Mathematics - 2021-02-01 (page 1 of 3)

Astyx

00:25

Algebra is a bubble that's going to burst any time now

I'd short it if I were you

Thorgott

people should do lots of algebra now so that it bursts faster

BigSocks

trying

Astyx

Imagine being the last guy with all the algebra but all of it being worthless

BigSocks

"left holding the 'bra"

Wasabi

01:01

Simple Question: How do I say "center of circle" in a mathematical equation?

I can't just put "O is the center of my circle"

Likely an equation

Ted Shifrin

@Wasabi You mean the equation of your circle? Are you using vectors or just usual $(x,y)$ equations?

I don't know what's wrong with saying $O$ is the center of your circle. Depends on context.

Wasabi

No I mean simply how to say "I is the center of my circle" but in a mathematical equation

BigSocks

needs more info

Ted Shifrin

Is this in the plane? What's the radius of the circle?

BigSocks

otherwise, like Ted said, you already gave a good enough description

Ted Shifrin

01:06

You didn't answer my question about vectors.

Mike Miller

One thing that people don't recognize --- I blame high school geometry --- is that any description which is sufficiently clear that anyone will understand you is a perfectly good mathematical description of something.

Ted Shifrin

Yeah, but it really depends on context and what Wasabi is doing, and he ain't tellin'.

Wasabi

it isn't that complicated. All I'm asking is how to say "O is the center of this circle" in an algebraic way.

Mike Miller

Some descriptions are more useful than others. Sometimes it's good to have a coordinate representation of a circle. Sometimes it's nice to say "the circle centered at the origin with radius 5".

condescension gets you nowhere, good luck!

Ted Shifrin

Don't get impatient with us, @Wasabi. I've asked you TWICE to give answers to my questions, and you refuse.

BigSocks

01:07

@MikeMiller same

Wasabi

I have no idea the vectors of my circle.

Ted Shifrin

Great answer.

NOT.

One last try, and then I'm leaving. What is your context? What are you actually doing with this circle? What is its radius?

Wasabi

Okay. Lemme rephrase: What is an equation to prove that O is the center of my circle?

It's a different kind of problem that basically only needs me to say that O is the center without using words.

Ted Shifrin

How are you given the circle?

Wasabi

I hate it

In a math problem with 2 triangles in it

BigSocks

01:10

that sounds like information

Wasabi

And I need a proof using AAS to prove the triangles are congruent.

But in this proof, I am given that O is the center of the circle.

Ted Shifrin

Therefore it is equidistant to points on the circle.

That's what a circle is ...

Wasabi

How in my proof do I say that O is the center? It's given, but I need to rephrase "O is the center of my circle" into an equation.

...that proves O as the center.

It's a proof, so I can't put words in it.

Only equations.

Ted Shifrin

Wrong. Proofs have lots of words.

BigSocks

@Wasabi oof

Wasabi

01:12

on the reasons side

Astyx

bubble is starting to burst

Ted Shifrin

Words go on the reasons side.

BigSocks

lmoa

Wasabi

the statements side in my case is not supposed to be having any words

So I can't simply say "O is the center of my circle"

Astyx

I'm gonna go ahead and ask for context as did three people before me

Wasabi

01:13

Mmhmm...

Ted Shifrin

We have epsilon more context now.

BigSocks

sounds like highschool geometry problem

Ted Shifrin

Why should it be a statement to say $O$ is the center of the circle?

Yes, @BigSocks, with the thing I hate most about high school mathematics — two-column proofs.

BigSocks

shudders

Wasabi

yep

Ted Shifrin

01:14

So answer my question. Why should this be a statement rather than a reason?

Wasabi

Because it's given, @TedShifrin

It's a given, which needs to be in my proof, at the top, of course.

Fargle

@TedShifrin You hating these is even in video form.

Ted Shifrin

That should be a reason, not a statement.

Wasabi

no, "O is the center" is the statement, and the reason is "given".

Ted Shifrin

LOL, oh, is it @Fargle? Did I rant in one of my lectures?

Fargle

01:15

@TedShifrin Very briefly, when explaining that you don't want that on your homework.

Ted Shifrin

So (a) I disagree: It can be a statement.

(b) You want to conclude something from that, evidently ... like some points are equidistant from $O$. So make that your statement, and, as I suggested, the given fact about $O$ is the reason.

Wasabi

what?

Ted Shifrin

There is no appropriate algebraic equation in this setting.

Wasabi

I'm not given points.

