i do remember getting to that limit when i first took calculus and thinking things were getting unusual. it had been all polynomials and factoring and canceling x - a and suddenly we're drawing pictures of a circle. the book did say something like, this is because we want to use sin x before we have more algebraic ways of expressing it.
@user863565 If you don't use area, you have to deal with arclength of the arc of the circle. Area is much more basic, going back rigorously to the early Greeks.
i hope it's just on the vaccine compounds themselves. they should leave a backdoor, like a patent on a method of preparing the compound, or making a machine that prepares the compound
there isn't a huge amount of law on trusts. it will surprise you to learn that california has a civil code on this subject, and judges just cut-paste lines from that and try to do whatever seems right. not a lot of 'precedent'
your honor, by the case of hearst v. hearst, where the trustee had to give over the castle and the personal zoo and the yachts to the other hearst, we have to ---
one of my ancestors mowed the lawn near a castle in killarney
there's probably a way of setting up an LLC or something and having people contract with it in a way that achieves the same end as a trust, but that seems like a tightrope act
you're a minimum if you're in the society and below everybody else in the society. think of being the only serf in a society of noblemen. you're minimal if you're in the society and not better than anyone else in the society. think of people under communism.
the poset pictures here would be a dot connected via lines to a bunch of dots above it (the serf) and a long row of millions of dots, all on the same level, nobody connected to anybody else. forget about party members for a moment.
extreme examples are helpful. if you're not comparable with any other member of the poset, you're minimal by definition and will fail to be a minimum unless you're the last person on earth.
If $\le$ is less or equal in its usual meaning and let’s stay in the set of reals for example then minimal element of a subset of R is same as minimum element of the subset. Is this statement correct?
note that any two elements of R are comparable to one another. this is a big part of what makes R and other "linearly ordered' sets different from general partially ordered sets.
Suppose $F$ is a field. Suppose $\sigma_1, \sigma_2 \dots \sigma_n$ are distinct elements of finite group of automorphisms of $K$, then lang proves that they are algebraically independent when $K$ is infinite. What about finite field $K$? I cannot find any counterexample. Can anyone help?
if \sqrt is the principal square root function on C, then sqrt(e^z) should be an entire function, because e^z does not vanish anywhere, and so we have no branch points. However, in the function plot on wolframalpha there appears to be a cut at z=x+iy for y=pi; see the contour plots here: wolframalpha.com/input/?i=plot+sqrt%28exp%28z%29%29
can someone explain this to me please?
maybe wolframalpha just messes something up... but from my understanding sqrt(e^z) should be analytic everywhere
I suppose wolframalpha doesn't understand that there exists an analytic continuation, and thus messes up
@courge9 there is no principal square root function on all of C, at least not a continuous - let alone holomorphic - one, regardless of whether you remove 0 or not
@robjohn I don't think I'm mixing that up... z\mapsto e^z is a holomorphic function on all of C, which takes values in C\{0}. Hence this function has a holomorphic logarithm, and thus holomorphic n-th roots for all n.
courge9 the principal square root in many books is given in polar form by sending r e^(it) to r^(1/2) e^(it/2). the problem with this is that it treats points near the negative real axis in very different ways.
points slightly above i get pulled back toward the positive y axis. points slightly below i get pulled down toward the negative y axis
there's no way of defining this at i to make it continuous let alone differentiable there
and you encounter a similar problem if you use other formulae
If $X$ is path connected, locally path connected, semilocally simply connected then there is a covering space $\tilde{X}$ such that $p:\tilde{X}\to X$, $p_*(\pi_1(\tilde{X})) = H$ for given $H<\pi_1(X)$. But we still don't know such $\tilde{X}$ is path connected right?
