akmather

I think a better way of thinking of that (in the context of hyperbolic geometry) is to think of it as "putting a boundary on the universe" (or the plane) But any point on that boundary is a "point at infinity" and so what value are you assigning? There are still "infinitely" many of them!

For problem 2: If every vector is of the form [math] \vec{x} = \begin{bmatrix}a\\a-b\\2a+3b\end{bmatrix}[/math], then [math] \vec{x} = \begin{bmatrix}a\\a-b\\2a+3b\end{bmatrix}=a\begin{bmatrix}1\\1\\2\end{bmatrix}+b\begin{bmatrix}0\\-1\\3\end{bmatrix}[/math] Hence a basis for your subspace is: [math] \left\{ \begin{bmatrix}1\\1\\2\end{bmatrix},\begin{bmatrix}0\\-1\\3\end{bmatrix} \right\}[/math] As the two vectors are certainly independent, the dimension of the space would be two. I know this is an old problem, but I didn't see any harm in putting down a solution at this point and I needed the latex refresher...

Anybody else ever taken any of the subject GRE's? If so, what did you think of them? I took the Math Subject last November and thought that it was pretty hard. most other students that I know who have taken the Math also thought it quite hard. Especially in comparison to the general GRE. Which is very easy. Just curious if the other subjects were similar, as I have never had the opportunity to talk with someone who had taken one.

So I am working through Hoffman and Kunze, and in the chapter on canonical forms I am having some difficulty due to the lack of examples. Using another book, I figured out how to calculate the Rational and Jordan canonical forms for a given matrix (linear operator). And I assume that the so called "rational decomposition" has something to do with the Rational Canonical Form? But what exactly is the relationship? For a finite dimensional vector space V and linear operator T, There exists r non-zero vectors in V and r respective T-Annihilators so that the direct sum of the cyclic subspace generated by the vectors with respect to T equals V AND the T Annihilator for each vector divides the next one. Hard to say in words, but I couldn't find the LaTex stuff. I haven't been on here in a while, did that go away? Anyways, like I said, I am sure it has something to do with the canonical form, just hoping someone can point me in the right direction. I am currently trying to figure out from the proof of the rational decomposition theorem where he is getting the vectors, but haven't yet seen the connection to the canonical form. Any help would be greatly appreciated.

At my university, it isn't. The only difference is that it takes two semester to take the college algebra - trig sequence, while the pre-calculus course is a one semester intensive course.

There also is the whole thing of context. If there is an exponential in the expression, you can pretty much assume that "log" is base e. Even at lower levels. Like in the expression that started this whole thing. [imath] a^x = e^{x\log a} [/imath] It is pretty obvious what is meant. The clarification is pretty much unnecessary, but why bother confusing a Calc 1 student over it? Incidentally, in my opinion most of the reason that math books don't bother writing "ln" is because in straight math I do not think that you will hardly ever see any log but base e after you get into calculus. In the few astronomy classes that I took though, they used both base 10 (written as log) and base e (written as ln).

Yes, but in the context of the original post, which did appear to be in a calculus class, it probably was a good idea to clarify it. I mean, just by the question being asked it was fairly apparent that this was not a discussion in higher math. It was a calculus discussion. So I think it much better to denote [math] log_e(x) [/math] by [math]ln(x) [/math]

Don't mind me, I am just trying to write my previous stuff in this LaTex and see if I can make it work. [math]y = a^x[/math] [math]\ln{y} = \ln{a^x}[/math] [math]\ln{y} = x\ln{a}[/math] now, differentiate implicitly [math]\tfrac{dy}{dx}(\tfrac{1}{y}) = \ln{a}[/math] [math]\tfrac{dy}{dx} = y\ln{a}[/math] now, substitute for y [math]\tfrac{dy}{dx} = a^x\ln{a}[/math] so, [math]\tfrac{d}{dx}a^x = a^x\ln{a}[/math] Allright, I think I got it. Thanks!

Ok, I am new here so I do not know what standard convention is, and if this is an inappropriate response feel free to delete it; but in answer to the original question, to differentiate a^x you can use impicit differentiation. (The other idea posted is probably somewhat easier, but someone seemed not to understand it so this is another way to look at it) That is, if y = a^x ln(y) = ln(a^x) ln(y) = xln(a) (now differentiate impicitly) y'/y = ln(a) y' = y ln(a) (substitute the original equation for y) y' = (a^x)ln(a) so hopefully that makes sense. Incidentally, forgive the new guy here, but how are people writing their mathematical stuff so nicely as I do not see the proper tools here in the reply box, and if I try to paste in from somewhere else it does not work? Edit: I also now notice that this is a very old post. Sorry....