Ted Shifrin

What is the point of knowing that $O$ is the center of the circle? I still have no idea what circle.

Wasabi

01:16

I just need to put "O IS THE CENTER" as "Given" in my proof, but it needs to be an equation.

Ted Shifrin

I give up. Bye.

Wasabi

And I don't know either.

So in my given, I need "O = ..." in which "..." is some equation that would make O the center.

I really don't understand how this is so complicated.

Astyx

Which is why we're asking for more context

BigSocks

you seem like a highschool kid looking for help, and I get that, but you should reread this conversation and learn 1 thing today- this isn't how you ask for help.

Astyx

Because you don't know why we're having trouble answering your question and we don't know what you seek as an answer

Wasabi

01:19

It is like this: "Problem: Need to find out how to "say" the given "O is the center of the circle" as a MATHEMATICAL EQUATION. Rules: Cannot use real "words" in statement, can only use mathematical equation to substitute for simple words."

BigSocks

that alone is a much better lesson than figuring out where this circle is centered

Astyx

How is a circle defined?

Thorgott

horrible rules

Astyx

^

Wasabi

So in other words, how to say "O is center of circle" in a universal mathematical equation.

Astyx

01:20

There isn't one

Wasabi

no>

*?

there's now way to say "O is the center of the circle" in an equation?

Like, how do I find the center of a circle using a formula? I could use that formula to spell out that O is the center.

Astyx

I'll repeat

How is a circle defined?

Wasabi

c=pi*d

Astyx

huh?

Thorgott

what does that mean

Wasabi

01:23

circumference = pi * diameter

That's a circle.

Or, "r = 5"

Astyx

Well, no

Wasabi

but really 5 can be anything but 0 or infinity

huh?

Thorgott

that's a property of a circle, not a definition

Wasabi

What I mean is that "B is half of T" would be "B = T/2" in my statement. Well, "O is the center of my circle" would be what as an equation?

Astyx

Which is why I'm asking you how a circle is defined

Wasabi

01:25

a circle? I don't get it.

You mean my circle? Or are you testing how I would define a circle?

I don't understand

Astyx

The latter

user3733558

$\forall P\in C,\ ( x_{P} -x_{O})^{2} +( y_{p} -y_{O})^{2} =r^{2}$ where $C$ is the circle and $r$ its radius?

BigSocks

yeesh

Wasabi

Uh, "Perfectly round polygon"?

BigSocks

@user3733558 way to blow it

Wasabi

01:28

enclosed polygon with no vertices?

why does that matter, @Astyx

user3733558

sorry, newbie here, btw how do I change my nickname?

Thorgott

a circle isn't a polygon, so that doesn't quite work

Wasabi

eh, shape.

Thorgott

well, how are we supposed to tell you how to describe the center of a circle when you can't tell us how you define a circle

copper.hat

how would you describe a circle in a mon mathematical context?

BigSocks

01:37

how nonmathematical? also probably shouldn't matter for this context

Ted Shifrin

01:49

Scroll up. I said this twice earlier.

runs back away

Astyx

@user3733558 On main, click on your profile, then go to the "Edit profile and settings" tab

Anyway I'm going to bed

'night

Thorgott

night

user3733558

@Astyx thanks (looks like I'm stuck waiting till March 1st to be able to change it, oh well), and g'night

Nike Dattani

02:50

Hi, I wonder if anyone knows much about "simultaneous" diagonalization? I cited two Mathematics.SE questions here, but both of them are not quite talking about this type of eigenvector matching:

0

I have two matrices for which part of the eigenspectrum of one matrix, very closely resembles the eigenspectrum of another matrix, but the only way I'm (currently) able to verify this is quite inelegant. I am open to any solution, but to present examples here of what I seek, I find it easiest to ...

Amin Idelhaj

03:17

@BalarkaSen that could be good! Class is using this book: springer.com/us/book/9783030276430

Amin Idelhaj

03:32

Hey Ted!

Ted Shifrin

Hi, Demonark.

Amin Idelhaj

How's everything going?

Ted Shifrin

Just finished cooking :)

I'm not going to read all that, @Nike, but it's standard that two matrices are standardly diagonalizable if and only if one of them is and the two matrices commute. More generally, you can simultaneously put two matrices in Jordan canonical form if and only if they commute.

BigSocks

@TedShifrin bon appetit

I always disliked that there's no proper counterpart to that in english

Nike Dattani

@TedShifrin That fact about commuting matrices sharing an eigenvector is already stated in the question. The issue for me is that (1) there seems to be no easy way to match the eigenvectors without doing some sort of brute force search and matching, (2) I still want the (normalized) eigenvectors even if their inner product is not exactly 1 but also if they are 1 - epsilon for some threshold epsilon.

