And in general, we do talk about Math. If it doesn't interrupt another conversation (which is common decency), a conversation about math is perfectly fine.
in other words, it's like if I am sitting at a random point and I swing a large bar into a mountain parallel-style, and I want to know where the collision point is where it first makes contact
@robjohn Hi. Just dropped in quickly to see what was happening, sparked by the recent meta post. Did similarly on sci.math yesterday and it has really deteriorated now that most of the non-cranks have migrated here. Ah for the good ole days.
@t.b. Yes, that is how I wanted it. I have been thinking about this because it gives 5 decimals in common with the the decimals of the first Riemann zeta zero.
If $A$ and $B$ are square matrices such that $AB = I$ where $I$ is identity matrix. Show that $BA = I$. I do not understand anything more than the following.
Elementary row operations.
Linear dependence.
Row reduced forms and their relations with the original matrix.
If the entries of the mat...
@QED Oh, this one. We discussed this here of course. It's interesting that Bill could identify which proof you were referring to even with the sparse hint. =)
@JohnSmith Could you try putting your "huge disgusting function" in PasteBin or some similar service? I'm now really curious as to how you obtained this function in the first place...
@Srivatsan As I said, I thought about it far too much. It's a FAQ in math forums at this level. Every time I revisit such questions I try to inject something fresh into them. Be sure follow the link to sci.math to the literature references if you want to understand it more generally.
So, there's a really barebones question here, and it's exactly the same notation and essential problem as this much more fleshed-out question here. I wonder where it comes from.
@BillDubuque Yes, your answers are in general thoughful -- and the ones QED mentioned are no exception. I try to follow your answers and subsequent pointers to the extent that I understand. :-) (And hi!)
So if you imagine the "guy" sitting in the air next to one of those contours, and he swings his "bar" into it, there is a particular contact point against the surface
Does anyone know how to edit BIT.LY links? Alas, some of my sci.math links in said thread have rotted, they need to change from WWW.GOOGLE.... to GROUPS.GOOGLE...
wolfram alpha adds a level of indirection, when it doesn't give you what you want you don't know if you stated the problem in a way it doesn't interpret
@J.M. Here on such rotted link from my comment in Pete L. Clark's post in that thread bit.ly/Shift1-1notOnto If your replace "www" by "groups" it resolves fine.
@Srivatsan I'm happy to hear that you found those posts helpful. Sometimes I never know since my posts often get few votes since they appear later than the FGITWs. If there is something that you don't understand then please do ask questions since it will help me to clarify the post, which will help many other readers. It's difficult to "go backward" and remember what the conceptual difficulties were when one was learning topics.
@J.M. The problem is that the short syntax I used was never "officially" supported. If I had a good contact at google perhaps we could get a fix. But it seems that google doesn't care about usenet groups generally, since the support has gone downhill as of late.
JM: Is there an iterative approach I could take, perhaps? Like finding the equation of the line through (x0,y0) at z0 and see where it best intersects the function at a particular range if I have a rough idea where it should intersect?
@robjohn Indeed, it seems that google was more interested in using usenet newgroups to gather users for its own "groups" than it was in supporting usenet newsgroups and their archives.
@Cam I wouldn't worry too much about it. But it would be helpful to readers to add a remark, e.g. "Note to readers: this is essentially the same as Matt's answer, except...." Even if you think they are equivalent, they may not prove equivalent to students, since one "trivial" difference might make the answer more comprehensible to a student. What's trivial to experts may not be so to students.
@robjohn Yes, the search has gone seriously downhill. It seems that the many of the servers have latency issues, so the search results often timeout (the DB search is apparently distributed across many servers). I often cannot find old posts that I know are in the archive. Dejanews never had that problem.
@J.M. I tried all the obvious permutations long ago with no success. The good thing about the groups search engine is that you can search on various fields, e.g. I usually use author:dubuque when I want to search only my posts.
@robjohn Ha! Luckily for you it didn't also improve your flaming skills. Many of us had to make great efforts to tone it down when moving here after hanging out in the wild, wild west for far too long. This is like a church compared to sci.math.
technically NSolve[{(y-100)/(x-100) = -(D(f,x))/(D(f,y)), f=(my huge disgusting function)},{x,y}] assumes the tangent is a true tangent, right? As opposed to a best-approximation
@robjohn I don't flame in real life either, but if you hang out long enough in a forum like sci.math the cranks and charlatans can get to you. E.g. see some of Tim Golden's threads, which seriously riled even Arturo.
@JM Oh, you won't see them with that intensity of color with the naked eye, even with a telescope. The low-light receptors (rods) are not very sensitive to color.
The camera can pick up the color as if the intensity of light were greatly magnified.
@BillDubuque Riling Arturo sounds to be a pretty hard thing to do. I don't know if I've actually been in too many "discussions" with him here or on sci.math, but from what I've seen, he is pretty level headed.
@BillDubuque :-) My son is going to college in Chicago, but if we decide to make our way further east, I'd like to visit Princeton once more and see Boston. I've never been to Boston, but I have a friend here who went to MIT. He tells me horror stories about the traffic there.
@Tim I wouldn't say "preferred". I've only used PARI/GP thrice, and Sage once. What I'm saying is that if you still don't feel comfy after a few hours of poking around an environment, you could try switching...
@BillDubuque He was mainly a pedestrian/bicyclist, so he viewed things from a pedestrian vs car standpoint. Perhaps that has improved since he was there.
@tb Thanks. I think I displayed that mostly it was a matter of not being detailed enough.
@tb Didier pointed out a missing ' and suggested changing $x$ to $x(u)$ in a few places, which I did as it makes the fact that $x$ is being considered a function of $u$ clearer.
@robjohn Yes, certainly. Sorry for not following up on my promise from yesterday. It is a bit sad that the neat computation is now a bit hidden in clutter... Anyway: all is well that ends well :)
@JohnSmith This came up yesterday in here. You might want to try the main site. The involvement of z seems superficial. You have an implicit equation f(x, y) = 0 for a curve in the plane and you want to calculate arclength along it. To me it seems like you need to find parametrizations for the curve. I don't know of a general procedure for doing this. In examples it probably isn't so hard. But it sounds like you want something general. That would be an interesting question to ask.
@JohnSmith This theorem can give you an answer to "Onto which axis can I project the curve (in a small neighborhood) to get a parametrization?" but it's not so clear to me how you can find this neighborhood.
The proof uses the inverse function theorem, and the proof of that which I remember relies on this. Which is some iterative process. Maybe you could explore that?
like if my arbitrary origin is at point px, py and my radius is R, landing on point x,y, and I sweep counterclockwise theta degrees -- being able to translate that into cartesian coords isn't simply rcostheta rsintheta etc