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The following really gets confusing:

As a simple example:

Is the function

  • $f(x) = 1$ the same function as $f(x) = \frac{x - 1}{x-1}$?
  • $f(x) = \frac{(x-1)(x+1)}{x-1}$ the same function as $f(x)= x + 1$?

Basically: does resolving and simplifying a function expression alter the function 'identity/essence'? I think it does because the domain changes ($1$ is put into the domain) and the only thing that matters are the exact values in the domain set and the range set, an algebraic expression is just a description.

So is it correct to say that one expression is an approximation of another function expressions? But that they are completely different functions, they just overlap?

Is there a bigger philosophy to be found somewhere?

Aiden Chow
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MuseOfMusic
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3 Answers3

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I think that this depends on what you're working with. Sometimes if you get an algebraic expression and it makes sense to simplify it and then you have the function that you will continue working with. However, if we are being strictly formal, you are right. The functions

$$x \mapsto 1 \text{ and } x \mapsto \frac{x-1}{x-1}$$ are different because their domains are different. The first function has $\mathbb{R}$ as its domain, whereas the second one has $\mathbb{R} \setminus \{1 \}$. Formally a function is a triplet $(A, G, B)$, where $A$ and $B$ are the domain and codomain of the function and $G$ is what we call its graph, by which we do not refer to its graphical representation as a 2D plot in the $(x,y)$-plane, but rather its set of ordered pairs of arguments and function values. Formal function equality requires the domain, codomain and graphs to be equal for two functions to be equal.

In elementary calculus however, there are very few situations where this actually has a significance. When you study complex analysis, these kinds of questions become more cogent and the answer is also more interesting.

Thusle Gadelankz
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By definition, it is necessary that for two functions to be the same, both functions must share the same domain as well as codomain. Your examples are not the same since $1$ is in the domain of one, but not the other. Although, you might be able to say they are "the same thing" in the sense that you are able take the restriction of $f$ to $\mathbb R - \{1\}$ so that $f|_{\mathbb R - \{1\}} = f_1$ where $f_{1}$ represents the function undefined at $x=1$.

Derek Luna
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$f(x) = 1$ and $g(x) = \frac{x - 1}{x - 1}$ are NOT same functions at all. They do not share the same domain set. The function $f$ can take values from all of $\mathbb{R}$ , whereas $g$ takes values only from $\mathbb{R} \setminus \{1\}$ .

When we we say that two functions are equal (or, identical), we mean that they take values from the same set, and they take two same value at each point in the domain set. So, in your question, if it's specified that $\text{domain}(f) = \mathbb{R}$ , then $f \not\equiv g$ .

However, in order to make $f \equiv g'$ forcefully, we may define $g' : \mathbb{R} \to \{1\}$ like : $$g'(x) = \begin{cases} g(x)&\quad\quad\text{if}~ x \neq 1\\ 1 &\quad\quad\text{if}~ x = 1\\ \end{cases}$$ In that case, we can surely say that $f \equiv g'$ .

JRC
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