Mathematics - 2023-03-25 (page 1 of 2)

12:11 AM

@robjohn Hello !

feel free to dig in and help me with this dilemma :D

No, you’re messed up.

Complex roots are one thing. They come in conjugate pairs, so we get the right number of real solutions. Multiplicity is another thing. Where are you getting your $2k$?

Jack Ohara

That might be the case that I messed things up

let me rephrase it so it will be proper

so if the char equation give us complex solutions each of multilplicity k

so from that we do have 2k lin indep solutions right?

since each root and its conjugate are solutions

how would we proceed from here? @TedShifrin

is it that we have n roots

so we have 2n lin indep solutions

and n of them are real?

using the euler formula e^ix

to split real and complex parts

leslie townes

12:27 AM

it would help if you wrote in more complete sentences, and were clearer about, solutions to what, roots of what. if the characteristic polynomial has a pair of complex roots, both of them contribute to the degree of the polynomial. there's no mismatch between the number of complex roots, counted according to multiplicity, and the degree of the polynomial.

Ted Shifrin

No, no. $n$ roots total means $n$ indep solns. Stop doubling.

leslie townes

you're using both n and k here and it isn't clear what either of them is supposed to be, but it also seems pretty clear that they're not going to be the same, so the switch from 2k to 2n is confusing.

or, is one of the confusing things.

Ted Shifrin

Reread what I wrote before..

Jack Ohara

Okay I agree that I am mixing things up with my terrible notation

Ted Shifrin

You’re mixing yourself up.

Jack Ohara

12:29 AM

yes that is what I meant :D

Ted Shifrin

Each pair of conjugate complex roots contributes 2 real solutions. Multiplicities or not.

Jack Ohara

could you say how it will go about , to solve an ODE of degree n with constant coefficients , the homogenous case

leslie townes

y'' + y = 0 might be an instructive example. before anybody blasts off to degree n.

Jack Ohara

haha okay we work with that!

the first thing i would do here is to get the char equation

x^2+1 = 0

and we do find i and -i as solutions

Ted Shifrin

$x$ might be a bad letter here.

What is your independent variable for the ODE?

Jack Ohara

12:33 AM

it differs from problem to problem

sometimes t and other times x

y(x) , x(t) , u(phi) etc

Ted Shifrin

Right, so avoid those letters for the char eqn.

Jack Ohara

but lets agree here to use t

okay

Ted Shifrin

Anyhow, go on with leslie’s exercise.

Jack Ohara

yes after that i would just have e^it and e^-it as solutions

cost and sint would be the real ones

Ted Shifrin

OK. Adding and subtracting gives $\sin t$ and $\cos t$.

Jack Ohara

12:36 AM

yepp

Ted Shifrin

So 2 roots and 2 solutions. That’s it. No big deal.

No doubling.

Jack Ohara

it seems to be no problems when i do exercices but when i try to understand the general theory

i get mixed up :D

i mean if we had this case that i for example had multiplicity 3

AMDG

Could someone check my work for correctness? I think I managed to introduce a useful asymmetry into the equation I've been after.

Jack Ohara

then the solution would have t* e^it and t^2 e^it

and e^it

Ted Shifrin

And their conjugates. Six functions, six roots.

Jack Ohara

12:40 AM

oh yeah that is true

AMDG

Jack Ohara

thanks Ted and Leslie !

Ted Shifrin

You’re welcome.

AMDG

1:02 AM

Checking myself, it seems wholly correct, and the transform is entirely possible within these constraints.

Really only needed something to introduce some asymmetry into the exponents such that the factors would be possible to extract.

Franklin

2:59 AM

Find the number of points in $(-2,2)$ where $f(x)=[x^2-1]$ is not continuous.

Can anyone please help me with this question?

I dont have a clue how to solve it 😅😅😅

Ted Shifrin

3:13 AM

Graph the function.

Franklin

@TedShifrin I could graph $y=x^2-1$

But how to graph, $[x^2-1]$ ?

Ted Shifrin

Look at your graph. What does the [] mean?

Franklin

@TedShifrin Greatest Integer Function -[]

Ted Shifrin

so what happens to your graph? Where does it hit integers?

Franklin

@TedShifrin that's the thing I am struggling with! 😩

Ted Shifrin

3:23 AM

Stop whining and think about it. Where is the function $0$? Where is it $1$?

Franklin

@TedShifrin ohh... if x\in [0,1), then f(x) is -1, if x\in [-1,-2), then f(x)=0, if x\in [-2,-3), f(x) is 3 and so on isn't it?

