Mathematics - 2023-02-13 (page 2 of 2)

Kumar Shuvam

21:07

A straight line passing through P (3, 1) meets the
coordinate axes at A and B. It is given that the
distance of this straight line from the origin O is
maximum. The area of triangle OAB is equal to... Can anybody help me with question..one way could be calculate the length of perpendicular from orgin and maximise it..but that would be quite lengthy...is there any other way

ペガサス Seiya

@KumarShuvam have you tried asking on the main site?

Kumar Shuvam

@ペガサスSeiya Umm! I can do that

But I thought it would be good to discuss here

ペガサス Seiya

@KumarShuvam if you do ask there, I suggest including what you tried so far within the question. Every detail helps

Kumar Shuvam

Umm! Okay

Can I include a image here?

ペガサス Seiya

@KumarShuvam sure, send the imgur link

Sine of the Time

21:16

@KumarShuvam the key word is "maximum". what can you conclude?

Kumar Shuvam

imgur.com/a/41ifsSH

@SineoftheTime well my mind says...calculate the derivative maximise that's it lol

Sine of the Time

@KumarShuvam sounds good

Kumar Shuvam

@SineoftheTime is there any other way?

Sine of the Time

21:31

@KumarShuvam probably, I think you can show somehow that the line must be perpendicular to OP, but I've not done this geometry problems for years

ペガサス Seiya

@KumarShuvam yeah. Shown that line passing through O and P is perpendicular to line passing through A and B

Kumar Shuvam

@ペガサスSeiya how?

ペガサス Seiya

@KumarShuvam when are two lines on a plane perpendicular?

Kumar Shuvam

@ペガサスSeiya when the product of their slopes is -1 assuming none of the two lines are vertical...

ペガサス Seiya

@KumarShuvam correct. Then, if OP is perpendicular to AB where O is the origin and P(3,1), what does that tell you about AB?

Kumar Shuvam

21:39

That slope of AB is "-3"?

ペガサス Seiya

@KumarShuvam exactly. Now you can easily find the equation for AB

Ted Shifrin

Anyhow, @Sine, with $p=[1,0,0,0]$ and the $\Bbb P^2$ given by $x_0=0$, you should find that the retraction $\Bbb P^3 - \{p\}\to\Bbb P^2$ is given by $r([z_0,z_1,z_2,z_3]) = [0,z_1,z_2,z_3]$. Now the homotopy to the identity map should be clear.

Sine of the Time

@TedShifrin yes, thank you. I've understood the key idea and built the homotopy

Sine of the Time

22:04

@TedShifrin now I'm doing this problem: let $B=\{\emptyset,[a,b]\}$ with $a\le b \in \Bbb{R}$ a base for the topology, and let $X=(\Bbb{R},U_B)$. show that $\pi_1(X,0)=1$. My idea is to show that the identity map is homotopy equivalent to the constant map

Is there a faster way?

Obliv

22:17

for $y' y = -x$ given $y(0) = r$ I did $\frac{y^2}{2} = -\frac{x^2}{2} + C$ as the implicit solution, how do you get $x^2 + y(x)^2 = r^2$ using the initial value

$y(0) = r$ means $y = \sqrt{-x^2 + 2C} = r$

Semiclassical

@Obliv what you did rearranges to $x^2+y^2=2C$, so if you can take $2C=r^2$ you're done

also, that's not what $y(0)=r$ means. you need to actually plug in $x=0$

Obliv

oh derp

$y(0) = \sqrt{2C} = r$

and $C = r^2 / 2$ then you just multiple by 2 on both sides to get $x^2 + y^2 = r^2$

Ted Shifrin

22:31

@SineoftheTime I don’t understand the topology.

Sine of the Time

@TedShifrin open sets are $[a,b]$ with $a\leq b$

Ted Shifrin

No, that’s not right.

If those sets are open, then so is every open interval, too.

Sine of the Time

@TedShifrin the base of the topology is formed by these intervals, but obviously also the countable union is open

@TedShifrin yes, I've proved that it's a base for a topology on $\Bbb{R}$

Ted Shifrin

So actually what is the topology?

Sine of the Time

I think every subset of $\Bbb{R}$ is open

Rithaniel

22:35

That'd be the discrete topology

Ted Shifrin

Yes, the discrete topology.

Sine of the Time

@Rithaniel yes

Ted Shifrin

So what are the continuous maps to it?

