is $x^2 + y^2 = 1$ a function?

Question

I understand that implicit functions are functions where $y$ isn't isolated and isn't immediately expressed in terms of $x$. But I wonder if $x^2 + y^2 = 1$ can even be considered a function since it violates the vertical line test since its graph is a circle. Is the definition of an implicit function slightly different from an explicit function where it can allow this?

Additional question, since some implicit functions can be expressed as explicit functions, can they still be considered the same function?

Answer 1

No, $x^2+y^2=1$ is not a function, it's an equation, and its solution set is a curve. Locally on this curve we have a single $y$ for each $x.$ Locally this equation therefore defines a function $x\mapsto y.$

There is also another function, $F(x,y)=x^2+y^2-1.$ When learning about implicit functions it's common to let $F:\mathbb R^2\to \mathbb R$ be some function and consider the equation $F(x,y)=0$ which locally around a point $P$ on its solution set defines a function $y=f_P(x).$

Answer 2

No, the full circle isn't a graph of a function.

There is no difference between "implicit" and "explicit" functions, function is a function. It would perhaps be better to say implicitly defined function.

In general, it is not enough to just write an equation like $x^2+y^2 = 1$ to implicitly define function, because there are infinitely many functions $y(x)$ on $[-1,1]$ that satisfy that equation, two of which are continous on $[-1,1]$, namely $y(x) = \pm\sqrt{1-x^2}$ (and their graphs correspond to upper and lower half-circles).

To uniquely implicitly define function, what we usually use is known as Implicit function theorem. You can take a look at First example section.

You can read precisely what the theorem states, but let me just explain how it works in your example. Let $f(x,y) = x^2+y^2-1$ and pick a point $(x_0,y_0)$ on the unit circle, that is, a point such that $f(x_0,y_0) = 0$. We want to define unique function $y$ such that $y(x_0) = y_0$ and $f(x,y(x)) = 0$, i.e. $x^2 + y(x)^2 = 1$ which is continuously differentiable. There is a condition for which points we can do it and that is $$\frac{\partial f}{\partial y}(x_0,y_0)\neq 0$$ (see this section). In our case, $\partial_y f(x,y) = 2y$, so by the Implicit function theorem, there is such a function whenever $y_0 \neq 0$. Explicitly, if $y_0 > 0$, then $y(x) = \sqrt{1-x^2}$ and if $y_0 < 0$, then $y(x) = -\sqrt{1-x^2}$.

Finally, what's the significance of the condition $y_0 \neq 0$? Well, take a look at the unit circle and the two points for which $y_0 = 0$, i.e. $(\pm 1,0)$. If you look at any arc containing any of those points, it can't be a graph of a function by vertical line test.

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To summarize, when we talk about implicit functions, it is often in the context of Implicit function theorem and to understand implicit functions, you should study that theorem.

Answer 3

It's not the graph of a real one variable function, for example, if we take $x=0$, the dependent variable $y=\pm1$ so this is uncompatible with the definition of function.

Anyway, if you write the equation as

$$y^2=1-x^2\Longrightarrow y=\pm \sqrt{1-x^2}$$

and you choose the expression with the plus or with the minus, then, both of them represent the explicit formula of a real one variable function.

On the other hand, if you consider the real two variable function $f(x,y)=x^2+y^2$ that equation represent a level set of $f$, in particular, the level set obtained "cutting" the 3d graph of $f$ (is a paraboloid) with the plane $z=1$.