Using this, I sketched the graph...

D Left Adjoint to U

That $-1$ was harsh

whoever graded that curve

Ted Shifrin

You’re not doing algebra right, but you’re thinking along the right lines.

D Left Adjoint to U

"right lines"?

What about the left lines

:D

Franklin

3:40 AM

@TedShifrin I made a mistake in my assertion literally! Yeah, I have sketched it now, intuitively, but wait, let me verify it...

@TedShifrin Yes, I have sketched the graph, correctly!(I checked it using desmos)...But then ?

Ted Shifrin

Then what?

Franklin

@TedShifrin Then the function is discontinuous at all non-integer points ?

No wait

The points which have breaks,

Or kink whatever we may say...

Ted Shifrin

So list the points where the function is not continuous.

Franklin

This was an mcq, as we see that it has breaks at infinite number of points, so we can say, it is discontinuous at an infinite number of points, and that statement "discontinuous at an infinite number of points" is indeed given as an option !

@TedShifrin I guess this option or statement, whatever, is the correct one ?

Ted Shifrin

Read the question carefully.

D.C. the III

3:57 AM

and look at the picture....as Ted would say again

Franklin

4:23 AM

@TedShifrin Silly me! My bad! Six kinks and hence ssix points of discontinuities, right ?

I meant six kinks in (-2,2)

Ted Shifrin

I don’t think kink is the right word, but yes.

Franklin

@TedShifrin what terminology do you prefer, calling the non-continuous regions, characterized by sharp bends or holes or kinks , etc as the case may be ?

Ted Shifrin

These are simple jumps

Franklin

@TedShifrin Okay, that's a good one. I accept it. So, I might say, six jumps hence six discontinuities....but jumps are typically for this particular case, isn't it ?

@TedShifrin Can we call any point of discontinuiy in a graph, a jump ? I am in two minds here...

A strange dilemma to me!

Ted Shifrin

4:42 AM

No. There are basically three different kinds of discontinuity.

Franklin

@TedShifrin discontinuity of 1st kind and 2nd kind. The discontuity of first kind has two subdivisions jump discountinuity and removable discontinuity. While the discontinuity of the 2nd kind is the case when either side limit do not exist. Is this what you mean ?

Ted Shifrin

Yes

Cotton Headed Ninnymuggins

5:02 AM

I'm trying to show the relation ~ on $V$ that $v~w$ iff there's a walk in graph $G=(V,E)$ from $v$ to $w$... is an equivalence relation. To prove reflexivity, do I need to show that any vertex $v_i$ has a walk to itself? How do I show that? Does this involve anything like connectedness since "walk from ___ to ___" seems to imply vertices are connected?

Franklin

@TedShifrin Thanks a lot!!!! I get it now! But, actually I saw in some books, whenever, one is trying to conclude discontinuity from a graph, like we did in this case, they use the phrase "the holes/kinks clearly denote discontinuities" , though. I think that's valid too...

Franklin

5:37 AM

Can anyone please help me with this: math.stackexchange.com/questions/4665964/… ?

leslie townes

someone just answered it

Franklin

@leslietownes yeah and that's a good one indeed!

one potato two potato

6:06 AM

Let $\mathcal{C}$ be any category with terminal object $*$. Then to show the product $X\times *$ exists in $\mathcal{C}$ for any $X\in\mathcal{C}$, do I need to construct an explicit $X\times *$? $\mathcal{C}$ is just an arbitrary category so I'm not sure it's possible.

I don't assume the pullback exists in $\mathcal{C}$.

Alessandro Codenotti

@onepotatotwopotato what is $X\times\ast$ in some explicit categories you know to have a terminal object?

one potato two potato

@AlessandroCodenotti So if $\mathcal{C}$ is a category of abelian groups, then the trivial group $0$ is a terminal object. $G\in\mathcal{C}$, $G\times 0$ then?

Alessandro Codenotti

Sure, give me another example

one potato two potato

$G\times 1$ if $\mathcal{C}$ is just a group category, $V\times 0$ if $\mathcal{C}$ is a $k$-vector space category...

Alessandro Codenotti

What are $G\times 0$, $V\times 0$ isomorphic to?

one potato two potato

6:18 AM

$G$ and $V$ respectively.

Alessandro Codenotti

Right

one potato two potato

But I already know the product exists in these explicit categories.

leslie townes

maybe X has the categorical properties required of the product "X times *" in general?

