# Why do reciprocal functions work for undefined values?

I have a certain problem: When you have a function $f(x)$ and it can have some undefined values, the reciprocal function does some things I don't understand. For example:

i) $$f(x)=\tan(x)$$

or another one: ii) $$f(x)=\frac {5}{x-3}$$

Clearly you can see that for ii) $x=3$ will be undefined and for i) $\pi/2, 3\pi/2,\ldots$

When I now work with the reciprocal functions my problem appears :

i) $$(f(x))^{-1}=\frac{1}{\tan x}=\cot x$$

ii) $$(f(x))^{-1}=\frac{1}{\frac {5}{x-3}}$$

Coming up to my questions:

i) The function of $\cot x$ is $0$ at those $x$-values where $\tan x$ is undefined. Why is $1/\text{undefined}=0$ and not undefined too?

ii) The same here; $f(x)$ is undefined at $x=3$, but depending on which graphic calculator you use it defines $$(f(3))^{-1}=0$$ or as a discontinuity.

*What happens at those "magical" points when $f(x)=\text{undefined}$ and $(f(x))^{-1}= 1/\text{undefined?}$

What I got so far:

"If y = f (x) = 0 for some value of x, then 1/f (x) is undefined. There is a jump or discontinuity in its graph for this value of x. This means that, as f (x) gets close to 0, 1/f (x) will become very large in value. Equally, if there is a jump or discontinuity in the graph of y = f (x) for some value of x, then y = 1/f (x) = 0 for that value of x." That's a definition I have from "Jenny Olive: Math a student's survival guide"

and of course : Why is cot(x)=0 instead of undefined

and my own thesis:

Since you can convert $$y= \frac {1}{\frac {5}{x-3}}=1\cdot\frac{x-3}{5}$$ the value for x=3 is defined

Same for $\cot(x) = \cos(x) / \sin(x)$ , where $x= \pi/2$ is defined.

Is my thesis working or am I hurting mathematics at this point?

• Your hypothesis is absolutely correct (at least as it seems to me). In fact, I like your question. One way of seeing the reciprocal functions is: as completely different functions. 1/f(x) is just a notation. Whereas, it becomes a completely different function as you have written in the last part of your question. So, the domain and all for 1/f(x) will depend on what expression we will get afterwards. – Aniruddha Deshmukh Jul 5 '17 at 6:28
• $\dfrac{1}{f(x)}$ is not discontinuous where $f(x)=0$. Simply the roots of $f(x)=0$ are not in the domain and continuity is defined only for values which are elements of the domain. For example $\dfrac{1}{x}$ is continuous on all its domain $\mathbb{R}-\{0\}$ – Raffaele Jul 5 '17 at 10:07

If you want to be technical, you can think of only defining the reciprocal on the domain of the original function, which means that at any point $f(x)$ is undefined, $(f(x))^{-1}$ is undefined... but what's more commonly done is to try to define a function everywhere that function makes sense, and in the two functions you have mentioned, since they are continuous functions around these "bad points" and the limits exist, it seems reasonable to extend the definition of $(f(x))^{-1}$ to include these "bad points," setting the value to be the limit at the previously undefined point.

The first idea makes sense if you want to define the reciprocal as "the function $g(x)$ such that $g(x)f(x)=1$ for all $x$ such that $f(x)$ is defined." The latter idea makes more sense if you want to think of the reciprocal in its own right. For an example of the latter, there's no reason to hold back from defining $\cot(\pi/2)=\cos(\pi/2)/\sin(\pi/2)=0,$ since the middle expression is perfectly well-defined there, and $\cot(x)$ agrees with this function everywhere in a neighborhood of $\pi/2.$

The reciprocal of $5/(x-3)$ is $(x-3)/5$ except at $x=3$ where it is undefined.
So you look near $x=3$, and you find it approaches $0$ both when $x<3$ and when $x>3$.
That is called a 'removable discontinuity'.
You might as well define $g(x)$ to be $(x-3)/5$ except at $x=3$, and $0$ at $x=3$. But that is just $g(x)=(x-3)/5$ for all $x$.

If $f(x)$ is undefined for some specific value $x^*$, so is $(f(x))^{-1}$. However, if $x^*$ is a value for which we have $\lim_{x\to x^*} f(x)=\pm\infty$ (undefined in your terms), then one usually uses this other (but not actually correct) reciprocal function:

$$g(x):=\begin{cases} (f(x))^{-1} &\text{for x\ne x^*}\\ 0 &\text{for x=x^*} \end{cases}.$$

This is because this function $g(x)$ equals $(f(x))^{-1}$ whereever it is defined, and otherwise extends it continuously. It is in some sense a natural extension of $(f(x))^{-1}$. It is easier to deal with this $g(x)$ than always having to keep in mind to exclude $x^*$.

