Partial Derivatives of a Two-Variable Dirichlet Function

[teleport]

Homework Statement

Does the function f(x,y) has any partial derivatives?

f(x,y) = (x^2 + y^2)^2, if both x and y are rational

= 0 , otherwise

Homework Equations

The Attempt at a Solution

Not sure what to do. I know this function has only one limit at (0,0) and is also continuous there, and nowhere else. So it is not differentiable for (x,y) =/= (0,0). However, this doesn't say anything about existence of the partials. Thanks for any help.

 

[HallsofIvy]

Have you written out the expression for the partial derivatives? In particular, if you look of
$$\frac{f(x_0+ h,y_0)- f(x0,y0)}{h}$$
and
$$\frac{f(x_0,y_0+ h)- f(x0,y0)}{h}$$
for irrational x₀, y₀, and h, what do you get? What about for rational values. I would recommend that you look at (0,0) especially carefully.

 

[teleport]

And you can do this because of irrationals and rationals being dense in R. OK but what is it so special about (0,0)? I'm not seeing it. thanks

 

[teleport]

I mean after all, with some of x, y, h, irrational we get 0/h and when you take the limit as h goes to 0, then what now?

 

[teleport]

Wow forgot l'hospital for a sec there. Anyways lim 0/h =0. Ok so I got for the irrationals the partials are 0, while for the rationals the partials are non-zero only for (x,y)=/= (0,0). So the conclusion is that the partials exist only at (0,0)? Thanks

 

[teleport]

Oh wait now I checked my work and is giving me that the x partial exists not only at
(0,0) but also at the y-axis. Is that right?

 

[Dick]

There are actually partial derivatives at a lot of places, if I read your problem correctly. Take the line y=sqrt(2). Then f(x,y)=0 for any x-value along this line since y is irrational. So an x partial exists anywhere along this line.

 

[teleport]

True! I considered the partials through the rationals and through the irrationals, and looked at where these two were equal. What I got was that they were equal only (for the x partial) at the y-axis. So what I did must be insufficient, and I am missing a major method of finding the answer. Can anyone share it? Thanks

 

[Dick]

There no real 'method'. You just have to think about it. I think you are right that the only rational (x,y) pts where partials exist is (0,0). Now we know if y is irrational then there is an x partial. What about if x is irrational? What about if they are both irrational? Is that all of the cases?

 

[teleport]

Oh, so by symmetry if x is irrational then the y partial exists symilarly. I said that the partials exist through at the y-axis (for the x partial) including (0,0), because I found that this was the only place where these two partials are equal. So would this be the answer for the both rational or irrational cases?

 

[teleport]

Oh now I see that the "both irrational" case is included in the "one irrational " case. So now my only concern is about the "both rational" case.

 

[Dick]

Right, but I don't see anything special about the y-axis. And, why should the two partials be equal?

 

[teleport]

Oh I see the mistake I was doing. I think I got it now. Check this out:
Take x,y rational. Letting h -> 0 through the irrational we get

f_x = lim (h -> 0) [-(x^2 + y^2)^2/h], which exists only for (x,y) = (0,0), where it equals 0.

Letting h->0 through the rationals we get

f_x = (some function with x and y, so that if x = 0, then f_x = 0)

Since Q and R\Q are dense in R, these two limits should be equal if they exist. Since they both work only if (x,y) = (0,0) then the partial exists only at (0,0) as you say? Thanks

 

[Dick]

That's a little confusing, but if you understand it... BTW in the case (x,y) rational and not (0,0), the function is not even continuous at (x,y) even when approached along x constant or y constant lines. This being the case, there really isn't any need to write down a difference quotient.

 

[teleport]

I'm done with this Diricklet! :) Great help! Thanks a lot.