Franklin

0

Q: Let $ f:(-1,1)\to(-1,1)$ be continuous, $f(x)=f(x^2)$ for every $x$ then the value of $f(\frac 1 4)$

Let $f:(-1,1)\to(-1,1)$ be continuous, $f(x)=f(x^2)$ for every $x$ and $f(0)=\frac 1 2$. Then $f(\frac 1 4)$ is $A)\frac{1}{16}\quad B)\frac{1}{4}\quad C)\frac{1}{2}\quad$ $D)$ can't be determined. I don't know how to start with this problem? Please give me hints.

Alessandro Codenotti

@onepotatotwopotato if the product of $X$ and the terminal object is $X$ in all the explicit categories you know, then...

Franklin

6:23 AM

I had a question about this.

The answerer assumes $\lim_{x\to c}f(x)=f(\lim_{x\to c} f(x)$ .

Is this always valid ?

Also is it valid in this case ?

If so, then I want to ask how?

one potato two potato

@AlessandroCodenotti Then I just declare $X\times *$ is $X$ with two projections $1_X :X\to X$ and $X\to *$?

leslie townes

one: a mentally reordered version of what you just said would be to prove that X with those maps satisfies the requirements of the categorical product X times *

you don't get to "declare" what a product of two things is, either something is a product of two things in the categorical sense or it isn't

maybe ignore the \times notation entirely. a product of A and B is an object E with a pair of morphisms from E into A and B respectively such that blah blahblah

in this case, can X with [some specified] pair of morphisms from X into X and X into * play the role of that E, or can't it

franklin: there are unbalanced parentheses in your "the answerer assumes . . . " but lim_{x to c} f(x) = f(c) is the definition of "f is continuous at c" (or equivalent to it, depending on how you define continuity)

one potato two potato

Right right. So $X$ with two projections satisfies the universal property of product so $X = X\times *$ is the correct expression.

@AlessandroCodenotti Thank you

Franklin

6:42 AM

@leslietownes I am sorry. I actually meant the identity $\lim_{x\to c}f(g(x))=f((\lim_{x\to c} g(x)))$ is essentially used in the solution of that post, isn't it? But is this identity always valid , as long as ,L=\lim_{x\to c} g(x) is not something indeterminate and $f(L)$ is not indeterminate as well ?

leslie townes

continuity of squaring and f is being used here

i don't think the "indeterminate" concept helps here. you have x^2, a continuous function, and f, a continuous function

Franklin

@leslietownes If we go, by this identity $\lim_{x\to c}f(g(x))=f((\lim_{x\to c} g(x)))$ then in this case, as it's a sequence the variable is n, and thus, $\lim_{x\to c}f(g(n))=f((\lim_{x\to c} g(n)))$

leslie townes

that's a little garbled

Franklin

@leslietownes This is essentially what is done, isn't it? Since $\frac{1}{4^{2n}}$ converges to 0 and again, f(0) is \frac 14 ....

leslie townes

you can notationally distinguish lim n->infty h(x_n), where x_n is some sequence of numbers, which is a sequential limit, from lim x to c h(x), where h is some function defined in a neighborhood of c, which is not a sequential limit

you were blending them

which is fine, i don't care, but that's what i meant up above when i said it was a little garbled

Franklin

6:48 AM

@leslietownes Ah...I see...

leslie townes

but yes, if f is continuous at 0 and y_n is a sequence that goes to 0, then lim n to infty f(y_n) is f(0), or, if you like, f(lim n to infty y_n)

and if y_n has the form h(x_n) for some function h and convergent sequence x_n, and h is continuous at the limit of x_n, you can put that limit inside there

but really, here we have a specific sequence going to 0, there is no need to work at that level of generality

there's no useful "if and only if" characterization of the continuity of a composite function in terms of its components. e.g. if f is 0 on rationals and 1 on irrationals, then f is not continuous at any real number, but f composed with f is identically 0

this is why i am pulling you back from basically trying to formulate some general thing where, if f(g(x)) has some limit at some point, then you can deduce things about limits of f and g

the useful characterizations are if f and g are nice, then f composed with g will also be nice

and that's what you have here

and maybe if any of this is at all confusing, you should stop worrying about general g and focus on the specific sequence in this problem

Franklin

@leslietownes Exactly, the identity that generalizes this or the lemma that generalizes the argument in the solution is : $\lim_{x\to c}f(g(x))=f((\lim_{x\to c} g(x)))$ is true, as long as $L=\lim_{x\to c} g(x)$ is not something indeterminate and $f(L)$ is not indeterminate as well.

