Mathematics - 2023-04-01

AMDG

12:15 AM

shintuku

hope that's not aimed at me because i'm not saying truth is relative

AMDG

Yet many think it. Anyways, no time for chitchat. I need to get back to my studies so I can get back to working.

shintuku

also lowtier argument vs relativists

who is this

shintuku

1:05 AM

what's the fastest way to show $|c_1||z|^n(1-\frac{2|c_2|}{|c_1||z|})$ diverges, where $c_i \in \mathbb C$ is a constant and $z \in \mathbb C$

Ted Shifrin

It needn’t?

shintuku

forgot some bars there

D.C. the III

Ted.....in asking us to graph something such as $r = 4\cos(\theta)$, I suppose you mean in the $r - \theta$ plane?

Ted Shifrin

Are you talking sequence or series?

shintuku

it's a continuous function

Ted Shifrin

1:08 AM

huh? @shin

no, I meant in cartesian, surely.

D.C. the III

oh..........well that is bit more demanding, but makes sense in this context

shintuku

it's a function $f:\mathbb C \to \mathbb R$, i think it's probably continuous

D.C. the III

well you did say "sketch the curve" in the instructions

shintuku

it clearly diverges because $|z|^n$ explodes and the stuff in the parenthesis goes to $1$

but i'm perusing my divergence tests and they're not nice looking

Ted Shifrin

You’re still not making sense. Are you talking about divergence of a sequence? You seem to be thinking about behavior at infinity.

shintuku

1:17 AM

right, as $z$ goes to any infinity, complex or not, the function $f$ goes to infinity

it seems to me

so, $f: \mathbb C \to \mathbb R$ with $f(z) = |c_1||z|^n(1-\frac{2|c_2|}{|c_1||z|})$ has $\lim \limits_{|z| \to \infty} f(z) = \infty$

but I'm trying to figure out if I'll need to prove that $f \to \infty$ and $g \to L$ implies $fg \to \infty$ first, or if I can prove divergence directly

Ted Shifrin

Why are you talking about divergence tests?

shintuku

i was thinking that if there was a function I knew diverged, I could compare it against it, but clearly misuse of terminology by me, my bad

shintuku

1:49 AM

how about: there is $z_0$ s.t. $z > z_0 \implies 1-\frac{2|c_2|}{|c_1||z|} > \frac{1}{2}$, and so, $z > z_0 \implies |c_1||z|^n(1-\frac{2|c_2|}{|c_1||z|}) > \frac{|c_1||z|^n}{2}$. Let $\epsilon$ be given, and make $\frac{|c_1||z|}{2} > \epsilon$. would this conclude the proof?

Ted Shifrin

2:00 AM

Remember that > makes no sense with complex.

shintuku

ohhhhh right thank you. i'll say instead: there is a $z_0$ such that $|z| > |z_0|$ implies the above

Ted Shifrin

You don’t customarily use $\epsilon$ for an arbitrarily large number, either.

shintuku

hm, what's the usual notation?

Ted Shifrin

Say $M$ :)

D.C. the III

When I graph $r = 3sec(\theta)$ there will be a trouble point at $\theta = \pi/2$, do we qualify it as a removable discontinuity? other than that it is just the straight ray

shintuku

2:07 AM

thanks!

D.C. the III

discontinuity might not be the right term for this though

unless I flip axes.

Ted Shifrin

No, you typically specify the domain as $|\theta|<\pi/2$.

D.C. the III

Ah.....that would be reasonable. I drew an open point around it, but specifying domain is more convenient.

leslie townes

3:13 AM

we say that a sequence {n_z}_{z=1}^{infty} of complex numbers converges to infty if for all delta > 0 there is a positive integer epsilon such that |n_z| > delta whenever z > epsilon.

Ted Shifrin

Go to your room.

3

shintuku

what's a cool way to show invertible implies nonsingular? rank-nullity theorem is banned

no determinant facts either

invertible: has inverse; nonsingular: full rank

shintuku

3:45 AM

suppose invertible. recall it implies only trivial solution in nullspace. suppose not full rank. so rref has a row of zeroes. so the augmented system $Ax = 0$ has a nontrivial solution, contradiction

Ted Shifrin

4:00 AM

Nonsingular is equiv (for square) to saying the equation $Ax=0$ has only the trivial solution.

shintuku

hm, how do you get there

Ted Shifrin

Think.

No free variables because …

shintuku

number of pivots equal to number of variables?

Ted Shifrin

Right.

shintuku

hm

I'm not sure how to follow that thread. i've meanwhile come up with: suppose $A$ invertible. Then $A \vec x = 0$ has only the trivial solution, and so does $U\vex = 0$, therefore the columns of $U$ are linearly independent, so the column rank is $n$, and therefore so is the matrix rank

but, how would we do it your way instead Ted? i don't know where to go from having the number of pivots equal the number of variables

hm nevermind, figured out what you meant

we would say: suppose $A$ invertible. Then $A \vec x = 0$ has only the trivial solution, and so does $U \vec x = 0$, but this implies $U = I$, so $A$ is nonsingular. would this be what you meant?

one potato two potato

4:29 AM

where does this number come from...

