12:01 AM
@robjohn Oooh. Nice. And I love envelopes.

@TedShifrin Yeah, I added the envelopes, but the circles have centers at $(0,\tan(\theta))$ and radius $2-\sec(\theta)$. Not the most natural family of circles.

Seems lovely to me!

12:39 AM
@TedShifrin I added the specs for the family of circles to the image.

Perhaps a different letter for the parameter would be less confusing (since you're using polar/spherical coordinates elsewhere).

@TedShifrin done

Thanks, Sire. :)

12:56 AM
@TedShifrin The $\phi,\theta$ in the parametric equations are actually the angles from the Gauss map. That is, they measure the normal to the surface, not actual latitude (the longitude is actual longitude since it is a surface of revolution).
off to the park

just curious, can you decompose a mathematical object into rings? (In manifold decomposition you can "break down" a manifold into smaller pieces)

1:16 AM
you don't decompose a manifold into rings, though

1:31 AM
I know, just wondering what kind of objects can be built out of rings, as if rings were the pieces that make up the object so to speak

In mathematics, a ringed space is a family of (commutative) rings parametrized by open subsets of a topological space together with ring homomorphisms that play roles of restrictions. Precisely, it is a topological space equipped with a sheaf of rings called a structure sheaf. It is an abstraction of the concept of the rings of continuous (scalar-valued) functions on open subsets. Among ringed spaces, especially important and prominent is a locally ringed space: a ringed space in which the analogy of a germ of a function is valid. Ringed spaces appear in analysis as well as complex algebraic geometry...

1 hour later…
3:05 AM
What special property does this type of sequence have : ($a$ is any integer) $$a , 0, a , 0, a, 0 \cdots$$ that is each term is either $a$ or 0 and they are alternate.
(By special property I mean one of the obvious property, I’m solving a problem and got this so I thought of asking if this sequence have something obvious in it)

I have a question on elementary inequality.
Given positive $x,y$ with $x+y=1$, prove that $x^{\frac 1 x}+y^{\frac 1 y}\leq 1$.

$$x \lt 1 \implies x^{1/x} \lt x \lt 1~~~~~~~~~(1) \\ y\lt 1 \implies y^{1/y} \lt y\lt 1 ~~~~~~~~~(2)$$
@WilliamSun Just add the inequalities (1) and (2)
What you need to focus on is that $1/x$ and $1/y$ are greater than 1 and any number less than one, when gets squared, cubed or alike gets even smaller than itself.

3:25 AM
Oh thank you yes actually the problem I am doing is this: for $x+y=2$ positive how can we prove that $(2x)^{\frac 1 x}+(2y)^{\frac 1 y}\leq 2$
And I am thinking of trying to deduce this from the inequality above
This weighted one cause much more difficulty since using Jensen’s inequality cannot ensure the part from $x=0$ to $x=1$.
If we use Cauchy’s inequality on this weighted one naturally we would be led to proving the inequality $2^{\frac 1 x}+2^{\frac 1 y}\leq 4$ for $x+y=2$.
But obviously this is wrong

@WilliamSun $1+1=2$ and $2^{1/1} + 2^{1/1} > 2$

And does this imply that the LHS is less or equal to $2$?

what?

I don’t understand how I deduce the weighted one from your statement above...sorry

I asked you to put $x=1$ and $y=1$ in your "weighted one"

3:35 AM
Oh I am sorry it should be $x+y=1$
Let me write down the question again: for positive $x+y=1$ prove that $(2x)^{\frac 1 x}+(2y)^{\frac 1 y}\leq 2$.

3:52 AM
@MikeMiller Thanks. That's what I needed. Suppose if $F = F_j + F_k$ where $F_j$ and $F_k$ are non-zero and homogeneous, then does $F$ necessarily have roots which are roots of neither $F_j$ nor $F_k$?
Seems believable to me, but I can't write a proof.

Hey is it possible to create a rational function with asymptotes $y=0$ and $y=x$?

Hi everyone
Please someone look at this question

4:08 AM
@ronakjain $cos(\theta)=-1$ satisfies the equation.

2 hours later…
6:26 AM
@TedShifrin I just couldn't leave well enough alone; I added a 3-D envelope animation ;-)

4 hours later…
10:18 AM
how do prove p(x,y)dx+q(x,y)dy=0 where both functions are same order homogeneous function can be written as y'=f(x/y) ?

user434058
What's up with reckless starring, man?