Ted Shifrin

03:39

I told you I'm not going to read the posts.

If matrices are slightly off, one might be diagonalizable and the other not.

So, from a numerical perspective, this is NOT a stable situation.

BigSocks

@TedShifrin right so I guess it's like complimentary variables in quantum mechanics

Ted Shifrin

Poisson bracket zero is analogous.

BigSocks

you can't simultaneously diagonalize bc they don't commute

copper.hat

you could try multiplying the matrices of eigenvectors as in $U_1^{-1} U_2$ and see what it looks like.

Ted Shifrin

Copper will have better numerical advice than Ted.

copper.hat

03:42

Its been a while, I've gone digital :-).

Ted Shifrin

But if they only "almost" commute, there's not much you can say. One can be diagonalizable and the other most definitely not.

copper.hat

Also, detecting matching eigenspaces (greater than one dimension) is more work.

Ted Shifrin

Well, my point is that some may become generalized eigenspace in the other matrix.

Anyhow, I don't intend to read two long posts, so unless there's a direct question forthcoming, I'm going to wash dishes.

BigSocks

wait

nah nvm probably left

copper.hat

finished my dish washing :-)

Ted Shifrin

03:49

Well, come do mine, @copper. I did stir fry, so wok and air fryer to wash.

copper.hat

04:13

@TedShifrin :-).

robjohn

04:35

@TedShifrin Just finished my dishwashing.

RewCie

04:51

It's my birthday today!

Let's make this room great today :)

XD :)

I got a reply from Guido van Rossum, inventor of python programming language... Also, he wished me on 1st feb :)

copper.hat

python is the next best thing to scheme.

happy birthday!

RewCie

thanks :-)

yes :-)

user15072279

06:10

1

Q: How is this possible regarding theta?

Here, $\theta = d\theta$. Therefore, we can say $s$ is almost like a line. Then , that makes it a triangle. I have read that to find $\theta$, we say it is $\theta = S / H$. Now , let us say theta is a final velocity at of particle $P$ at $A$ and initial velocity of particle $P$ at $B$. 1]: http...

geometry trigonometry vectors triangles physics

Guys , I am having confusion in this whether it is right or wrong

Some say wrong and some right.

polite proofs

07:45

@TedShifrin What do you mean?

porridgemathematics

08:14

hi, can someone give me a hint on this? suppose we know that for any $2L$-periodic continuous function on $[-L,L]$ f, there is a sequence ${S_n(x)}$ defined on$ [-L,L] $of trigonometric polynomials, i.e.$ S_n $is a finite linear combination of ${1,cos((k(pi)/L) x),sin( (k(pi)/L )x)}_{k>=1} $converging uniformly to f on$ [-L,L]$.

How can I show that for any only continuous function on $[-pi,pi]$, there is a sequence of trigonometric polynomials that are linear combinations of ${1,cos(kx),sin(kx)}_{k >= 1}$ that converge to $f $in $L_2$ ?

oh nvm , i figured it out

polite proofs

11:22

Define $a_n = \tfrac{n!}{n^n}$. I want to prove that $\inf(a_n) = \lim_{n \to \infty} a_n = 0.$ If we write out $a_n$: $a_n = \frac{1}{n} \cdot \frac{2}{n} \cdots \frac{n-1}{n} \cdot \frac{n}{n}$. We then have $0 < a_n \le \left( \frac{1}{2} \right)^{\frac{n-1}{2}}$ which makes sense, but what is the next step?

Astyx

What happens to $(1/2)^{(n-1)/2}$ as $n\to \infty$ ?

polite proofs

It goes to 0, but I don't know how to prove it.

Because then we'd have 0 < 0 ≤ 0, which makes no sense!

Astyx

Strict inequalities don't go through limits

if $a_n<b_n$ are two sequences that converge, you can only say that $\lim (a_n) \leq \lim (b_n)$

polite proofs

Oh, I see

Astyx

For instance $0<1/n$ but both sequences converge to 0

polite proofs

11:33

However, why $(1/2)^{\frac{n-1}{2}}$ and not $(1/2)^{\frac{n}{2}}$ Or what about $(1/2)^{\frac{n+6}{2}}$?

Astyx

That was your bound

polite proofs

I meant, are either of the other two bounds good as well?