Cotton Headed Ninnymuggins

If an undirected graph $G$ has $1$ edge connecting vertices $v_1$ and $v_2$, can a subgraph of $G$ have multiple edges connecting $v_1$ and $v_2$ or only $1$?

one potato two potato

4:52 AM

0

Q: Farkas & Kra Riemann surfaces Proposition III.5.9

The question is from Farkas & Kra Riemann surfaces. I can't make the question self-contained because there're too many definitions, and propositions I need to type. So I'm asking for people who have the book (p.87). But if someone asks for some details and notations via comment, I'll add them. P...

I'm not sure if this question not violates MSE guidelines...

Ted Shifrin

5:16 AM

You’re not thinking about what a Wronskian is, are you?

user 7269591

5:29 AM

...and not thinking violates all math guidelines

one potato two potato

@TedShifrin Ahhh it's just a sum of $q+(q+1)+\cdots+(q+d-1)$. I'll delete the post soon

user 7269591

Why delete? Answer it yourself so others,
with the same query, may benefit.

one potato two potato

Ok, I'll undelete and wait for one day. After that, I'll either delete or answer by myself.

user 726941

6:41 AM

Shortest chapter ever in a physics book.
From Why String Theory?
by Joseph Conlon

one potato two potato

7:49 AM

One more... math.stackexchange.com/q/4670577/668308

Franklin

10:52 AM

Can we make this assertion:" If $f(x)$ and $g(x)$ are two functions that have same derivatives, then $f(x)-g(x)=k$ , where $k$ is a fixed real number " ?

robjohn

11:09 AM

@Franklin If $f$ and $g$ are real functions.

Ajay

11:55 AM

@robjohn Do you think it would be worthwhile to create a hints tag?

robjohn

To what end? Hints are not accepted as good answers

Besides, the tag is supposed to qualify the question, not the requested answer

Ajay

oh, I didn't think of it like that.

PM 2Ring

12:44 PM

Here's a lovely sweet acoustic guitar song from a couple of 60s Mod guitarists: Pete Townsend from The Who, and Ronnie Lane from Small Faces.

April fool - Ronnie Lane

►

Franklin

1:03 PM

@robjohn Ok, so you mean : "If $f(x)$ and $g(x)$ are real functions that have same derivatives, then $f(x)-g(x)=k$ , where $k$ is a fixed real number ", this is true, right ?(I am just confirming)

PM 2Ring

Hmm. Maybe Pete isn't playing on that track. One YouTube comment claims that Eric Clapton played the Dobro part.

@Franklin Yes. If the functions are complex, then $k$ may be complex.

Franklin

1:51 PM

@PM2Ring ohh..thanks a lot foy your validations!

@PM2Ring I see, you are interested in classics. Have you ever listened to Nat King Cole? He is one of my favorite American singers....You might give it a try if you haven't:)

Or you might want to listen to Glenn Miller. He tuned a beautiful composition called "Moonlight Seranade" , it's one of greatest melodies I heard. It's a nice orchestra in general!

@PM2Ring They did a wonderful symphony, no doubt !

PM 2Ring

2:26 PM

@Franklin Nat was a great singer, and a great pianist. Here's a clip from a Glenn Miller movie that you'll probably enjoy. It has some excellent vocal harmonies. It's great that we have such an excellent "video" with high quality audio from that era.

Glenn Miller - Chattanooga Choo Choo - Sun Valley Serenade (1941) HQ

►

PM 2Ring

2:43 PM

One of Nat's early big hits was Route 66, written by Bobby Troup. Here's Bobby performing a song he co-wrote with Neal Hefti, Girl Talk. Sorry, the audio quality is a bit muddy.

GIRL TALK by Bobby Troup

►

PM 2Ring

2:54 PM

Here's an acapella version of another Neal Hefti tune.

The Real Group - Li'l Darling (Live at "Go'morron Sverige" -91)

►

PM 2Ring

3:08 PM

More vocal harmonies, but a bit less schmaltzy. The Blues and the Abstract Truth...

NEW YORK VOICES - Stolen Moments (1991)

►

Sourav Ghosh

3:48 PM

How to find area of the image of $\{z :|z|<r \} under the map $f(z) =\sin z$ ?

Franklin

@PM2Ring I have to admit you have a good collection of wonderful music. Sometimes, you know, I avoid listening to 50s,60s,70s songs because I think, that's a lost era. Yeah, there were much much more hardships than today, but something enchanting was there. When I listen to these, I sometimes wonder if I could go back to the time, where I wasn't even there, but it gives me nostalgic, melancholic and an exciting feeling at the same time and that can't be explained

@PM2Ring On the other hand, if I listen 80s and 90s pop and soft rock music, I get a satisfied feeling. Again, it was a time when I wasn't even born. But the 80s era music hits me in a different place than the 70s which hits me in a different way! They have different moods, but nonetheless they provoke a much deeper unimaginable satisfaction and a strong yet a comical and funny desire, to go back to the 80s.

