Question

Use Implicit Function theorem to show that \(F(x,y)= y^2+2xy +x-1=0\) can be solved for y in terms of x near all points \((x_0,y_0)\) with \(y_0\neq x_0\).
Check the result directly. Also determine if the equation F(x,y) =0 is locally solvable near any point \(x_0,-x_0)\)
Please help correcting my solution.

Answer 1

To consider F can be solved for y in terms of x, we need partial \(F_y\neq 0\) . And we got it, \(F_y = 2y-2x \neq 0\)

Answer 2

2y+2x?

Answer 3

To check the result directly, I replace and get \(y=-x\pm \sqrt{x^2-x+1}\) but my prof commented :"This is not the function of x"
How to fix?

Answer 4

oh yea, 2y + 2x

Answer 5

because of the +/-

Answer 6

if it was just + then it would be a function

or if it was just - then it would be a function

but it can't be both
like for each x we want one unique y
but for each x here we would have at max 2 y

Answer 7

So? I just separate them ?

Answer 8

For the last one, I find \(F(x_0, -x_0)= x_0^2-x_0+1\) and it has no solution. Hence, \(F(x,y) \)is not solvable near any point \((x_0,-x_0)\) . My prof's comment: " Lucky you"
It means it is not correct, just lucky. I don't want that luck. I need logic. Please, correct me

Answer 9

can I asked how you got the first relation?

Answer 10

that is the relation $y=-x\pm \sqrt{x^2-x+1}$

Answer 11

\[y=-x\pm \sqrt{x^2-x+1}\]

Answer 12

Just solve F(x,y)=0, that is \(y^2 +2xy +x-1 =0\)

Answer 13

oops just found out

Answer 14

I worked backwards
still trying to familiarize myself with the theorem

Answer 15

\(y^2 +2xy+x^2-x^2+x-1=0\\(y+x)^2 =x^2+x-1\)

Answer 16

\[y^2+2xy+x^2-x^2+x-1=0 \\ (x+y)^2=x^2-x+1 \\ \text{ now we have } y \neq -x \\ \text{ so \let's assume } y>-x \text{ then } y+x>0 \\ \text{ so } x+y=\sqrt{x^2-x+1} \\ y=-x +\sqrt{x^2-x+1} \text{ when } y>-x \\ \text{ now } \\ \text{ assume } y<-x \text{ then } y+x<0 \text{ so we have } \\ x+y=-\sqrt{x^2-x+1} \\ y=-x-\sqrt{x^2-x+1} \]

Answer 17

does that make since to you?

Answer 18

sense*

Answer 19

Yes, it is

Answer 20

now the second question you say I kinda don't get it yet
are you asking to solve x^2-x+1=0?

or is something else meant by solution/

Answer 21

That is when \(y_0= -x_0\), we have \(F(x_0, -x_0)= (-x_0)^2 + 2 (x_0)(-x_0)+x_0-1=0\\-x_0^2+x_0-1=0\)

Answer 22

oh we are dealing with same function

Answer 23

And this equation has no solution, that is at \((x_0,-x_0)\), \(F(x,y) \neq 0\), Hence it is not solvable near there

Answer 24

i mean relation
\[F(x,y)=x^2+2xy+x-1\]
\[F(x,-x)=x^2+2x(-x)+x-1 \\ F(x,-x)=x^2-2x^2+x-1 \\ F(x,-x)=-x^2+x-1\]
so yes I see how you got this

Answer 25

\[x^2-x+1=0 \\ x=\frac{1 \pm \sqrt{1 -4(1)}}{2}\]
so you don't want to include imaginary ( I mean complex) solutions?

Answer 26

I think we are working on R^2, not C

Answer 27

I don't know about that one

by the way did you asked this on mathexhange someone had your exact same solutions

Answer 28

http://math.stackexchange.com/questions/1227592/implicit-function-theorem-checking
is that you?

Answer 29

Yes, :)

Answer 30

I didn't really a straight forward answer for the second question
it seemed like they were agreeing with you

Answer 31

I make a lot of question there since from here, not many people help me. :)

Answer 32

My prof took of 5 credits out of 10 for that mistake. I ..... frustrated.

Answer 33

what would happen if we looked at dF/dx instead of dF/dy
and solved for x instead

Answer 34

And I got mad with his comment : Lucky you, Ha!!

Answer 35

Nope, I typed it right.

Answer 36

lol made he was trying to bring humor to it all :p

Answer 37

\[F(x,y)=y^2+2xy+x-1=0 \\ \frac{dF}{dx}=2y+1 \neq 0 \\ \text{ means } y \neq \frac{-1}{2} \\ y^2+2xy+x-1=0 \\ y^2+x(2y+1)-1=0 \\ y^2-1=-x(2y+1) \\ x=\frac{y^2-1}{-(2y+1)} \\ \]
now what would happen if we use this for our y=-x

\[x=\frac{x^2-1}{-(-2x+1)} \\ -x(-2x+1)=x^2-1 \\ 2x^2-x=x^2-1 \\ x^2-x+1=0\]
same equation :(

Answer 38

which makes sense I guess

Answer 39

Ok, I will come to see him tomorrow. I need make it clear for my upcoming final. Thanks for the help. I have a bunch of thing need check. Willing to help?

Answer 40

I can only try. I can not guarantee I will be able to help. You have really exceeded my knowledge but I will look at it.

Answer 41

@Michele_Laino we are still trying to figure out why @Loser66 reply to F(x,y)=0 isn't solvable for (x,-x) is wrong

Answer 42

Or I'm still trying to figure it out :p