Sine of the Time

to show that $\pi_1(X,0)$ is trivial, I considered: $id_X$ (identity map) and $c_y$ (costant map) and then build a homotopy between them

and checked that the function is continuous

Specifically, I defined: $F:X\times I\to X$ with $F(x,t)=x $ if $0\leq t <1$ and $F(x,t)=y$ if $t=1$

Ted Shifrin

Forget the question and answer mine.

Sine of the Time

22:40

@TedShifrin I think every well-defined map is continuous

Ted Shifrin

Really? I’ll be back in a while.

Obliv

for $y' = \frac{1+2x}{y^2 + y^2x^2} \to y'y^2(1+x^2) = 1+2x \to y^2 dy = \frac{1+2x}{1+x^2}dx$ is integration by parts the way to go or can I do a u-sub

actually u sub should work

actually no

Sine of the Time

@Obliv just split the fraction

Obliv

that makes more sense lol

how would you classify $y' = x^2y - y + x^2 - 1$? it doesn't seem to be in the linear form

$\frac{dy}{dx} - x^2(y + 1) + y - 1 = 0$

Sine of the Time

you can separate variables

Obliv

22:49

I don't think so

nvm

Sine of the Time

$y'=x^2y-y+x^2-1$ is equal to $y'=(x^2-1)(y+1)$

Obliv

Oh I was gonna say $\frac{dy}{dx} = x^2(y + 1) - y - 1 \to \frac{dy}{dx} + y + 1 = x^2(y+1)$ then yeah

Thorgott

@Ted I finally understood the whole geometric explanation of Nakayama's lemma being an inverse function theorem just now

in proving that two curves intersect with multiplicity 1 iff they are smooth at the point and the intersection is transverse

Sine of the Time

@Thorgott thank you for your help before :)

Thorgott

glad you got it

Sine of the Time

22:56

I'd like to examine in depth projective geometry, hope that will be such course next year

Node.JS

What is a pre-order that is not reflexive nor irreflexive? so just transitive and total. What is called?

Ted Shifrin

23:17

@SineoftheTime You have a revised answer for me now?

@Thorgott I never saw that.

Sine of the Time

@TedShifrin in order to be continuous, the preimage of an open set must be open

ペガサス Seiya

Molecular geometry is weird

Sine of the Time

and every subset is open

Ted Shifrin

In the codomain, not in the domain

For example, if the domain is a circle.

Sine of the Time

oh I thought you were referring to functions from the space to itself

@TedShifrin it's to be open for the induced topology

Ted Shifrin

23:22

And closed?

Sine of the Time

same for open sets but you consider a close set instead

Ted Shifrin

So, if $X$ is connected, what are the continuous maps from $X$ to your discrete space?

Sine of the Time

must be a constant map, since the image of a connected space is connected

Thorgott

the Nakayama lemma says that if $(R,\mathfrak{m})$ is a local ring and the residue classes of $f_1,\dots,f_n\in\mathfrak{m}$ generate the cotangent space $\mathfrak{m}/\mathfrak{m}^2$ as $k=R/\mathfrak{m}$-vector space, then $f_1,\dotsc,f_n$ generate $\mathfrak{m}$ as $R$-module.
in the concrete example, say $R=\mathcal{O}_{\mathbb{A}^2,0}$. this is the local ring of stalks of regular functions near the origin of the affine plane. its maximal ideal consists of those stalks vanishing at $0$ and its cotangent space consists precisely of the differentials of stalks of regular functions at the …

Ted Shifrin

@SineoftheTime Right. Now answer your question.

Sine of the Time

23:31

@TedShifrin every continuous function $f:I\to X$ must be constant for the aforementioned reasoning

Ted Shifrin

@Thorgott I see. Infinitesimal normal crossings implies normal crossings.

@Sine So what about $\pi_1$?

Thorgott

yeah, locally

Sine of the Time

It must be trivial because all paths are equivalent to the constant path

don't know if path is the appropriate word

Ted Shifrin

Not true, is it?

@Thorgott well, globally if true for every intersection.

Interesting that Artin, who taught me commutative algebra, never taught me your observation. I would never have forgotten.

Sine of the Time

@TedShifrin we have only the constant map no?

Ted Shifrin

23:37

Why only one constant map?

Sine of the Time

in the case $\pi_1(X,0)$ or in general?

Ted Shifrin

In our case.

Sine of the Time

if $f:I\to X$, we have $f(0)=f(1)=0$

Ted Shifrin

OK, so we have only one map because of the basepoint. So no homotopy discussion at all.

Sine of the Time

@TedShifrin yes, this is definitely faster

Thank you again :)

Ted Shifrin

23:43

Sure :)

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$Mathematics$ Mathematics