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f(x,y) = lxl^k*lyl^k/(x^2+y^2) is convergent, find min k

Posted: Wed Oct 12, 2011 6:26 am
by cathy_liping
how to find the mininum of k to make f(x,y) = lxl^k*lyl^k/(x^2+y^2) convergent?

in practice book, there is a question which is similar but simple than this one, like xy/(x^2+y^2) is convergent or not, i know this is not convergent, as when keep y constant, if y=x, lim(x->0)=1/2, while if y=0, lim(x->0)=0, so it is not convergent.

what about this one? tks in advance!

Re: f(x,y) = lxl^k*lyl^k/(x^2+y^2) is convergent, find min k

Posted: Wed Oct 12, 2011 11:18 am
by owlpride
You mean continuous?

Any k > 1 will do. If you consider the straight line y = ax through the origin, then the limit of f(x,ax) as x ->0 along this line is 0 as long as 2k > 2.

Re: f(x,y) = lxl^k*lyl^k/(x^2+y^2) is convergent, find min k

Posted: Thu Oct 13, 2011 1:50 am
by cathy_liping
owlpride wrote:You mean continuous?

Any k > 1 will do. If you consider the straight line y = ax through the origin, then the limit of f(x,ax) as x ->0 along this line is 0 as long as 2k > 2.
thank you owlpride for your reply :D and sorry for my wrong info... yes i mean continuous... but when k=2/3 will make f(x,y) continuous, which is <1... am i right?

Re: f(x,y) = lxl^k*lyl^k/(x^2+y^2) is convergent, find min k

Posted: Thu Oct 13, 2011 2:59 am
by blitzer6266
Nope. k=2/3 will make the limit at 0 not exist: If you go along y=x, the limit will be infinity, where as if you go along y=0, the limit will be zero (since the function is identically 0 along the x and y axis).

Re: f(x,y) = lxl^k*lyl^k/(x^2+y^2) is convergent, find min k

Posted: Mon Oct 17, 2011 5:01 am
by cathy_liping
blitzer6266 wrote:Nope. k=2/3 will make the limit at 0 not exist: If you go along y=x, the limit will be infinity, where as if you go along y=0, the limit will be zero (since the function is identically 0 along the x and y axis).
oh... ok, I see, my mistake :oops: tks for pointing out!

Re: f(x,y) = lxl^k*lyl^k/(x^2+y^2) is convergent, find min k

Posted: Fri Aug 22, 2014 5:54 pm
by DDswife
I have a question.

For k = 1, lim |x|*|y| / (x^2+y^2) = 0 for both x and y approaching 0
For x=y, the limit is 1/2.

So the limit doens't exist. I get that.

Now, if we use polar coordinates, we get:

lim (r cos a)^k (r sin a)^k / r^2, as both, r and a go to 0

lim r^2k [sin (2a) / 2]^k / r^2 = lim r^(2k-2) [sin (2a) / 2]^k

2k - 2 = 0, the limit is 0 since lim sin (2a) / 2 = 0 as r, a both approach 0

2k - 2 > 0, then both approach 0

So I get that f continuous if k >= 1

Can someone explain to me where the problem is? Thanks

Re: f(x,y) = lxl^k*lyl^k/(x^2+y^2) is convergent, find min k

Posted: Sun Aug 24, 2014 12:58 am
by Austin
DDswife: I could be wrong, but I believe your error is sending both r and a to 0. If you just send r to 0, you should get the limit at (0,0). By also sending a to 0, I think you're evaluating the limit by approaching (0,0) along the line y=0.

So, for k > 1, lim [r^(2k-2) sin(2a)/2 , r->0] = 0. For k=1, lim [r^(2k-2) sin(2a)/2 , r->0] = lim[sin(2a)/2, r->0] = sin(2a)/2.

Note that this agrees with your y=x example, because in that case a=pi/2, so sin(2a)/2=1/2.

Hope this helps!

Re: f(x,y) = lxl^k*lyl^k/(x^2+y^2) is convergent, find min k

Posted: Sun Aug 24, 2014 1:10 am
by DDswife
That might be the problem. Funny, when I took Multivariable Calculus, I watched some videos and learnt this trick and, if I am not confused, this is the way it was done. But I will recheck on that.

Thanks for your reply. BTW, I think that you meant a = pi/4, then 2a would be pi/2. But I got your point.

Good to know I am not by myself! This forum seemed to be abandonned lately. Are you planning to takie the MGRE at some point?

Re: f(x,y) = lxl^k*lyl^k/(x^2+y^2) is convergent, find min k

Posted: Sun Aug 24, 2014 1:25 am
by DDswife
You were right. I just found this link and they say that r approaches 0 and the rest needs to be independent of the angle. Good that I corrected one of my mistakes! Thanks again.

http://www.math.ucla.edu/~skalyanswamy/ ... -polar.pdf

Re: f(x,y) = lxl^k*lyl^k/(x^2+y^2) is convergent, find min k

Posted: Sun Aug 24, 2014 2:15 pm
by Austin
Great! Glad I could help. Yeah, it should have said a=pi/4...I also left off an exponent inside the limit.

Yeah, the forum does seem pretty dead, but that could have to do with the time of year. I applied to grad schools a couple of years ago and the peak time for this forum was the October-April stretch when the admission process is in full swing.

I am hoping to make another run at grad school this year; if I do, I'll take the MGRE in October. Are you also aiming for Fall 2015?

Re: f(x,y) = lxl^k*lyl^k/(x^2+y^2) is convergent, find min k

Posted: Sun Aug 24, 2014 2:28 pm
by DDswife
I am far away yet. Taking Linear Algebra next semester in college and, if the University finally accepts me (I am not from the US and they keep requesting stuff I don't even think I will be able to submit! But who know...), I will be taking Abstract Algebra. Watching some Topology videos on these days.

I am a Math teacher but my Abstract Algebra class was completely different from the classes here. So different, that I don't even understand the questions from the GRE! I didn't know what a symmetric group was till I googled for that. I didn't know what Z/Z2 was. I used the notation Z/R, were R is a relation, or Z_2, [0], etc. Now I see that these things are sort of the same. I didn't know what a coset was, either. My class and the ones I see you take here are as similar as a hand and a foot!

My Topology class wa lousy (half a year without a teacher to begin with!), and it was only about Metric Spaces. I didn't even know what a topology was!

And I need to take Complex Analysis. Never took this one.

But I am already preparing the GRe to take it at some point (whenever I think that I am finally ready!). And I will be around this forum.

Re: f(x,y) = lxl^k*lyl^k/(x^2+y^2) is convergent, find min k

Posted: Sun Aug 24, 2014 2:35 pm
by DDswife
I have posted several messages lately with solutions to the exercises. If you find mistakes, or if you have better ways to solve them, let me know.