@robjohn I don't. I will install it now.

you will see that $2^13$ does not show the same as $2^{13}$.

I have to go walk the dog. BBL

@robjohn I don't. I will install it now.

cure, there are some arithmetic errors in there, but it would work in principle

I have a question, is there a simple step to calculate s in the last pic?

The book said it can be found with "simple arithmetic". I know I could calculate it, but through a relatively lengthy process

@leslietownes Thank you. I will check it out again for mistakes :)

cure: i agree that it's simple arithmetic, but if you had to do it by hand, maybe not that fun? using the result from 5., (g^m)^n = 39^31 mod 97, and you could then do that similarly as the above thing with bigger numbers, with the repeated squaring bit

i.e. compute 39^2, ^4, ^8,^16 by squaring, and then multiply all of them together to get 39^(1+2+4+8+16)

the point of the squaring method is to control the size of the numbers you have to deal with in intermediate steps (reduce mod 97 at each step) while also reducing the total number of multiplications

if you have a calculating device that doesn't barf on large inputs, just feeding it in can work

cure, did you know that yesterday was robert smith's birthday?

Yes! He turned 63 :)

Do you like The Cure?

yes

awesome

do you have a favorite album?

or favorite songs? :)

i think faith is my favorite album, followed by wish, which is a little weird in that they're opposites. but they're all good.

Is this a PSQ math.stackexchange.com/questions/4434058/…?

@XanderHenderson

Let $f:X\to X$ be bijective continuous map between metric spaces $X,Y$. Let $g$ be the set theoretical inverse $g:X\to X$ of $f$. Let $\{x_n\}$ be a convergent sequence in $X$ (assume $x_n\to x$). Then by continuity of $f$, $f(\lim_{n\to\infty} g(x_n)) = \lim_{n\to\infty} fg(x_n) = \lim_{n\to\infty} x_n = x$. Hence, $\lim_{n\to\infty} g(x_n) = g(x)$. So $g$ is continuous.

But in general, bijective continuous map need not to be a homeomorphism. Is my argument wrong?

@onepotatotwopotato I think that it pre-assumes existence of $\lim g(x_n)$.

@Koro Right. Thanks

Hello everyone. I was wondering that in the Riemannian sphere, would a divergent sum be considered as infinity, or would it be undefined?

@MathGeek a divergent sum is any sum that does not converge. Diverging to infinity is just one way to diverge. 1-1+1-1+1-1+... is another

@robjohn Ok, so what if it diverges to infinity?

@MathGeek Think about the topology of the Riemann sphere. What are the neighborhoods of infinity? Given a neighborhood, do the partial sums eventually end up in that neighborhood?

So it would be equal to infinity, right?

yes

Ok, thanks!

Can I ask a very simply slightly-physics related question? It's mostly math, anyway.

I've been trying to solve this thing since last night, but to no avail:

Of course, $W=F_{fr}*d$

And we know $F_{fr}=\mu * F_N = \mu * M * g$

This is all fine and good, since I know $\mu, M, g$, but what about $d$?

I tried so many ways to get $d$ -- I tried to find the acceleration of the system (which I found to be $1$), applying the Work-Energy Theorem, and Conservation of Energy. But I just can't figure it out.

What am I doing wrong?

If anyone could help me figure out where I'm making a mistake, that would be much appreciated.

@rb3652 Does the rope stretch? How far does $m_2$ move? So how far does $m_1$ move?

Not closed does not mean open.

@onepotatotwopotato wow, so my misunderstanding runs deep. Are there any necessary and sufficient conditions for openness and closedness involving the containment of limit points?

One can formulate it using definition but I can't find the usefulness of stating openness in terms of limit points.

It would help me with proofs in Rudin lmao

Looks like (at least) my second statement is false; the set $E = \{1/n \mid n\in\Bbb N\}\subset \Bbb R^1$ has the single limit point $0$ which isn't contained in $E$, yet the set is closed.

@dsillman2000 It's not closed in $\mathbb{R}^1$

@Jakobian Really? Isn't its complement open?

closure of $E$ is $E\cup \{0\}$

@Jakobian This is what is confusing to me. This makes me think that a set is closed iff it contains all of its limit points, so $\bar{E} = E$. But I thought there was something wrong with that intuition

well that's true

it's your understanding of what an open set is that's wrong

Ah you're right. My intuition ignores the gray area that a set can be both closed and open (containing all of its limit points, but also such that every point has an $\epsilon$-ball contained in the set)

Is this akin to how $\Bbb R$ is both closed and open? (i.e. it has the $\epsilon$-ball property, but also contains all of its limit points)

Clopen sets make the space disconnected, so that's why you don't see a lot of them, but they exist

hello, does anyone know if there's a program that will allow me to graph 3d shapes defined by multiple equations? e.g. $x^2-y^3=y^2-z^3=0$

this deleting of questions is such a waste of every bodies time math.stackexchange.com/questions/4410639/…

The issue is compactness not being closed + bounded in general. The OP was too unresponsive.

true, but the answer may have helped someone.

Hi! How to draw the graph of $f(x)=\ln (1+x^2)$ manually?

mm, the usual calculus curve sketching techniques will work for the qualitative features (local extrema, concavity, inflection points, limits at +/- infty). that's about the best you can do manually.

it's an even function, which cuts the work in half.

it looks a lot like x^2 around 0, and a lot like 2 ln |x| for large |x|

@leslietownes I haven't studied those yet :-/ It's from a function transformation exercise.

huh. that's a little baffling to me, because i'm not sure that having a graph of f(x) for example would allow you to accurately graph f(x^2) or ln(f(x)) without a bit of thought that goes somewhat beyond the usual 'transformation' amount of thought.

e.g. if you have a graph of ln(1+x) (which you can get by translating the graph of ln(x)), then you get the graph of ln(1+x^2) for positive x by squishing that curve horizontally, because x^2 is going to travel 'faster' than x. but i don't know that the change in concavity present in the actual graph is easy to 'see' from that picture.

and same with starting with the graph of 1+x^2 (an easy transformation of the graph of x^2) and then applying ln to it. maybe someone else in the chat has a better idea.

I wondered if it can simplified to a simple equation like another question of the exercise $f(x)=\ln(ex^2)$

Anyways thanks for the inputs :)

Nope, not like that.

@rb3652 $d$ is simply y, no?

there's a change in concavity that will be missing. those graphs do work out via simpel transformations, from the properties of logs. we don't quite have that here.

we do have ln(1+x^2) is approximately ln(x^2) for large x, but this misses the key flavoring near to 0, but not exactly at 0. as does ln(1+x^2) being approximately x^2 for small x.

and both of those approximations are out of the realm of 'transformation' based graph stuff. so i'm still baffled.

Yeah, this seems a hard problem other than to observe a few things.

it's certainly a nice calc 1 curve sketching problem. all of the numbers work out nice.

we were at a cafe today and the non censored version of a rap song came on. the cafe people changed to another song after about 20 seconds, i think because a toddler was present.