@Astyx The reasoning my book gave me for finding a bound of (n-1)/2 is because half of the terms are less than or equal to 1/2, but (n-1)/2 seems kind of random?

Astyx

It means that $a_n = \left(\prod_{1\le k\le (n-1)/2}k/n\right)\times\left(\prod_{(n-1)/2\le k\le n}k/n\right)$

The first term is lower than your bound, the second is lower than 1

polite proofs

11:50

But all of the terms are lower than one except the very last one, which is 1

Astyx

Yes, but 1 isn't a good enough bound

because 1 doesn't go to 0 as n goes to infinity

polite proofs

Are you sure your product of products is correct? I think you have (n-1)/2/n twice?

Astyx

that is less than 1/2

And you get issues with n=1 otherwise

polite proofs

I meant that your product is $1/n \cdot 2/n \cdot 3/n \cdots (n-1)/2/n \times (n-1)/2/n \cdot ((n-1)/2+1)/n + \cdots n/n$

Is that right?

Astyx

Ah yes you're right

polite proofs

11:59

But it should be $\prod_{1 \le k \le (n-1)/2} k/n \times \prod_{((n-1)/2 + 1)/n \le k \le n} k/n$?

Or maybe better notation would be $\prod_{1 \le k < (n-1)/2} k/n \times \prod_{(n-1)2/n \le k \le n} k/n$

Astyx

No, because you don't know that (n-1)/2 is an integer

Now that I think of it it should be $\prod_{1 \le k \le n/2} k/n \times \prod_{n/2 < k \le n} k/n$

polite proofs

But why is $n/2$ necessarily an integer?

Astyx

It isn't

polite proofs

Why does it not matter?

Astyx

So you want to bound from below the number of terms in the first term

Let's say that bound is m

Then you know that your product is lower than $1/2^m$

polite proofs

12:06

@Astyx Sorry, what is first term here?

The first product?

Astyx

yes

polite proofs

Sorry, I still don't understand why we care for (n-1)/2 has to be an integer, but n/2 doesn't.

Astyx

We do care if it's an integer, because that changes what we can take as a bound for m

polite proofs

But what changes for n/2?

Astyx

Nothing

You just get the correct bound (the one you mentioned above)

But you could do it with any other number less than n

polite proofs

12:10

Hmm, but you said that $\prod_{1 \le k \le (n-1)/2} k/n$ is incorrect because that (n-1)/2 is not necessarily an integer, but the same could be said for $\prod_{1 \le k \le n/2} k/n$

T_01

For lebesgue integrals we know that $f \leq g$ a.e. implies $\int f \leq \int g$. Is my intuition correct, that this does not in general hold for integrals over submanifolds (because the inequivality can be destroyed by the parametrizations) ?

Thorgott

No, it's still true

T_01

Oh, okay... thank you. Is it hard to proove? I do'nt see how I would

Astyx

@politeproofs Look at the example when n=4. With what you said above, the first product get the term $1\le(4-1)/2$ while the second product gets the terms $(4-1)/2 +1\le 3, 4$ but we're missing 2

(sorry for the multiple pings)

T_01

Ahh no I think I got that wrong in my mind. We have $f \circ \psi$ (if $f$ is the function i want to integrate and $\psi$ the parametrization) in the integral so the inequivality just carries over, right?

love_sodam

12:14

If $a\otimed b=0$ then $a=0$ or $b=0$? $a\in M,b\in N$ where $M,N$ are A-modules. A is commutative ring with unity

Thorgott

A reparametrizarion cannot destroy an inequality. If $f\le g$, then $f\circ\varphi\le g\circ\varphi$

T_01

Yep, okay. Thanks!

Thorgott

@T_01 yeah, something like that

@love_sodam far from it, try some examples with torsion over Z

polite proofs

Of course it's no problem (pings). But if we use yours for $n = 3,$ then $\prod_{1 \le k \le 3/2} k/3 = 1/3 \cdot 2/3$, also $\prod_{3/2 < k \le 3} k/3 = 2/3 \cdot 3/3$

So we still repeat a factor?

Astyx

No, because 2 is not less than 3/2

polite proofs

12:17

Oops.

Astyx

(note I changed the first inequality in the second product to be a strict one)

love_sodam

@Thorgott tough life

polite proofs

I did note that, but my other product does satisfy the strict inequality?

