What makes elementary functions elementary?

Is there a mathematical reason (or possibly a historical one) that the “elementary” functions are what they are? As I’m learning calculus, I seem to focus most of my attention on trigonometric, logarithmic, exponential, and nth roots, and solving problems that have solutions which are elementary functions. I’ve been curious why these functions are called elementary, as opposed to some other functions that turn up rather naturally in mathematics. What is the reason that these functions take up most of our attention, and is there a reason that some additional functions are not included amongst the elementary functions? In other words, what property or properties do these functions possess that separates them from non-elementary functions (if there is one)?

Answer

As Sivaram Ambikasaran mentioned the description on Wikipedia is fine.

I believe the class of elementary functions, E, is commonly thought of as a construction of the form

  1. All polynomials are in E

  2. The exponential and the logarithm function is in E

  3. The sine and cosine functions are in E.

  4. E is closed under addition, subtraction, multiplication, division and composition (finitely many operations of these).

  5. E is the smallest set with the properties 1-4.

This applies to both real or complex valued functions.


Edit 1:

Some examples of functions that are elementary

  • f(x)=1-x^2
  • s(x) = \sqrt{x} (see addendum below)
  • g(x)=\arctan x (see addendum below)
  • U(x)=\sin\frac{1}{\log(1+x^2)}
  • A(x)=|x| = \sqrt{x^2}
  • _2F_1(1,1,2,x) = \log(1-x) (a Gauss Hypergeometric notation that ends up in a elementary function)

Some examples of functions that are not elementary

  • The Sine integral \operatorname{si}(x)=\int_0^x\frac{\sin t}{t}dt
  • The Error function \operatorname{erf}(x)=\frac{2}{\sqrt{\pi}}\int_0^x e^{-t^2}dt
  • The Cantor function
  • The characteristic function of an interval.

Edit 2:

The nomenclature elementary function is of course used since the functions that are elementary can be deduced using finitely many applications of elementary operations on basic “high school functions”.


Edit 3:

Convinced by comments – removed closed under inversion in step 5. Added log in step 2.


Addendum (2013-02-28):

To see that the square root function s(x)=\sqrt{x} is elementary using the above definition just note that
x\mapsto \frac{1}{2}\log x = \log x^{1/2}=\log\sqrt{x}
and hence s(x)=\sqrt{x}=\exp(\log\sqrt{x}) is elementary too.
Perhaps \arctan is a bit harder to deduce from the other steps in the construction of E, first note that
\begin{eqnarray}
\tan x=\frac{\sin x}{\cos x}= \frac{e^{ix}-e^{-ix}}{i(e^{ix}+e^{-ix})}=
-i\frac{e^{2ix}-1}{e^{2ix}+1}=\\
-i\frac{e^{2ix}+1 -2}{e^{2ix}+1} = -i\frac{e^{2ix}+1 }{e^{2ix}+1}-i\frac{-2}{e^{2ix}+1}=\\-i +i\frac{2}{e^{2ix}+1}
\end{eqnarray}

Hence if we solve for, e^{2ix} in the above identity we get
\begin{eqnarray}
e^{2ix}=\frac{2}{1-i\tan x}-1= \frac{2-(1-i\tan x)}{1-i\tan x}= \frac{1+i\tan x}{1-i\tan x}
\end{eqnarray}

and taking the complex logarithm (more precisely the principal branch of the logarithm) we get x=\frac{1}{2i}\log\frac{1+i\tan x}{1-i\tan x}
That is t\mapsto \arctan t =\frac{1}{2i}\log\frac{1+it}{1-it}
belongs to E.

Attribution
Source : Link , Question Author : Community , Answer Author : AD.

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