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...