory
If the function $f$ is bounded and continuous a.e. on $[a,b]$, then $f$ is Riemann integrable, and any such sequence of sums over equally-spaced partitions will have to converge to the Riemann integral.
1d
comment Two conflicting answers: Problem in linear algebra involving quotient spaces and T-invariant subspaces
@AbeerAbassi : That answer is gone. But keep in mind that if you start with a two dimensional space $\mathbb{C}^{2}$ and you have a one dimensional subspace $U$, then $\mathbb{C}^{2}/U$ is one dimensional and, hence, any linear operator on this space is a constant multiple of the identity.
1d
comment Is taking the real part required in vector orthogonality and projection?
Why would it not be good to define orthogonality to mean that the real part of the inner product is $0$? Because that definition would lead to $e^{i\theta}x \perp x$ for some $\theta \in [0,\pi)$.
1d
answered Two conflicting answers: Problem in linear algebra involving quotient spaces and T-invariant subspaces
1d
answered The minimal poly of $T_w$ divides the minimal poly for $T$
1d
awarded  Organizer
1d
comment How Kriging, Bochner theorem and Positive definite (PD) function are related?
I added a tag. There may be a better one, but you need something to do with Statistics I think. Wish I had time to look into this more. Very interesting topic and a good question.
1d
revised How Kriging, Bochner theorem and Positive definite (PD) function are related?
edited tags
2d
comment How to show $\Im\{z\cot(z)\}$ is not $0$ in the first quadrant?
@DavidC.Ullrich : Maybe this is the only reasonable way to prove such a thing, but it sure seems strange. You get a nice representation using Herglotz' theorem that results in a classical sum expansion of $\cot$ from just knowing the residues. You can, of course get it from the sum expansion, but that's worse than the indirect method I mentioned above!
2d
comment How to show $\Im\{z\cot(z)\}$ is not $0$ in the first quadrant?
No luck trying identities. I gave up.
2d
comment Is there a deep reason why replacing $\cos(x)$ with $e^{ix}$ and taking the real part often makes a contour integral work out?
@DanielMcLaury : Or, you may apply the construction to the $\cos$ function only, and allow the rest to come along as a multiplier. Lots of choices.
2d
answered Differentiation and punctured neighbourhoods
2d
revised Is there a deep reason why replacing $\cos(x)$ with $e^{ix}$ and taking the real part often makes a contour integral work out?
deleted 4 characters in body
2d
comment Is there a deep reason why replacing $\cos(x)$ with $e^{ix}$ and taking the real part often makes a contour integral work out?
@DanielMcLaury : I've added more to explain why this sort of thing is to be expected, and some conditions under which it does.
2d
revised Is there a deep reason why replacing $\cos(x)$ with $e^{ix}$ and taking the real part often makes a contour integral work out?
added 1123 characters in body
2d
answered Is there a deep reason why replacing $\cos(x)$ with $e^{ix}$ and taking the real part often makes a contour integral work out?
Jul
30
comment “Every function can be represented as a Fourier series”?
If you're going to sample the signal over an interval $T$ of time, then the Fourier series with exponentials $e^{i2n\pi/T}$ for $n=0,\pm 1,\pm 2,\cdots$ will represent the signal over $[0,T]$. Obviously outside of that interval $[0,T]$, the Fourier series extends periodically, even though the signal may do something else.