For example, @geocalc, if you take the curve for parameter $t$ to be the line $(\cos t)x+(\sin t)y = 1$, draw the pictures and tell me what you think the envelope is.
@TP_1: Yes, and there are lots of answers there ...
@TP_1: It's not a named theorem. But it's a correct result. As the answers posted indicate, if the function were positive somewhere, because it's continuous, it would have to be positive near that point and hence would have positive integral.
Yeah, a former student of mine is riding his motorcycle from Arizona to here tomorrow, planning to go on to Baja on Saturday. I think he's insane, but he claims he has appropriate equipment to be safe.
You're student is a crazy person. Also, it looks like there is a digital version, so I just have to log into the VPN, and won't have to brave the 115°F heat!
@Fargle: It's a cool exercise with the implicit function theorem to see what the sufficient condition for existence of the envelope is ... and to apply the theorem.
Oh, interesting, @Xander. Does UCR digitize old books?
@TedShifrin I guess I'd need to go back to your book, then, lol. I just abused the fact that the unit tangent vector on the parametrization of the circle by (cos t, sin t) is (-sin t, cos t), and then found that the line spanned by that tangent vector at a given point (cos t, sin t) of the circle is exactly characterized by $(\cos t)x + (\sin t)y = 1$.
Fair enough, Fargle. You'd see it from the picture, too. The neat thing is that there's a recipe for the envelope of a family of curves $f(\vec x,t)=0$. It's given by the equations $f(\vec x,t) = \frac{\partial f}{\partial t}(\vec x,t) = 0$. That's the point of the exercise.
@TedShifrin Ah, I see. The envelope of the family can be gotten by finding a curve which locally "stays on" whatever curve in the family it's tangent to at that point, hence the partial derivative condition.
I would be fine with remembering every proof and statement in an algebra course, but in the modular forms course, a lot of proofs were just huge computations that I don't really find intuitive
The original Collatz function is defined as the following:
$$f(x)=
\begin{cases}
3x+1, & \text{if $x$ is odd} \\
x/2, & \text{if $x$ is even}
\end{cases}$$
However, I am more interested if the following modified rule can be proven to sometimes wander off to infinity or not:
$$f(x)=
\begin{cas...
@TedShifrin the prof is an expert on modular forms (his papers are refered to by Shimura, among other things) and he gave most lectures without any notes (no idea how he does that)
I only had one oral exam by him before, that was on complex analysis 2. He asked vague questions like "What do you know about the Gamma function?", then I mentioned some stuff and sometimes he asked for proofs but he was happy with me just giving the proof ideas
but there are a lot of proofs in this course that I can't really distill down to some ideas. Some proofs are just "well, you work it out by 2 pages of computations"
Unless by the end of the course a "main idea" has become clear. There are some technical results in every field, where one doesn't ever see an important main idea. To me, those aren't good things for exams.
I'm going to cook. Good luck on your exam. I'm sure you'll do fine.
The original Collatz function is defined as the following:
$$f(x)=
\begin{cases}
3x+1, & \text{if $x$ is odd} \\
x/2, & \text{if $x$ is even}
\end{cases}$$
However, I am more interested if the following modified rule can be proven to sometimes wander off to infinity or not:
$$f(x)=
\begin{cas...