The premise of the question isn't true. The reciprocal of an undefined expression is still an undefined expression.

$$(f(x))^{-1}=\frac{1}{\tan x}=\cot x$$

...

The function of $\cot x$ is $0$ at those $x$-values where $\tan x$ is undefined. Why is $1/\text{undefined}=0$ and not undefined too?

The function $\cot x$ is $\frac{\cos x}{\sin x}$, not $\frac{1}{\tan x}$. The expression $\frac{1}{\tan x}$ is undefined when $\tan x$ is undefined, but this doesn't affect $\cot x$, because $\cot x$ and $\frac{1}{\tan x}$ are different.

$$f(x)=\frac {5}{x-3}$$

$$(f(x))^{-1}=\frac{1}{\frac {5}{x-3}}$$

The same here; $f(x)$ is undefined at $x=3$, but depending on which graphic calculator you use it defines $$(f(3))^{-1}=0$$ or as a discontinuity.

Well, regardless of what your graphic calculator says, $(f(3))^{-1}$ is undefined, because $f(3)$ is undefined.

and my own thesis:

Since you can convert $$y= \frac {1}{\frac {5}{x-3}}=1\cdot\frac{x-3}{5}$$ the value for x=3 is defined

Same for $\cot(x) = \cos(x) / \sin(x)$ , where $x= \pi/2$ is defined.

Is my thesis working or am I hurting mathematics at this point?

It's not correct to simplify $\frac {1}{\frac {5}{x-3}}$ to $\frac{x-3}{5}$, because the denominator of $\frac {1}{\frac {5}{x-3}}$ is undefined when $x = 3$.

It may be worth noting that "undefined" isn't a mathematical value. The word "undefined" is an adjective, and it means "not having a definition". This is why mathematicians always write "is undefined", never "= undefined".

Some definitions are conditional, meaning that they only "work" under certain conditions. For example, the definition of division over the real numbers is:

If $x$ and $y$ are real numbers, and $y \ne 0$, then $\frac{x}{y}$ is that real number such that $\frac{x}{y} \cdot y = x$.

This definition tells us what, say, $\frac23$ is, but it does not give a definition for $\frac10$, so the expression $\frac10$ still does not have a definition.

The definition also doesn't give a result when either $x$ or $y$ is not defined as being a real number. That's why a fraction is undefined whenever its numerator or denominator is undefined.

• The fact that "undefined" is an adjective is also why I cringe whenever I think about how JavaScript has a value called undefined. The name undefined is, of course, defined. It's possible for a name in JavaScript to be undefined, or for it to have the value undefined, but not both at the same time! – Tanner Swett Jul 5 '17 at 15:32

In grade school mathematics, functions like: $$f(x) = \frac{5}{x-3}$$ . . . are indeed said to be "$\text{undefined}$" at points like $3$. This is because these functions are implicitly over the real numbers.

In my opinion, it's usually more intuitive to instead consider real-valued functions over the projective reals (i.e., the "one-point compactification" of the reals). This is basically the same real numbers we all know and love, plus the number "$\infty$". You treat $\infty$ (nearly) like any other number, with some common-sense rules:\begin{align} c \mp \infty &= \infty, \text{ all } c ~\neq \infty\\ c \cdot \infty &= \infty, \text{ all } c ~\neq ~0\\ c ~/~ \infty &= ~0, ~\text{ all } c ~\neq \infty\\ c ~/~ 0 &= \infty, \text{ all } c ~\neq ~0 \end{align}

This makes everything cleaner and lets you define a broader class of functions. Long story short: $$f(3) = \frac{5}{3-3} = \frac{5}{0} = \infty$$ $f(3)$ is no longer undefined. It's defined as $\infty$. So now when you want to calculate the reciprocal, it works out either way:\begin{align} (f(3))^{-1} &= \frac{1}{(f(3))} = \frac{1}{(\frac{5}{3-3})} = \frac{1}{(\frac{5}{0})} = \frac{1}{(\infty)} = 0\\ (f(3))^{-1} &= \frac{3 - 3}{5} = \frac{0}{5} = 0 \end{align} Easy!

This sort of analysis works for many "$\text{undefined}$" functions. E.g.:$$\log(0) = \tan(\pi/2) = 1/0 = -8/0 = \infty$$

The projective reals can't fix everything. For example, consider: $$f(x,y) = \frac{x}{y}$$ We're still stuck for $x=0, y=0$ (we can't apply any of the rules above). So $f(0,0)$ must still be $\text{undefined}$. However, observe that $(f(x,y))^{-1}=\frac{y}{x}$ is also $\text{undefined}$ at $x=0, y=0$. So your intuition about $1/\text{undefined}=\text{undefined}$ is satisfied. I haven't proved this will always be true for the projective reals, but it seems to be at least more common, anyway.