@leslietownes is that correct? so the problem here is that while 0 is not a branch point of sqrt(e^z), we still have a branch cut on the negative real axis in the image of e^z (that is, at z=i) where sqrt(e^z) becomes discontinuous
Follow $\sqrt{e^z}$ along the line from $z=-\pi i$ to $z=\pi i$. That goes around the unit circle counter-clockwise from $-i$ to $i$. We cannot continue along this path since it would take us out of the range of the principal square root, so we need to jump back to $-i$. Thus, the discontinuity
The range of the principal branch of the square root is the right half plane
and did I put it right with what I've written above: " so the problem here is that while 0 is not a branch point of sqrt(e^z), we still have a branch cut on the negative real axis in the image of e^z (that is, at z=i) where sqrt(e^z) becomes discontinuous"
okay, so the misconception on my side was that I thought it suffices that 0 is not a branch point to conclude that the function is entire, while in reality one still has to consider the cut along the negative real axis
This isn't analysis, it's pure topology. $\mathbb{C}\setminus\{0\}\rightarrow\mathbb{C}\setminus\{0\},z\mapsto z^2$ is a non-trivial, connected cover, so it doesn't admit a global section, i.e. there is no global square root. The image of $\pi_1(\mathbb{C}\setminus\{0\})\cong\mathbb{Z}$ under this covering map is identified with $2\mathbb{Z}$, so any map $X\rightarrow\mathbb{C}\setminus\{0\}$ which maps $\pi_1(X)$ into $2\mathbb{Z}$ admits a lift through the covering. The covering map is holomorphic, whence a local biholomorphism, so a lift of a holomorphic map through it will be holomorphi…
@Thorgott : but $z\mapsto e^z$ is a nowhere vanishing holomorphic function $\mathbb{C}\rightarrow\mathbb{C}\setminus\{0\}$, so why doesn't it admit a holomorphic square root?
@robjohn but if we take the principal branch, is it still true that $\sqrt{e^z}$ becomes discontinuous at $\pi i$? How does that fit together with what you said before?
Follow $\sqrt{e^z}$ along the line from $z=-\pi i$ to $z=\pi i$. That goes around the unit circle counter-clockwise from $-i$ to $i$. We cannot continue along this path since it would take us out of the range of the principal square root, so we need to jump back to $-i$. Thus, the discontinuity
yes, it is discontinuous, or rather it cannot be continued across $\pi i$ continuously
it's a bit like why i don't like the classical notation sqrt(-1) for i. it invites "power arithmetic" like -1 = (sqrt(-1))^2 = sqrt((-1)^2) = sqrt(1) = 1
@courge9 Wolfram Alpha sees that you explicitly wrote sqrt(e^z), so it assumes you want the ugly version. If you really wanted $e^{z/2}$ then you would have writtten e^(z/2). It's kind of similar to $\sqrt{x^2}$ vs $\left(\sqrt x\right)^2$ for real $x$.
I have a quick follow-up question regarding what Thorgott wrote: any nowhere vanishing holomorphic function $f:\mathbb{C}\rightarrow\mathbb{C}\setminus\{0\}$ admits a holomorphic square root. I understand that this "square root" is (in general) not literally $\sqrt{f}$, but a function whose square gives the original function. But is it unambiguous to write $f^{1/2}$ for that function, in particular when trying to define an analytic continuation of $\sqrt{f}$ from the reals?
put differently: is it meaningful to regard this "holomorphic square root" as the proper analytic continuation of $\sqrt{f}$ on the reals?
I think it is still ambiguous because fractional powers of complex numbers are (usually) defined in terms of taking a branch of the logarithm and expressing the fractional power in terms of exp and log.
e.g., if you write $z^{1/2}$, usually this means that you need to choose a region in which to define $\text{Log}$ (normally $\Omega=\mathbb{C}\backslash (-\infty,0]$) and then define the branch $\text{Log}_\Omega$ to be the inverse of exp on $\Omega$. Then you can write that $z^{1/2}=e^{1/2\text{Log}_\Omega(z)}$ in this region.
but all this is still local and what thorgott was saying is that there are theorems which allow you to avoid this choice of branch under certain circumstances (choosing a branch is similar in my mind to choosing a basis!)
yeah, it's just gonna be nicer to work with $e^{z/2}$ as an extension of $e^{x/2}$
i wouldn't think of e^(z/2) as an exception to the case of f^(1/2) being a problem. post composing with ^{1/2} is the problem because you need to know what ^{1/2} means.
it's clear what z/2 means, so no trouble in precomposing it with anything.
i may be reversing post and pre because i'm left handed but we all know what i mean.
but there is notational ambiguity when you do fractional powers of complex numbers unless you really nail down stuff explicitly. its annoying but such is life
i had to deal with this once and i forget the details what i did. i think i really needed a log, the formula really only made intuitive sense as a log, but i got rid of it by differentiating and then working in the world of rational functions of what had been stuffed into the log. the formulas weren't intuitive but i didn't have to worry about nightmare worlds.
Considering additive groups R (set of reals) and Q(set of rationals), I know that they can’t be isomorphic because R is uncountable and Q is countable so no bijection will exist between the two. But I wonder if there’s any element in R which has an order that no element in Q posseses.