10:42 AM
1

I would say that $$A[i_1]\cdots[i_k] = B + W f(i_1-L_1, \ldots, i_k-L_k)$$ where for $k>1$ and $0\le i'_p < N_p$ $$f(i'_1,\ldots, i'_k) = f(i'_1,\ldots, i'_{k-1})N_k + i'_k$$ and $$f(i'_1) = i'_1$$ Edit: to answer...

Any ideas here?

@feynhat I didn't see how to end the argument.

1 hour later…
11:46 AM
0

Consider the twin primes. More specific the first of the smallest of the pairs. So we consider $3,5,11,17,29,...$ BUT not $7,13,19,...$ Call the $n$th element in the sequence $t(n)$. Now consider the Sophie Germain primes ( if $p$ is prime, then so is $2p + 1$. ONLY count the $p$ not the $2p+1$ )...

12:10 PM
I just learned that the homeomorphism type of a compact Hausdorff space $X$ is determined by the ring isomorphism type of $C(X)$. Can these conditions on $X$ (compactness, in particular) be relaxed?

12:47 PM
If you want to keep $C(X)$ I'm not sure, because for example $C(X)$ cannot tell $\Bbb N$ and $\beta\Bbb N$ apart, so it is already false for locally compact spaces, but if you're willing to move to $C_b(X)$ or similar, then there are generalizations
See chapter 10 in "rings of continuous functions" by Gillman and Jerison, the whole book deals with the question "which topological properties are recoverable by rings of functions on a space" basically

If you want an example you can take the long line or something
Every continuous function is eventually constant
So $C(L) \cong C(L^*)$ where $L^*$ is the one-point compactification of the long line $L$ I think
But that's not really different from Alessandro's example because $\beta L \cong L^*$

Ah no wait, my example is actually backwards, $C(X)$ distinguishes between $\Bbb N$ and $\beta\Bbb N$, while $C_b(X)$ does not
There are more general theorems like "if $C(X)\simeq C(Y)$ and singletons are $G_\delta$ in $X$ and $Y$, then $X\simeq Y$
Also $C_b(X)\simeq C_b(Y)$ iff $X\simeq Y$ for metrizable $X,Y$
There's plenty of such theorems, as I said above there's at least a whole book on this topic

1:13 PM
nice, thanks for the reference

1:52 PM
@Thorgott The nicest statement (Equivalence of categories) holds in the case of compact Hausdorff. If you're willing to lose niceness a bit, then $C_0(X)$ recovers the homeomorphism type of a locally compact Hausdorff space, but you don't have the same equivalence of categories statement
And that's clear by passing to the 1-point compactification, resp adjoining a unit

2:09 PM
@TedShifrin In going through G&P again I see that they orient normal bundles incorrectly. To them, $NS \oplus TS = TX$ is an oriented splitting.
They have the factors backwards! $NS = TX/TS$, so should come second.

3:03 PM
is a semi-group the "simplest" algebraic structure above a set?
one set, one operation, one axiom

@Charlie Well, you have an axiom, so it could be simpler

oh yeah I guess so
I've been searching just out of curiosity but can't find anything that's literally just a set and an operation, is there a name for something that is just 1 set 1 operation?

that is usually called a magma
it used to be called a groupoid, but that term means something else nowadays

ahh ok thank you :)

3:53 PM
@MikeMiller I get your point. However, for Stokes's Theorem to work, you need to have the signs right for boundary orientation. This correct sign is consistent with orienting the normal bundle of the boundary as they say, not as you wished they did.

4:34 PM
Hi everyone
I have a question

Just post it, @satan.

in many situations (such as solving differential equations, or u-substitutions in integrals), we treat $dy/dx$ as a fraction, and treat dy and dx analogus to simple variables
for instance

That's not quite right, but go on.

Consider the equation: $dy/dx = 2x/y$
$dyy=2xdx$
then integrate to yield $y^2/2 = x^2 +c$

@satan29 Such differential equations are called separable

4:38 PM
the Multiplying with dx part gets me...

First, there are things called differential forms (which you typically learn about in more advanced analysis/geometry courses).

If you think about more precisely what you do when you integrate you will see how little sense that step really makes

They put all of this in a rigorous, consistent setting.

in the context of "dy/dx NOT being a fraction, and rather, being an operator" i.e d/dx()

Second, if I may finish ...

4:39 PM
@TedShifrin i see
yes, sorry.

I mean, you integrate one side with respect to dy and the other with respect to dx. Why would we expect that to preserve equality

The chain rule justifies what you're talking about here with separating variables.

@TedShifrin can you elaborate this bit?

