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There's something I can't quite wrap my head around, and I am not satisfied with my professor's "that's just defined that way" answer. So suppose we have:

$$f_1(x) = x - 5$$

$$f_2(x) = \frac{1}{x - 5}$$

$$f_3(x) = 1$$

By simple algebra, we can see that $f4(x) = f_1(x) \cdot f_2(x) = f_3(x)$

What I mean is the rational function we get from dividing $x - 5$ by $x - 5$ is the same as the constant function $y = 1$

Now I like to look at functions graphically to understand them. So we can clearly see that $f_1(5) = 1$ and $f_2(5)$ is not defined. So how come, when we multiply these two functions, we end up with a function (or a curve) that is defined at point $x = 5$. By logic, on $x = 5$, we should have $0/0$, which is undefined. But here it just takes the value $1$. How come?

Visuals to better understand whats bugging me: The three functions visualized

I would appreciate if your answer addresses this logically (or rather conceptually), not as a plain mathematical proof using algebra. I too, know, that algebraically speaking $(x - 5)/(x - 5)$ is just $1$, but thinking of functions visually rather than algebraically, this does not look convincing at all.

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  • $\begingroup$ Sorry for not using the standard math syntax and formatting for the forum, I'm quite new here, I hope someone edits out the awful looking formatting. $\endgroup$
    – Silidrone
    Mar 9 at 10:24
  • $\begingroup$ One idea to convince yourself of it, is to show that $$\lim_{x\to 5^+}f_1(x)\,f_2(x)=\lim_{x\to 5^-}f_1(x)\,f_2(x)=1$$ $\endgroup$
    – Davide
    Mar 9 at 10:31
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    $\begingroup$ Your computer visuals might be misleading. If you use a program like GeoGebra to plot $\frac{x}{x}$ it only shows the ‘hole’ of a discontinuity when you click on the line, otherwise it looks continuous. In one sense this is accurate as a mere point is absent, but by hand we would mark the absence with a fat little circle normally. $\endgroup$ Mar 9 at 14:27
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    $\begingroup$ This has nothing to do with division-algebras, so I removed the tag. Read the tag description to get an idea for the reason. $\endgroup$ Mar 13 at 4:12
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    $\begingroup$ Those screenshots look like they're from the Desmos app for Android, right? If you plot the function $f(x) = \frac{x-5}{x-5}$ and place your finger directly on the point $(5,1)$, your phone would say "(5,undefined)" because Desmos is defining that function as $f: \mathbb R\setminus\{5\}\to \mathbb R$. I don't think there's any continuous extension going on. Visually, Desmos would just graph that function as a straight line without making a visual like some kind of "puncture" or "hole". $\endgroup$ Mar 13 at 7:09

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In fact, $f_1(x)\cdot f_2(x)=\dfrac{x-5}{x-5}$ is not the same function as $f_3(x)=1$.

One has that $f_1(x)\cdot f_2(x)=1=f_3(x)$ for every $x\neq 5$, but in $x=5$ $f_1\cdot f_2$ is not defined as the denominator vanishes.

Nonetheless, $f_1\cdot f_2$ can be extended continuously to another function which is defined in $5$, namely the constant function $1$ (which would now be $f_3$).

This happens because the numerator has a zero of at least the order of the zero of the denominator.

For example, if we defined $f_4(x)=\dfrac{x-5}{(x-5)^2}$, you can check there is no way to continuously extend $f_4$ to $x=5$, because the denominator has a zero of order $2$ and the numerator has a zero of order $1$ (one gets a vertical asymptote).

However, $f_5(x)=\dfrac{(x-5)^2(x-3)}{(x-5)(x-2)}$ may be continuously extended to $x=5$.

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  • $\begingroup$ Can you elaborate what you mean by continously extended? $\endgroup$
    – Silidrone
    Mar 9 at 10:54
  • $\begingroup$ Take a continuous function $g: D\backslash\{x_0\}\to \mathbb{R}$. Then $h:D\to \mathbb{R}$ is a continuous extension of $g$ if $h(x)=g(x)$ for all $x\in D\backslash\{x_0\}$ and $h$ is continuous in all $D$. Intuitively, $h$ is a function which agrees with $g$ and fixes the problem that $g$ has at $x_0$ (doing so continuously, of course) $\endgroup$ Mar 9 at 11:02
  • $\begingroup$ I understand now, but I am confused as to why this is the default thing we do? To me, continous extension sounds like we are assuming things for sake of practicality, and we discard and not care altogether about the problematic point. This is fine to me, as a separate special procedure you can do, but why is it the default? For example, if someone asked you to sketch x - 5 / x - 5, the default should be to leave the x = 5 as undefined, not continously extend it to what it can be algebraically simplified. Does that make sense? $\endgroup$
    – Silidrone
    Mar 9 at 14:49
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    $\begingroup$ @Silidrone: You might be misreading the comment. That comment is stating the definition of continuous extension. It is not "the default thing we do". Instead, as this answer nicely explains, sometimes continuous extensions exist and sometimes they don't exist. $\endgroup$
    – Lee Mosher
    Mar 9 at 19:34
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    $\begingroup$ Well, assuming that a continuous extension exists, it is a tool you can use. Sometimes there is a reason to apply that tool in one's mathematical investigations, and sometimes there isn't. There is no "default" to either "do it" or to not "do it". I would not make any assertion such as "there really isn't a reason to do continuous extension", because I can indeed think of situations where there is a very good reason to do it. $\endgroup$
    – Lee Mosher
    Mar 10 at 16:00
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Another way of looking at it other than Julio Puerta's excellent answer is to realize that the domain of function is an essential part of the definition of a function and can not be ignored.

So the definitions of your functions have to be:

  1. $f_1:\mathbb R \to \mathbb R$ and $f_1(x)= x-5$.
  2. $f_2:\mathbb R\setminus\{5\}\to \mathbb R$ and $f_2(x)=\frac 1{x-5}$
  3. $f_3:\mathbb R \to \mathbb R$ and $f_3(x) = 1$

Now the definition of multiplying functions is not just figuring that $f\cdot g(x)=f(x)\cdot g(x)$ but it is also a case of figuring out the domain.

The proper definition should be If $f:A\to B$ and $g:C\to B$ then $f\cdot g$ will be defined as: $f\cdot g: A\cap C \to B$ and $f\cdot g(x) = f(x)\cdot g(x)$.

With this definition we get:

That the domain of $f_1\cdot f_2$ is $\mathbb R\cap (\mathbb R\setminus \{5\}) = \mathbb R\setminus \{5\}$.

Meanwhile, of course, $f_1(x)\cdot f_2(x)=\frac {x-5}{x-5}=1$ so......(That was your "simple algebra" and it was perfectly correct.)

If $f_1\cdot f_2=f_4$ the $f_4$ is: $f_4:\mathbb R\setminus\{5\}\to \mathbb R$ and $f_4(x)=1$.

And that is a different function than $f_3:\mathbb R \to \mathbb R$ and $f_3(x)=1$.

They are different functions simply because they are defined to have different domains. And that's the thing "simple algebra" gave us the formula for the mapping, but it didn't take calculations for domain into account at all. "Simple set intersection" can do that though.

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