@PM2Ring Each of these eras 50s,60s,70s,80s were characterized by many hardships, some of which I maybe aware of, while some I maybe unaware, but listening to them gives me an unexplainable feeling which I feel can't be described in words maybe !

user4539917

4:30 PM

Qualia

In philosophy of mind, qualia ( or ; singular form: quale) are defined as individual instances of subjective, conscious experience. The term qualia derives from the Latin neuter plural form (qualia) of the Latin adjective quālis (Latin pronunciation: [ˈkʷaːlɪs]) meaning "of what sort" or "of what kind" in a specific instance, such as "what it is like to taste a specific apple — this particular apple now". Examples of qualia include the perceived sensation of pain of a headache, the taste of wine, and the redness of an evening sky. As qualitative characters of sensation, qualia stand in contrast...

Sourav Ghosh

Probably you need a different chatroom.

user4539917

Why is that.

One room for quality, one room for quantity, one room for quanti, and one room for qualia...

Franklin

4:45 PM

0

Q: Let $1,4,...$ and $9,14,...$ be two arithmetic progressions. Then the number of distinct integers in the collection of first $500$ terms of each of p

Let $1,4,...$ and $9,14,...$ be two arithmetic progressions. Then the number of distinct integers in the collection of first $500$ terms of each of progressions is $$A. 833,\space B. 835,\space C.837 ,\space D. 901$$ My solution goes like this: The two A.P.'s general term might be represented a...

sequences-and-series solution-verification arithmetic-progressions

Can anyone please help me with this ?

Ted Shifrin

4:56 PM

Ah, @Franklin. I figured it out just as someone posted the answer. You're misinterpreting "distinct."

They want you to count the numbers that appear, without double-counting. They did not ask you how many numbers appear on only one list.

@SouravGhosh Do you know the change of variables formula for multiple (double) integrals? You need that, but this seems to be a yucky thing to compute for this function.

Franklin

5:15 PM

@TedShifrin yeah, what a stupid error! Thanks a lot!

Franklin

5:50 PM

Let f be an onto and differentiable function defined on [0, 1]
to [0, T], such that f(0) = 0. Which of the following statements is
necessarily true?
(a) f‟(x) is greater than or equal to T for all x
(b) f‟(x) is smaller than T for all x
(c)f‟(x) is greater than or equal to T for some x
(d) f‟(x) is smaller than T for some x.

Can someone please help me with question ?

I don't have a clue where to start :?)

Ted Shifrin

Where are you getting all these multiple choice things?

Think about this one in terms of physics. If you accelerate a lot, what distance do you travel?

Franklin

@TedShifrin I had a different approach, I was thinking about, say, taking y=arcsinx, from [0,1] to [0,\pi/2]$

Ted Shifrin

Thinking about one specific function may not make it clear.

Franklin

@TedShifrin hmm...that's true but here if we take y as arcsin, then we see option c is automatically satisfied

Ted Shifrin

So what?

Franklin

5:58 PM

@TedShifrin But, I think that here, I am just being lucky...

@TedShifrin so C is the answer

Ted Shifrin

No.

Both c and d could be the answer.

If you're correct about your example, all that it does is rule out b (and maybe a, too).

Franklin

@TedShifrin You are valid in your point, but this a one-answer correct type question, so I take it as the fallacy in the question as well.

Ted Shifrin

No, it's a fallacy in your logic.

It asks which is necessarily true? Not, which might be true depending on your function?

You must pay attention.

Franklin

@TedShifrin now, that's a good observation and I overlooked it. Ok, but I want to know, how would you approach this question this question in an elementary way?

Ted Shifrin

I already told you.

Franklin

6:04 PM

To be honest, I am not good in physics

My bad:(

But is there any other way?

Ted Shifrin

This is just elementary school physics. What does $f''$ mean?

Every Calc class does what I'm talking about.

Franklin

@TedShifrin umm...double derivative,😂😂😂 but I think, you are talking about it's significance here...

Ted Shifrin

And what does double derivative mean? What information does it give you?

Franklin

@TedShifrin rate of change of f' ?

Ted Shifrin

OK. And can you determine $f'$ if you know $f''$?

Franklin

6:07 PM

@TedShifrin Yes, by performing definite integrals

Ted Shifrin

And then can you determine $f$?

Franklin

@TedShifrin Yeah sure, by proceeding in the same way...

Ted Shifrin

So think about doing this with the information you have about $f$.

D.C. the III

Ted you asked us to evaluate $\int_{S} y^2 dA$ over the annular region, you also asked in the same question to evaluate it by calculating $\int_{S} x^2 + y^2 dA$. ...Why? What was I supposed to glean from doing it in the second form?

It is easier clearly, but how/ where would the idea of using the second form come into play?

Ted Shifrin

Symmetry.