Astyx

(and either $k\le x$ or $x< k$, there is no in between, and each k is in one of those products)

which other product?

polite proofs

$\prod_{3/2 < k \le 3} k/3 = 2/3 \cdot 3/3$

Astyx

12:21

Yes, that works

I agree with what you said, except that 2/3 is not part of the first product

polite proofs

Yes I understand that now

Could you explain the philosophy for showing a sequence like this has an infimum of $0$?

Astyx

If you know that geometric sequences converge to zero (when the reason is <1), then it's about seeing that there are an ever growing number of number less than a number strictly less than 1 (here it's 1/2)

polite proofs

And we care about $1/2$ because it's a convenient number or is there any deeper reason?

Astyx

It's convenient, the same argument works for any number <1

You can try to do it with 1/3, or 2/3, or 145673/39182823

polite proofs

Okay

Astyx

12:29

So if $N_n$ is the number of terms $\le r$ for some $r<1$ in $a_n$, then $a_n \le r^{N_n}$, which goes to zero if $N_n$ diverges to infinity

polite proofs

Okay, fantastic, thanks.

Got it.

Astyx

So do it for 1/3

polite proofs

$a_n \le (1/3)^{(n-1)/3}$, as $n \to \infty, (1/3)^{(n-1)/3} \to 0,$ and so $a_n \to 0$

Astyx

What about n=2?

polite proofs

Well, you could replace all $3$'s by $2$, so $a_n \le (1/2)^{(n-1)/2},$ or we could do $a_n \le (1/2)^{n/2}$

Astyx

12:34

That's not quite true

polite proofs

Sorry?

Astyx

Write out $a_2$ explicitely

polite proofs

$a_2 = \frac{1}{2} \cdot \frac{2}{2}$

Astyx

And you'll see that there are no terms less than 1/3

So your bound (although it may be correct) is not justified

Not with this reasoning at least

polite proofs

For $a_n \le (1/3)^{(n-1)/3}$ I'm taking no more than a third of the terms of $a_n$

The rest of the terms are at least as large as that, but still less than $1$

Astyx

12:37

Do you rememeber how we got the $(n-1)/2$ exponent with the $r=1/2$ case?

It doesn't work exactly the same way with $r=1/3$

polite proofs

We wanted to bound from below

I don't see why not :(

We have that $a_n = \prod_{1 \le k \le n/3} k/n \cdot \prod_{n/3 < k \le n} k/n.$ But then, observe that $\prod_{1 \le k \le n/3} k/n \cdot \prod_{n/3 < k \le n} k/n \le (1/3)^{(n-1)/3} \cdot 1/3 \prod_{n/3 < k \le n} k/n$

Hmm

And then each term in $1/3 \prod_{n/3 < k \le n} k/n$ is less than $1$ but positive, so we have that $(1/3)^{n-1)/3} \cdot 1/3 \prod_{n/3 < k \le n} k/n \le (1/3)^{(n-1)/3} \cdot 1$

So $a_n \le (1/3)^{(n-1)/3}$

Astyx

Where does the (n-1)/3 come from?

polite proofs

Well it's arbitrary, I am taking no more than a third of all the terms of $a_n$

Astyx

The case $n=2$ disproves that

polite proofs

But what's wrong with this then...

Astyx

12:48

In the n=2 case, the first product is empty, so you're just bounding by 1, but somehow you get that $1/3^{1/3}$

Because you're not thinking about where that $(n-1)/2$ comes from in the r=1/2 case

polite proofs

But $1/3$ of the terms in $a_n$ are not greater than $1/3$

Astyx

That would be true if n could be divided by 3, but you don't know that

for instance, not 1/3 of the elements of $\{1/2, 2/2\}$ are less than 1/3

polite proofs

Well

Astyx

I'll let you ponder on it, I have to go

polite proofs

In that case, there are no terms which are less than 1/3... sigh

Okay

Well I think we need casework

That's because there are less than 3 terms, so there can't be $1/3$ of the terms which are less than that

love_sodam

13:30

Why is $0\to k\otimes N\to k\otimes F\to k\otimes F\to 0$ is exact?

Edward Evans

because $F$ is flat

love_sodam

How does flatness of $F$ imply that?

