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It is said for two functions $f,g$ to be equal they must have same domain and codomain and for each $x\in X$, $f(x)=g(x)$.

But shouldn't functions such as $f:\Bbb R \to \Bbb C$ where $f(x)=x^2$ and $g:\Bbb R \to \Bbb R$ where $g(x)=x^2$ still be considered equal functions for example? Even if codomain is different.

Xander Henderson
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Rivaldo
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  • Rivaldo $f:R --> C$ is wrong, if it was the case than $f(x) = u(x)+iv(x)$ – James Apr 24 '18 at 12:36
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    Equality of functions is a definition. There is no "should." – Paul Apr 24 '18 at 12:36
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    @JimmySabater This is fine, in the example given just take $u(x) = x^2$ and $v(x) = 0$. – JackR Apr 24 '18 at 12:37
  • @JimmySabater but isn't R subset of C? – Rivaldo Apr 24 '18 at 12:39
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    Functions are defined as subsets of the Cartesian product of the domain and the co-domain (with certain properties). In your case $f$ is a subset of $\mathbb{R}\times\mathbb{C}$ and $g$ is a subset of $\mathbb{R}\times\mathbb{R}$. And those subsets are the same (well, if $\mathbb{R}$ is seen as a subset of $\mathbb{C}$). In general, from the definition of function you see that the function doesn't really depend on the co-domain. But there are contexts in which it is useful to consider the co-domain, part of what a function is. This happens when the sets are the subject matter of the study. –  Apr 24 '18 at 12:39
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    On the other end of things, you could bite the bullet and do away with the horrible ambiguity that is the codomain. Let $i:\mathbb{R}\to\mathbb{C}$ be the function $a\mapsto a+0i$. Then $f=i\circ g$. – J. Moeller Apr 24 '18 at 12:39
  • im confused now my friends this makes no sense to me – James Apr 24 '18 at 12:42
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    @Paul Actually, definitions should serve their usage. Moreover, the particular case of equality of functions does have two different versions, depending on the context, one in which his two functions are the same and one in which they are not. –  Apr 24 '18 at 12:56

2 Answers2

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It might be worth noting that from a set theoretic standpoint, the copy of the reals contained in the complex numbers is not the same set as the reals on their own. They are isomorphic, but distinct as sets.

We usually construct $\mathbb{C}$ as an ordered pair $(x,y) \in \mathbb{R} \times \mathbb{R}$, and define multiplication on these pairs. Here $x$ is the real part of the complex number and $y$ is the imaginary part. We have a natural isometric embedding of $\mathbb{R}$ into $\mathbb{C}$ by $x \mapsto (x,0)$. Thus, if we're talking about the real number "$2$" in $\mathbb{C}$, we're really talking about the ordered pair $(2,0)$.

To bring it back, the two functions you described: $$f:\mathbb{R} \rightarrow \mathbb{R}, \quad f(x) = x^2$$ $$g:\mathbb{R} \rightarrow \mathbb{C}, \quad g(x) = x^2$$ Obviously $f$ and $g$ give the "same information" in some sense, but the objects in the image are set theoretically distinct, even if we interact with them in exactly the same way.

Joe
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Some authors define function equality differently: if the domains are the same, and the rules of association are the same, the functions are the same. If $f(x)=x^2$, but you also write $f:\mathbb{R}\to\mathbb{C}$, you're not really gaining a whole lot. Also recall what a function actually is, at a fundamental mathematical level: it's a set where the elements look like ordered pairs $(x,f(x))$. If the sets corresponding to two functions are the same, the functions are the same.

Adrian Keister
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  • I've been trying to understand analytic continuation of the Riemann zeta function. Something similar comes up with Dirichlet eta alternating function and the Riemann zeta sum(1/n^s). Since these two have different domains (eta: Re(s)>0 and zeta Re(s)>1) can these two functions be said to be equivalent? For instance, Wolfram/Mathworld https://mathworld.wolfram.com/RiemannHypothesis.html gives the "Euler continuation" formula (eqn. 1) and than says they are equilvalent. Thanks for any comments. – zeynel Feb 07 '24 at 15:41
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    If two functions have the same rule of association, but one has a domain contained entirely within the domain of the second, then the second function is said to be a continuation of the first. The functions would not be said to be "equivalent" - not sure that term even applies to functions. The two functions would be equal on the first functions's domain. – Adrian Keister Feb 07 '24 at 19:59
  • Thanks. If I understand correctly, you describe a situation where one domain is a subset of the other. So if $f1$ has domain $D1$ and $f2$ has domain $D2$ then $D2$ is the continuation of $D1$ if $D1 \subseteq D2$. This makes sense to me. But usually, I read that "analytic continuation" only requires that $D1$ and $D2$ intersect at some region. So if $D1 \cap D2$ then it is said that $D2$ is the "continuation" of $D1$. I don't understand the mathematical justification that makes $D2$ a continuation of $D1$ just because they intersect. Even if $D1 \cap D2$, $D1$ still has no value at $D2$. – zeynel Feb 08 '24 at 08:20
  • I read about the intersection issue that I mentioned here https://www1.phys.vt.edu/~ersharpe/spec-fn/ancont.pdf – zeynel Feb 08 '24 at 08:22
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    Analytic continuation has a nice MathWorld article here: https://mathworld.wolfram.com/AnalyticContinuation.html. It's similar to what you linked to. The whole point of a continuation is that you have one function $f_1$ that is NOT defined on $D_2,$ but you would like it to be. So if you can find a function $f_2$ defined on $D_2$ such that $f_1=f_2$ on $D_1\cap D_2,$ and everything in sight is analytic, then you've got a unique extension or continuation of $f_1,$ like you originally wanted. – Adrian Keister Feb 09 '24 at 16:48