And in our question we had f(x_n) where x_n is a sequence

@leslietownes Thus, $x_n$ converges at an L and f(L) is not indeterminate. That's why, it's working, right ?

@leslietownes this is so true

@leslietownes yes

But I was saying this was the core of it, ig

leslie townes

x_n converges to L and f is continuous at L

please stop saying "indeterminate" it's not helping

Franklin

@leslietownes f is continuous at L , is this a necessary condition?

leslie townes

necessary for what? for lim f(x_n) to exist if x_n goes to L? in this case, no, it's not necessary. but it's sufficient. (and as i said up above, it isn't fruitful to think in terms of "if and only if" characterizations here)

Franklin

6:59 AM

It should be f(L) even if f is not continuous at L, I this might be the case

Rather $\lim_{n\to \infty} f(x_n)=f(L)

leslie townes

again blending the real variable limit with the sequential limit

Franklin

@leslietownes that was a typo...fixed it

Here $L=lim x_n$

leslie townes

"if x_n converges to L then f(x_n) converges to f(L)" will be true if f is continuous at L [where here the background assumption is just, say, that f is defined on some neighborhood of L]

Franklin

That's why I am saying: f is continuous at L, this condition is irrelevant here?

leslie townes

if f is not continuous at L then the statement in quotes will fail for at least one sequence x_n converging to L, but might hold for other sequences x_n converging to L

Franklin

7:19 AM

@leslietownes why is that ?

leslie townes

i said two things, which one is "that"

Franklin

"will fail for at least one sequence x_n converging to L,"

leslie townes

the first one is either the definition of "f is not continuous at L" or equivalent to it

approximately depending on whether you use limits of sequences or real variable limits to define "f is continuous at L"

Franklin

@leslietownes In short,
"if x_n converges to L then f(x_n) converges to f(L)" is valid if f is continuous at L and if f is not continuous at L, then this is not true. But my question, is: if f is not continuous in L, then why the statement in quotes will not hold true for all subsequences converging to L

leslie townes

what is "this" in "this is not true"

if f is not continuous at L, then the limiting behavior of f(x_n) as n goes to infinity may depend on the sequence x_n, and cannot be deduced simply from the fact that x_n goes to L

depending on the sequence x_n, it is entirely possible for lim f(x_n) to still exist and be f(L) even if f is not contnuous at L

example: if f is 0 on rationals and 1 on irrationals then lim n to infinity f(1/n) = 0 = f(0) even though f is not continuous at 0

the second part of what you said above appears to be a restatement of "if f is not continuous at L then the statement in quotes will fail for at least one sequence x_n converging to L," which is true

i feel like we've repeated essentially the same thing several times

Franklin

7:37 AM

@leslietownes Yeah that clears things.

leslie townes

ah OK

Franklin

@leslietownes Thanks a lot!

Maybe, the reason for these, I will get to know, after I have gained a good lot of knowledge on real analysis:) I anticipate, after that I will be able to prove them rigorously

For now, I take that as a theorem

By "that" in my previous comment, I mean : if x_n converges to L then f(x_n) converges to f(L)" is valid only if f is continuous at L...

I take the above assertion as a theorem for now...

leslie townes

that's missing quantifiers, it needs something like "for all sequences x_n," at the front

Franklin

@leslietownes Ah ok, trying to fix it...

if x_n converges to L then f(x_n) converges to f(L) for all subsequence $x_n$ converging to L " is valid only if f is continuous at L...

leslie townes

that's a little jumbled up because "x_n" is beginning an "if...then" statement before the thing that quantifies it

put the for all at the front

Franklin

7:45 AM

Ok

leslie townes

and you do mean "sequences" there, not "subsequences"

Franklin

For all possible sequences $x_n$ converging to $L$, $f(x_n)$ converges to f(L) if and only if f is continuous at L

@leslietownes Should the word "iff" be there ?

leslie townes

yes, if you rewrite the thing in some order so that it makes slightly more sense. f is continuous at L if and only if lim f(x_n) = f(L) for all sequences (x_n) that converge to L

Franklin

@leslietownes Thanks a lot! Indeed, writing this mess up like: f is continuous at L if and only if lim f(x_n) = f(L) for all sequences (x_n) that converge to L , looks good and exciting...

leslie townes

some books take this as the definition of "f is continuous at L" because it requires only sequential limits (i.e. no need for the full 'epsilon-delta,' just 'epsilon-N')

i think the formal simplicity of that approach is undermined somewhat by the fact that "the set of all real sequences convergent to L" is a very large collection that is not particularly easy to "visualize" unless you have some comfortability with very abstract concepts

the epsilon-delta has two greek letters instead of one, but you're dealing with things that range over sets that feel "smaller" and you can capture more of the idea in a single picture

Franklin

7:56 AM

@leslietownes I completely agree with you. By the theorem, you wrote out, magically, a new definition of continuity appears before my eyes!