Balarka Sen

13:46

The obstruction to that sequence not being exact lies in Tor_A(k, F) = Tor_A(F, k) (symmetry of Tor), which vanishes by flatness of F

Mike Miller

Looks like A&M, doubt the has Tor yet

Balarka Sen

You can translate that to an elementary proof but I am not gonna

Sorry if that doesn't help

love_sodam

Tor is introduced before that exercise which I skipped

Balarka Sen

Ah, that's great, so now you have to do that exercise :)

Mike Miller

@BalarkaSen Exactly my feeling lol

love_sodam

13:56

The book assumed I'm familiar with Tor

which is not true

I skipped many newly introduced topics in exercises and as I go on, I realize I need to go back

Mike Miller

Time to learn it!

love_sodam

show time

Balarka Sen

@MikeMiller Finished reading Smale's proof of Diff(S^2) ~ O(3)

Surprisingly by-hand

How did no one before Smale do it

Mike Miller

The trick to reduce from diffeomorphisms to vector fields is not obvious IMO

That's the clever part

Once you have that first step, the rest is not obvious, but a patient worker will figure it out.

Balarka Sen

The details are not that clear actually but it sounds like someone from the 40's would know this

Mike Miller

14:02

Yeah I don't mean it's easy to do the rest but if you have the patience you will do the rest.

You know what I mean?

Balarka Sen

Yeah

I gotchu

Mike Miller

And Smale is nothing if not a patient man

Balarka Sen

That makes sense yeah!

Alessandro Codenotti

@love_sodam half of A&M is in the exercises, maybe more than that

There is a pdf somewhere from some guy that did every exercise in AM, the pdf has more pages than the book

Thorgott

15:00

Sounds excessive

love_sodam

@AlessandroCodenotti yeah I know that but not all of his solution is good

Mike Miller

Good

That means you have to do it yourself

Thorgott

I only worked through the first couple chapters diligently, but I think most exercises were ultimately pretty straight-forward

Sha Vuklia

15:16

Guys, shouldn't there be $I\Omega^1_A$ in the denominator? That is, $\Omega^1_{A/I}=\Omega^1_A/(I\Omega^1_A+Adf_1+\dots+Adf_n)$.

Here $I=(f_1,\dots,f_r)$

and $\Omega^1_A$ is the module of Kähler differentials of $A$

and likewise for $A/I$

Mike Miller

In your picture Omega^1_{A/I} never appears

Sha Vuklia

Oh well eh

can you read it as

A=k[x_1,\dots,x_n]

in the picture

wait

I'll make it match

Mike Miller

Anyway dunno

What's there seems right

Sha Vuklia

ah I think that explains why things didn't work out

I read $A=k[x_1,\dots,x_n]$ in the text

A-Level Student

Sorry to interrupt. I have

Sha Vuklia

15:31

ah ya, now it's fine

A-Level Student

a question about inverse functions

Mike Miller

@A-LevelStudent No such thing as interrupting

Whether or not someone will respond is always a different question but you can always say whatever whenever

A-Level Student

Thanks :) If I have a curve ( I don't say function as it won't necessarily be a function; eg it could be one to many) then if it intersects with its inverse- so where solutions EXIST -must they be on both of the lines $y=\pm x$, or only one of them, or none?

Mike Miller

15:44

I unfortunately don't think most of that parses. "It intersects with its inverse" is not a phrase that parses, nor is there such a thing as the inverse of a curve. Then it's not clear what is meant by solutions existing.

I suspect that when you make the question make sense (to me) the answer will be no. But I don't understand the question as it is.

Balarka Sen

15:56

@A-LevelStudent You're asking a good question so you should try to make it communicable.

I won't do it for you, just wanted to give a word of encouragement.

A-Level Student

16:23

I will try to explain what I mean. Suppose I have something (a function in this case) like $x^2+y=5$ and I graph it on the Cartesian plane. Now suppose I also graphed the function $y^2+x=5$ on the Cartesian plane, ie the inverse function of $x^2+y=5$. We can see that the intersection points of the two curves lie on the line $y=x$. Similarly if I graph $x^4+y^2=20$ and $y^4+x^2=20$ on the Cartesian plane the curves intersect, both on the line $y=x$ and $y=-x$.

Although these two curves cannot be called inverse functions of each other as they are one to many, we have still interchanged the $x$'s and $y$'s so I will refer to them as inverse curves. My question is, if inverse curves do intersect, must they intersect only on one of the lines $y=\pm x$ or are there other lines they can intersect on?

@MikeMiller and @BalarkaSen is that any clearer?

user3733558

Think of what it means to swap $x$ and $y$ in the equations. Also, consider $x^2 + y^2 = 1$, what is the intersection in this case?

123

17:05

Hello Guys...

Astyx

17:27

@politeproofs can you give the exact number of integers less than n/3

copper.hat

@A-LevelStudent plot something like $x \mapsto \sin (20 \pi x)$ on $[-1,1]$.

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$Mathematics$ Mathematics