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The wikipedia entry on elementary functions describes them to be "of a single variable (typically real or complex) that [are] defined as taking sums, products, and compositions of finitely many polynomial, rational, trigonometric, hyperbolic, and exponential functions, including possibly their inverse functions". Piecewise functions do not fit this description, and I believe these functions are continuous for the regions for which they are defined, as well as their derivatives for the regions along which those derivatives are defined.

Taking two functions which are composed of elementary functions and are unequal for the majority of the interval along which they can be defined for the independent variable, can these functions be equal (meaning they contain all the same points) over an interval with nonzero width, such as being defined and equal for (2,3) or (0, inf)? If this is impossible, why is it impossible?

Thanks!

Snaw
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FriendlyFriend
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    What do you mean by "elementary functions"? You could have $1$ and $\frac{|x|}{x}$. – user816709 Jan 25 '22 at 01:30
  • @user816709 those two functions don’t work. They only intersect at $x=1$. – Radial Arm Saw Jan 25 '22 at 01:38
  • By elementary function, I mean what the Wikipedia page on Elementary functions describes, which are continuous functions "of a single variable (typically real or complex) that is defined as taking sums, products, and compositions of finitely many polynomial, rational, trigonometric, hyperbolic, and exponential functions" and sometimes their inverses. I'm about to update my question to reflect this. – FriendlyFriend Jan 25 '22 at 01:44
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    @FriendlyFriend $;f(x)=x$ and $g(x)=\sqrt{x^2}$ are equal on $\mathbb R^+$, different on $\mathbb R^-$. – dxiv Jan 25 '22 at 02:00
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    @dxiv $g$ doesn’t have a continuous derivative, though. – Radial Arm Saw Jan 25 '22 at 02:05
  • @RadialArmSaw Missed that part, but then $,f(x)=x^3,$ and $,g(x)=\sqrt{x^6},$ will do. – dxiv Jan 25 '22 at 02:34
  • @RadialArmSaw That was not part of the question. The question only stated that elementary functions have a continuous derivative wherever that derivative is defined, which is correct, and the example of $f(x)=x$ and $g(x)=\sqrt{x^2}$ certainly satisfies that property. – Snaw Jan 25 '22 at 02:43
  • @Snaw oops- my bad. Sorry about that. – Radial Arm Saw Jan 25 '22 at 02:44

1 Answers1

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Extending the ideas already present in the comments, pick any $a<b$.

Let $f(x)=b-a$ and $g(x)=\sqrt{(x-a)^2}+\sqrt{(x-b)^2}$. Both $f(x)$ and $g(x)$ are elementary functions. $g(x)$ is just $|x-a|+|x-b|$ in disguise so that it would be clear that it is an elementary function according to all definitions.

$f(x)$ is identically equal to $g(x)$ for any $x\in[a,b]$.

Snaw
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  • Fantastic! Cool to see that it can cover any closed interval. Do you think there are other types of solutions which are not f_inverse(f(x)) type functions? – FriendlyFriend Jan 25 '22 at 02:50
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    @FriendlyFriend There's the example by dxiv of $f(x)=x^3$ and $g(x)=\sqrt{x^6}$ which are even differentiable. That's not exactly the $f^{-1}(f(x))$ type thing, but it's still close enough and certainly the same idea. You're asking if there's a different type of examples. That's a good question, I don't know. – Snaw Jan 25 '22 at 02:59
  • I did some investigating and I found some criteria which a function and its inverse likely need to fit in order to create a semi-equal imitating function.
    1. It must be be continuous along a region for which the derivative is discontinuous at some point in the neighborhood of a.
    2. For an originating function, h, for the inverse to the surrounding function, g, and for the surrounding function, f, there must be the property of g(f(x)) that g(f(h(a))) = h(a) (or lim(n->+/- infinity)g(f(h(n)))=h(n).)
    3. The derivative must be nonzero in the neighborhood of a,

    (By the inverse function thm.)

     – FriendlyFriend Jan 27 '22 at 23:13