0
$\begingroup$

$x,y \in \mathbb{R}$ and $F: D\subset \mathbb{R}^2 \to \mathbb{R}$ where

$$ F =\sqrt{x^2 - y^2} + \arccos\Big(\frac{x}{y}\Big) = 0 \tag{1}$$

Noticing that the square-root and inverse cosine functions only return positive or $0$ values, the only way for this statement to work is if both output $0$. Therefore $(1)$ defines an implicit function $y = g(x) = x$ excluding the point $(0,0)$. Unfortunately, you cannot conclude that $(1)$ implies an implicit function from the implicit function theorem. Consider the point $(2,2)$. We apply the implicit function theorem around this point. The partial derivatives $F_x$ and $F_y$ evaluated at $(2,2)$ or any point $(x, y=x)$ where $F(x,y) = 0$ is undefined and so we cannot use the theorem, which only gives a sufficient condition for the existence of an implicit function. So when it doesn't apply, who knows if a function exists. In this case, we were able to find one.

If the implicit function theorem holds for a function $F$, guaranteeing the existence of some function, we can apply implicit differentiation to the equation. If the implicit function theorem doesn't hold, but we know an implicit function exists, you'd think (or maybe hope) 'well the function exists so I might as well try implicit differentiation.' In $(1)$, we know that $g'(x) = 1$. Yet, implicit differentiation gives you junk (you get a square-root in the denominator). How can an implicit function $y(x)$ exist but implicit differentiation not work? Differentiation is just a operation. If you have an object to work on, shouldn't differentiation give something reasonable when the object is reasonable? $(1)$ must be much different than $y = x$ excluding the origin

$\endgroup$
  • $\begingroup$ So, $F$ is a number? And what is $D$? An arbitrary set? $\endgroup$ – amsmath Aug 21 '18 at 15:09
  • $\begingroup$ $F$ I try to denote as a function set equal to $0$. $D$ is the domain of the function where $y \neq 0$, $|x| \leq |y|$, and $x^2 - y^2 \geq 0$ $\endgroup$ – DWade64 Aug 21 '18 at 15:11
  • 1
    $\begingroup$ Again, $F_x$ and $F_y$ don't exist. How would you invoke implicit differentiation if the conditions are not met at all? $\endgroup$ – amsmath Aug 21 '18 at 16:00
  • 1
    $\begingroup$ "It can't be done any other way." That's correct. Your question becomes more interesting though if you replace $x^2-y^2$ under the square root by $y^2-x^2$ and let $D := \{(x,y) : |x|\le|y|,\,y\neq 0\}$. Then still the implicit function theorem cannot be invoked because your manifold $\{(x,\pm x)\}$ is the boundary of $D$. However, $F_x$ and $F_y$ exist in the interior of $D$ and $-F_x/F_y$ tends to the "correct" value $1$ on the boundary. $\endgroup$ – amsmath Aug 21 '18 at 16:17
  • 1
    $\begingroup$ You're welcome! $\endgroup$ – amsmath Aug 21 '18 at 16:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.