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Jul
29
comment Does an $n\times n$ matrix $A$ only have an inverse if $rank(A)=n$? If so, why?
of course, by "general" I mean general invertible $3 \times 3$ case.
Jul
29
answered Does an $n\times n$ matrix $A$ only have an inverse if $rank(A)=n$? If so, why?
Jul
29
comment Proving Hyperbolic Equation
btw, $\cosh x = \frac{1}{2}(e^x+e^{-x})$ and $\sinh x = \frac{1}{2}(e^x-e^{-x})$ can be seen as the even+odd decomposition of the function $e^x$ as it is manifestly clear that $e^x = \frac{1}{2}(e^x+e^{-x})+\frac{1}{2}(e^x-e^{-x})$.
Jul
29
answered Proving Hyperbolic Equation
Jul
29
comment harmonic functions on the disk which agree on the real are identical?
Ah, there was an error in my answer somewhere, nice.
Jul
28
awarded  Enlightened
Jul
28
awarded  Nice Answer
Jul
28
reviewed Approve About the boundary conditions of the Black-Scholes-Merton PDE
Jul
28
comment reoder basis vectors to get 'more diagonal' representation of NxM matrix
I see, well, I guess you need to decide what is first then sort through your columns and select all the first columns and put them at the start of your new list. Then, look for what comes next according to the order as indicated by the pixelation scheme. Put all those second.... keep sorting until you have them all in a nice tidy list.
Jul
27
comment Eigenvalues of Matrix Product.
$det(AB) = det(A)det(B)$ and $det(M) = \lambda_1 \cdots \lambda_n$ so there is a relationship. If $AB=BA$ then they share e-vectors, with possibly different e-values... there is much to say here.
Jul
27
comment reoder basis vectors to get 'more diagonal' representation of NxM matrix
see pages 174-175 of supermath.info/LinearNotes2015.pdf essentially the idea is to choose the inverse image of the range basis to start your domain basis then stick the basis for the kernel at the end. This gives you an identity matrix padded with zeros.
Jul
27
reviewed Reject Is this a valid example of a non-euclidean Sierpinski attractor?
Jul
27
comment Does a mathematical construct exists which explains all theories?
@sasha category theory is very general, if I understand correctly, it includes just about everything you can think of, or, at least gives a language in which those things can be reformulated. As to complexity theory, I don't know details... but, I'd be surprised if could not be fit... I'll leave the details to those who have studied these things properly :)
Jul
27
comment Does a mathematical construct exists which explains all theories?
As the years go on, more and more physics is decided in computer simulations. Some even go so far as to say physics should be computer simulations (this is roughly what Wolfram says in the big book). This much we can say, physics is incomplete thus any existence theorem about "where" physics fits has to deal with the inevitable ignorance of physics.
Jul
26
comment Calculus of Variations transformation
I usually assume this is because we are implicitly working on some jet-space where $y'$ is not really the derivative of $y$ with respect to $x$. This is relevant to variational calculus where we often take derivatives with respect to $y'$ and for example $f(x,y,y') = y$ would have partial derivative w.r.t. $y'$ of zero; $\frac{\partial f}{\partial y'} = 0$. Sort of like, but even more frustratingly in my experience, the independence of $z$ and $\bar{z}$ when those are used as a complex notation for real derivatives...
Jul
25
awarded  Yearling
Jul
25
comment Sanity check: smooth structure of tangent bundle
The set $TU$ described below is open in the bundle. I can't see using smaller sets than $T_pM$ since over a given point $p \in M$ you want to allow for all possible $v \in T_pM$ in $(p,v)$. Using an open set smaller than all of $T_pM$ would put some tangent vectors outside the chart of the tangent bundle. That would defeat much of the purpose of studying $TM$. So, we use adapted charts.
Jul
25
comment Sanity check: smooth structure of tangent bundle
good news, you have not lost your mind.
Jul
25
answered Sanity check: smooth structure of tangent bundle
Jul
25
comment Finding a Gradient Vector given only a derivative and direction
@Vlad just the point in question, but, since we are given the direction and magnitude for $(\nabla f)(P)$ neither the formula for $f$ nor the details of $P$ particularly matter. That was my knee-jerk read of this.