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#1 2006-04-15 04:59:41

Chemist
Member
Registered: 2005-12-12
Posts: 35

Domain of f(x,y)

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

f(x,y) = Ln [ y / ( x[sup]2[/sup] + y[sup]2[/sup] -1 ) ]

D[sub]f[/sub] = { (x,y) belong to R[sup]2[/sup] : y > 0 , (1-y[sup]2[/sup] ) [sup]1/2[/sup] < x < - (1-y[sup]2[/sup] ) [sup]1/2[/sup] } U { (x,y) belong to R[sup]2[/sup] :

y < 0 , - (1-y[sup]2[/sup] ) [sup]1/2[/sup] < x < (1-y[sup]2[/sup] ) [sup]1/2[/sup] }

Thanks,


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

Werner Heisenberg

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#2 2006-04-17 16:27:24

George,Y
Member
Registered: 2006-03-12
Posts: 1,306

Re: Domain of f(x,y)

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

Last edited by George,Y (2006-04-17 16:30:00)


X'(y-Xβ)=0

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#3 2006-04-18 09:29:38

Chemist
Member
Registered: 2005-12-12
Posts: 35

Re: Domain of f(x,y)

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

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

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

Last edited by Chemist (2006-04-18 09:30:31)


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

Werner Heisenberg

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#4 2006-04-27 11:20:18

Chemist
Member
Registered: 2005-12-12
Posts: 35

Re: Domain of f(x,y)

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

D[sub]f[/sub] = { (x,y) belong to R[sup]2[/sup] : y > 0 , (1-y[sup]2[/sup] ) [sup]1/2[/sup] < x < - (1-y[sup]2[/sup] ) [sup]1/2[/sup] } U { (x,y) belong to R[sup]2[/sup] :

y < 0 , - (1-y[sup]2[/sup] ) [sup]1/2[/sup] < x < (1-y[sup]2[/sup] ) [sup]1/2[/sup] }

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

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