$r(r^5 - 2r^3 - 719r + 864) = 720$

@mo-_- I tried to do your exercise, but I got here

Does anyone know how I can solve this? :/

Using the rational root theorem, I found that r = 5 is a solution

But then how do I know that there are no other r that satisfy the equation?

i.e this: $r^6 - 2r^4 - 719r^2 + 864r - 720 = 0$

@Pizza There's another solution near r=-5.55

sagecell.sagemath.org/…

But I guess we can ignore that, since $r = \sqrt{x^2+y^2}$

im wondering if this answer makes sense to you: Since we're integrating a circle of radius $R$, it makes sense to parametrize the integral as such. A circle of radius $R$ going from $0\to 2\pi$ in the counter-clockwise direction is given as $f(\theta) = R\cos\theta\hat{i} + R\sin\theta\hat{j}$. Since we want the contour of this circle, we take $\vec{A}(f(\theta))\cdot f'd\theta$

over the interval. $f'(\theta) = -R\sin\theta\hat{i} +R\cos\theta\hat{j}$ $\int_0^{2\pi}\vec{A}(f(\theta))\cdot f'(\theta)d\theta$ becomes $$\int_0^{2\pi} R^2\sin^2\theta + R^2\cos^2\theta d\theta = \int_0^{2\pi} R^2d\theta = 2\pi R^2$$

if there is a simpler way of answering this, i'd like to know

i think u can also convert to polar unit vectors

i think it makes more sense to do that instead. idk what i've written here like what even is $\vec{A}(f(\theta))$

While expressing points in the Cantor set we usually make use of their ternary representation. As we are removing middle third interval in each step, it follows that all such points have only $0$'s and $2$'s as digits in their ternary representation. But after some trials, it seems to me that binary representation of points in the Cantor set does not have two consecutive $1$'s. Although I cannot able to show prove it. Do anyone have any idea on it?

anacardium: have you looked at the binary representation of 2/9?

@PM2Ring Yes, the problem is that I wanted to check manually, do you know a method to check if r=5 is the only solution?

@PM2Ring yes

if you factor r - 5 out of that sixth degree polynomial, you're left with r^5 + 5 r^4 + 23 r^3 + 115 r^2 - 144 r + 144 = r^5 + 5 r^4 + 23 r^3 + 115 r^2 + 144 (1 - r). if r is in [0,1) all of these terms are positive. if r >= 1 then r^5 + 5 r^2 + 23 r^3 + 115 r^2 >= r + 5r + 23 r + 115 r = 144r so the polynomial is >= 144r + 144(1 - r) = 144

so that polynomial has no nonnegative roots except r = 5

How did you conclude that it is always positive?

Between 0 and 1 the term to the right of 144 has a decreasing phase that reverses at a relative minimum, how can i obtain the value of the minimum?

by "it," do you mean the quintic polynomial i wrote above? i guess i meant nonnegative. i say above, it's [nonnegative] because all of the terms in that expression for it turn out to be, and a sum of nonnegative things is nonnegative.

e.g. if r >= 0 then r^5, r^4, r^3, and r^2 are >= 0, and they stay >= 0 if you multiply them by positive numbers. and 1 - r is positive because i assumed r was less than 1 in that case.

and the sum isn't 0 because, well, you fill in the rest. maybe r = 0 is a special case.

i don't know what you mean about "decreasing phase" but that might be a translation issue. i separate into two cases above, and give different reasoning for the two cases.

Yes, I didn't understand if it is possible to assume r≥0 in this exercise

in that exercise, "r" was sqrt(x^2 + y^2) with x and y real numbers

so even if you don't assume r >= 0, it happens that r >= 0 :)

@leslietownes right! But instead, to find the solution r=5, how would you do it?