This is a common sort of argument.

D.C. the III

6:13 PM

symmetry?....so the second is the circle, and the first is a parabola.......I guess a perfectly drawn parabola is "half" a circle?

Ted Shifrin

No.

Franklin

@TedShifrin All I know is f is onto and differentiable in [0,1]. And not to forget f(0)=0. But then, f(x) is monotonic (I guess this intutively from the fact, that f is continuous since it's differentiable)

Ted Shifrin

We're integrating functions. The region isn't changing.

D.C. the III

Right. I got the region. but the change in function here is getting me.

Franklin

@TedShifrin But I think I am wrong in this argument. But where's the off-beam portion ? Is it not monotonic? I can't think of a rigorous explanation 😕

Ted Shifrin

6:16 PM

I don't know why $f$ has to be monotonic.

Since we don't know anything about $f'(0)$, actually, this problem is not so easy.

Franklin

@TedShifrin that's for sure, but I think, in an mcq test my first approach, with arcsin is the way to do these things in a jiffy!

Ted Shifrin

But if you integrate twice you can certainly eliminate option b, for example.

Well, I am not fond of your approach, but do it if you want.

I don't see how one example does it.

It can eliminate a few, which is good, but you can't be sure.

Franklin

@TedShifrin That might be the case really...

Ted Shifrin

As I pointed out, the answer might be d or c and you can't tell.

Franklin

@TedShifrin but then you said necessarily true ?

Weiiird question in my opinion

Ted Shifrin

6:19 PM

That was my point. Doesn't your example satisfy both c and d?

Franklin

@TedShifrin let me check for d in my eg

Yes, it satisfies both @TedShifrin :)

Ted Shifrin

So it is not helping.

I can give the simplest possible example that shows a and c are wrong. Think about that.

My integral approach will then show that b is wrong.

Franklin

@TedShifrin I am not convinced with this, as the answer key suggests c as the correct choice!

Ted Shifrin

It's garbage.

What is the simplest possible function you could use?

You need to think more critically and not learn from answer keys.

Franklin

@TedShifrin you are so inspiring (I am not joking really!)!!!!

@TedShifrin these hit and trial testing is soo frustating

Is it the experience that counts?

Ted Shifrin

6:29 PM

Well, multiple choice is often testing eliminating certain answers in an easy way.

Anything with "for all $x$" in it begs for counterexamples. But then you have to work a bit harder.

Franklin

For quickly judging the functions(like in this case) and coming up with the counter examples like you did, isn't it because you have a whole lot of experience ?

Ted Shifrin

No.

Everyone should think to try the simplest possible function.

Franklin

For quickly judging the functions(like in this case) and coming up with the counter examples like you did, isn't it because you have a whole lot of experience ?( I think, that's what permitting you)

Ted Shifrin

I would never have thought of arcsin.

Franklin

@TedShifrin Should I ask: Why?

Is it your priori intuition ? My approach here seems filled with fallacies and fallacies...

Ted Shifrin

6:32 PM

Because once I get rid of the easiest example, my tendency is to think about integrating. I may be wrong, but that's what I do.

You can draw pictures, too, rather than writing integrals. To me thinking physically is most natural. I can certainly accelerate very hard (much bigger than $T$) on a certain interval, and then decelerate hard to make sure I don't go farther than distance $T$.

Franklin

@TedShifrin this indeed eliminates a and b

Ted Shifrin

What is the easiest example? I'm still waiting.

Franklin

If I go readily, with your reasoning, I think c and d are the possible answers ...

@TedShifrin f(x)=x. But this isn't helping much ?

Ted Shifrin

You need to be more careful. But, yes, linear.

Franklin

@TedShifrin Maybe, $f(x)=ax+b$ ? But then, we arrive, again at options c and d

PNDas

6:42 PM

I don't understand. $f(x)=x$ implies that possible answers are $b,d$ and $f(x)=x^2$ implies possible answers are $a,c$.

May be I am missing something.

Ted Shifrin

$f(x)=x$ does not satisfy the hypotheses.

Franklin

@TedShifrin yes, I see now...

I was too quick...

PNDas

@TedShifrin Why?

Ted Shifrin

@D.C.theIII Why is $x^2+y^2$ easier to integrate? Once you realize that, how does it answer the question?

@PNDas Because we need to surject to $[0,T]$.

PNDas

Here my T=1

Ted Shifrin

6:45 PM

Well, mine isn't.

If you can argue that fixing $T=1$ answers the question for general $T$, I'll accept that. On multiple choice, you don't need to defend it, but you'd better have an argument ready in your head that justifies it.

D.C. the III

Well that converts to $r^2$ once turned to polar. I saw that part. In just doing $y^2$, I will end up with a $\sin \theta$ in there. I was thinking for a second before you pinged me and was conjecturing because $\sin \theta$ is an odd function over that interval it disappears.

PNDas

What I'm saying is that the question is for general T so it should work for T=1. But when T=1, then like I said $x,x^2$ show that no options hold for all functions.

