@JingWeng I do have a technique but it is not for everyone in every situation. You need to know that 'it's not your fault'. Here is a video for you. youtube.com/watch?v=GtkST5-ZFHw
Hi @ShaVuklia I thought you left because you were distracted lol
If I understand right we have hollow sphere and outside may be a electric field, if the sphere is perfect conductive obviously the electric field is null inside...but the potential is defined as the way integral from the reference point to your point so the potential inside the sphere is the same as on it since there is no voltage between any points a&b being inside or on the sphere
Musing: does the term "to figure", meaning "to believe/understand", come straight from the idea of constructing a figure to help one reason through things?
@ShaVuklia Oh, then it's rather easy to explain. When it's spherically symmetric, the field created by each point on the sphere is cancelled by the field created by the opposite point
@ShaVuklia Symmetry implies that both the shape and the charges are symmetrical. Whether they're positive or negative doesn't matter, as long as they're all positive or all negative.
@ShaVuklia Nah, for instance if you take a randomly charged sphere (as in, the charge changes randomly accross the surface), the field inside won't be constant
@Semi I understand that we can deduce the field from the flux when we have this symmetry, but how do we know for sure there won't be a net electric field if there is no symmetry.
For any Gaussian surface which contains the dipole, there will be no net flux. But thr dipole certainly produces a nonzero field (almost) everywhere in space
If I have a series of non-empty, bounded, closed sets $(C_i) \subset \mathbb{R}^n$ such that $C_{n+1} \subseteq C_n \forall n \in \mathbb{N}$, I'd say, without proof, that the limes, I'll call it $C \subseteq C_n \forall n \in \mathbb{N}$, is this true? Are there any exotic examples where this is not true? Perhaps for others spaces?
Point I was going to make is that while the entire surface is equipotential, the local surface charge density is higher where the curvature is greatest
@AlessandroCodenotti Ok is that so? I was never really sure about how to deal with limits of series of sets... @Fargle No, it wasn't what I meant, I've seen a proof that it isn't empty
@Semi yea, because Gauss' law only tells you something about the flux, so you should be able to write your flux in terms of the electric field, to say something about the electric field, right? In the case of the hollow cylinder for instance, we can choose a Gaussian surface such that $E$ is constant, and therefore a zero at the RHS will give us that $E$ is zero.
@JingWeng Because here you want $|x - a|$ to be less than both $\delta_1$ and $\delta_2$, so that you have $|f(x) - K|$ and $|g(x) - L|$ less than $\epsilon$.
We already have our chosen $\delta_1 $ and $\delta_2$, and we choose $\delta=min\{\delta_1, \delta_2\}$, yet how is it possible that we can say $|x-a|<\delta \leq \delta_2$ if we say that $\delta_1 <\delta_2$ ?
Because you found $\delta_1$ and $\delta_2$ to be greater than $|x - a|$. They're the numbers that put $f(x)$ and $g(x)$ (respectively) within $\epsilon$ of $L$.
@JingWeng Draw some pictures on the real line. Visualisation helps to motivate the proofs, and then after that the rigorous proofs are independent of the pictures.
You can't understand the proof by staring at a bunch of inequalities. You need to see pics to really get it.
Of course, the actual way some people do it is to manipulate the pics in their heads, but since they can't tell you how they do that they tell you to draw.
@Ted: I'm having trouble with one of your exercises, the one where you express a vector in terms of the dot product and cross product of that vector with another.
@TedShifrin Oh, well I can do that. There will be a component of $x$ along $a$, with magnitude $x \cdot a$/$\|a\|$, and a component along $(a \times x) \times a$ with magnitude $\|a \times x\|/\|a\|$.
Oh, two different courses. The first is an introduction to differential geometry. The second is some serious analysis (so there are surely Lebesgue integration prereqs for that).
Well, we have a powerpoint that sums up the prereqs for each class, and there's nothing special for math classes, so I'm fairly confident they'll start from what we already know
We have math classes this year... but they're a bit special, we didn't chose them. They're supposed to make sure everyone starts next year with solid bases.
Demonark: To make sure you satisfy university general requirements, etc., a non-faculty member is fine. To guide you through your math curriculum and give you guidance for grad school, etc., a faculty member is highly desirable. But you told me you had such a faculty adviser.
Oh, OK, @Hippa. I don't know what you've done this year, because you've hardly been around here at all.
@TedShifrin The solution set of the dot product equation is exactly the plane perpendicular to $a$ containing $(x \cdot a/\|a\|^2)a$, and the solution of the cross product equation...is more subtle.
That or maybe I mentioned "my adviser" at one point, who wasn't actually faculty? I dunno, but anyway, unless there's a surprise next year we don't have any formal faculty advisers.
I mean, faculty can't be bothered keeping up with rules and regulations, but some of them should care enough to guide you through a decent major, warn about teachers, etc. Maybe that's something Paul Sally did and no one has replaced him.
@Julius: Unless you're particularly working in the plane $A^2$, I don't see why. If I'm talking about multiplying $x,y\in\Bbb R$, I'm going to say that and not $(x,y)\in\Bbb R^2$.
Hmm, I know Fefferman is in the position that Sally used to be, Boller is probably the one best suited to take an advising role, though there's nothing official or scheduled, just that a lot of people tend to ask him things. Though various professors will say very different things
I know it's a one-dimensional curve, but I don't know how I'd describe it without knowing that the magnitude of the cross product involves $\sin \theta$.
