Reputation
6,617
Top tag
Next privilege 10,000 Rep.
Access moderator tools
Badges
6 29
Newest
 Yearling
Impact
~112k people reached

2d
comment Prove this result relating to the sign of a permutation
Can you prove that sign is well-defined? i.e. given 2 representations of $\phi$ as a product of transpositions, can you prove they have the same sign?
2d
comment Determine whether the following set is a vector space
What are the axioms for a vector space?
May
24
comment A really basic integration question concerning differentials
This is an example of treating differentials as if they were ordinary real numbers: If we have the equation $3x=9$ then we divide by $3$ to get $x=3$. This is not defined for differentials, but if we pretend it is then we get $dv=adt \implies \frac{dv}{dt}=a\frac{dt}{dt}=a$, for example.
May
24
comment A really basic integration question concerning differentials
NB: When I use the word incorrect, I don't mean that it gives the wrong answer. Just that it's completely unjustified. Lots of things in physics are unjustified, and that shouldn't alarm you.
May
24
answered A really basic integration question concerning differentials
May
23
comment Vector calculus identity for $\nabla\times(\vec{b}\cdot\nabla)\vec{b}$
Thanks, I look forward to it!
May
22
comment Vector calculus identity for $\nabla\times(\vec{b}\cdot\nabla)\vec{b}$
EDIT: It turns out the identity is false. See the above for a counterexample (I leave verifying it as an exercise to the reader!)
May
22
revised Vector calculus identity for $\nabla\times(\vec{b}\cdot\nabla)\vec{b}$
added 146 characters in body
May
22
comment Taylor series in two variables?
No, it's fine. Notice that $\frac{dg}{du}=\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}$
May
22
answered Taylor series in two variables?
May
22
reviewed Approve Taylor series in two variables?
May
22
comment Vector calculus identity for $\nabla\times(\vec{b}\cdot\nabla)\vec{b}$
Perhaps I was being a bit hopeful in saying this hopefully isn't too hard... I can't seem to verify it! I'll leave this up for the time being and come back to it when I work it out.
May
22
revised Vector calculus identity for $\nabla\times(\vec{b}\cdot\nabla)\vec{b}$
deleted 66 characters in body
May
22
comment Vector calculus identity for $\nabla\times(\vec{b}\cdot\nabla)\vec{b}$
@MathNewbie That is not the case, since $\nabla$ is an operator.
May
22
answered Vector calculus identity for $\nabla\times(\vec{b}\cdot\nabla)\vec{b}$
May
16
comment What is the density of $(X,X)$
This is only one random variable, do you mean the vector $(X,Y)$ where $X$ and $Y$ have the same distribution?
May
16
comment solve a fairly simple equation
Indeed, if r=0 then we can't deduce x either (it can be anything provided d=0)
May
16
comment solve a fairly simple equation
There's still a good deal of simplification to be made here.
May
16
comment solve a fairly simple equation
(NB: If $d=0$ then your equation becomes $x=x-r$, from which we cannot deduce the value of $x$.
May
16
answered solve a fairly simple equation