I wrote ±(1,2,3,4,5 etc...) , so by inserting 5 I found it was a solution

But I don't know if it's the best way to do an exercise

sure, some version of what is sometimes called the rational root theorem (or i guess integer root theorem in this case) would do it

Yes i used the rational root theorem

i mean, part of it is a kind of metamathematical principle, which is, if this is coming up on some random person's homework, it is going to have a 'nice' root

there's nothing out there that would be like "oh, i'll just do this general method that would work for absolutely any polynomial" that i know of

you can do some specific fiddling with inequalities to maybe rule out the larger factors of 720, right? that r^6 is going to get very big very quickly

but i dont think its particularly illuminating to focus on any one person's way of individually doing that. the most generally applicable thing worth remembering is "aha, the rational root theorem reduces a search for rational roots to checking a finite number of integers"

hi :-)

@leslietownes @Pizza hi

$r^6 - 2r^4 - 719r^2 + 864r - 720 = 0$

It's not clear to me what I should do here

$r = \sqrt{x^2+y^2}$ so its $\geq 0$

@leslietownes so like $r^6 \approx 720 ?$

How do you show that the "complex cone" $\{(x, y, z) ∈\mathbb C^3 : x^2 + y^2 - z^2 = 0\}$ is not a manifold?

need to analyze a neighborhood of the x=y=z=0 point

I mean, real manifold

I think it has real dimension =4. Can we prove a neighborhood B of (0,0,0) is not homeomorphic to R^4 by removing one single point and connectedness?

@leslietownes hi, sorry i have the last question

if r >= 1 then r^5 + 5 r^2 + 23 r^3 + 115 r^2 >= r + 5r + 23 r + 115 r = 144r so the polynomial is >= 144r + 144(1 - r) = 144

@leslietownes what do you mean here ?

its not clear why you write r + 5r + 23 r + 115 r

@mo-_- When r >= 1 then r^2 >= r and r^3 >= r, etc.

Does anyone know of any introductory books on elliptic functions, which can be read by someone who knows only calculus and are of a more historical flavour? By "historical flavour" I mean that the book should go according the historic development of the subject.

is there a reason why when we define things like irreducible elements of integral domains, we require them to be non units?

I have a silly question, but I'm working a challenging exercise in Rudin's PMA I think. It's basically Tietze extension theorem on $\mathbb R$. I'm stuck on a small detail and a confirmation would be nice. We have a set $E\subset \mathbb R$ and a point $x\in E$. $E^c$ is open and hence a countable union of $(a_k,b_k)$ with possibly $(b,\infty)$ and $(-\infty,a)$. Let $\delta>0$ be fixed.

Then in what ways can $E\cap (x,x+\delta)$ fail to be nonempty? I'm thinking, this happens if $x=a_k$ (with $b_k>a_k+\delta$) or $x=b$? Is this a correct observation?

$E$ is closed, by the way.

@SoumikMukherjee thank you

@hbghlyj I would assume it's locally connected at $(0,0,0)$

@Thorgott I wonder if it is a manifold

Just wanted to ask a quick question: The set {x in Q: x^2 < 2} does not contain its own upper bound right? Furthermore, I am right in claiming that {x in Q: x^2 < 2} does have a supremum in {x in Q: x^2 < 2} U {2} and the sup is equal to 2?

Its own upper bound in which ordered set?

psie: you might think about "cantor sets" in relation to this. i maybe wouldn't think of them as prototypical closed subsets of R, but they are closed sets, and sometimes informative examples. the usual construction (e.g. en.wikipedia.org/wiki/Cantor_set but there are probably much better pages, at a glance this looks pretty bad) explicitly involves complementation and open intervals. the ternary expansion viewpoint is also helpful for this example.

i would add, while there is a reasonably concrete-sounding description of an arbitrary open subset of R, i would not take this as evidence that there will be an accessible characterization (i.e., not just a definition) of what an arbitrary closed subset of R will "look like." people mislead themselves into a lot of wrong stuff that way even in R and it gets worse in R^2 let alone R^d.

and as even cantor sets show a countable union of open intervals can be pretty goofy. it's all about where you put those intervals :)

@leslietownes Goofy?

GOOFY?!

Why I oughta...!

Dr. Goofy has entered the chat

?

@leslietownes anyway thanks for before, I managed to understand :-)

If a complex number raised to an irrational number gives a real number, is there a real number which, elevated to a complex number, gives a pure imaginary number?

@Binky I'm not sure what one has to do with the other, but every point on the unit circle can be expressed as $\mathrm{e}^{i x}$ for some real $x$.

