Implicit and inverse theorems

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xoninhas

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May 25, 2008
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you have this system of two equations and 3 variables:

x^2 + y^2 = e^(2t) + 1
x sen( (?/2) x) = t

Show that this system defines, in a neighborhood, of the point (t0, x0, y0) = (1, 1, e) a C1 function, given by:

?(t) = (x(t), y(t))

Since it is kinda very hard to invert the function to xy = t than I think that what you do is get:

F(t, x, y) = (x^2 + y^2 - e^(2t) -1, x sen( (?/2) x) - t)

and made the d^2F/dxdy (don't know if it should be to those variables) on point (1, 1, e) which I got:

[-2e^(2t)]
[ 1 ]

well and now I can't get the determinant so to say that it can be inversed thus ruining everything else...

I don't know exactly what I'm doing very well... I know I'm proving first that the function is invertible and if it is I have no idea why you can proceed to say that it can be said that t can be written in function of x, y.

Thanks for every help!
 
You could investigate it more simply to some benefit.

What happens when you substitite t = x = 1 and solve for y?

What happens when you substitite t = 1 and y = e and solve for x?

What happens when you substitite x = 1 and y = e and solve for t?

Now think about the defintion of "function".
 
Ok, i did those substitutions and you can indeed with only two variables always get to the other one. so in fact we could write each of them in function of the other two.

But I still don't get ow I'll build the a(t) = (x(t), y(t))... hmmm unless I have to isolate x=blablat and y=blalat

which in fact i dont know how to do with that sin there... Thanks so far!

EDIT:

Well I was thinking, if they say in the neighborhood of (t0, x0, y0) = (1,1,e) so we can say that:

x(t) = t
y(t) = sqrt[ -t^2 + e^(2t) +1 ] and thus:
a(t) = ( t , sqrt[ -t^2 + e^(2t) +1 ] )

Is that it??
 
xoninhas said:
Ok, i did those substitutions and you can indeed with only two variables always get to the other one. so in fact we could write each of them in function of the other two.
Click to expand...
You didn't find it significant that two were multi-valued and one was single-valued?
 
What do you mean by multi-valued and single valued?

for t = x = 1:
y = +/- e
1 = 1

for t=1 y=e:

x = +/- 1
x * sen(x*pi/2) = 1 I suppose its x = 1

for x = 1 y=e:
t=1
t=1

and what does it help that t=1 in both systems? And by the way is that what you meant multi-valued and single-valued, I'm sorry but my math was given in Portuguese.

Thank you again for the help.
 
Basic Definition of "Function". When you use "+/-" you are not saying there is a single value. This violates the premise, doesn't it?

Okay, for real. You have (t,x,y)=(1,1,e).

Substitute t = 1. This gives two functions in x and y.

Create the Jacobian with the four partial derivatives. Where is this thing zero? Anywhere near x = 1 and y = e?
 
Ok I did the matrix and i got:

[ 2x 2y ]
[ sen(x*pi/2) + x*cos(x*pi/2) 0 ]

well I don't know what you meant when this is zero... but obviously the matrix is 0 when x = y = 0, if we do the determinant it's the same thing...

so next move is? :| I'm getting kinda confused....
 
Isn't that the point? As long as we stay away from x = 0 and y = 0, the thing is invertible. Since we're at x = 1 and y = e, I think we're done.
 
I'm kind of slow with this I'm so sorry... so let me see if I got everything...

First what we did was see in the neighborhood of the point given what was the variable that could be written in function of the other two... only t gave only one value, which means that we have F(x,y) = t... so we did the determinant of the Jacobian of the F(x,y) so that we can prove that there is the inverse a(t) = (x,y) in the neighborhood of that point.

is that it?? damn it would be cool if it was... cause it almost makes sense! :p Thanks so much so far!
 
Just one thing. The Jacobian IS the determinant.
 
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