Anyway I came here ask a complex analysis question.

Ted Shifrin

@D.C.theIII Not when you square. Now think back to the word I said when you asked at the beginning.

@Franklin @PNDas I got to d as the only possible answer, by eliminating a,c,b. That doesn't yet prove that d is correct.

D.C. the III

True, when you square it turns even.....and there is a property of even functions when integrating which I have forgotten at this moment...which applies to symmetry.........let me go fetch my SPivak....:)

PNDas

@TedShifrin $x^2$ eliminates d, no?

Ted Shifrin

6:50 PM

Nothing to do with Spivak, DC.

How are the integrals of $x^2$ and $y^2$ related on this region?

PNDas

In theorem 10.21 of papa Rudin, we have to prove that If f is holomorphic except a point then one of the following cases must occur: 1) f has removable singularity at a(2) f has a pole at a (3) f has essential singularity.

Ted Shifrin

@PNDas, yes $Tx^2$ maps onto $[0,T]$ and the second derivative is $2T$ everywhere. So all the answers are wrong. The answer key said c but it's clearly wrong.

PNDas

But Rudin proved that if 3 fails then either 1 happens or 2 happens

Ted Shifrin

I don't know Rudin's definitions. To me that's almost definition. Look at the Laurent series and decide if it has (a) no negative terms, (b) finitely many negative terms, (c) infinitely many negative terms.

PNDas

But does it imply that if 1 and 2 both fail then 3 must happe? He didn't show it.

Ted Shifrin

6:53 PM

You're making a logic error. That is not needed.

Oh, I guess we need to know the list is exhaustive. So I just did that.

PNDas

@TedShifrin This is the easiest definition in my opinion, But Rudin follows different definitions.

Ted Shifrin

You need to know such a function has a Laurent series on some annulus of radius $0$. I presume he already proved that with those hypotheses.

Well, I do not have the book.

PNDas

@TedShifrin But I got the idea. Thanks

Ted Shifrin

Rudin is so hyper-rigorous. I have confidence he is correct.

@Franklin The conclusion is that the question is just plain wrong. PNDas has a natural suggestion easier than my approach. Think about linear and quadratic functions.

D.C. the III

The integrals of $x^2$ and $y^2$ are related on this region because they are equal...but that involves me actually having evaluated them

Ted Shifrin

6:55 PM

You should not need to evaluate to see that.

What symmetry does the region have that switches $x$ and $y$? I suppose you'll want the change of variables theorem to make this rigorous, but it's obvious.

PNDas

Actually, in the theorem, (3) is If r>0 and the deleted neighbourhood $D'(a,r)\subset \Omega$ then $f(D'(a,r))$ is dense in $\mathbb C$.

D.C. the III

I'll get to the change of variable theorem when you said...in a few sections time. For now just intuition building

PNDas

And this is the definition of essential singularity for him.

Similarly pole has a different definition so I think it's difficult to argue that the list is exhaustive.

Ted Shifrin

That's Casorati-Weierstrass, which is one of my all-time favorite theorems.

D.C. the III

What do you mean by the question "what symmetry does the region have?"? As in it "is symmetric along such and such an axis?"....this is now a language thing for me.

Ted Shifrin

7:00 PM

He's never been my favorite author, so I have no idea what he does in there, PNDas. But I am confident it's correct.

What symmetry switches $x$ and $y$ and happens to map the annulus to itself (by an isometry)?

Franklin

@TedShifrin What a waste of time then! Like seriously!!!!

Ted Shifrin

Well, it's educational to learn about how to think about examples carefully. Learning is never a waste of time.

But you need to be learning in a way that isn't just "look at the answer key and get that answer."

I don't know if you're in India, but we've seen plenty of examples in here of tests in India that are full of mistakes and wrong answers.

PNDas

@TedShifrin Hmm I saw the proof of that theorem from Wiki. The proof is exactly same as Rudin. What they did is: start with $z_0$ a essential singularity then conclude that $z_0$ is either pole or removable singularity so we get a contradiction. But in Rudin's setup I don't know why a pole can't be essential singularity. same for removable case.

Ted Shifrin

So the definition of an essential singularity is that the image is dense?

PNDas

Yes

Ted Shifrin

7:09 PM

Well, we can see easily that for a removable singularity the image is bounded and that for a pole the image is a neighborhood of infinity.

PNDas

Then he said this "dense" definition is equivalent to: for each complex number $w$ there exists a sequence $z_n\to a$ such that $f(z_n)\to w$.

Sourav Ghosh

No. Essential singularity means non removable, non pole. Equivalently Laurent development has infinitely many negative coefficients .

PNDas

@SouravGhosh Rudin doesn't follow this.

So I think this sequence definition means that removable or pole implies not essential.

Because he earlier said that pole implies $|f(z)|\to\infty$ as $z\to a$

So I just need to prove the equivalent definition.

Ted Shifrin

Of course, the density is equivalent to that.