Including $\pm i$ (with $x = \pm \pi /2$).

Your answer is interesting, but it doesn't seem directly related to my question. I was asking whether there exists a real number that, when raised to a complex exponent, results in a purely imaginary number. The fact that every point on the unit circle can be written as $e^{ix}$ is useful, but I don't immediately see the connection to my question. Could you clarify the link?

$e$ is real, $i\pi/2$ is complex and $\exp(i\pi/2)=i$

Thank you for the clarification!

@SineoftheTime That's what I said!

X(

>:(

:'(

@XanderHenderson There was no example :(

Thanks for the clarification! At first, I didn’t fully understand because the example wasn’t clearly explained, but now I see that $( e^{i\pi/2} = i )$ is exactly the answer I was looking for.

Okay... sure. I generally expect people to do at least a little thinking on their own when I give an example. :/

$\frac 13 = 0.\overline{33}$ Is it finite or not?

I have no idea what "finished" means.

I corrected

Don't use pronouns. What is "it"?

Could i say $1 + 1 = 2.\overline{000}$?

If "it" is $1/3$, then yes, "it" is finite.

@Binky You could even say that $2 = 2.\overline{0}$.

I am not sure what you gain from that, but it is true.

so more on maths including infinite sequences

You've just taken a leap, and I don't know how you got there. There are no sequences involved at this point.

What sequence are you talking about?

like take the taylor series

we can never really have an end

just an approximation

Again, you are jumping from place to place, and I don't see how you are connecting the ideas that you are trying to connect.

Slow down.

You've gone from the decimal representation of $1/3$ to sequences to series, and I have no idea how you are connecting these things.

I feel silly for asking, but in the Tietze extension theorem for $\mathbb R$, can the closed subset $A$ from which we extend the function be the empty set? In the statements I've been looking at, I haven't seen it explicitly stated that $A$ should be nonempty.

@psie What does a function with empty domain look like? Can you extend it?

Is a constant function an extension of the empty function?

@XanderHenderson why does maths sometimes gets to infinity? Like 1/3

@Binky I don't understand... $1/3$ is not infinite.

Not 1/3 but 0.3333333333....

@XanderHenderson ah ok, so any extension would work. In particular, a constant function.

The latter is not infinite, either. It is just a representation of $1/3$.

i don't learn this stuff but it's good to know

these*

@psie yes. The statement would then simply be "there is a continuous real-valued function on the space" which is true

ah ok 👍

For example a constant function

:P

I'm currently reading a proof of this theorem for $\mathbb R$, using the fact that $A^c$ is the disjoint union of open intervals (including possibly semi-infinite intervals). The hard part is verifying continuity of the constructed function on $A$.

Actually, I'm watching a YouTube video :)

It's kind of weird. We have $f:A\to\mathbb R$ and we construct an extension $g:\mathbb R\to\mathbb R$. For checking continuity of $g$ on $A$, the author proceeds by contradiction. Assume $g$ is not continuous at $x_0\in A$. Then there exists a sequence $(x_n)\subset\mathbb R$ such that $x_n\to x_0$ but $g(x_n)\not\to g(x_0)$. Then they say if $x_n\in A$ for $n\geq N$, then $g$ is not continuous on $A$, and since $g=f$ on $A$, this is a contradiction.

So $x_n\in A^c$ for infinitely many $n$...and the proof continues. How can one assume that $g$ is not continuous on $A$ and then say that it is a contradiction that $g$ is not continuous on $A$?

This is the video. One only needs to watch until 7 minutes and 15 seconds or so.

i.e. from the beginning until 07:15 or so.

see Carrie from Brian De Palma

it is a horror movie

@psie a function $g$ is continuous on $A$ if and only if for every sequence $(x_n)\subseteq A$ with limit $x_0$ also in $A$, $g(x_n)\to g(x_0)$. Here you built a sequence as in the hypothesis, but which does not converge to $g(x_0)$, so $g$ is not continuous on $A$

And this is a contradiction because by construction $g$ agrees with $f$ on $A$, and $f$ is known by hypothesis to be continuous on $A$+

ok, thanks 👍