At this level, I take it as a theorem or rather as a definition. This will probably become obvious once the real analysis course progresses much :)

Franklin

8:21 AM

@leslietownes An off topic question: I understand $f(x_n)$ converges to $1/2$ , thus it's a liitle or rather juuust a sliiight inaccurate to say f(1/4)=1/2 it's better to say, $f(1/4) \sim 1/2$

@leslietownes An off topic question: I understand $f(x_n)$ converges to $1/2$ , thus it's a liitle or rather juuust a sliiight inaccurate to say f(1/4)=1/2 it's better to say, $f(1/4) \approx 1/2$

leslie townes

in the problem above, f(1/4) = 1/2. it would be slightly inaccurate to suggest otherwise

Franklin

But then again f(x_k)=f(x_{k+1}) , for all natural k , so, the conclusion is f(1/4)=1/2

leslie townes

the fact that you can deduce the fact from a statement about limits doesn't mean that it's an approximation

Franklin

" the fact that you can deduce the fact from a statement about limits doesn't mean that it's an approximation" is valid here only because f(x_k)=f(x_{k+1}) is true for all natural k

Otherwise, I would've doubted the accuracy

leslie townes

you can certainly write statements about limits that look inaccurate, or are at least informal, but f(1/4) is definitely 1/2

well, under some other set of hypotheses, there would be some other answer :D basically

it doesn't change the fact that in this problem, f(1/4) = 1/2

in analysis, it's sometimes helpful to keep in mind that very often, the hypotheses you get to work with are much stronger than would be absolutely necessary

Franklin

8:27 AM

@leslietownes you are more than correct in this comment, but as I suggest the equality is valid only because of this: f(x_k)=f(x_{k+1}) is true for all natural k

leslie townes

often because the "if and only if" statement that describes what would be both both necessary and sufficient for something to be true is (a) really complicated and beyond the reach of an intro to analysis, or (b) not any easier to work with

so yes, it's certainly useful here that we've got this sequence x_n going to 0 for which f(x_n) is constant, and we should not be ashamed to use this fact

Franklin

@leslietownes But can you really deny my previous comment ?

leslie townes

even if you could envision more general problems where f(x_n) wouldn't be constant and yet you could still deduce some other thing

i don't know what it means to say that f(1/4) = 1/2 is valid "only because" of some consequence of the hypotheses of the problem

i don't think this is a useful way of thinking about this problem or other problems

this is a bit like my analogy above, about there being no useful "if and only if" for continuity of f o g in terms of f and g

the solution on that web page certainly uses the fact that for this f, there is a sequence x_n going to 0 for which the sequence f(x_n) is constant

under other hypotheses you could still conceivably deduce things about values of f from milder hypotheses

Franklin

@leslietownes hmm...

leslie townes

this is just some random exercise on MSE, i wouldn't pay too close attention to why its hypotheses take the form they do

D Left Adjoint to U

8:32 AM

I can prove to you that for $n \lt m$ arithmetic in $\Bbb{Z}/n$ relates polynomially ($f \in \Bbb{Z}[X])$ to arithmetic in $\Bbb{Z}/m$.

If $n = m$ the polynomial decomposes to just $f(x) = x$.