It's the same thing, as you pointed out, that goes on with $u$-substitution in integrals.
So, let's ignore $dy/dx$. Let's write your equation as $f'(x)= f(x)/x$.
Now I write $$\frac{f'(x)}{f(x)} = \frac 1x$$ and I antidifferentiate with respect to $x$ only.
So the LHS is the derivative of $\ln|f(x)|$, of course.
Differential forms, as I said, make this completely rigorous, but the formula $du = \dfrac{du}{dx}dx$ makes integrals work because this is the chain rule:
$$\int f(g(x))g'(x)\,dx = \int f(u)\,du = F(u) = F(g(x))$$ if we substitute $u=g(x)$ and $F'(u)=f(u)$. Why? Because the derivative of $F(g(x))$ is $F'(g(x))g'(x)=f(g(x))g'(x)$, as desired.
OK, I'm done. :)

Right,
One more question

4:48 PM
So the method that we teach when we do separable DE is "engineering" style, but this discussion I just gave justifies it completely.
Sure.

@TedShifrin is "antidifferentiating" both sides ...rigorous?
because what i picture while "anti-differentiating" the right side for example, we are "integrating after multiplying by dx" and this multiplying by dx is what bothered me in the first place

Why?
If $f'(x)=g'(x)$, then $f(x)=g(x)+C$, so this is why, of course, we always put the $+C$ when we do antiderivatives.

this might seem a bit silly but:

@TedShifrin: we can also do a rigorous treatment of $dy/dx=y/x$ by rewriting it as $(1/y)(dy/dx)=1/x$. Assuming $y$ as a function of $x$ both sides can be integrated to get $\log y=\log x+\log c$ and thus $y=cx$. I see separation of variables as using chain rule but with more ease.
I am not really sure if we need to invoke differential forms to justify this

starting from $f'(x)=g'(x)$, what exactly do we do to obtain $f(x)=g(x)+c$? just trying to be rigorous here

4:55 PM
@satan29: well you use mean value theorem which implies that any function with zero derivative on an interval is necessarily a constant on that interval

Sure. Rigor comes from applying the Mean Value Theorem. Let $h(x)=f(x)-h(x)$. If they're continuous/differentiable on a closed interval $[a,b]$, then $h'(x)=0$ on that closed interval. By the mean value theorem, $h(b)-h(a)=h'(c)$ for some $c$ in the interval, which we know must be $0$. Therefore $h(b)=h(a)$, so $h$ is constant on the interval. This means $f-g$ is constant on the interval.

@ParamanandSingh But you are not integrating the two sides with respect to the same variable. So that does not seem like a good justification

@Tobias: I already explained this carefully.
You do in fact integrate with respect to the same variable. You just think you don't.

Ahh, right

@TedShifrin I see!

4:57 PM
@TobiasKildetoft: I am integrating both sides wrt $x$. The derivative of $\log y$ wrt $x$ is $(1/y)(dy/dx)$.

Does it make sense to you now, @satan? I agree that 95% of calculus classes don't explain this carefully.
Yes, @ParamanandSingh, you're right. I explained all this above.

@TedShifrin: that's the unfortunate part about calculus classes. Why do most of them ignore fundamentals and focus on computational aspects

So i guess "multiplying" by dx (or dy or any differential) is simply not rigorous, it just...seems to work (mostly due to the chain-rule).
@TedShifrin yes, somewhat. Thanks a lot sir!!

Until you learn about differential forms, you should just think of $\int \dots dx$ as a symbol that means "find a function whose derivative is $\dots$." Then the symbolic manipulation $dy = \frac{dy}{dx}dx$ is building in the chain rule as I explained.

right. Thank you everyone for the help!

5:02 PM
I wanted to post a question but it tells me that I can post only once every 40 minutes. I did not post anything for weeks ... Is that a bug?

@ParamanandSingh: Because the majority of students are science/engineering students, not math majors who "want" proofs. But even when I lectured a big lecture with mostly science/engineering students, I gave derivations provided the derivation gave understanding of what was going on. I didn't belabor proofs the way I would in a theoretical course.

Hmm, yeah

@СССР did you ask and then deleted a question in the last 40 minutes?

nope
nothing

@TedShifrin There are good arguments that the irritating signs surrounding lots of things in topology can be chased to the boundary orientation being normal-first instead of normal-last, but I do agree (and did forget!) that this makes Stokes more irritating.

