Mathematics - 2020-05-26 (page 3 of 3)

« first day (3583 days earlier) ← previous day next day → last day (1733 days later) »

BAYMAX

22:31

I have two functions $f(x,y), g(x,y)$

I wan to check for the x and y values at which they intersect

How about setting $f(x,y) = g(x,y) ?$

Thorgott

oh lord, the mods are flocking

Semiclassical

that'll do it, yes

Ted Shifrin

What does it mean for two functions to intersect?

BAYMAX

I meant curves

sorry

Ted Shifrin

Well, how are they curves?

Alessandro Codenotti

22:35

Uh nice I didn't know I get pinged when another room owner kicks somebody @Semi

quid

Is there an issue or is there just inappropriate banter?

Semiclassical

huh, didn't know that

Alessandro Codenotti

There had never been the need to do so in more than a year apparently

Ted Shifrin

I just came in to find 6 removeds.

doppelgreener

@quid There was; see the deleted messages.

Semiclassical

22:35

it's (hopefully) been dealt with

Ted Shifrin

How come you get pinged and I don't, @Alessandro?

Semiclassical

weird

BAYMAX

$f = 1 - y - \frac{y}{\lambda} |-y + Rx|$

$g = \delta (1-x) - \frac{x}{\lambda} |-y + Rx|$

Ted Shifrin

@Baymax, you're missing the point. If I set $f(x,y) = x^2+y^2$, how is that a curve?

Alessandro Codenotti

Dunno @Ted, it made the same sound as when I get pinged and I had that notification

doppelgreener

22:37

@TedShifrin I think it happens only if you're in the room at the time and on desktop. (Like you're present or the room's in your room list in the sidebar.) A banner comes down on the top mentioning someone got kicked in a room you're in.

Ted Shifrin

Maybe I've been demoted after being the "original" new room-owner.

Ah, I came in right after that. Of course. Thanks.

Alessandro Codenotti

Ah makes sense then, I had the chat open in a side tab as I always do

BAYMAX

Circle is a curve in $\Bbb{R}^2$

Ted Shifrin

We don't have a circle, @Baymax.

Semiclassical

which circle?

BAYMAX

22:38

I got that

Ted Shifrin

You have to specify a value of $f(x,y)$ to get a curve. Same for $g$.

BAYMAX

yup

$f=0, g= 0$

Ted Shifrin

In other words, it takes an EQUATION, not a function, to define a curve.

Aha.

Alessandro Codenotti

Anyway looks like we can go back to maths. Thanks mods for the quick intervention

Ted Shifrin

Do you see that solving $f(x,y)=g(x,y)$ does not give you what you want? Why?

doppelgreener

22:39

Thanks to the room for bringing it to our attention!

BAYMAX

@TedShifrin I wan to do it analytically and not graphically. Thanks for correcting me!

Semiclassical

first time i've used the kick-mute power, hopefully won't need to do so again any time soon

Ted Shifrin

Mazltov, @Semiclassic.

Alessandro Codenotti

Do the mods get automatically notified when somebody is kick-muted or did you need to flag too?

BAYMAX

that gives me $1 - y - \frac{y}{|\lambda|} |-y + Rx| = \delta (1 -x) - \frac{x}{\lambda}|-y + Rx|$

Ted Shifrin

22:40

There are no algorithms, Baymax. If you can solve one equation for one variable and plug it into the other, you're fortunate. Of course, if they're both linear equations, there are well-known algorithms.

No, you've thrown away information when you remove the VALUE.

BAYMAX

okay

If I solve both

Semiclassical

I think the message on my end said that mods may be notified, but I flagged as well to be sure

Ted Shifrin

If I want to solve $x^2+y^2 = 4$ and $y-2x = 4$, setting $x^2+y^2=y-2x$ does NOT give me the answer.

BAYMAX

yup I see

Ted Shifrin

Since you have $|-y+Rx|$ in both equations, I would solve each for that expression and then see what you have.

BAYMAX

22:42

yeah

that gives me

doppelgreener

@AlessandroCodenotti Nothing special there. Messages got flagged for us though.

BAYMAX

$y = \frac{x}{x + \delta(1-x)}$

but not able to get $x$ now

Semiclassical

@ted i see what you mean re: that griffiths book, btw. the table of contents makes the relevance immediate

Is $\delta$ a number? If so, that's certainly solvable for $x$

Ted Shifrin

It's a very classical approach, @Semi. It should fit your interests and skills.

Semiclassical

it's a little tedious but quite doable

@ted nice

Ted Shifrin

22:44

You then have to substitute that (I didn't check to see if you're correct) into one of the original equations, @Baymax.

BAYMAX

Poh ok

that gives me

$\delta(1-x)(x + \delta(1-x)) - x^2 = 0$

but they have mentioned that one would get $y$ as a cubic in terms of $x$

But I get a quadratic

Ted Shifrin

I haven't checked your algebra. Have you rechecked it?

BAYMAX

I wan to show that there is a chance that I will get three points

where $f=0,g=0$

Ted Shifrin

The $|-y+Rx|$ has to come back into it.

Semiclassical

the presence of absolute values there makes me suspicious about getting a cubic (though I could certainly imagine 3 solutions)

BAYMAX

22:52

gotcha!

there will be cubic equation in $x$

Semiclassical

oh, is this restricted to real x and y

BAYMAX

no actually

Semiclassical

hmm

BAYMAX

@Semiclassical fixed points

Semiclassical

then I remain suspicious

BAYMAX

22:54

graphically I can see there are three points of intersections

but of course, they will depend on the parameter values

Semiclassical

how are you seeing that graphically---by plotting inn a plane of (x,y)?

BAYMAX

yes

Semiclassical

the real (x,y) plane?

BAYMAX

yup why

Semiclassical

then why are you saying it's not restricted to real x,y?

BAYMAX

22:55

oh ok

Semiclassical

i mean, if you're only interested in real x,y, that's fine

it definitely makes the problem moreo tractable

BAYMAX

oh yes yes

only real $x$ and $y$

Thank you!

Semiclassical

« first day (3583 days earlier) ← previous day next day → last day (1733 days later) »

all rooms

Transcript for

May25

May '2026

May27

$Mathematics$ Mathematics

Associated with Math.SE; for both general discussion & math qu...

join this room
about this room

00:00

06:00

12:00

18:00

all times are UTC

$Mathematics$