I wrote a complicated pro-existentialist surreal neo-Dadaist analysis of Octavio Paz. Not sure how the examiner would take it but I really wanted to pull that stunt.
Ever since reading this NewScientist issue on how even if for some beliefs that despite being based on falsehoods, can have actual impacts to the world via cultural influences, I started to wonder about the nature of belief, and by extension, unreal entities or concepts that have real impacts.
C...
Hey guys I have complex analysis class at university and we currently look at integrals, unfortunately we don't proof the theorems, since this is more like a crash course for electrical engineers.
I read through some papers about the proofs for the Cauchy-Integral theorem, but it's so much stuff and there seem to be a vast number of interpretations of the Cauchy theorem.
Oups, not finished yet
When we want to calculate the circle integral of a holomorphic function $f$, but there is a singularity inside the circle, why do we have to use the cauchy fromula? Since when we evaluate $f$ in a circle around the singularity we never touch that "bad" point actually no? Same with wholes in sets, e.g. unconnected sets...is this due to missing homotopy of our circle function?
@Felix.C There may be bad points inside the circle, is the point. Eg, $f(z) = 1/z$ is a holomorphic function which has no singularities on the unit circle $|z| = 1$. But $\displaystyle{ \int_{|z| = 1} \frac{dz}{z} }$ is not zero.
@Felix.C It's because holomorphic functions on $D\subseteq\Bbb C$ are not guaranteed to have global antiderivatives. If $D$ is simply connected, this becomes true, but that's kind of a miracle
For functions with antiderivatives, the fact that their integrals on closed loops equals zero follows from the Fundamental Theorem of Calculus.
What's true is that every holomorphic function has local antiderivatives around any point. If $D$ is simply connected, you can patch it all up to a global antiderivative. If it has singularities (like if it's $\Bbb C^*$) then everything can't be patched up.
So if I understand this right, it boils down to the problem that holomorphic functions $f:D \subseteq \mathbb{C} \mapsto \mathbb{C}$ with $D$ not being simply connected, aren't guaranteed to have a global antiderivative $F:D \mapsto \mathbb{C}$, therefor we can't use the fundamental theorem of differential calculus on them, but rather have to watch local antiderivates...such that the integral won't be necessarily 0.
Then is there a theorem about the situation where you have a whole in your set, and you want to calculate the integral around it?
It may be true that Tobias can not answer your question. That does not mean he doesn't do "modern stuff". He's a serious rep theorist. That's all I am saying
Hmmm... If $\exp \sum_{n=1}^\infty \frac{a_n}{n}z^n$ and $\exp \sum_{n=1}^\infty \frac{b_n}{n}z^n$ are both rational functions, is there a reason to assume that $\exp \sum_{n=1}^\infty \frac{a_nb_n}{n}z^n$ will be rational?
Started a Discrete Math assignment 9 am in the morning. Spent all day working on it with minimal breaks until 4 am. I managed to wake up after a few hours of sleep, made my way to Uni, and walked into class only to hear my professor announce that the assignment is not due Monday, minutes before the deadline. Terrrrrrrrrrrrrrrrrific.
I was sleep deprived and stress-induced for nothing :)
Assuming $\lim_{R\to \infty} a^{(R)}(n)$ exists and is bound. Then $\lim_{R\to \infty}\int a^{(R)}(n) u^{-R}e^{nu}dn=0$ and hence the terms in the series goes to zero as $R \to \infty$
@MikeMiller I have a quick question. So if we have $p : E \rightarrow B$ where B is path connected. Then I am trying to prove that the fibers are actually all homotopy equivalent.