5:05 PM
You don't seem to have a math.stackexchange account, only a quantum computing one, where were you trying to ask a question?

the first time i tried i got a validation error, since I've forgotten to the tags and after I did so I now get this error
i do have math.stackexchange

@MikeMiller LOL @ irritating. I think that if you do it your way (which I agree is reasonable), then you have to make a special case for boundary orientation. I find that more annoying :P
I should look in Hirsch and see what he says (although he doesn't do integration, so perhaps he doesn't care).

This mainly happens in Morse theory. The boundary orientation being defined the way it is leads to a lot of extra signs showing up in various formulae, that disappear (or simplify) when you have the boundary orientation as outward-normal-last instead.
But in the end more conventions makes math harder to do.

Oh, Hirsch defines boundary orientation with it at the end, but pointing in. Oy.
Oh, he's doing this because of collaring.
But this will mess up Stokes's Theorem if ever he did it.
So he's off by $(-1)^n$ compared to G&P, I guess.
Spivak gets around it by doing chains and only needing to get the boundary of a cube oriented correctly.
Ah, on p. 260 he does boundary orientation. He does it the way G&P do it. Whew.
Conclusion: If you're doing just topology and don't care about integrating, then your remark is good. :) But otherwise it's bad. :)

1 hour later…
6:29 PM
Do everything with $\Bbb Z/2\Bbb Z$ coefficients

That's always good with integrals.
Howdy, demonic.

6:46 PM
Hi Ted

7:10 PM
Hello
Can I get some help understanding a math book?

You need to ask a specific question. No one can answer that question.

Ah I am reading a book on Functional Analysis. And I found an example which is a bit difficult to realize and it's so abstract

Well, you're still not telling us much. Ask the question.

Can you understand this example clearly?

Yes, this is a standard example.
It's generalizing the norm $\max_{1\le i\le n} |x_i-y_i|$ on $\Bbb R^n$.

7:17 PM
X is a set. Good. Each element of that is a sequence. Now they are showing one such element x. And then they are saying each sequence element has magnitude <= c_x

This is what it means to say the sequence $x=(\xi_i)$ is a bounded sequence. No big deal.

Now what is this metric doing?
taking two such sequences and then it does a crazy thing
I mean ok ...I can tell you my dilemma. So you have a sequence and I have a sequence. Let's say x and y and then suddenly you pick a element and I pick exactly that number of element from my sequence.
Now we take a difference . And we do it many many many times and now we say ..what is the number that is just bigger than all of these? And that's a metric?

Huh? You measure the maximum distance the individual elements of the sequence are apart. Just like doing this with two functions on $[a,b]$. You look to see the greatest distance (vertically) between the graphs.
Why don't you start by understanding what I wrote up 10 lines about that metric on $\Bbb R^n$. Work that out first.

Wait it's not rendering proerly...Idk why.
I see what you said
So it's basically doing that right?
Max distance between sequence elements?
I see.

Yes, that's precisely what it's doing.

7:25 PM
Why didnt it say it clearly like you did?
These authors try to make it look so complex

They do assume some basic mathematical maturity, I would say.

No, I think they wrote it fine. I like to believe that as a professor I added value to the books, even if I wrote them.
I thought it was odd that they explained sup (lub). I would have assumed that was well-known by this level.

But it is not lowest upper bound...It is just max.
@SayanChattopadhyay yes I know. I am just a common man trying to understand this stuff.

They are using the supremum and not max.

Ok i get it
supremum is indeed LUB
Geez I talked so stupid.

7:31 PM
No, it is sup or lub. There need not be an actual max, because we have infinitely many numbers.

Yes I see
Asking both of you...I just read 1.5 page today. The book has 700 pages. There is no way I can finish this book in a month or so. Do you people know any children book on Functional analysis?
Like it would explain everything like it is talking to a dumb person

What is your background with analysis, algebra, topology, proofs?
Functional analysis should only come after very serious real analysis (including Lebesgue theory), and you need solid linear algebra for sure.

I got an annoying calculation question. The Res( \frac{z^a}{(z^2 + 1)^2},i ) = \frac{(1-a)exp(aipi/2)}{4i} is supposed to be this. Since i is a residue of order 2, there is a derivative involved. Now according to mathematica, wolframalpha.com/input/?i=D%5Bx%5Ea%2F%28x+%2B+i%29%5E2%2Cx%5D , how does i^a disappear?

7:49 PM
@TedShifrin I know Lin Alg. But rest I did not study.

You can't do functional analysis without solid foundation in analysis, topology, and some Lebesgue integrals. What gave you the idea that this was reasonable?

