Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °

You are not logged in.

## #1 2006-04-16 02:59:41

Chemist
Member

Offline

### Domain of f(x,y)

Can someone verify if this is the domain of def of the following function?

f(x,y) = Ln [ y / ( x2 + y2 -1 ) ]

Df = { (x,y) belong to R2 : y > 0 , (1-y2 ) 1/2 < x < - (1-y2 ) 1/2 } U { (x,y) belong to R2 :

y < 0 , - (1-y2 ) 1/2 < x < (1-y2 ) 1/2 }

Thanks,

"Fundamentally one will never be able to renounce abstraction."

Werner Heisenberg

## #2 2006-04-18 14:27:24

George,Y
Super Member

Offline

### Re: Domain of f(x,y)

Except for  (1-y²)1/2< x < -(1-y²)1/2 should be
x < -(1-y²)1/2 or x>(1-y²)1/2 instead, or you may use three sets.

Last edited by George,Y (2006-04-18 14:30:00)

X'(y-Xβ)=0

## #3 2006-04-19 07:29:38

Chemist
Member

Offline

### Re: Domain of f(x,y)

Yes, (1-y²)1/2< x < -(1-y²)1/2 doesn't make sense, in the solutions guide, it's written :

- (1-y²)1/2< x < (1-y²)1/2 , which is definitely wrong!

Perhaps three sets , as you suggested, is more logical.

Last edited by Chemist (2006-04-19 07:30:31)

"Fundamentally one will never be able to renounce abstraction."

Werner Heisenberg

## #4 2006-04-28 09:20:18

Chemist
Member

Offline

### Re: Domain of f(x,y)

Hey, I've rechecked this : Plotting y=x , y=-x would show that

Df = { (x,y) belong to R2 : y > 0 , (1-y2 ) 1/2 < x < - (1-y2 ) 1/2 } U { (x,y) belong to R2 :

y < 0 , - (1-y2 ) 1/2 < x < (1-y2 ) 1/2 }

The domain is on the right bounded by y=x , y=-x. y changes from x to -x. The left part of the domain is bounded by y=x, y=-x, y changes from -x to x.

"Fundamentally one will never be able to renounce abstraction."

Werner Heisenberg