@JessyCat It looks like you're going down the partial fractions route in the work you added. I don't have much to say there, on the grounds that I don't think it gives nearly as a simple a route as what I sketched.
@Semiclassical it gives a route that is more connected with what I've learned about the Laplace transform pairs. I just wish that that Kitty person understood what I was asking about. She keeps on thinking I'm talking about what happens for $e^{s-b}$, and I'm actually talking about $e^{-bs}$.
Fair enough. For me it boils down to the three functional relations that I indicated earlier (though now I'm repeating myself). Might add those to my answer, since that's about all that makes sense for me to add. (She's got the partial fractions route covered, so I'm not inclined to have my answer overlap with that.) @JessyCat
Hello, I have a question regarding spectra (in homotopy theory). Let X be an omega-spectrum. The notation Ω^∞(X) is usually meant to mean the 0th term of the spectrum, is this correct? Am I then right to interpret Ω^(∞-1)(X) as the -1th term? This notation is used in the statement of the Madsen-Weiss theorem, where X is a certain spectrum MTO(d). In this setting, is MTO(d) an omega-spectrum? If not, can we interpret Ω^∞(X) for any spectrum X (not necessarily an omega-spectrum)?
@Fargle i'm trying to understand why i can preserve spatial relationships if i shift the x- and y-values of two points by the same amount -- but i can't preserve them if i shift the angle and eccentricity of the two points by the same amount.
@Adrien Nope. Remember that the point of Omega spectra (other than being fibrant) is that the map X_i -> Omega X_{i+1} is a homeomorphism. So if you want to undo an Omega, you go one further up the tower, not down.
If I remember right, we can say there's a space TO(n,d), a certain Thom space. There's a natural map TO(n,d) -> Omega TO(n+1,d). Then the zeroth degree space of MTO(d) is the limit of Omega^n TO(n,d), and the kth degree space is Omega^{n-k} TO(n,d). It is indeed an omega spectrum.
@Fargle i have points A, B, and C that fall roughly on a circle.
@Fargle i want to know whether B is systematically closer to A than C is.
@Fargle but from observation to observation, A varies randomly and B and C also vary, independently of A.
@Fargle what i want to do is move all my points A so that they are all at the origin, and then look at how far the Bs are from this origin relative to how far the Cs are.
@Fargle i don't see how that works. if i multiply the eccentricity of all my As by 0 -- to get them at the origin -- and then multiply the eccentricity of all my Bs by the same amount, this doesn't preserve distance.
@Fargle and because (0, 0) is on average closer to all points on a circle centered at 0 and than any individual point on the circle, this means that my Bs are closer to my As than my Cs are just by virtue of being closer to the origin.
What's the name of the property of complex functions that the image of am intersection of contours under f preserves the angle between the tangent lines?
because B is closer to the origin than C -- and B and C are otherwise always about at the same angle -- B will on average be closer to the corresponding A
@dbliss On half the circle, A is closer to C. On half the circle, A is closer to B. This holds true even if you allow things to lie "roughly" on the circle (i.e. on some annulus).
@Fargle i'm trying to talk about minimizing average distance from all points on the circumference of a circle.
@Fargle i think (0, 0) would minimize that, for the unit circle.
@Fargle mean distance = 1.
@Fargle i don't know what the mean distance for (1/2, 1/2) is, but it doesn't seem obviously shorter.
user228700
Hi, everyone :-) Is anybody interested to help me to find the domain and range of compositions of two piece-wise defined functions? Do ping me if u are...
@MikeMiller @Mike Miller Thanks for your help. MTO(d) is as you described, except that in my reference, the zeroth degree space of MTO(d), Omega^{infinity}MTO(d), is defined as colim Omega^{n+d}TO(n,d), note the +d.
I am looking at a solved exercise but I got stuck at a point...
We have tht each continuous function on $D$ is bounded. We suppose that $D$ is not closed. Then there is a sequence $(x_n)$ in $D$ with $x_n\rightarrow b\notin D$. Let $f: D\to \mathbf R$ with $f(x)=1/(x-b)$. Then $f$ is an unbounded continuous function on $D$. This is a contradiction.
