This is so obvious, that $\frac{1}{x}$ is elementary function. But how this can be proven? I've been searching for information, and have found a whole list of elementary functions, and $\frac{1}{x}$ is one of them. It is like an axiom, which is always true. There are many ways to prove that different compound functions are elementary, but $\frac{1}{x}$ is always considered to be elementary.

We were give a list of properties, based on which we should prove that $\frac{1}{x}$ is also elementary. I'm quite confused, because I have no idea how to start proving. Would appreciate any kind of help.

Base Cases.

  • Identity function, $id(x) = x$ is in EF.
  • Any constant function is in EF.
  • The sine function $sin(x)$ is in EF

Constructor Cases. If $f,g \in EF$, then so are

  • $f+g$, $fg$, $2^g$
  • The inverse function $f^{-1}$;
  • The composition $f \circ g$.

Original: Given properties

  • $\begingroup$ The question is a bit confusing, the title and first paragraph seems totally unrelated to the question itself - deriving that given function based on specific properties given. I would suggest to perhaps get rid of the first paragraph at all, and also add some your thoughts on the problem, people generally dont like when someone posts homework problems here without showing any thoughts on the solution... In other words, what have you tried? $\endgroup$ – Sil Oct 29 '17 at 16:06
  • $\begingroup$ Well you probably need to think a bit as follows. First, is $\frac 1x$ a Base Case? Can you say it is not? Which of the Constructor Cases look unpromising? What is left to try? Give it a go, and report back what you find. $\endgroup$ – Mark Bennet Oct 29 '17 at 16:18
  • $\begingroup$ Where does the definition of EF come from? Its different and more difficult to apply than the definition usually given for example mathworld.wolfram.com/ElementaryFunction.html and WIkipedia. $\endgroup$ – Bernard Massé Oct 29 '17 at 16:31
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    $\begingroup$ The second constructor case ("inverse function") seems hard for me to understand - does it only apply to invertible functions? If not, exactly which function does it give back when we apply it to a noninvertible function? For example, if we apply it to $\sin(x)$, what function do we get back? When we apply it to $x^2$? $\endgroup$ – Carl Mummert Oct 29 '17 at 21:10

Recall that $\cot x=\tan(\pi/2-x)$, so $$ \frac{1}{x}=\cot\arctan x $$ More precisely, for $x>0$ you have $$ \arctan x+\arctan\frac{1}{x}=\frac{\pi}{2} $$ so $$ \frac{1}{x}=\tan\left(\frac{\pi}{2}-\arctan x\right)=\cot\arctan x $$ and, for $x<0$, $$ \arctan x+\arctan\frac{1}{x}=-\frac{\pi}{2} $$ so $$ \frac{1}{x}=\tan\left(-\frac{\pi}{2}-\arctan x\right)= -\cot(-\arctan x)=\cot\arctan x $$ The tangent and the cotangent are elementary, because so are the sine and the cosine.

The exponential function is elementary, because $e^x=2^{x/\!\log 2}$. Therefore also the natural logarithm is elementary. Thus $$ \frac{1}{\cos^2x}=\exp(-\log(\cos^2x)) $$ is elementary and $$ \tan x=\sin x\cos x\frac{1}{\cos^2x} $$ Similarly for the cotangent.

  • $\begingroup$ @CarlMummert Now it seems to work, doesn't it? $\endgroup$ – egreg Oct 29 '17 at 21:26
  • $\begingroup$ I think so. Is there are reason that the natural exponentiation is required instead of just $2^x$ and $\log_2(x)$? $\endgroup$ – Carl Mummert Oct 29 '17 at 21:30
  • $\begingroup$ @CarlMummert Just because I prefer them notation-wise. ;-) Anyway, I agree with you that the problem is not well-posed. $\endgroup$ – egreg Oct 29 '17 at 21:30
  • $\begingroup$ @CarlMummert Well, I use the arctangent… $\endgroup$ – egreg Oct 29 '17 at 21:32
  • $\begingroup$ I see. I'd need to think about that (but I do suspect that it would still work regardless of the branch of the inverse that you get back). $\endgroup$ – Carl Mummert Oct 29 '17 at 21:33

Since $2^x$ is elementary and elementary functions are closed under inverses, $\log_2(x)$ is elementary. Then: $$\frac{1}{x} = x^{-1} = 2^{\log_2(x^{-1})} = 2^{(-1)\log_2(x)}.$$

Edit: As noted in the comments, this expression is only defined for $x>0$. Following egreg's suggestion:

$$\frac{1}{x} = x(x^{-2}) = x(2^{\log_2(x^{-2})}) = x(2^{(-1)\log_2(x^2)}).$$

This is defined for all $x\neq 0$, since then $x^2>0$.

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    $\begingroup$ This just holds for $x>0$. $\endgroup$ – egreg Oct 29 '17 at 16:20
  • $\begingroup$ However, I think this can be easily fixed $\endgroup$ – egreg Oct 29 '17 at 21:37
  • $\begingroup$ @egreg , can you please clarify, how this can be proven completely using this example? What should be added in order to make this proof favorable for x < 0 $\endgroup$ – Nk Dibz Oct 30 '17 at 8:25
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    $\begingroup$ @NkDibz $x^{-1}=x\cdot x^{-2}$ $\endgroup$ – egreg Oct 30 '17 at 8:41

Wikipedia says that

In mathematics, an elementary function is a function of one variable which is the composition of a finite number of arithmetic operations (+ – × ÷), exponentials, logarithms, constants, and solutions of algebraic equations (a generalization of nth roots).

Clearly, going by the above definition, $\frac{1}{x}$ is elementary.

That pretty much answers your question, right?

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    $\begingroup$ I am not sure this answers the question, the OP gave list of properties from which it should be proven, for example division is not one of them. $\endgroup$ – Sil Oct 29 '17 at 15:55
  • $\begingroup$ I agree, that by definition it should be elementary. But the task is to prove that not by using definition, but by using given properties, i.e. f(x)=x, f(x)=c and f(x)=sin(x) are elemntary functions. If the defition was enough, I wouldn't ask for help. This is the trick. But anyway appreciate your answer $\endgroup$ – Nk Dibz Oct 29 '17 at 15:57

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