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Jul
14
comment $O\in M_{3}(\mathbb{R})$ is orthogonal and $\det O=-1$ then $\lambda=-1$ is an eigenvalue of $O$
The eigenvalues of a general orthogonal matrix may be complex. Do you already know the properties of unitary matrices? Then you can obtain the properties of real orthogonal matrices by noting that they are just unitary matrices with all entries real.
Jul
14
comment $O\in M_{3}(\mathbb{R})$ is orthogonal and $\det O=-1$ then $\lambda=-1$ is an eigenvalue of $O$
The determinant is the product of the eigenvalues. What do you know about the eigenvalues of orthogonal matrices?
Jul
14
comment Explicitly writing out a differential 2-form
Note: The equality sign in my last comment of course only applies if it is an alternating form, of course.
Jul
14
comment Explicitly writing out a differential 2-form
@AndrewLedesma: If you defined it with $\wedge$ it automatically is alternating. If you defined it with $\otimes$ you have to test it, but then you sum over all $ij$ anyway: $\sum_{ij}\omega(e_i,e_j) dx^i\color{red}\otimes dx^j = \sum_{i\color{blue}\le j}\omega(e_i,e_j)dx^i \color{blue}{\wedge} dx^j$.
Jul
14
comment Explicitly writing out a differential 2-form
Does it fail? As far as I can see, $\omega(e_1,e_1) = \omega(e_2,e_2) = \omega(e_3,e_3) = 0$, $\omega(e_1,e_2) = -\omega(e_2,e_1) = p^3$, $\omega(e_1,e_3)=-\omega(e_3,e_1)=0$, $\omega(e_2,e_3)=-\omega(e_3,e_2)=0$. So why do you think it fails?
Jul
14
comment $\nabla \cdot \color{green}{(\mathbf{F} {\times} \mathbf{G})} $ with Einstein Summation Notation [Stewart P1068 16.5.27]
Also note that $\epsilon_{ijk}=\epsilon_{jki}=\epsilon_{kij}$. You want to rotate them in a way that the indices for the derivatives come in front of the indices for what they are applied to.
Jul
14
comment $\nabla \cdot \color{green}{(\mathbf{F} {\times} \mathbf{G})} $ with Einstein Summation Notation [Stewart P1068 16.5.27]
@LePressentiment: Now your calculation is correct. In your first two questions you still name the results from the incorrect calculation, though. Now, how do you check that the actual terms are $G\cdot(\nabla\times F)$ and $F\cdot(\nabla\times G)$? Well, the easiest is to just write those two vector expressions in components and compare. Alternatively, whenever you have an $\epsilon_{ijk}$ where $j$ and $k$ are contracted with vector components $X_j$ and $Y_k$, that's the $i$-th component of the vector product of $X$ and $Y$, $(X\times Y)_i$. Basically you apply the rewrite rules backwards.
Jul
14
answered When does it make sense to say that something is almost infinite?
Jul
14
comment When does it make sense to say that something is almost infinite?
@GerryMyerson: No, it isn't. There's absolutely nothing quantum in twin primes. Quantum superposition is a well-defined concept which is not identical with "we do not know" (indeed, it is in some sense the antithesis of it).
Jul
13
comment Zech's logarithms - Why are they called “Zech”?
The only mathematician with that name I could find is Julius August Christoph Zech (1821–1864). Unfortunately the internet doesn't seem to have any more information about him. The dates and the fact that he was mathemnatician and astronomers are all I could find (he's listed in the German Wikipedia page for the name Zech, but unfortunately there's no dedicated Wikipedia page about him). Is is of course possible that this is just a name coincidence.
Jul
13
comment Zech's logarithms - Why are they called “Zech”?
From the "'s" part, I'd guess they were invented or studied in detail by some person whose last name was "Zech"
Jul
13
answered Logic/Intuition behind the Uniqueness Theorem
Jul
13
comment Do De Morgan's laws hold in propositional intuitionistic logic?
Do you mean that as "it's another way to write $\subseteq$", or as "while it is something different, in this context the differences are not important"?
Jul
13
comment why 0=0 is not possible??
@JMCF125: Thank you.
Jul
13
awarded  Nice Answer
Jul
13
comment Do De Morgan's laws hold in propositional intuitionistic logic?
What is $\le$ in $W\cap U\le V$?
Jul
13
answered Is it possible to eliminate $g(x)$ in the formula $\dfrac{\int g(x)a(x)}{\int g(x)b(x)}$
Jul
12
comment what does $|x-2| < 1$ mean?
Ah, that way it makes sense.
Jul
12
comment what does $|x-2| < 1$ mean?
"it means that x is at a distance of at most 1 from 2 and it never reaches 2." Of course it reaches 2. It won't reach 1 or 3, though.
Jul
12
answered Reference a single element within a set