What do users here follow, wrt questions: New tab, or Active tab. I have always followed "new" questions. But it seems like many follow "active" questions, and hence the concern about bumping posts via edits or a community bot. Poll? It seems that a lot of users follow "active questions". I'd consider changing my preference for new, if I understood better, the value of "active". Or at least I'd consider using both, in any given day. @TedShifrin What are your thoughts?
koro's problem also a classic. great stuff. happy thanksgiving, everybody.
stuff like that feels harder if you have to give 'formulas' to see it. fairly intuitive that a curve can bounce between two curves that share a common non-horizontal tangent at a point without 'settling down' enough to make the curve 1-1 on a neighborhood of the point.
a lot of that stuff is in the 'after a bit of thought, no reason to expect it to be true' ballpark and then you just have to find formulas if you want them. which tend to suck and involve lots of arbitrary choices.
i will hit the wine shortly. my consumption has been unusually light lately. something is even more wrong than usual. my usual accomplice was traveling for a while which induced the break.
i like a drink every now and then, but its much more enjoyable with company.
i mean company that pretends to listen.
Reading Terrence Tao is depressing for me. I know he's a wunderkind, but still, it reminds me of a comment Norbert Wiener made
yeah, if you're trying to fit it into a program of study, more detail might help those with opinions. i don't have strong opinions, because to me most of those books (interpreting those topics broadly) are kind of the same
in particular, I liked how $\sin x $ was approximated by a polynomial of degree $5$ using inner products with better accuracy than Taylor's polynomial of the same degree on $[-\pi, \pi]$
Leslie: Actually I was referring to a solved example in Axler's wherein he approximates $\sin x$ by projecting it on space of polynomials of degree 5 using an inner product.
Yes, you are right. But in the exercise, he also tells to use a symbolic calculator to retain $\pi$'s etc. which in the solved example he replaced by numerical approximations.
but i blew my chance by snorting when the high school principle announced to the council that they were removing the model of a ww 1 fokker from the local studio
i asked if they realised that ww 1 & ww 2 were separated by a few years in addition to some other minor details?
just one simple question, Does ln(x) continuous at infinity? According to the condition that states a function is continuous if both the limit and its value at that point equal and exist
@EnthusiastiC no, infty is not real, nor is log defined there (since it is not real). the limits are equal, but that does not extend the log function to non-reals.
@EnthusiastiC You can define an extension to the log function anyway you want, as long as you don't expect it to obey the relations that log obeys.
if you want to talk about an extension to the log function that is defined on an extension of the reals, you need to define those extensions. You can work with the two-point compactification of the positive reals, and define $\log(0)=-\infty$ and $\log(\infty)=\infty$. Then $\log$ is continuous on the compactification.
So with the proper extensions, the answer to your question is yes.
Suppose I write one of 1~7 on each tile on infinite tile. Is it possible to write 1~7 so that every time I see 3X3 subtile, all numbers 1 to 7 appear on the boundary 8 tiles.
@AdilMohammed oh, that is not really a short cut, as you are computing all the same products and adding them. It is just a way to remember the signs used. And it is only for 3x3 matrices.
Define numerically stable. From the perspective of mathematics, the semantics are the same, and the result is the same. I assume you're referring to some limited subset of the model of numbers such as floating point arithmetic.
Anyways, so has it been considered whether or not division by zero manifests a number line perpendicular to the reals as well as the imaginary numbers?
In mathematics, in general topology, compactification is the process or result of making a topological space into a compact space. A compact space is a space in which every open cover of the space contains a finite subcover. The methods of compactification are various, but each is a way of controlling points from "going off to infinity" by in some way adding "points at infinity" or preventing such an "escape".
== An example ==
Consider the real line with its ordinary topology. This space is not compact; in a sense, points can go off to infinity to the left or to the right. It is possible to turn...
So compactification is just another way of saying that a topological space is converted from one of infinite space for every dimension to one of bounded indefinite space?
Regardless, I don't know enough about complex analysis to give a proper answer to your question. I'm just considering division by zero as an operation and its immediate conclusion as a result of further contemplation.
If we go back to first principles and consider division as asking "how many times can I subtract one number from another before it equals zero?", then for $\frac{0}{0}$, we could say that there are infinitely many answers, and generalize or build upon this as a foundation for numerators other than zero such that you do not end up with the contradiction that every number is equal to every number.
So any real number greater than or equal to zero must intuitively be a valid solution to $\frac{0}{0}$ since I can substract zero from itself any arbitrary quantity and end up once more at zero, including a total of zero subtractions. I suppose zero would in some way here be considered a "minimal" or "optimal" answer, but we would have to extend the idea of functions to operations on sets which yields sets and state that all reals greater than or equal to zero is the answer to $\frac{0}{0}$.
@robjohn Haha maybe at your level it may not be a shortcut but at my level it's like using l'hopital for differentiation to save time😂
Also Rob, I have a HS doubt, if f(x)=4x³-18x²+27x-7 and f'(x)= 3(2x-3)² would you say the function is increasing throughout the domain? Or throughout the domain except 3/2?
Why is there no good way to handle division by zero?
I've seen all the videos on it, heard all the usual things about it, but it just feels like all the avenues haven't been considered.
Or at least, if they have, something's been missed because someone thought it was interesting for a second and then dismissed a possibility out of bias.
No matter how you try to define it, you get contradictions. That is why in almost all, if not all, computer languages, division by zero raises an exception.
In mathematics, a well-defined expression or unambiguous expression is an expression whose definition assigns it a unique interpretation or value. Otherwise, the expression is said to be not well defined, ill defined or ambiguous. A function is well defined if it gives the same result when the representation of the input is changed without changing the value of the input. For instance, if f takes real numbers as input, and if f(0.5) does not equal f(1/2) then f is not well defined (and thus not a function). The term well defined can also be used to indicate that a logical expression is unambiguous...
Bruh, zero is an anomaly, all things considered. All this time and it still has no definition for division by zero. You would think something so fundamental to the set of reals would have a definition under the operation of division.
Correct, @Lobic. As I say, I came in in the middle. If you want the domain to be all of $X$, then you have to add that condition (for all $x$, there exists $(x,y)\in f$).
@robjohn I got a mini iPad long before I ever got an iPhone.
The phone is too small for reading and typing. I use it only when I'm out and about.
@TedShifrin At first you can see there is a line connecting from point a to point b where point both a and b contains 0 charge and c got charge $Q^+$ but c is not connected to any of them. You can distribute charge from the line.
At second all of the position is connected by line which means charge can be distributed.
C will do something more at A than at B so you shouldn't have equal charges but I don't remember the details. I think I did these kind of problems in my class 12.
With one wire joining two point charges, at equilibrium they have the same charge.
@Koro: My claim was that A had to balance B+C, and by symmetry B,C should be equal. So A=Q/2, B=C=Q/4. But that may be wrong. Does it matter if all the charge starts off at C or at A?