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 Yearling
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  • 87 votes cast
Feb
9
accepted Calculus over integers for derivative/integral of factorial?
Feb
9
accepted Why do mathematicans care so much about the incompressible Navier-Stokes equations?
Feb
9
accepted If $(-1) \cdot (-1) = +1$ shouldn't $(+1) \cdot (+1) = -1$?
Feb
9
accepted Why is there only a complex conjugate, but no real conjugate?
Feb
8
asked Is a random variate a functional?
Jan
15
comment Is every square matrix a tensor of 2nd order?
"The total number of indices required to uniquely select each component is equal to the dimension of the array, and is called the order, degree or rank of the tensor." (see en.wikipedia.org/wiki/Tensor#As_multidimensional_arrays). But searching through the article the term "order" seems to be the most used one.
Jan
14
asked Is every square matrix a tensor of 2nd order?
Jan
12
comment Why is there only a complex conjugate, but no real conjugate?
Aren't the roots of $X^2 -1 = 0$ also symmetric, so $X = 1$ and $X=-1$ is an arbitrary choice?
Jan
12
comment Why is there only a complex conjugate, but no real conjugate?
@null Yes. I corrected the text.
Jan
12
revised Why is there only a complex conjugate, but no real conjugate?
added 2 characters in body
Jan
12
asked Why is there only a complex conjugate, but no real conjugate?
Dec
23
comment If $(-1) \cdot (-1) = +1$ shouldn't $(+1) \cdot (+1) = -1$?
Could one not assume Positive × Negative == Zero and Negative × Positive == Zero ?
Dec
23
comment If $(-1) \cdot (-1) = +1$ shouldn't $(+1) \cdot (+1) = -1$?
@Pierre-GuyPlamondon The most symmetric (commutative) solution would be $(+1) \cdot (-1) = 0$.
Dec
23
asked If $(-1) \cdot (-1) = +1$ shouldn't $(+1) \cdot (+1) = -1$?
Dec
22
accepted How to generalize the Thue-Morse sequence to more than two symbols?
Dec
18
awarded  Yearling
Dec
9
comment What is the advantage of the Fourier Transform over the Hartley Transform?
I don't think so. A cosine transform is simply the real part of a Fourier transform. But the real part of the Fourier transform can be computed from a Hartley transform like $Re[F(\omega)] = (H(\omega) + H(-\omega))/2$. So only the even part of the Hartley transform is equivalent to a cosine transform. This is because the Hartley transform kernel is a shifted cosine function, which is not symmetric around the origin.
Dec
7
asked What is the advantage of the Fourier Transform over the Hartley Transform?
Nov
15
awarded  Tumbleweed
Nov
8
revised Why do mathematicans care so much about the incompressible Navier-Stokes equations?
added 332 characters in body