I didnt think it was reasonable.
Just wondering.
@TedShifrin But this is the thing. I have to say this, even if it sounds bad, if I want to study all the 3 things today, there is not a single author who has shown everything from scratch so that a beginner can study. It seems none of the math people had time to explain stuff

8:12 PM
It's extremely hard to just jump into functional analysis like this @mathuser001. You need to cover whatever Ted has mentioned to get some idea of the kind of techniques functional Analysis employes. Not really a book on functional analysis, but a very good book nonetheless would be Introduction to Topology and Modern analysis by Simmons
That should atleast give you a starting point

Yes, Simmons's book is wonderful. But one should still do some baby real analysis (metric spaces and $\delta$-$\epsilon$) before learning topology. Mathematical maturity is a real thing. One could start with Spivak's Calculus. @mathuser said he "knows" linear algebra. Does that mean lots of proofs or just matrix computations?

Ah I see. There is a book on Linear Algebra of Gilbert Strang. I know that book.

Which one?
He is not good at making students write proofs.
And reading books isn't good enough to learn mathematics. One must write lots of exercises to learn to do/write proofs.

That book has almost no proofs at all.

8:20 PM
-_-

I taught out of an earlier edition, gave up, and wrote my own textbook.

Strang has a wonderful view of linear algebra, but his books are not going to teach one how to do theoretical mathematics.
I have four books, @mathuser, yes.

I see. But those are advanced book I guess.
Google says you have 3 books

The linear algebra book is at the level of Strang's book, but emphasizes proofs and learning to understand/write proofs.
Yes, the fourth is a free .pdf (on my webpage) on differential geometry.
I decided to keep it free.

8:25 PM
I see.
I see you are a famous professor.
I have to read a book like this ...and that's why I am trying to look into these kind of things
Idk what to do

Quick question, if I want to show the ideal generated by I = (19,x^2 +1) in Z[x] is maximal, I have to show that I is irreducible. The only obvious thing I can see if that the generators 19 and x^2 + 1 are relatively prime. But that's not enough.

@mathuser: What is your background in multivariable calculus? You know it computationally?
@Hawk: Why does it mean for an ideal to be irreducible?
I would recommend a different approach.

@TedShifrin Not much. I forgot a lot of stuff. -_-!

you could also show that the quotient by this ideal in a field

Better to say now that I dont know anything

8:33 PM
i meant to say the polynomial generating the ideal is irreducible.

Yeah, maybe you should start by watching my YouTube videos on multivariable calc/analysis and linear algebra. That doesn't give you exercises to do, but the lectures will be easier to learn from than reading books for you, I think.

but the ideal is generated by two polynomials

@Hawk: This is not $F[x]$. That doesn't make sense here.

Ok I will check that. But you have video on functional analysis as well?
@TedShifrin

No functional analysis. But some of the things you definitely need to know are covered (e.g., working with open/closed sets, sequences, convergence, compactness). It's all in $\Bbb R^n$, though. But lots of proofs with linear algebra, as well, and you can practice by trying to do them for yourself.

8:35 PM
Ok I will check that. Thanks a lot

@Thorgott how do we show the quotient is a field in this case?

What's the best way to mod out by the ideal generated by $2$ things?

@Hawk The hint is in one of the ring isomorphism theorems.

@Hawk: My suggestion is to reduce to the theorem you're trying to apply. Can you see how to turn your question into a question in $F[x]$ for some field $F$?

I just know it is not the 2nd theorem. I thought Z isn't a field here
actually i think i know what to do, to reduce it to F_19

9:08 PM
indeed, that's what Ted was hinting at
then adapt the usual criterion over PIDs that you tried using earlier

9:42 PM
@TedShifrin Good afternoon. It is very hot and humid today.

hi all
3

Consider the twin primes. So we consider $3,5,7,11,13,17,19,29,31,...$ Call the $n$-th element in the sequence $t(n)$. Now consider the Sophie Germain primes ( if $p$ is prime, then so is $2p + 1$. ONLY count the $p$ not the $2p+1$ ) Call the $n$-th element of that sequence $s(n)$. Now for \$1000 <...

any ideas ?

10:08 PM
@robjohn Hi sir.

10:24 PM
@robjohn Yup. It's summer. Whether the postal service can deliver or not.

11:09 PM
If I have a description of two 2d shapes on a grid is there a shortcut to detect if they're touching or overlapping without comparing every point in the first shape to every point in the second shape?
And if so, is there a way to cheap in a similar way with 3d shapes?

11:27 PM
@TedShifrin we need to protect ponies in case the government forces us to revert to the Pony Express.
@Binod how are you today?