Franklin

"the fact that you can deduce the fact from a statement about limits doesn't mean that it's an approximation"- I would add a phrase "in some cases" at the end of this sentence.

leslie townes

there's no "in some cases" about "the fact" i'm talking about, which is that f(1/4) is 1/2

i'm not talking about all statements about all limits, or about limits at all, i'm talking about the equality f(1/4) = 1/2 for the f in this problem

Franklin

@leslietownes I am not convinced about this as of yet. Maybe, it is only because I repeatedly mention as f(x_n) do not change at all for this sequence, and hence, it is a constant sequence and as it converges to 1/2, the f(x_n)=1/2 for this sequence

leslie townes

there are two things potentially worth thinking about, (1) the problem itself about this function f, and (2) general facts about limits and continuity. i'm focused on (1)

you definitely are using the hypothesis about f(x) (and not just the value of f(0)) to reach a conclusion about f(1/4), and if that hypothesis on f(x) were changed, you might not be able to solve the problem in the same way

or be able to determine f(1/4) at all from the given information

but again, all that really means is that if you ask a different question you might get a different answer. not that f(1/4) isn't 1/2 in this specific case

changing the subject, it's worth noting that there's nothing special about 1/4 here. the hypothesis implies that f(x) = 1/2 for all x in (-1,1)

as such, it's a very strong hypothesis

one potato two potato

9:08 AM

If $G_i$ is abelian for each $i\in I$, then $\coprod_{i\in I}G_i\to \bigoplus_{i\in I}G_i$ by $(g_i)\mapsto \sum g_i$ is a well-defined map? I mean the sum could be infinite sum.

Word should be finite. hehe...

Franklin

@leslietownes exactly, this is what I mean, we could conclude f(1/4)=1/2, only because f(x)=f(x^2)

The complete equality about f(1/4) was only possible because of the hypothesis...

@leslietownes yeah...maybe that's true, but we are sure f(x) is same for all x=1/2^{2n}

D Left Adjoint to U

9:29 AM

@onepotatotwopotato, it's well-defined by UMP of direct sum isn't it?

You have that the disjoint union has injections into it. just like the direct sum UMP

one potato two potato

@DLeftAdjointtoU Yes. I just confused at the moment.

I'm hoping the kernel of that map is a commutator subgroup

D Left Adjoint to U

Is the one on the left a group?

Isn't it just the same thing as $\prod_{i \in I} G_i$ since abelian categories

one potato two potato

coproduct of $G_i$ in Group category

D Left Adjoint to U

Oh, well, then restrict to $\textbf{AbGrp}$ since all your morphisms will be from there anyway

Basically the well defined map is piece-wise, right?

one potato two potato

What do you mean by restricting to AbGrp? $\coprod G_i$ is not abelian.

D Left Adjoint to U

9:34 AM

Why not?

Abeliean group coproduct coincides with product

:|

Oh crap, don't trust what I'm saying pls

one potato two potato

$\coprod_i G_i$ is a coproduct in group category not abelian group category. $\Bbb Z\coprod \Bbb Z = \Bbb Z*\Bbb Z\neq \Bbb Z\oplus\Bbb Z$.

D Left Adjoint to U

What is the disjoint union of groups?

Looking like

one potato two potato

Free product

D Left Adjoint to U

$\prod$ then?

one potato two potato

Just a product of groups

D Left Adjoint to U

9:40 AM

I see now it's en.wikipedia.org/wiki/Free_product

Franklin

Check if the polynomial p(x)=x^4-4x^2+1 has exactly 2 roots in the interval [0,3]

Can anyone please help me with this ?

I have proved that there is atleast one root in the interval [0,1) and another root in the interval (1,3]

Thus, atleast 2 roots are there in the interval [0,3]

Nevermind, I found another one in the interval, [-1,0).

Problem solved! Thanks you!!!

nickbros123

10:18 AM

When converting multiple integrals from x,y to polar coordinates, it's not obvious to my why we can't apply the formula for $ \del x $ and $ \del y $ in terms of r and $\theta$ and doing the integral. I've read a few answers on SE, but I'm not able to understand them frankly.

What i got was that it was an abuse of notation, which i can agree, there possibly is some deeper limiting process happening that the notation dx dy simply doesn't convey

But there seems to be no answer online that drives home the point whilst also not redirecting me to higher math like diff geometry or differential forms

robjohn

@Franklin There is always Sturm's Theorem.

You can also note that $p(x)=(x^2-2)^2-3$

The roots are $\frac{\pm1\pm\sqrt3}{\sqrt2}$

Sturm's Theorem says that there is one root in $[0,1]$ and one root in $[1,2]$.

user2236

11:05 AM

How goes the recovery pal @robjohn

Franklin

11:36 AM

Let $l_1$ and $l_2$ be a pair of intersecting lines in the plane. Then the locus of the points P such that the distance of P from $l_1$ is twice the distance of $P$ from $l_2$ is : A. Ellipse B. A parabola C. A hyperbola D. A pair of straight lines

Can anyone please help me with this mcq ?