@MikeMiller We are taking the limit of the sequence X_0 --> Omega(X_1) --> Omega^2(X_2) --> ..., right? So we want a universal object X with maps from each Omega^i(X_i) into X, i.e. a colimit.
Let $f(x)=\arcsin (x)$ To calculate the interval $f([-1,1])$ we have to check if $f$ is descresing or incresing or not? Or can we say that since the $\sin(x) : \mathbb{R}\rightarrow [-1,1]$ then for the inverse function it holds that $f([-1,1])=\mathbb{R}$ ?
@MaryStar Regarding your previous questions: yes, x_n is arbitrarily close to b by definition of what it means for the sequence (x_n) to converge to b.
let be $N(x) = \sum\limits_{d|p - 1} {\mu \left( d \right)} {F_x}(d)$ that $p$ is a prime number and ${F_x}(d) = \frac{{{x^2}}}{{4d}} + O(x{p^{\frac{1}{2}}})$
that $x+1=g(p)$ and $g(p)$ is the least primitive root modulo $p$. Applying Brun's method to $N(x)$ in conjunction with ${F_x}(d)$ in orde...
1) Took it with my phone 2) Transferred it to my laptop using an app called xender, its really efficient 3) Uploaded it with the "upload" button near chat bar
@Adrien Let $g=\sin (x) : \mathbb{R}\rightarrow [-1,1]$. We have that $g^{-1}([-1,1])=\{x\in \mathbb{R} \mid g(x)\in [-1,1]\}$. Since $[-1,1]$ is the range of $g$ we have that $g(x)\in [-1,1]$ for all $x\in \mathbb{R}$. Therefore $g^{-1}([-1,1])=\mathbb{R}$. Is this justifiction correct?
@Adrien So, we have the function $f(x)=\frac{1}{x-b}$ and since $x_n\rightarrow b$ there are $x\in D$ arbitrarily closed to $b$, so we conclude that it is unbounded. Is this correct?
[Abstract algebra] a(bc)=(ab)c Interpretation: Given two maps $a,b$ with some given domain, the image of the two maps acting in sequence to $c$ is the same as their composition. In other words $a \circ b= (a \circ b)$ Therefore nonassociativity can mean there exists some $c$ such that the map $ab$ is not decomposable into a composition of maps $a$ and $b$
More generally, we can treat a binary operation together with some element $x \in M$ in a magma $(M,\cdot)$ as an action from the underlying set $M$ to itself. Then $x \cdot : M \rightarrow M$ represents a left magma action by $x$ and $\cdot x: M \rightarrow M$ represents a right magma action by $x$.
Then for some elements $a,b,c \in M$ such that $(a\cdot b)\cdot c=a \cdot (b \cdot c)$, it means the left magma action $ab\cdot$ can be decomposed into a left magma action by $b$ followed by $a$
Equivalently, we can equally say that the right magma action of $bc$ can be decomposed into right magma actions of $b$ followed by $c$
therefore, nonassociativity will mean there is some element $c$ such that $a\cdot b \neq ab$ and likewise there is some element $a$ such that $b\cdot c\neq bc$
Further discussion requires more in depth understanding of nonassociative algebras, which I currently don't have yet
There's also one more interpretation of associativity that reminds of Light's associativity test: We can say that if for all $b$, $a\cdot (b\cdot c)=(a\cdot b)\cdot c$, then $b$ act like some kind of center of the algebra wrt the order of binary operation, that is, the brackets () commute under $b$.
As for the distributive law $a(b+c)=ab+bc, \forall b,c$ it means the left magma action of $a$ is a homomorphism of the $+$ structure. Likewise, $(b+c)a=ba+ca, \forall a,c$ means the right magma action of $a$ is a homomorphism of the + structure.
Therefore if an algebra is distributive, then multiplication is a homomorphism of addition
Now that we knew distributive law implies multiplication is a homomorphism of addition, it is now easy to see that if some element is an additive identity, so are their products
@Semiclassical I have a piece of rubberband, I hold a small object with it, and pull the whole thing. I want to compute the velocity of the object once I release the pull. The framework should be that the kinetic energy of the object after the release should be equal to the potential energy of the rubberband, yes?
