I have seen many times here on math.SE that some users write the word application instead of the word function. Does this come from some language where the word for map/function is similar to the English word application?
You might be right. French Wikipedia article says: "En théorie des ensembles, une fonction, ou application, est une relation entre deux ensembles pour laquelle chaque élément du premier peut être mis en relation avec un unique élément du second."
I do not really understand what it says, but "une fonction, ou application" is " a function or application" according to google translate. So French probably use these two words for function.
There's not much pollution in NYC. It's more common in industrial centers. LA is in principle more polluted, but I live closer to the ocean and farther from the factories.
That's thrice the farthest I have gone from my home, I think, and it was going approximately from one edge to the other of India. Sometime I forget how big US is.
How can you express the volume of a cylinder in spherical co-ordinates? Would you have to split up the polar angle into two cases? i.e. If $0 \leq \phi \leq \frac{\pi}{4}$ then $0 < r < h \sec(\phi) $ and when $\frac{\pi}{4} \leq \phi \leq \frac{pi}{2}$ then $0 < r < a \csc(\phi) $ where $h$ is the height of the cylinder and $a$ is the radius?
that's not a definition. in any case, the word you want is "with radicals". I.e, just using +, -,division, multiplication and taking nth roots for whatever n you want.
Abel proved there is no solution for the general quintic in terms of just those.
if there is no solution for the general quintic in terms of the operations in $\mathbb{Z}$ doesn't that mean there is no solution for the general quintic in $\mathbb{Z}$?
Hi. I just posted my first question. I am nervous as my question can be "dumb" or "unformatted" So I wonder if I could get some advice on "asking appropriate question" before I get downvoted, link
@BlueBag It's a perfectly fine and not dumb question. Formatting looks about alright. To answer the question: yes, there are functions $f$ such that $f(a)$ and $\lim_{x \to a} f(x)$ are not the same. Example: $f(x) = \text{sign}(x)$ where $\sign(x)$ is the sign of the real number, i.e., $f(x) = -1$ if $x < 0$ and $f(x) = 1$ if $x > 0$ with the convention that $f(0) = 1$. (i.e., $0$ is assumed to be positive).
symplectomorphic's comment below your post is an example where the limit is wrong no matter which direction you approach it from, if you have seen it, @BlueBug.
What I am being confused is that if I approach from negative I get $ x *( -(x-1) ) $ but if I approach from positive then I get $x(x-1)$ the equation looks different because of the absolute sign
I mean... I think the value comes out to be the same but the equation looks different
@BlueBug: you're overthinking it. take a calculator and compute $x |x-1|$ for some value of $x$ a bit below $1$. now take a different value $x'$ that is a bit closer to $1$, but still below. do it again and again. what happens? do the same from above.
notice how the Riemann integral of $\int_0^1 \ln(x)dx$ is $\sum_{k=1}^n \frac{1}{n}\ln(k/n)$. in general, $\int_0^1 f(x)dx$ will have Riemann sums $\sum_{k=1}^n \frac{1}{n}f(k/n)$.
now with $\sum_{k=1}^n\frac{1}{n}\frac{1}{1+k/n}$, you need to think backwards to figure out what integral this is a Riemann sum of. specifically, what function is being integrated? hint: $x=k/n$.
are the sets $\mathbb{Q},\mathbb{C},\mathbb{R}$ all constructed from $\mathbb{Z}$? they all seem to share that their associativity under + is a consequence of the associativity of $\mathbb{Z}$ under +
@JoeStavitsky I've never seen the usage on that page before. Just think of them as thought bubbles in the middle of equations to record indeterminate forms, that's apparently how the author is using them there.
Take the double earring, subspace of R^2, consisting of circles of radius $1/n$ with center at $(1/n, 0)$, union circles of radius $n$ with center at $(n, 0)$ (for all $n$). Denote this by $X$. Consider the map $X \to X$ which sends $i$th circle to $i+1$-th.
Mapping cone of this map is the same as mapping cone of the identity map, nope?
But I am pretty sure this map is not homotopic to the identity map.
So, hmm, here you go: take unit circles centered at each integer on the real line. Give it a subspace topology. This guy is homotopy equivalent to wedge of countably many circles. Now take a map $X \to X$ which takes $i$th circle to $i+1$th circle.
Does this sort of result/property have a name?: Let $q$ be prime. $f(1) \equiv 0 \pmod q$ if and only if $x-1 \mid f(x)$ in $(\mathbb Z/q\mathbb Z)[x]$.
i don't think there's anything stopping you from doing $n<0$, but i also don't see any reason it'd give you something new i.e i expect $\mathbb{Z}/(-n)\mathbb{Z}=\mathbb{Z}/n\mathbb{Z}$
Hello :) Someone here good with function in set theory?
During the lecture we learned this phrase: "If F function then $F^{-1}$ function iff F function injective (one-to-one)." But why? what with onto? F don`t need to be Surjective (onto)?
If you're hell-bent on leaving $B$ is the codomain, then correct. But if $f$ is injective, it can be made to be invertible by replacing the codomain with the image.
no I'm saying even if you leave the codomain as the image, the map is $f: A\to B$ not $f: A \to f(A)$ if it were the latter then it would be surjective, too, by definition.
@MikeMiller Great, so all three of my examples work :P Good point about the identity and reflection, much simpler than the automorphism of wedge of two circles.
@Balarka: A more interesting question: Find an example of mapping tori that are homeomorphic but such that their defining homeomorphisms f, g are conjugate (so there's no homeomorphism h, inverse h', with hfh'=g)
@Tobias To me they are equivalent (at least in the context I'm usually in as I don't have to deal much with complements of things with a group structure).
I'm asking because in my exercise group I told them to use $-$ and when they asked why, all I could say was "it formats nicer" and "you dont mix up quotient with difference"
@Tobias $(A-B)/(C-D)$ vs $(A\setminus B)/(C\setminus D)$, although I admit taht is a very deliberately constructed example. But also $A^C \cup B^C= (X-A) \cup (X-B) = X-(A\cap B)$ sort of looks nicer
looks nicer because there is more space to seperate the larger expression somehow dont know
Speaking of conjugacy. If $A$ is a square matrix then $AA^T$ and $A^TA$ have the same characteristic polynomial (being conjugate). But if $A$ is not square it seems like these still have the same characteristic polynomial up to a power of the variable. Does anyone know if this is true?
@TobiasKildetoft a weaker version of that: Taking $A$ to be $m\times n$ with $m\geq n$, must the characteristic polynomial of $A^T A$ divide that of $AA^T$?
@Semiclassical Certainly, evaluating the characteristic polynomial of $A^TA$ in $AA^T$ gives something that annihilates $A^T$ from the right (though not necessarily $0$)
My motivation towards this question is literally most manifolds are hyperbolic and I don't think the classification is known in any case (which is kind of crazy).
It's also (well-known) open if every hyperbolic 3-manifold admits a tight contact str.
@Semiclassical Hmm, actually what I just said also says something about the power of the variable in the larger polynomial (since we get a bunch of eigenvectors of value $0$ from it)
@MikeMiller Essentially they consider for a knot $K$ which is hyperbolic they can prove that if the HFK for it is nice enough with respect to the contact invariant they can look at $p/q$ surgery on the knot as $0$ surgery on $L(q,p)$ (maybe a framing error), and show the contact invariant is still nonzero for a suitable contact surgery.
So I think for suff. large surgeries on k(=P(2,-3,2m+1)?) they can show they have a tight contact str.
