robjohn

similarly. I am now retired, but still dealing with some administrative issues. I hope your new hip does not fight back.

@copper.hat hello. How are things?

@skullpatrol Thanks!

@DanielSank $A-x$ is constant in $y$, so yes.

@XanderHenderson I have the left box. I could never take the lies from the right box.

@XanderHenderson skeptical does not seem to be in the Crayola colors.

Yeah, I've been busy at work. I'm retiring soon, so things have been a bit hectic.

Sorry, I am on a Zoom meeting, so I may be a bit slow.

@SineoftheTime How are you?

@onepotatotwopotato shhh! People are sleeping here!

@Thorgott Arrow envy?

it's raining or it isn't.

Suppose neither has a subsequence converging to $0$

that is sufficient, but not necessary.

@ILikeMathematics Have you checked if the derivative is bounded?

@skullpatrol Dealing with a lot of stuff irl, but I hope I can spend more time online soon.

@skullpatrol hey, there. How are you doing?

The wind...

@leslietownes You require too much!

@XanderHenderson We are very thankful. That keeps the fire away from us.

The town of Topanga has suffered some, but some places have escaped. The southern part of Topanga Canyon is pretty crispy.

@handan_toddler Thanks. We are, for the time being, safe. We can see flames and see smoke, however.

This conversation can go on for days.

@copper.hat This is why the room description says, "Just ask; don't ask to ask."

@leslietownes Thanks. We are doing better, though my wife fell and has hurt her ankle.

@think_meaning_builds Thanks, though I have been checking in here several times a day. We've had some medical issues here recently.

@ILikeMathematics I can't paste the javascript into the address bar. I need to have it as a link on the shortcut bar. I don't know if Chrome has that. There is a suggestion on the installation page for Android and Chrome.

@ILikeMathematics It works for me. What OS and browser are you using?

@Jakobian Also the integer closest to $\frac{n!}{e}$ for $n\ge1$.

To see a representation of a Riemann Surface for the cube root, see the answer to this question.

You ask about the cube root function. There is indeed only one branch cut, for the same reason as the square root or the log. The Riemann Surface for cube root would have three sheets, as the one for square root would have two, and the one for log would have infinitely many.

It doesn't matter where we put the branch cut, as long as it prevents us from circling the origin, which would be the branch point if we were to create a Riemann surface for these functions.

If one travels along the unit circle, the square root function comes back to being negative of what it was one revolution prior. If one travels along the unit circle counterclockwise, the log function comes back to being $2\pi i$ different from what it was one revolution earlier.

@LuckyChouhan Since you mentioned branch cuts, which are used to make analytic functions on $\mathbb{C}$, I was assuming that's where we were looking. Yes, branch points are used when looking at Riemann Surfaces, which are a bit more complicated than branch cuts. Branch cuts are used when a function would be multi-valued on all of $\mathbb{C}$, such as square root or log.

@LuckyChouhan I have seen singular points and branch cuts, but not branch points.

@SineoftheTime Ah, I see you noted this.

@SineoftheTime it probably has to do with the bounties offered.

@Jakobian Belated birthday wishes, etc are a common sentiment in greeting cards here.

@Sahaj hello, belatedly.

@Pizza I fixed it

I hop in once in a while, but things have been slow. I’ve been occupied IRL for quite a while. Now we’re in Mammoth Lakes on vacation.

@Pizza $-\frac12$

@Sahaj did you get a closed form? I can show how to get it.

@leslietownes Dreamer

Ack, I will look later, but I think this doesn't work.

Look at it as $\frac1{i\sqrt3}\left(\frac1{n-\frac{-1+i\sqrt{3}}2}+\frac1{-n+\frac{-1-i\sqrt{3}}2}\right)$

I think this might bear something. I'm looking.

$\frac1{i\sqrt3}\left(\frac1{n-\frac{-1+i\sqrt{3}}2}-\frac1{n-\frac{-1-i\sqrt{3}}2}\right)$

I don't see a contour that would work.

Note that in this sum, $n=0$ cancels $n=-1$, $n=1$ cancels $n=-2$, etc. So the sum is $0$.