Question

If we know f(x,y)=f(y,x) what else do we know about the function? And as an extra, if we know g(x,y)=-g(y,x) what do we know about this function?

Answer 1

some kind of symmetry / antisymmetry

Answer 2

Well for the first one it seems like we can safely say x=y so we can just see that \[\LARGE f(x,x)=f(y,y)=f(t)\] it's really just a one variable function disguised as a multivariable function I guess.

But for g(x,y) it's not so obvious and this is where I'm stuck in my own little question. We can try to limit ourselves to certain functions and it will work. \[\LARGE g(x,y)=\sum a_n(x-y)^{2n+1}\] so basically we can throw on any constant coefficient and have a bunch of odd power terms so that the negative will come out for the identity g(x,y)=-g(y,x). But past that I'm not sure, it just doesn't feel as clean.

Answer 3

Or maybe somehow my reasoning that f(x,y) is really a single variable function is wrong, I'm not sure.

Answer 4

Let f(x,y)=e^x+e^y
does e^x+e^y=e^y+e^x
imply x must be y ?

Answer 5

or like f(x,y)=x+y
x+y=y+x
this doesn't imply x=y, does it?

Answer 6

I think f(x,y)=f(y,x) implies it since it means x=y by comparing their corresponding parts. f(x=y, y=x) if that makes it make sense.

Answer 7

f(x,y)=x+y
f(2,3)=2+3 and f(3,2)=3+2
but 3 doesn't equal 2
Maybe I'm misunderstanding

Answer 8

if f is a one to one
and f(x,y)=f(y,x)
then (x,y)=(y,x)
then y=x

Answer 9

for the case that f(x,y) =x+y
t = (x+y)/2

Answer 10

is this symmetric\[f(x,y) = (x+y)^{y+x}\]

Answer 11

Hmm well this is interesting because it seems like we can replace all cases of x and y with just a single variable now but you are right @myininaya that x=y is sort of false. Hmmm. I'd really like to see some kind of counter example I guess.

Answer 12

@ganeshie8 that is symmetric.

Answer 13

kainui , why 3+3=8 :P ?

Answer 14

i still cant see why x=y xD
why if i have function like this :3
x^2+y^2 ?!

Answer 15

Yeah it's not true what I said, I don't know but it makes sense though in a way I guess I'm not sure.

Answer 16

hmm but it might make sense to say
|x|=|y| ?
idk if it match all cases , if we have angles for example it might not work
lets check properties of function

Answer 17

Nah, we shouldn't say any of this I think because it puts up false restrictions. Since f(x,y)=f(y,x) we are saying that they are interchangeable like f(1,2)=f(2,1) so it's a surface reflected over the line f(t,t) I guess.

Answer 18

yes !
f(x,y)=f(y,z)=z
make sense but hmmm im not thinking of anything else

Answer 19

I think we can make quite a few functions that satisfy this property f(x,y)=f(y,x) with a single valued function if we choose f(u) with u=x+y

and we can make some functions that satisfy g(x,y)=-g(y,x) with g(v) when we use v=x-y

Answer 20

i don't know if this is interesting but
say we have f(x,y)=x^2+y
f(x,y)=f(y,x) could give us
x=y or x=-y+1

x^2+y=y^2+x
x^2-x-y^2+y=0
\[x=\frac{1 \pm \sqrt{1-4(-y^2+y)}}{2}=\frac{1 \pm \sqrt{(2y-1)^2}}{2}=\frac{1 \pm (2y-1)}{2} \\ \text{ so } x=y \text{ or } x=-y+1\]

Answer 21

That's because if you have f(x,y)=x^2+y that's not equal to f(y,x)=y^2+x.

Although this makes me realize that we can make any function that satisfy these situations quite easily!
\[\LARGE p(x,y)\] p is just any arbitrary function. Now we can make f(x,y)
\[\LARGE f(x,y)=p(x,y)+p(y,x)\]

so we see that \[\LARGE f(y,x)=p(y,x)+p(x,y)=f(x,y)\] is satisfied!

We could have also multiplied them instead.

Answer 22

oh yeah I forgot what we were talking about for a sec

Answer 23

Kainui what was wrong with my counterexample above

Answer 24

f(x,y)=f(y,x)
when f(x,y)=x+y

Answer 25

f(3,2)=f(2,3)=5
but 2 doesn't equal 3

Answer 26

f(x,y) is only a surface around Z hmm
if we tried to check traces we will get always same trace what ever Z was , right ?

Answer 27

You're right, x=y is wrong, but you can always rearrange the function so that it looks the same as it did before you replaced them. So where x was "held" is the same. The spots holding your choice of x and y are identical is what I meant to say but I don't know how to say that.

Answer 28

For example, \[\LARGE x+y+xy+y^x+x^y=y+x+yx+x^y+y^x\]

Answer 29

this is the type of symmetry that it should satisfy

Answer 30

I don't think this line is a line of symmetry but I might be mistaken.