Let $Exp_t^{[y]} (x) $ denote the $y$ th iteration of the exponential function with base $t$ : $t^x.$
For example $Exp_t^{[1]} (x) = t^x. $
Let ~~ denote best fit.
Now as $x$ Goes to positive infinity and a pair $(a,b)$ with $e<a<b$ Is given , I wonder how to find the best fit base value $C$ ...
for a 4 year old mathematics might be counting to 10(or 100), and that would ineed be something to be proud of. Nontheless is counting not all, neither is proofing or implementing.
I don't know if "art" is quite right either. Sure, there is a palpable aesthetic nature to it, but that isn't the primary function of math, at least to me.
@saturate I didn't just say numbers. Generally the kind of math I like is described moderately well as "study of pictures". But of course it's hard to explain, let alone define all of mathematics in one line.
this is because mathematical objects can be the subject of inquiry, a toolbox to investigate the subject of inquiry, and "atoms" and "laws" that govern an abstract world
music is a combination of soundwaves given a subject meaning, which is why it is one of the few things that lies at the intersection of art, linguistic and science
@Fargle The obvious trouble is that you then have to define when a point belongs to a line. Equivalently, define when you'd consider two lines to be "the same".
So, at least one of $\pi+e$ and $\pi-e$ is transcendental; at least one of $\pi+e$ and $\pi e$ is transcendental; at least one of $\pi-e$ and $-\pi e$ is transcendental.
Is it true that $a$ is transcendental iff $-a$ is transcendental? I think so because you can always substitute $x$ with $-x$ in the generating polynomial...
@MeesdeVries Something like that. But it'd not be a proof by contradiction argument.
@DHMO Yes, its false in $\Bbb R$.
Complex analytic functions have a bit more rigidity to them than real analytic ones. Not every real analytic function on $\Bbb R$ extends to a complex analytic one, precisely for the reason you said.
@saturatedexpo Every function $\Bbb R \to \Bbb R$ extends to a function $\Bbb C \to \Bbb C$. But not every analytic functions extends to an analytic function.
No worries. In general, a surprising theorem in analysis says every continuous function on a closed interval can be approximated, uniformly, by polynomials.
I'm trying to get a big cross which I can subscript in order to denote a generalized cartesian product (much like how \bigcup works for generalized unions). How can I accomplish this?
my tutorers used $(M^{},\circ )$ if they wanted to notate the set M without those Elements that would "destroy" the Group. Like 0 in multiplication. So $(\mathbb{R}^{},\cdot )$ would be simply $(\mathbb{R}\backslash {0},\cdot )$. But one could find bigger exclusions easily. My question is: is there a more international (yet short) notation for "exclude all elements that would destroy this group"
@saturatedexpo you can't destroy a group you haven't defined yet, so maybe that's not the best way to phrase it. if M is a monoid, then $M^\times$ might be used to denote the group of invertible elements. this applies in particular with the multiplication operation in any (unital associative) ring.
$M$ can be different types of things in different contexts, but I can't off the top of my head think of another usage for $M^\times$. I've even seen $V^\times=V\setminus 0$ for vector spaces $V$, which is a slight generalization.
perhaps you want $(M,\circ)$ to be a monoid, then $M^\times$ can be the largest subset for which $(M^\times,\circ)$ is a group sharing the same identity element as $M$.
note the last condition is necessary. for instance there are subsets of $M_n(\Bbb R)$ (that's $n\times n$ real matrices) which are groups under multiplication but do not have the identity matrix, e.g. $\{{\rm diag}(1,0,\cdots,0)\}$
Am I right to say that a homeomorphism $f \colon M \to N$ of orientable topological manifolds with orientations $(\mu_x)_{x \in M}$ and $(\nu_y)_{y \in N}$ is orientation-preserving iff $f$ maps $\mu_x$ to $\nu_{f(x)}$ in homology?
also is there a relation to the elements $\mu_K \in H_n(M, M \setminus K)$ which restrict to $\mu_x$ for all $x \in K$ for compact sets $K \subseteq M$?
