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Mar
5
comment Show that the linear function f(x)=Ax is differentiable at a
To begin, the fundamental theorem of linear algebra tells you that there exists $A$ an $n \times n$ matrix for which $f(x)=Ax$. Of course, $A$ is just the standard matrix of $f$ as we know and love from linear algebra...
Mar
5
answered Show that the linear function f(x)=Ax is differentiable at a
Mar
1
comment Expand $(\vec{A}\times \nabla)\times \vec{B}$ using tensorial notation
I'm sorry, I have trouble with this one. I need to stew on it a while, perhaps someone will swoop in and solve it before I finish this.... I haven't found the right way to write the operator expression in the Einstein notation just yet.
Mar
1
comment Expand $(\vec{A}\times \nabla)\times \vec{B}$ using tensorial notation
I think we need to antisymmetrize on the $\vec{A} \times \nabla$ term as to pick up the product rule. Unfortunately the usual formula $\vec{A} \times \vec{D}_k = \epsilon_{ijk}A_iD_j$ assumes that $\vec{A}$ and $\vec{D}$ have commuting components. Of course in this context that is false as $\vec{D}= \nabla$ is an operator so... we have to respect that.
Mar
1
comment Expand $(\vec{A}\times \nabla)\times \vec{B}$ using tensorial notation
I'm not yet sure how to do it with the tensorial approach, I need to ponder it a bit, but, I hope you can see from my current half-answer why there are three terms. You get one from raw derivatives acting on $\vec{B}$ then the terms like $\partial_2A_3$ act on $\vec{B}$ under a product rule hence producing two terms; $2+1=3$.
Mar
1
answered Expand $(\vec{A}\times \nabla)\times \vec{B}$ using tensorial notation
Feb
27
comment Multi-complex arithmetic in MATLAB?
I should talk to you more about what you are doing... let me digest that paper a bit.
Feb
27
comment Multi-complex arithmetic in MATLAB?
I don't know about any built-in commands, but, I could derive a matrix representation that you could build in matlab if interested.
Feb
26
comment Goursat's Theorem and Real Differentiability
you should read math.stackexchange.com/questions/833471/…
Feb
26
comment Goursat's Theorem and Real Differentiability
as the comment above might lead you... note that the difference between real and complex differentiability in the plane is the fact that the complex case allows the differential to be written as a multiplication of a complex number. In contrast, real differentiable allows many other possibilities, you need two complex numbers and the product of $w-z$ as well as the conjugate to capture real differentiability... sorry if this comment is cryptic, I'll try to find something to link.
Feb
25
comment Is it ever easier to show differentiability than continuity?
for example: math.stackexchange.com/q/447104/36530
Feb
25
answered Is it ever easier to show differentiability than continuity?
Feb
23
comment Reduction of order and lost in arithmetic
I haven't worked out this problem recently, but, perhaps multiplying by $\sqrt{x}$ then differentiating would make the calculation of $y'$ and $y''$ less annoying. That said, you are aware there is a formula for generating the second solution from the first for such a problem. However,the formula is not easy either. See page 86 of supermath.info/DifferentialEqns.pdf
Feb
8
comment Does $f(x,x)=0 \: \forall\; x\in \mathbb{R} \: \implies f(x,y)= g(x,y)(x-y) $?
Very nice, I just love calculus a bit too much...
Feb
8
comment Does $f(x,x)=0 \: \forall\; x\in \mathbb{R} \: \implies f(x,y)= g(x,y)(x-y) $?
@CameronWilliams here's an ugly question, for which set of functions is (2.) true?
Feb
8
answered Does $f(x,x)=0 \: \forall\; x\in \mathbb{R} \: \implies f(x,y)= g(x,y)(x-y) $?
Feb
8
comment Does $f(x,x)=0 \: \forall\; x\in \mathbb{R} \: \implies f(x,y)= g(x,y)(x-y) $?
I can prove $(1.)$. Just set $g = f(x,y)/(x-y)$ for $x \neq y$ then $f = (x-y)g$... I'll have to think about 2.
Jan
26
comment Matrix representation induced by quotient space
The quotient space identifies everything in the kernel as a single point. For your $A$ the kernel is one-dimensional, it follows that we lose one dimension. Another example, if $A$ was just a single vector glued together three times then the kernel would be two-dimensional and the quotient space would be just a one-dimensional space. More to the point, the set of vectors you list is not linearly independent in the quotient.
Jan
25
answered Matrix representation induced by quotient space
Jan
25
revised Matrix representation induced by quotient space
improved non-native English, I think.