I proved that we get vertical homotopy equivalence
as before yesterday
I considered the following diagram
Let $b_1$ and $b_2$ be points in B let $\phi : I \rightarrow B$ is a path between $b_1$ and $b_2$.
consider the following diagram
$p^{-1}(b_1)\times \{0\} \rightarrow E$
on the bottom square we have
$p^{-1}(\{b_1\}) \times I \rightarrow I \rightarrow B$ where the second map is $\phi$
now I am trying to prove we get homotopy equivalence given this square
I wouldn't really agree with that assessment. It's definitely not like before when going to college just kinda meant you were set for life, so for a lot of people, especially if you don't major in those subjects deemed "practical", it'll be stacking on debt without the guarantee of career security
But it's still better to have gone to college than to not, from a purely practical standpoint, and I find education to be an end in itself
(Also, I have financial aid which really helps)
Admittedly, it can be context dependent, but at least for me personally, this timeline is far better to one where I don't go to college
A very reasonable way of viewing it. I find myself in a similar situation too. Graduated from college, and currently in Uni. I find that University gives you a "reason" to keep moving forward. Sort of like a momentum, and a purpose.
Also, financial aid is nice and all, until you start paying it off :P
Sometimes down here a university often "contains" a college. Like the school Daminark and I go to, we're students in the college, which is a subset of the university. I don't know if other schools do this though.
Assuming separable, then take a sequence $w_n$ on the unit ball so that $\|Aw_n\| \to \|A\|$, and then use weak compactness to get that a subsequence $w_{n_j}$ converges weakly to some $w$.
Yeah this does more to make me suspicious of the axiom of choice than anything else. Well ordering, Zorn's, bases of vector spaces, that's all fine, and the statement of the axiom of choice seems obvious
This is the one thing I've thus far seen which is making me kind of iffy
yeah idk, axiom of choice was always fine with me, but some of it's equivalents seem like utter nonsense to me. But The same thing seems to be true of things like determinacy honestly
but honestly i dont care much about those kind of issues
Last summer during the REU I was in an apartment with 4 friends and toward the later parts of the summer, our conversations were mostly centered around, what I guess you can call "the philosophy of set theory"
Canyon was kind of a constructivist so we often had those debates as well
It's admittedly of questionable productivity, like when actually working in math, it's entirely fair to just say that regardless of what you believe a set actually is, you can just say that this is a model of something and roll with it, letting truth apply merely within that model
But I've found such conversations to be fun nonetheless, if only to kill some time with
@MeowMix Here's something real nice I solved a few days back. Prove that, for every $n$, there exists a circle (radius depending on $n$) on the plane which contains exactly $n$ lattice points - that is points of the form $(x, y)$ where $x$ and $y$ are integers.
Your proof should generalize to hyperspheres in $\Bbb R^k$.
Someone mentioned a type of graph to me the other day, but I forgot what it was called. It looks like a plane divided into four quadrants, of which any one of those quadrants might be divided into four more quadrants, and so on.
@BalarkaSen Ah, I think I've solved it. That's nice.
user228700
Is it that critical points also include those points at which the derivative is not defined (where as stationery points only include points at which the derivative is zero)
Or is it that critical points are points where there is a potential maximum/minimum (the sign of derivative changes as the function passes through this point)? If so, I have a follow-up question; isn't this the same as an inflection point?
All stationary points are critical points but not all critical points are stationary points.
A more accurate definition of the two:
Critical Point:
Let $f$ be defined at $c.$ Then, we have critical point wherever $f '(c)= 0$ or wherever $f(c)$ is not differentiable (or equivalently, $f '(c)$ i...
@Kaumudi They're saying, critical points are points where the function has zero derivative. The trick is, this also includes points where the derivative does not exist!
Because if the derivative doesn't exist, vacuously the function has zero derivative.
(I generally like to have an everywhere differentiable function whenever I talk about a critical point, but let's put that aside. All of this seems to be convention)
All stationary points are critical points but not all critical points are stationary points.
A more accurate definition of the two:
Critical Point:
Let $f$ be defined at $c.$ Then, we have critical point wherever $f '(c)= 0$ or wherever $f(c)$ is not differentiable (or equivalently, $f '(c)$ i...
