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I have the following task:

"State the maximal domain $D \subseteq \mathbb{R}^2$ of the following functions $f: D \rightarrow \mathbb{R}$

$\color{grey}{\textrm{and draw D in each case.}}$" (I'll do this bit).

  1. $f(x,y) = \dfrac{xy}{x-y}$
  2. $f(x,y) = x + \sqrt{x^2 + y^2}$
  3. $f(x,y) = \dfrac{\tan{(xy)}}{x^2 - y^2}$

Whilst I think that I understand what is being asked, I don't think I could articulate my solution with the appropriate rigour such that I could present it. Is my approach for determining the maximal domains valid? As such, have i accurately represented my maximal domains with the correct notation?


Progress so far:

Generally, I have tried to identify subdomains in my functions, and then wish to state the domain of each whole function as the intersection of those subdomains.

  1. Here I would argue that the maximal domain is the intersection of the numerator and denominator.

Thus, \begin{align*} D_1 &= \{x,y: x,y \in \mathbb{R}^2\} \ \cap \{ x,y: x,y \in \mathbb{R}^2 \cap (x \neq y)\} \\ D_1 &= \{ x,y: x,y \in \mathbb{R}^2 \cap (x \neq y)\} & \text{(Rule missing.)} \\ \end{align*}

  1. Here I would argue that the maximal domain is the intersection of the lone $x$ term and the term contained within the root function.

Thus,

\begin{align*} D_2 &= \{x: x \in \mathbb{R} \} \cap \{x,y: x^2 + y^2 \in \mathbb{R_{\geq 0}} \} \hspace{6.1cm} \\ D_2 &= \{x,y: (x \in \mathbb{R}) \cap (x^2 + y^2 \in \mathbb{R_{\geq 0}}) \} \\ \end{align*}

  1. Analogue to 1, except that another term needs to be introduced for the repeating asymptotes of the tangent function.

Thus,

\begin{align*} D_3 &= \{ x,y: x,y \neq n\cdot \frac{\pi}{2} \in \mathbb{R^2} \ (n \in \mathbb{N}) \} \ \cap \{ x,y: x^2 - y^2 \neq 0 \} \hspace{2.6cm} \\ D_3 &= \{ x,y: x,y \neq n\cdot \frac{\pi}{2} \in \mathbb{R^2} \ (n \in \mathbb{N}) \cap (x^2 - y^2 \neq 0) \} \\ \end{align*}

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    $\begingroup$ In $D_2$ the term in the sqaure root can be 0. So i think you should change $\mathbb{R}_{\gt 0}$ to $\mathbb{R}_{\ge 0}$. $\endgroup$ – Shervin Sorouri Jun 24 '18 at 12:05
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    $\begingroup$ And the union in the first equation for each of $D_i$ should change to an intersection. $\endgroup$ – Shervin Sorouri Jun 24 '18 at 12:07
  • $\begingroup$ @ShervinSorouri well spotted, I have made the changes. $\endgroup$ – Oscar Jun 24 '18 at 12:10
  • $\begingroup$ And you should change your notation. Meaning, when you say $D_i = \{x , y : ...$ it doesn't have any set theoretical meaning it should be changed to $D_i = \{(x, y) \in \mathbb{R}^2: ...$. $\endgroup$ – Shervin Sorouri Jun 24 '18 at 12:10
  • $\begingroup$ It would be better stated as "Find the natural domain of the expressions ..." $\endgroup$ – zhw. Jun 24 '18 at 15:16
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In 1. the only restricting factor is the division, as $(x,y) \to xy$ and $(x,y) \to x-y$ are defined everywhere. So the maximal domain is where the division is defined i.e. where $x-y \neq 0$.

So $D_1 = \{(x,y) \in \mathbb{R}^2: x \neq y\}$.

$D_2$ is just $\mathbb{R}^2$ because $x^2 + y^2 \ge 0$ always, so that the square root is always defined.

For the third we need both $\tan(xy)$ to be defined and $x^2 - y^2 \neq 0$ for the division to be defined. The latter is the case iff $x \neq y$ and $x \neq -y$, as $x^2 - y^2 = (x-y)(x+y)$. $\tan(t)$ is defined for all $t \notin \{\frac{\pi}{2} + k\pi: k \in \mathbb{Z}\}$. So the maximal natural domain for the third $f$ is the intersection of these two sets so $$D_3 = \{(x,y): y \neq x \land y \neq -x \land \forall k \in \mathbb{Z}: xy \neq \frac{\pi + 2k\pi}{2} \}$$

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