Q: If we have a weird (irregular)-shaped function, can a definite integral only be approximated or can it actually be precisely determined (by a human)
Anyhow: How about this. The minimal polynomial is the polynomial $m(x)$ with the property that it divides every polynomial $f(x)$ satisfying $f(A)=0$ ?
Oh so I've been wondering recently, how exactly does analytic continuation work? I've heard about it all the time as how you extend domains of functions, but like, what's the mechanism by which you do so?
So basically the functioning principle is identity theorem.
Take a holomorphic function $f$ on a disk $D$ around $0$
If it's not too wild, you can find a point $p$ on the boundary of the disk such that $f$ can be defined on the disk $D_1$ around $p$
i.e., the Taylor series of $f$ is going to define a $g$ on $D_1$ agree withing $f$ on $D_1 \cap D$
So you can patch it up and extend it to $D \cup D_1$.
You do the same thing with $D_1$, etc to extend $f$ disk-by-disk along a path on the complex plane
What goes wrong for some functions is that say you do this along a loop $\gamma$ from the origin back to itself
Then you'd return to $D$ with a different function $g$
For example, you can try to do this with $\sqrt{z}$ defined on a disk around $1$, and let the loop wind around the origin back to $1$. You'd return to $-\sqrt{z}$.
If you have a subset $U \subset V$ for which $f : U \rightarrow C^n$ is analytic
for specific U and V
this isn't Generalization of Riemannian contuition
So $U \subset V$ is open for open V. Then any function which is analytic on U can be extended to U.
the way it works geometrically is as follows so you consider the annulus V - U and slice it each chunk of it use Cauchy integral formula to extend each slice and we do it in uniform continous manner
and so that is how you do it
@Daminark category theory is important, but people sometime just keep generalizing everything for the sake of generalizing and don't even have a grasp about what they are actually studying in the end.
@Daminark You look at the full subdomain $\Omega$ of $\Bbb C$ where $f$ can be analytically continued by paths from the original point, say, $p$. Then, construct an object as follows: Take all the paths $\gamma$ starting at $p$, and patch together the disks of radius of convergence $D_1, \cdots, D_n$ covering $\gamma$ according to whether they intersect (this is just quotient space $\bigsqcup D_i/\sim$). Do this for all paths starting at $p$.
Then you'll end up constructing a Riemann surface $X$ where $f$ is actually a well-defined function $f : X \to \Bbb C$.
Sorry, I took ages to find a way to write that construction
@Adeek I don't have a particular view on abstraction for what it's worth. Some people care about it as a means to an end of understanding more concrete things, others seem to prefer to think about abstract things than concrete things. I'm just one of the "let people do what they want" types
And @Balarka so basically the idea would be that if you try to build a function along a circle and fail, the Riemann surface construction would simply not bring that path back to its starting point, right?
Notaton wise, if $M$ is a smooth manifold, $p \in M$ and $v_p \in T_p(M)$, $v_p(f)$ means the tangent vector $v_p$ acting on $f$, and it equals the directional derivative in the direction $v$ of $f$ at $p$, which notation-ally is $D_vf(p)$ correct?
@Perturbative In Ted's notation $Df(p)$ is the Jacobian matrix of $f$ at $p$, and $Df(p)v$ is then that matrix eating the vector $v \in T_p M$ which is the same as the directional derivative $D_v f(p)$.
@LeakyNun I think if you have "a logic" in the sense that you have some rules of inference for well-formed sentences (I don't want to make this super precise right now, I'm not an expert on this), then you can consider a category with morphisms = implications even without assigning actual truth-values to the sentences
@LeakyNun this is what my category book writes on logic "We can associate to every mathematical theory in the sense of mathematical logic a category. Objects are theorems of the theory. A morphism $A\to B$ is a proof of $A \to B$. Composition is the composition of proofs"
My favorite proof is where you say you'll prove something by contraction, give a direct proof, and be like "Our statement is true, contradicts our assumption that this was false"
Hello, who know this lemma on topology: let $X$ be a Hausdorff space, let $(x_n)$ be a sequence and $(x_{n_k})\subset (x_n)$ and $u\in X$, if there exists a sub sequence $(u_{n_{k_l}})\subset (u_{n_k}), u_{n_{k_l}}\to u$ then $u_n\to u$ ?
@TastyRomeo if a space is covered by a simply connected cover, then it has to be semilocally simply connected: for any point in the base space, you have an open set that is evenly covered, so if you have loop in that open set, you can lift it to the universal cover and there it will be entirely contained in one leaf, so it is a loop in the universal cover which can be contraced to a point in the cover as it is simpylyconnected, you can just push down the homotopy.
For a smooth manifold $M$ of dimension $n$ and a point $p \in M$ do y'all normally think of the tangent space $T_pM$ as the image of a parameterization $\varphi : U \subseteq \mathbb{R}^n \to M$ for some open set $U$ in $\mathbb{R}^n$ or as the set of all derivations of $C^{\infty}(M)$ at $p$?