That's beginning real analysis/topology. I would say "trivial" if I were prone to use that term.

PNDas

Yes.

Hmm

Sourav Ghosh

7:16 PM

Yes. Converse of Casorati-Weierstrass theorem also true.

PNDas

Actually we can say more by the Great Picard theorem

Ted Shifrin

That is beyond cheating.

That is a huge cannon.

PNDas

@SouravGhosh That's what I basically said.

Sourav Ghosh

Liouville's simplicity ( image of a non constant entire function is dense) Vs Piccard's rudeness (litte; can omit atmost one point)

PNDas

Rudin defines Laurent series in the exercise. I think one should keep less important things in the exercise. And IMO Laurent series is very important.

I'll just cover chapter 10 from Rudin. Then I'll shift to Greene, Krantz. I heard this book is very good.

Sourav Ghosh

7:26 PM

@TedShifrin In india, to challenge a question in provisional answer key, for each question it costs ₹700 ( $1=₹82.28)

PNDas

I don't think there may be 1 or 2 wrong questions in competitive exams.

schn

So I have the following series: $\sum_{k=1}^\infty \frac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k}}$. Grateful for any hint on how to determine its convergence or divergence.

PNDas

$a_k=\frac1{\sqrt k(\sqrt{k+1}+\sqrt k)}>\frac1{2(k+1)}$

leslie townes

schn: if you 'rationalize the numerator' you see sqrt(k+1) - sqrt(k) = 1/(sqrt(k+1) + sqrt(k)) which maybe suggests limit comparison of that series with sum 1/k. same result as what pndas does, but without inequalities.

schn

ok, thanks!

leslie townes

7:40 PM

note the point of what i call "rationalizing the numerator" here is not to make anything rational, but to rewrite a difference of things that individually go to infinity (which, if all you know about something is that it has that form, is not informative about limiting behavior) as something that goes to 0 like a recognizable power of k.

Ted Shifrin

@PNDas I would throw the book in the trash.

leslie townes

if you have f(n+1)-f(n) for a differentiable f(x) you can write that as f'(something) with something between n and n+1 to the same effect - sometimes making the limiting behavior clearer - even if it does not involve 'rationalizing' anything.

Ted Shifrin

@leslietownes No, limit comparison with $1/\sqrt k$!

leslie townes

ted: look at the original problem.

"that series" is not sqrt(k+1)-sqrt(k)

Ted Shifrin

Oh.

leslie townes

7:42 PM

haha!!

PNDas

@leslietownes I didn't know this technique in my BSc.

Ted Shifrin

My apologies.

leslie townes

ted: you'd like this. we're at the cafe this morning, eating bagels, and out of nowhere, munchkin turns to my wife and says "you're 40!" like it's a punchline.

@PNDas sometimes hidden in the exercises of the better calculus books :)

Ted Shifrin

@SouravGhosh Cool that they charge you for their errors. And they make so many.

PNDas

@TedShifrin They return the money if you are right.

Ted Shifrin

7:44 PM

I know a famous error on the college board math achievement test here (not SAT, I believe).

That is fair. It discourages noisy critics.

PNDas

If it was free then one can challenge all the questions. They'd have to spend time and money.

D.C. the III

So $y^2$ is symmetric around the origin, $x^2$ is symmetric around the origin, $x^2 + y ^2$ is symmetric about the origin....$x^2 + y^2$ is pretty much "hovering" over the region we want to integrate....

Ted Shifrin

You ignored what I typed an hour ago, DC.?

PNDas

Usually coaching institutes publish question banks. They have a lot of errors. I believe all the faulty questions you see must be from some random books and not from the national exams.

D.C. the III

I hope not....I was thinking about it while working on some other probelms....

the region is symmetric about the origin

Ted Shifrin

7:53 PM

No, you responded to part of what I typed but apparently ignored the substance of the remark.

The famous mistake in the US was a question like this: A coin of radius 1" rolls without slipping around the outside of a coin of radius 2". How many complete revolutions does the president's head make?

The story goes (I believe I read this in print) that a confident student contacted the ETS when he found out his score was not a perfect score and had to tell them they had the wrong answer to this question. Shameful.

D.C. the III

The substance being that we are integrating functions, the region stays the same.....I've heard this story.

Ted Shifrin

No, the substance being that there is a symmetry of the region that switches $x$ and $y$ ... namely ... ?

D.C. the III

Symmetry about the origin is the symmetry isn't it? Or are you looking for a particular function?

Ted Shifrin

Symmetry about the origin is what ?

D.C. the III

Symmetry about the origin would imply the function is odd

Ted Shifrin

8:08 PM

What are you talking about?

D.C. the III

You asked what symmetry about the origin is and if something is symmetric about the origin it would mean the function is an odd function.

I know what's coming next..........................THINK.....

Ted Shifrin

Sorry. This is all utter nonsense.

You’re not saying things carefully, and you’re saying stuff totally out of context from precalculus.

And you have yet to respond to what I said totally specifically.