@robjohn yeah, that's a method too. But Sturm function calculation is sooo lengthy!

@robjohn Thanks!

user2236

That kind of mcq is forcing you to draw a quick sketch.

I would use a ruler and set up a chart of points.

Franklin

@user2236 If we consider the l_1 as x-axis and l_2 a directrix, then can't we conclude that locus of P has a real valued eccentricity and it should be a conic?

Ohh , my bad!

In case of a conic, we should have fixed distance from a focus

Not a straight line as x-axis

user2236

Draw the picture plz

Franklin

@user2236 I think that's the only quick way for this...

user2236

Exactly 💯

Franklin

11:48 AM

@user2236 I got one straight line. Just a sec...

Excuse me for the bad sketch please

@user2236 I think this is a rough sketch

user2236

No problem 👍

Franklin

Option D suggest a pair of straight lines, nevermind I got only one, maybe if we take from other side, we might obtain another straight line. But D seems to be the only option which has the phrase " straight line" . So, I think, D is the correct option!

@user2236 what do you think?

user2236

I agree.

✅

Franklin

@user2236 Thanks for suggesting the witty trick!

user2236

Make drawing a sketch a method and you will go far.

Franklin

11:56 AM

@user2236 You talking only about these type of questions ?

user2236

✏️📈📉 in life

Franklin

@user2236 Nice one!

@user2236 drawing a sketch method, may sometimes become tedious. But in these problems of mcq, it's really helpful...

user2236

When it becomes tedious at least you know you're on the right track and have a better understanding of the situation.

Franklin

@user2236 Yeah...true...

Two subsets $A$ and $ B$ of the $(x,y)$ plane are said to be *equivalent* if there exists a function $f:A\to B$ which is both one-to-one and onto.

$(i)$ Show that two line segments in the plane are equivalent.
$(ii)$ Show that any two circles in the plane are equivalent.

Any ideas for this?

I have no clue how to proceed !

user2236

I gotta run TTL

Franklin

12:04 PM

@user2236 sure!

But can someone please help me with this ?

Ajay

A little experimenting seems good...

$(i)$ Let $L_1$ and $L_2$ be two line segments in the plane. Without loss of generality, assume that $L_1$ and $L_2$ have the same length, and let $P_1Q_1$ and $P_2Q_2$ be the respective endpoints of $L_1$ and $L_2$. Let $f:L_1\to L_2$ be defined by $f(P_1)=P_2$ and $f(Q_1)=Q_2$. Then $f$ is clearly one-to-one and onto, and hence $L_1$ and $L_2$ are equivalent.

$(ii)$ Let $C_1$ and $C_2$ be two circles in the plane. Without loss of generality, assume that $C_1$ and $C_2$ have the same radius, and let $O_1$ and $O_2$ be the respective centers of $C_1$ and $C_2$. Let $f:C_1\to C_2$ be define…

@Franklin Does this make sense?

I've been making short Kotlin code snippets to connect Chatgpt to the internet and i'm testing out how it fares.

Franklin

12:21 PM

@Ajay Wow! What's Kotlin ? A language ?

@Ajay what if L_1 and L_2 are not of same length ?

That might be a loophole

But wait

As a line-segment is made of infinite points, it might be valid

in a sense

But it looks a little weird...um....I would like some validation ....

@Ajay incidentally, I thought of the same solution

But as before, I felt it was somewhat offbeam ...

@Ajay But thanks! Although I did the first part like that, but I need some validation. I posted this question but not the solution of 1st part as I was lazy about typing that out( I know I shouldn't be). But then you posted the exact same thing in the latex format. That saved the whole burden of typing things out. I copy-pasted the first part for validation in my OP.

Let's see for validation. I hope you dont mind. After I got some clarifications, I will post the 2nd part as a "response from ChatGPT" and I bet, we might get some fruitful or at the most some desirable results...

Ajay

12:46 PM

Lol

I'm surprised it made some sense

Yeah, Kotlin is a programming language

Franklin

@Ajay math.stackexchange.com/questions/4666087/… -You can follow up here...

It made only a liiiiitle sense but not valid as you will see in this link

In my opinion ChatGPT is bad for mathematical purposes

Ajay

true, but that can certainly be improved

However, I would have liked to be able to answer this question myself...Sadly I cannot as I screwed up my math education and have to start from the beginning again...

I rushed the process.

Heavy sigh

Koro

2:45 PM

0

Q: The inclusion $A\hookrightarrow X$ induces isomorphisms on all homology groups iff $H_n(X,A)=0$ for all $n$.

