Search results

I am assuming that "cone" means a set like {(x,y,z): x >= 0, y^2+z^2=c^2x^2}, rather than the usual solid with a circular hat, otherwise the problem is trivial. Under this assumption, the answer is no. There is probably a nicer way to do this, but I will sketch my argument below. Pf: One face...

If you want to avoid using angles, an alternate method is as follows: Let d be the length of the diagonal, and let p be the length of the perpendicular from a vertex to the diagonal that does not pass through it. By considering the area of the triangles the diagonal bisects the rectangle into...

By invariance of the properties in question under rotation and dilation about the origin it suffices to consider disks of radius r < 1 centred at z0=1. Then for any fixed angle w in (-pi,pi], we can consider the ray z=t*cis(w). Finding the distance of this ray from z0 is optimising a single...

Although angles are dimensionless for the purposes of dimensional analysis, a choice of "angular unit" must be made in order for an angle to be considered as a real number, a necessary step in constructing trigonometric functions which we can use calculus on. (In fact when we do things...

A "function" in the usual sense of the word assigns an element of the codomain (real numbers here) to every element of the domain (real numbers also), there is no reference to any particular units in the function or the plane itself. That's why lengths/areas/etc are measured in units/units^2/etc...

What do you mean by this?

Why is that last sentence a reason to not care about how much you learn in your degree and how properly you learn it? It is up to the individual to weight his/her priorities with what he wants out of uni (replace uni by any aspect of life). This goes equally for He-Mann saying that that the...

Strongly disagree for higher math courses than the basic ones where you just do computations. Studying smart not hard is far more efficient*, and "rote learning" becomes utterly unfeasible eventually. (*I suppose this is a tautology based on your definition of studying smart, but I am factoring...

Idk what they want you to say with that wording and those assumptions really. Basically it just means that the image of f'(a) is a subspace of the kernel of g'(f(a)), or equivalently that the image of f'(a) is orthogonal to the gradient vector of g at f(a).

What was your end answer?

$Prove that there exists a unique \emph{non-negative} sequence $(x_n)$ with $x_0=0$ and $n=x_n(x_{n-1}+x_n+x_{n+1})$ for $n>0$. $

It really doesn't take that long, especially since to prove discontinuity you can just show that approaching the origin via different axes gives a different result, so most terms in the mixed partials drop out from x or y being zero. I would definitely go via this route rather than by restating...

They are, but so what? These are not the hypotheses of Clairaut's theorem.

Re: Several Variable Calculus The partial derivatives are all zero, but the directional derivatives are as you say for u1 nonzero, and zero if u1=0. This is one of those funny functions that you don't see the full weirdness of by just looking at 2d slices of its graph. Each partial derivative...

Re: Several Variable Calculus What book/notes are these? Anyway, yeah as I said, you need some kind of niceness condition on the domain for these things to be well-defined. You should be able to see what this condition is by going through the proof of Green's theorem, unless it is presented...

Re: Several Variable Calculus Well, what kind of regions was Green's theorem stated/proved for? For any domain A that is nice enough that Greens theorem works, you will get that the RHS boundary integral is equal to the integral of the indicator function 1_A of A. This can be taken as the...

Re: Several Variable Calculus Are you sure you are writing the question correctly? Is there an established convention in your book/notes for what "planar region" means? My guess is that it means things like the regions bounded between two lines x=a,b and g(x) < y < h(x). Otherwise your...

No worries, hope it helped :). I haven't actually worked with Brownian motion before but some aspects of it seem like they would be pretty intuitive.

Handwavy explanation using a particle moving with Brownian motion: Note B(t) is the particles position at some intermediate time t. If it ends up at B(s), then B(t) has a tendency to be in the same direction as B(s) from the origin (this is the dependence statement). This means B(s)+B(t) will...

Z not R but yes. a|b means that there exists an integer c with ac=b, i.e. a divides b.