D.C. the III

ok....So I wanted to find out why integrating $x^2 + y^2$ is easier than $y^2$ over the annular region. You said it was because of symmetry. The region is symmetric about the origin. Now it was also talked about that the integral of $x^2$ and $y^2$ are the same and I should be able to see that without having to calculate the value

Ted Shifrin

8:24 PM

What symmetry of the region switches $x$ and $y$?

D.C. the III

Symmetry along the line $y= x$ is the only other notion that comes to mind

Ted Shifrin

Reflection across that line, precisely.

So, without full rigor, since that switches the two functions and leaves the region and $dA$ the same, the integrals have to be the same. Symmetry is important and powerful throughout the rest of the course.

D.C. the III

Sigh....so the ideas are there, but the comfort in using them and being confident in using them is not....only cure is practice for that.

Hold on there is something interesting you just wrote......

how would a region or $dA$ change if I switched functions? The region I'm integrating over is still the same....

also as you mentioned, the integrals will be the same, but why would you add them?

Ted Shifrin

8:43 PM

Because $x^2+y^2$ is a function just of $r$.

D.C. the III

So is the reasoning since we are adding a function "to itself" (since $x^2$ and $y^2$ are the same) we are not doing anything to our region and $dA$ as you said?

Ted Shifrin

They are NOT the same!

D.C. the III

they are equal though. So I get why you shouted that

Ted Shifrin

9:04 PM

No. Their integrals are equal. Be correct and precise.

D.C. the III

Yes sir. Their integrals are equal......the language does matter. THanks for the virtual dress down

Gwyn

9:31 PM

Does the notation $(f^{-1})^{-1}(A)$ means the preimage of the set $A$ under the inverse function of $f$?

leslie townes

depending on intent, the notation would potentially make sense even if there is no inverse 'function'

Ted Shifrin

@leslie Any thoughts?

@leslietownes Is the preimage notation used with relations?

@Gwyn Context?

Gwyn

I am trying to prove that if a continuous, bijective function $f:X \to Y$ is open then it is an homeomorphism. So I'm trying to prove that for each closed set $C \subseteq X$ the preimage of $f^{-1}$ under $C$ is closed. However, I don't precisely know how to mean this without ambiguity. So I thought about $(f^{-1})^{-1}(C)$, because the parenthesis isolates $f^{-1}$ and so there is no confusion between the preimage of $f$ and the preimage of $f^{-1}$.

But this is the first time that I use the preimage of an inverse, of course I could call $g=f^{-1}$ and avoid this...

But I'm starting to think that in math I am a little masochist :]

Sorry, a typo: $f$ is closed and not open.

Ted Shifrin

You have a bijective function, so $f^{-1}$ is a function, no problem.

You mean the preimage of $C$ under $f^{-1}$.

Gwyn

Thanks Ted :) in these hypothesis, is it true that $f(C)=(f^{-1})^{-1}(C)$?

Ted Shifrin

9:43 PM

So figure out how the notation should simplify for you.

Precisely.

Just check element by element.

Gwyn

Ok. So maybe this works: let $C \subseteq X$ be an arbitrary closed set in $X$. Since $f$ is closed, $f(C)$ is closed in $Y$, but $f(C)=(f^{-1})^{-1}(C)$ and so $(f^{-1})^{-1}(C)$ is closed in $Y$. This means that $f^{-1}$ is continuous. Hence, since by hypothesis $f$ is already continuous and bijective, this means that $f$ is an homomorphism.

Ted Shifrin

Homeo … yes

Gwyn

Oh right, dropped and e! Thanks again Ted

Ted Shifrin

Sure.

mick

10:19 PM

this question is closed :/
I added more data and resulted in closed
meh

1

Q: Special numerical method for $\sqrt 2$ with rational functions

Numerical methods for approximating Pythagoras' constant $t =\sqrt 2$ by fractions. (This is an idea from my mentor while he was barely $13$ yo, as a response to a challenge). We all know Newton's method for finding $t$. It converges quadratically meaning like $o( C x^{2^n} )$ where $C$ is a cons...

sequences-and-series numerical-methods radicals rational-functions

shintuku

10:50 PM

what's a good movie

Xander Henderson

@shintuku Citizen Kane?

shintuku

already seen

Xander Henderson

Nausicaä of the Valley of the Wind is one of my favorites.

I like the original Tarkovsky Solaris.

The Princess Bride is a classic.

leslie townes

Out For Justice (1991)

Xander Henderson

Barry Lyndon is an interesting film. It gets panned a lot, but I think it's quite good.

shintuku

10:57 PM

only seen Solaris out of those, nausicaa of the valley of the wind is next watch

Xander Henderson

The Thin Man is a delight.

If you want to see Leslie Nielson before he became a comic actor, Forbidden Planet is a lot of fun.

shintuku

will be going through these this weekend

D.C. the III

11:16 PM

@shintuku Nonsense.......you have mathematics to do!

shintuku

brain can only sustain math 4 days a week

D.C. the III

only acceptable number is 7.

