Recent content by malicx

Homework Statement So, I have a set S = (x, y) \in \mathbb{R} | (x and y \in \mathbb{Q}) or (x and y \notin \mathbb{Q}) I want to find T = \mathbb{R} minus S so I am negating this and get T = (x, y) \in \mathbb{R} | (x or y \in \mathbb{Q}) and (x or y \notin \mathbb{Q})...

Looks like you'll probably have to use induction on n. 1. Show that the recurrence relation is true for n = 1. (base step) 2. Assume that, for some n in the natural numbers, your recurrence relation holds. Try to show that n + 1 holds as well. (inductive step). It's probably going to...

Perhaps you should try solving directly from the formula for Newton's Law of Cooling. i.e. dT/dt = -k(T - Tambient) where T = temp, t = time, k is a positive constant.

if you are considering the codomain to be (1/2, 1], then since your image is equal to the codomain it is both open AND closed by the very definition of a topology (the null set and the whole set are both open and closed)... but we aren't considering them in R, we are considering only (-1, 1) and...

1 is the endpoint of the set, there ARE no points greater than 1 in im(f). EDIT: the DOMAIN is R, the IMAGE is [-1, 1]

f(V) = (sqrt(3)/2, 1] f(V)c = {x in [-1, 1] | x not in f(V)}. sqrt(3)/2 is not in f(V), so it IS in f(V)c. Thus, [-1, sqrt(3)/2]. If that were the complement, then f(V) union f(V)c would not be the entire interval... which is a basic complement law. right? Call U the universe of discourse...

The complement of f(V) is [-1, sqrt(3)/2], obviously closed, so f(V) is open

I'd like to point out that (2n)! and 2(n!) are completely, completely different things...

So, what you really have is [A|B], where B is the column vector of solutions, right? So when you multiply by the matrix you have by its inverse what you really get is [I|A-1B]. Remember when you right it in the form [A|B], it is really just shorthand saying Ax = B. When A is invertible, you have...

Well... lim log(n)/sqrt(n) = lim log(e^n)/sqrt(e^n), since in either case, e^n and n both go to infinity (one just does it faster!) There are a lot of sequence theorems involved with sequences, functions, and combinations of both but I don't know which you have learned or are expected to know...

I was mistaken about f(V) before, here is what we have: x takes values in R, but f(X) = [-1, 1]. So, f(V) = (sqrt[3]/2, 1]. But that is still open in [-1, 1]. so I am still missing something I think...

I guess you could exponentiate your sequence, show that it goes to one, then take the log at the end.

It is a symmetric square matrix, so if it is invertible it satisfies A-1AT = I. Take a look here, http://en.wikipedia.org/wiki/Symmetric_matrix.

Have you learned about L'Hopital's rules for finding limits of indeterminate forms? (in this case, infty/infty)

Oh of course, if I would have thought about that for 5 seconds more I should have seen it (been a long night!).