Hi

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 $f(x,y) = g(x,y) ?$

oh lord, the mods are flocking

that'll do it, yes

What does it mean for two functions to intersect?

I meant curves

sorry

Well, how are they curves?

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

Is there an issue or is there just inappropriate banter?

huh, didn't know that

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

I just came in to find 6 removeds.

@quid There was; see the deleted messages.

it's (hopefully) been dealt with

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

weird

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

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

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

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

@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.

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

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

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

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

We don't have a circle, @Baymax.

which circle?

I got that

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

yup

$f=0, g= 0$

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

Aha.

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

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

Thanks to the room for bringing it to our attention!

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

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

Mazltov, @Semiclassic.

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

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

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.

okay

If I solve both

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

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.

yup I see

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

yeah

that gives me

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

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

but not able to get $x$ now

@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 $x$

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

it's a little tedious but quite doable

@ted nice

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

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

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

where $f=0,g=0$

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

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

gotcha!

there will be cubic equation in $x$

oh, is this restricted to real x and y

no actually

hmm

@Semiclassical fixed points

then I remain suspicious

graphically I can see there are three points of intersections

but of course, they will depend on the parameter values

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

yes

the real (x,y) plane?

yup why

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

oh ok

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

it definitely makes the problem moreo tractable

oh yes yes

only real $x$ and $y$

Thank you!

np