@TedShifrin I just finished the question of sketching and figuring out the volume of $\rho = \sin(\theta)$. I made it a lot harder than it had to be which brings up two questions:

1) to find the limits of integration on $\rho$, I was toiling away to try and get my function in terms of $\phi$, eventually I found no way of doing it and accepted using $\sin(\theta)$ as the upper bound. Is there a way to put it in terms of $\phi$?

2) Because I was reading fast I failed to notice that you specified bounds on $\theta$, specifically $0 \leq \theta \leq \pi$, prior to that I was stuck on trying…

Ted Shifrin

My convention is always that $\rho\ge 0$.

This is the doughnut with no hole?

D.C. the III

The pillow

Xander Henderson

@TedShifrin Don't be dirty!

D.C. the III

11:27 PM

says the man with a starred dirty comment............lol

Xander Henderson

@D.C.theIII I don't know what you are talking about.

Ted Shifrin

Oh right. I switched $\phi$ and $\theta$. This is the one with a picture in the errata.

Xander Henderson

@D.C.theIII Maybe you're being dirty, too!

D.C. the III

due to entonation not beinf a thing online...I don't know if you are being serious or not Xander....

Xander Henderson

@D.C.theIII :P

I am almost never serious. Unless I have to put on my moderator hat.

But that thing is itchy.

D.C. the III

11:30 PM

Yea I "sketched" the surface (well used geogebra) and saw it. That's why I'm curious about the bounds on $\theta$.

because they are not easy to determine algebraically. I had the "idea" of what I wanted, but to have precise bounds I saw no way of finding them

@XanderHenderson duly noted.

or if you get incessantly tagged.....

Xander Henderson

@D.C.theIII Yes. Then, too. :D

Cotton Headed Ninnymuggins

11:42 PM

Is $G^'$ a subgraph of $G$? I'm thinking not

Xander Henderson

@CottonHeadedNinnymuggins What is your definition of "subgraph"? And why are you thinking not?

leslie townes

what is your definition of 'graph,' even

Xander Henderson

That, too.

leslie townes

not asking to be annoying, but because it might change the answer. like you, "i'm thinking not," but as an inference from the fact that you probably wouldn't draw the thing on the left at all if an edge were determined by the vertices it connects, and not as a consequence of any definitions

Cotton Headed Ninnymuggins

My MathJax isn't working in here or when I ask a question right now, but a subgraph of $G=(V,E)$ is a graph $H=(V',E')$ where $V'$ is a subset of $V$ and $E'$ is a subset of $E$. I'm talking about unweighted, undirected graphs. Just started graph theory a few weeks ago

Xander Henderson

11:50 PM

tinyurl.com/cfqcvpc

shintuku

but what are the graphs made of

what are the subsets made of

Xander Henderson

Also, just V', not V^'

leslie townes

a key question is, what's an edge; is an edge determined by its vertices, or not

my guess is not

Thorgott

V^{\prime}

Ted Shifrin

@leslie Did you see that post about double-dual norms? I never had heard of this.

Cotton Headed Ninnymuggins

11:53 PM

I'm "thinking not" because the instructions to the HW question says "in other words, subgraphs are graphs obtainable by removing vertices/edges from $G$"

Ted Shifrin

@D.C.theIII Sure they are. It's where $\rho\ge 0$, as I done said.

Cotton Headed Ninnymuggins

@leslietownes edge is determined by its vertices yes. An edge connecting vertex 1 and 2 is not the same as an edge connecting vertex 2 and 3

Xander Henderson

The definition of a graph with which I am familiar is something like the following: a graph is a set $V$ (the set of vertices) together with a set of functions $E$, where each function in $E$ has a point in $V$ as its domain, and a point in $V$ as its codomain.

(Or something like that---I can't quite remember the formalism).

Cotton Headed Ninnymuggins

@leslietownes A graph $G$ is a pair $(V,E)$ of sets of vertices and edges. That's pretty much all I've got

Xander Henderson

So in the picture above, the first graph is something like $V = \{1,2,3,4\}$ and $E = \{e_1 : 1 \to 3, e_2 : 3 \to 4\}$, while the second has the same vertex set, but the edge set is $\{f_1 : 1 \to 3, f_2 : 1 \to 3 \}$ (I'm leaving off a lot of braces to keep things more simple.

Cotton Headed Ninnymuggins

11:57 PM

I also don't know what my bookmark bar is to fix MathJax through your link

@XanderHenderson Yes, that's why I think it isn't a subgraph, we don't have 2 edges connecting the 2 vertices that are connected in $G'$

leslie townes

cotton: as i was using the term, an edge being "determined by its vertices" would prevent there from being more than one edge between the same vertices. the vertices alone would tell you the edge. there would be no pair of dots with two distinguishable edges between them

ted: saw it, didn't care enough to think about it

i clicked into the previous post, saw a number of errors that looked fixable but annoying, and tapped out

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$Mathematics$ Mathematics