Hi WannabeNewton. Your explanation is very clear, but I don't think you got the point of my comment. Most students don't consider existence theorems to qualify as "explicit". So while the implicit function theorem shows that x and y are local coordinates, it does not give an explicit...
Because no matter how you word it, it is the number of ways to partition a set of size N into subsets with respective sizes n1, n2, ..., nk. And there is no "ordering" of the subsets, so you interpret it in any order you like.
Suppose you have the surface given by the equation
(x^5y^2+3x^3y^4z^2 - 5)^2 = 1.
At most points, a parametrization of this surface exists in principle, but that does not mean you can explicitly find it. And if someone comes on here and solves this equation to prove me wrong, then simply add a...
In the context of undergrad math, I think "rigor" refers to a focus on Definition/Theorem/Proof both in lecture and homework. Any course that is part of the standard undergraduate curriculum (past linear algebra) will be rigorous in this sense. That includes analysis, algebra, topology, and...
Both sides are positive so the inequality must hold if and only if the square of both sides satisfies the same inequality. That might simplify it a bit.
So, the point of dividing by 2pi is that instead of measuring in radians, your angle units go from 0 to 1. So you just have three variables:
Sa, Sb, Sc
These measure how far around each gear is. So, in these units, you have:
Sa = t,
Sb= pt+q,
Sc = rt+s,
where p, q, r, s are all...
Hi Bobs,
I would suggest ditching x and y coordinates and just focusing on the angles:
t,
ft+j,
gt+k,
You want to know whether there is a value of t for which these all differ by a multiple of 2pi. Let's simplify it more. Take j' = 2pi*j, k'=2pi*k, and t'=2pi*t. (In other words t',j',k'...
No, it is not quite valid.
The transformation is fine. The problem is that the rectangle in the xy plane does not transform into a rectangle in your new coordinates.
r and theta usually denote polar coordinates, which these are not, so it is clearer to change the notation a bit.
x = uv, y =...
Bigfoot, your solution is incorrect. The trajectory should have constant speed and yours decays exponentially. The solution should reach the origin in finite time.
Also, the equation that Halls wrote down is an example of a first order homogeneous equation. So you could solve it using another...
I reread the definition again just to clarify to myself whether strong implies weak. And it does. The norms are only used to restrict the class of variations. So the "stronger norm" is more restrictive on the allowed variations. Thus it produces a weaker form of extremum.
They aren't reversed. Strong is called strong because it allows a wider class of variations. To be a weak extremum you only have to check differentiable variations.
It would be nice to hear some book recommendations. I love Gelfand Fomin, but a more advanced presentation that assumes more...
Ok in polar coordinates, set up the origin as the center of the square. After any amount of time the ants will still form a square with the same center. So you can draw the velocity vector of which points inward at a forty five degree angle with the radial direction. And its length is 1. using...
f(x,y) = x^2 + 100xy +y^2
This has a saddle point at the origin. If you look at f(t,t) you get 102t^2, which is positive. If you look at f(t,-t) you get -98x^2 which is negative. So you have both positive and negative values near the origin.
Yeah, there are a lot of connections. Cool, isn't it.
Since you are talking in terms of cross products and parallel vectors and such, I thought I would mention one other thing in case you didn't know this already. If you have n vectors v1, ... vn in n dimensional space ( \mathbb{R}^n ), then...