Given that $f(x,y)=[x^2+y^2]\frac{x^2y^2}{x^4+y^4}$ and $g(x,y)=[y^2]\frac{xy}{x^2+y^2}$ then find the number of elements in the set E.

Question

Question : For $t\in \mathbb R$, let $[t]$ denote the greatest integer less than or equal to t. Let $D=\left\{(x,y)\in \mathbb R^2:x^2+y^2<4\right\}$. Let $f:D\to\mathbb R$ and $g:D\to \mathbb R$ be defined by $f(0,0)=g(0,0)=0$ and $f(x,y)=[x^2+y^2]\frac{x^2y^2}{x^4+y^4}$ ang $g(x,y)=[y^2]\frac{xy}{x^2+y^2}$ for $(x,y)\neq(0,0)$. Let E be the set of points of D at which both f and g are discontinuous. Find the number of elements in the set E.

Clearly the elements in the set D will look like $(\frac{1}{n},\frac{1}{n} )\forall n\in$$\mathbb R$-{$0$}. To check the continuity at "origin" normally I used to do it by choosing a path like $y=mx$ here I tried the same to examine something about $f$ and $g$. So I got $\lim\limits_{x\to 0}[x^2(m^2+1)]\frac{m^2}{m^4+1}$ and since $x^2$ is always positive so this limit will be $0$.Similarly I can check for $g(x,y)$. What should I do next? I know that the continuity is a property on the members of the domain and if a function is not continuous at a point then it will be termed as "discontinuous" at that point. What if to go more deeper I do this to look for the behaviour of $\lim\limits_{(x,y)\to(\frac{1}{n},\frac{1}{n})}[x^2+y^2]\frac{x^2y^2}{x^4+y^4}=\frac{1}{2}[\frac{2}{n^2}]$ so here I can choose $n$ specifically so that its value will come out to be non-zero so for $n=1,-1$, I got values for $(x,y)=(1,1),(1,-1),(-1,1),(-1,-1)$ i.e. $4$ elements in $E$ for the time being(same for $g(x,y)$).The total elements are $18$(The Answer).

Thanks.

Answer 1

Functions $\frac{x^2y^2}{x^4 + y^4}$ and $\frac{xy}{x^2 + y^2}$ are continuous everywhere except for in $(0, 0)$. However, inside disc $x^2 + y^2 < 1$ both $f$ and $g$ are constant $0$ (because both $[x^2 + y^2] = 0$ and $[y^2] = 0$ in that region), so that discontinuity is irrelevant.

Function $[x^2 + y^2]$ is continuous everywhere, except when $x^2+y^2 \in \{1, 2, 3, ...\}$. Similarly, $[y^2]$ is continuous everywhere, except when $y^2 \in \{1, 2, 3, ...\}$.

$f$ is discontinuous at points where $x^2 + y^2 \in \{1, 2, 3\}$, because in all such points $[x^2 + y^2]$ is discontinuous and $\frac{x^2y^2}{x^4 + y^4}$ is non-zero and continuous. The set of discontinuous points of $f$ is drawn as three concentric green circles in picture below.

Similarly $g$ is discontinuous, when $y^2 \in \{1, 2, 3\}$. The set of discontinuous points of $g$ is drawn as six parallel blue lines in picture below.

Counting all their intersections (red points), we get 18.

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Some comments on your attempt of solution

1.

To check the continuity at "origin" normally I used to do it by choosing a path like $y=mx$

This is incorrect. For some function you can prove that they're discontinuous this way, but you cannot use this method to prove that a function is continuous. For an example, see e.g. this question: Does there exist a function $f:\mathbb{R}^2\to\mathbb{R}$ that is discontinuous at a point but all of whose directional derivatives exist there? .

2.

Clearly the elements in the set D will look like $(\frac{1}{n},\frac{1}{n} )\forall n\in$$\mathbb R$-{$0$}.

This is obviously false, for example $(0.1, 0.3)$ is an element of $\mathrm D$.