The statement to be proven is the following: For a pair of spaces $(X,A)$, the inclusion $A\hookrightarrow X$ induces isomorphisms on all homology groups iff $H_n(X,A)=0$ for all $n$. I know that the inclusion induces the following l.e.s. (long exact sequence): $$...\to H_{n+1}(A)\to H_{n+1}(X)\t...

algebraic-topology

Ted Shifrin

4:15 PM

@Ajay Stop bullshitting. You certainly defined no function $f$.

Franklin

4:46 PM

@TedShifrin any ideas you wanna suggest?😂😂😂 Yeah, Chat GPT is just bullshit!!!

schn

6:17 PM

A simple one, maybe. For a convergent series it holds that $c\sum a_n=\sum ca_n$ for some constant $c$. How do you prove that if $\sum a_n$ diverges, so does $\sum ca_n$ for $c\neq 0$?

robjohn

@TedShifrin the definition just says they have the same cardinality. It does not mention continuity.

leslie townes

6:32 PM

schn: if such a thing were to converge, apply the result about convergent series with c^{-1} as the scalar to deduce...

schn

@leslietownes so something along the lines of; given that $\sum a_n$ diverges, suppose $\sum ca_n$ converges. Then $\sum ca_n=c\sum a_n$, but the right-hand side diverges, so the left-hand side diverges too. Contradiction. So $\sum ca_n$ diverges when $\sum a_n$ diverges.

leslie townes

well... yes, but i'd put in the part where you actually use the theorem about convergent series and the hypothesis that c is nonzero

schn

ok

leslie townes

that "but the right hand side diverges" is not the way you want to refer to that result

it diverges precisely because if converged, you could use the theorem about convergent series with c^{-1} in it to...

etc.

if you like you could also eliminate the reference to contradiction by just proving the contrapositive. "we prove the contrapositive, that if sum c a_n is convergent and c is nonzero them sum a_n is also convergent"

schn

6:54 PM

ok, thanks

Ted Shifrin

7:45 PM

@robjohn I grant that, but just defining a function on the endpoints of an interval does not define a function. Or did I miss something?

robjohn

8:17 PM

@TedShifrin No, you didn’t miss anything. The assumption might have been a linear function, but I sense a great dearth of specification has occurred.

Silly Goose

im having trouble seeing how every cauchy sequence does not converge given this definition. if we fix $p_m$, then don't we have a limit of $\{p_n\}$?

shintuku

yes all cauchy sequences converge

leslie townes

given epsilon, knowing that you can find some x (perhaps depending on epsilon) for which d(x, p_n) < epsilon holds for all sufficiently large n is very much not the same as knowing that {p_n} converges to x

Silly Goose

but isn't that precisely this definition

(perhaps this is where my misunderstanding lies)

leslie townes

that p doesn't depend on epsilon

that's one p that will work for all epsilon > 0

shintuku

8:30 PM

gotta prove cauchy sequence implies convergence

leslie townes

to "fix" your "p_m" (i assume you have in mind something like p_N), you need to specify an epsilon, with no general guarantee that the same "p_m" will satisfy the inequality for other values of epsilon

shin: goose is working in a general metric space, not necessarily a complete metric space

shintuku

oh ok

on normal R at least, every cauchy sequence converges

leslie townes

goose: consider something like X = (0,1) and x_n = 1/n, this is definitely a cauchy sequence in X, but all of the candidate "p_m"'s you might be thinking of are positive numbers that the sequence can't converge to

more generally i guess note that the structure of your argument would seem to prove that if a sequence is cauchy, then it actually must not only converge, but converge to one of the elements of the sequence

which hopefully feels wrong (and fails even in complete metric spaces)

Silly Goose

hm okay the counter example helps

im still having trouble seeing the dependence of epsilon on the points

to me the definition reads: "for every positive $\epsilon$ there exists an integer $N$ such that $d(p_n, p_m) < \epsilon$ if $n \geq N$ and $m \geq N$; including the particular situation when $n \geq N$ and $m = N$. So that: for every positive $\epsilon$ there exists an integer $N$ such that $d(p_n, p_N) < \epsilon$ if $n \geq N$.

Ted Shifrin

@shintuku Oh really?

shintuku

8:45 PM

i changed my mind

in $\mathbb R$

Ted Shifrin

It’s the lub property in all of $\Bbb R$ that makes this true.