Thanks, but I'm not really sure what to do with the information you just gave me. What does g_{\mu\nu} mean? Do I really need to compute the Christoffel symbols? If yes, how would I compute them from "ds^2"?
All I care about right now is the simplest way to get from "ds^2" to arc length. If I...
I am reading Thurston's book on the Geometry and Topology of 3-manifolds, and he describes the metric in the Poincare disk model of hyperbolic space as follows:
... the following formula for the hyperbolic metric ds^2 as a function of the Euclidean metric x^2:
ds^2 = \frac{4}{(1-r^2)^2} dx^2...
On a finite-dimensional vector space over R or C, is every norm induced by an inner product?
I know that this can fail for infinite-dimensional vector spaces. It just struck me that we never made a distinction between normed vector spaces and inner product spaces in my linear algebra course...
But can you move only two points? Since you are initially restricted to squares, it seems that you would need to move three points at a time.
Now you can see why the median score is so low :) Make an extra assumption and you are down from 10 points to 1. There isn't much partial credit.
That's an interesting approach. I am not sure you can take partial derivatives though because you don't even know that f is continuous, let alone differentiable.
The combinatorial argument is pretty straight forward. Suppose you want to show that f(pt) = 0. Draw a square centered at pt, and...
A1 and A4 have elementary combinatorial solutions. B2 doesn't require anything beyond Riemann sums and the intermediate value theorem, which should be covered in AP calculus.
But now you got me curious - how you would you solve A1 with line integrals?
Putnam problems are "shorter" than IMO problems. On the Putnam you only have half an hour to solve each problem, compared to one-and-a-half-hours on the IMO. That does not mean that the Putnam is easy. The Putnam is scored out of 120 points and in most years the median score is between 0 and 2...
These extra topics are not hard to learn. Most of them are covered in a single college course called Discrete Math.
You might find it enjoyable to take the Putnam recreationally, but I wouldn't waste too much time preparing to take it competitively. One former high school math olympiad...
College students would take multivariable calculus or linear algebra after single-variable calculus. If she really enjoys single-variable calculus, she could also work through a first course in real analysis. It might appeal to her preference for deriving math rather than memorizing it. (I avoid...
Calc 2 will actually use all that trig stuff you learned in pre-calc. Conceptually calc 2 shouldn't be any harder than calc 1, but the problems might be harder. Finding derivatives is straight forward; antiderivatives, not so much.
They might have guided several generations of students through the graduate admission process and observed patterns which students are accepted or rejected. They also know the content of their own letters, which will likely make or break your application :)
You could try to compare your own qualifications to the resumes of current graduate students. Or ask your references about your chances :)
Be careful with overall admission statistics because many programs have a huge discrepancy between their admission standards for domestic and international...
This definition of f^(-1) is usually called a "pre-image". That's the definition I was working with in my last post. It also agrees with your non-standard earlier definition
Let S = f(U). Both definitions allow me to say that f^(-1) (S) = Union[ f^(-1) (f(x)), for all x in U]. Here's an...
First off, all the DisjointUnions were meant to be regular unions (copied and pasted without thinking) but you probably figured that out yourself. There's no need for disjoint unions based on your definition of f^(-1).
No, that's just by the definition of f(U). Instead of indexing over y in...
Thanks for clarifying. I was confused because I was not reading your earlier posts careful enough. My apologies.
Your proof by contradiction is exactly what I would have done.
The obvious solution to your indexing problem is to note that f^(-1) o f (x) = {x} when f is 1-1. Then replace y with...
I think the product topology and box topology coincide for finite products. If you are looking at the space X x Y in particular, then the product topology is generated by the set
\{ O_X \times Y :O_X \in T_X \} \cup \{X \times O_Y:O_Y \in T_Y \}
Finite intersections of these sets are open as...
Can you state clearly how you define your inverse map? And what you are trying to show?
Given only the assumption that f is 1-1, your only shot at defining an inverse is as follows: let f^(-1) be the unique function that satisfies f^(-1) (f(x)) = x for all x in X. (f^(-1) exists thanks to f...
You should go from y is in f(U) to f^(-1) (y) is in f^(-1)(f (U)).
This only makes sense if f is onto. In general there might be points in Y that are not hit by points in X. If your only assumption is that f is 1-1, you can only get a "partial inverse" on the subset f(X) in Y.
I don't know if this is what you are confused about, but I would like to point out that the symbol dx^dy as a differential form has a very different meaning from the symbol dx dy in
\int f(x,y) \, dx dy.
Don't get them confused.
The differentials in an integral are just dummy symbols. They...
You have already seen signed areas in at least two different contexts:
1) Calculus: one interpretation of an integral is the "area under the curve." But it's the signed area. The integral of sin(x) over [0,2 Pi] is 0, not 4. If you computed integrals with absolute areas instead, you wouldn't...
I agree with the conclusion that
\frac{\partial F(x, y)}{\partial x} = 0
implies that F(x,y) = const under two assumptions:
(1) y is a function of x, not an independent variable
(2) \frac{\partial F(x, y)}{\partial x} means "take the partial derivative of F with respect to x" and not...
dx^dy = - dy^dx captures orientation. Minus signs for orientations show up all over the place: Changing the order of two columns of a square matrix changes the sign of its determinant. Changing the order of two vectors in a cross product changes the sign of the resulting vector. The definition...
Because it naturally introduces minus signs in places where you would otherwise need to specify them explicitly.
The most important operation on differential forms is exterior differentiation, denoted d. With the minus signs occurring naturally, you get the nice property that...
As adriank said, there are a *lot* of them. In general real numbers are defined as the limit of a "convergent" sequence of rational numbers, and most of these limits will be irrational. If you want to generate a random real number, just take an arbitrary infinite decimal expansion...
You don't need to learn all of the material ahead of time to breeze through the class. It's enough to familiarize yourself with the concepts and basic results. If you know the "big picture" going into the class, you can focus on filling in the details during lectures and on the homework.
The square picture Jamma points out is probably the easiest way to see non-orientability in dimension 2.
In general, RP^n is orientable for odd n and non-orientable for even n. That's because the antipodal map on S^n is orientation-preserving in odd dimensions and orientation-reversing in even...