I mean, if $f$ is orientation-preserving, does $f$ map $\mu_K$ to $\nu_{f(K)}$ for all compact subsets $K$ of $M$?
if you don't specify that $\circ$ is unital and associative to begin with, then crazy things can happen
for instance the loop of unit octonions has many subgroups isomorphic unit quaternions, all of which are "maximal" in the sense they can't be extended to bigger subgroups, but there's no reason to prefer one subgroup over another so there is no uniquely defined subgroup
I've read some pages (some of those that I can read) from Chen Nanxian, Mobius Inversion in Physics, World Scientific 2010, as a Google Book, and I recommend you, section 1.3.4.2 in pages 22 and 23 because seems to me incredible. Enjoy with it, @user1952009
@arctictern i guess what i wanted to say is: a field is not always better then a ring. But now you still can only guess what i mean with better. Got to go, but that will rustle my jimmies haha
Let $(V,\|\cdot\|)$ be a normed vector space. Is it true that, if $\{x_1,\dots,x_n\}$ is a linearly independent subset of $E$ then $\|x_i\|\leq \|x_1+\dots+x_n\|, \forall i=1,\dots,n$?
Given a commutative ring $R$ and an $R$-module $M$, can you call another $N$ module free with respect to $M$ if it is isomorphic to a direct sum of $M$'s and get a reasonable notion?
Given the function $f(x)= x^{\frac{p(p-1)}{2}}-1.$ Also let $p$ be an odd prime number. If $\epsilon$ is a number $\pmod{p}$ such that
$f(\epsilon) \equiv 0 \pmod{p}$. Then how do I prove ( prove because Hardy and Wright use it but don't prove it ) that
$$ f(\epsilon)\equiv 0 \pmod{p^2}$$
let z=a+bi. (a+2)^2=(a-2)^2 a=0 the equation becomes (2+bi)^n - (bi-2)^n = 0 ((2+bi)(bi-2))^n - ((bi-2)(bi-2))^n = 0 (-b^2-4)^n - (bi-2)^(2n) = 0 now 2+bi and bi-2 has the same radius. Let 2+bi=re^(iu). bi-2=re^(ipi/2-iu). niu = nipi/2-niu+2ikpi u = pi/4 + kpi/n
don't ask me how to continue; i have no idea
@teadawg1337 no. if you make two edits within 5 minutes without any comments interrupting, it will be registered as one edit.
"any comments interrupting" means that if you edit and then someone comments and then you edit again, it will be counted as two edits
I'm trying to reconcile two seemingly-contradictory statements re: holomorphic 1-forms on a Riemann surface
On one hand, I see statements like "Given a compact Riemann surface R of genus g, we have a basis $\{\omega_1,\cdots,\omega_g\}$ of holomorphic one-forms on R"
In the sequence with the Bockstein homomorphism, I just computed that the map $H_n(X;\mathbb{Z}) \to H_n(X;\mathbb{Z})$ induced by the multiplikation $\mathbb{Z} \to \mathbb{Z}, x \mapsto k\cdot x$ is just multiplication with $k$, is that correct?
@Semiclassical Yeah, you just caught it. There is a splitting of de Rham groups into Dolbeaut groups, and holomorphic forms live in H^{1,0}. There is a duality which says that the dimensions of H^{p,q} and H^{q,p} are the same.
@MikeMiller What I'm confused about then becomes why, in a certain context, it seems like one needs to consider the de Rham group H^1 instead of the Dolbeaut group H^{1,0}
Let $Exp_t^{[y]} (x) $ denote the $y$ th iteration of the exponential function with base $t$ : $t^x.$
For example $Exp_t^{[1]} (x) = t^x. $
Let ~~ denote best fit.
Now as $x$ Goes to positive infinity and a pair $(a,b)$ with $e<a<b$ Is given , I wonder how to find the best fit base value $C$ ...
I'm trying to get a big cross which I can subscript in order to denote a generalized cartesian product (much like how \bigcup works for generalized unions). How can I accomplish this?
Given the function $f(x)= x^{\frac{p(p-1)}{2}}-1.$ Also let $p$ be an odd prime number. If $\epsilon$ is a number $\pmod{p}$ such that
$f(\epsilon) \equiv 0 \pmod{p}$. Then how do I prove ( prove because Hardy and Wright use it but don't prove it ) that
$$ f(\epsilon)\equiv 0 \pmod{p^2}$$
