For writing a (german) article about the power with natural degree I have the following question:

In school one defines the power with natural degree via

$$n^k = \underbrace{n\cdot n\cdot \ldots \cdot n}_{k \text{ times}}$$

In calculus normally recursion is used to introduce the power:

$$\begin{align} n^0 & := 1 \\ n^{k+1} & := n \cdot n^k \end{align}$$

Question: What are the disadvantages of the expression $\underbrace{n\cdot n\cdot \ldots \cdot n}_{k \text{ times}}$ and why it is not used to define the power in advanced mathematics (although it is intuitive)?

Note: I'm sorry, I had an ambiguous question before. I do not want to know, why we define the power. I want to know, why we use recursion and not the expression $\underbrace{n\cdot n\cdot \ldots \cdot n}_{k \text{ times}}$ to define the power in advanced mathematics...

My ideas so far:

  • $\underbrace{\ldots}_{k \text{ times}}$ is no operator, which was introduced before.
  • definitions with recursion lead naturally to a scheme, how properties of these concepts can be proved via induction
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    $\begingroup$ It takes soooo much more effort to type (58 chars vs 3 for $n^k$) and takes approx 5 times as much space! $\endgroup$ – Winther Sep 15 '14 at 16:23
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    $\begingroup$ In my opinion and what I've experienced @Winther 's argument is the only one against it. What you said about induction is true but only if you want to be a pedant. I have seen the notation with 'k times' pretty often in lectures. $\endgroup$ – flawr Sep 15 '14 at 16:38
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    $\begingroup$ With the underbrace notation, it's not at all clear what $n^0$ is. $\endgroup$ – celtschk Sep 15 '14 at 16:38
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    $\begingroup$ Your new question can be asked more generally as why do we choose some notation over another? The answer to such a question usually falls into one of two categories. 1) Historical reasons. It just become standard and its hard to change it unless a new notation has some great advantage. 2) Simplicity and beauty. A simple/short notation has a greater chance of making it than a more cumbersome one. But in the end, this is really just a matter of taste. If you want to define $n^k$ for integer $k$ that way, there is nothing wrong with it and I'm sure there are textbooks that do this today. $\endgroup$ – Winther Sep 15 '14 at 16:43
  • $\begingroup$ Perhaps it's because you'll have calculus students doing the following: $\frac{d}{dx}(x^2)=\frac{d}{dx}\underbrace{(x+\cdots+x)}_{x}=1+\cdots+1=x$, :) $\endgroup$ – Alex R. Sep 15 '14 at 17:13

$n^k = \underbrace{n\cdot n \cdots n}_{k\ \textrm{times}}$ only works when $k$ is a positive integer. In order to define exponentiation more generally, we must refine the definition of exponentiation several times:

  • First, we must address a suitable definition for $k = 0$ and $k \in \{-1, -2, \ldots \}$, including operations of the form $n^{k_1}n^{k_2}$ that reconcile with the "repeated multiplication" definition when $k_1, k_2$ are positive integers;
  • Next, we must address a suitable definition for $k \in \{ \frac{1}{m} \mid m \in \mathbb{Z} \}$, including operations of the form $n^{k_1}n^{k_2}$ that reconcile with the similar operations defined above (and in doing so, we cover the more general case $k \in \mathbb{Q}$);
  • Next, we must address a suitable definition for $k \in \mathbb{R} \backslash \mathbb{Q}$. This turns out to be more complicated than one might expect, one such approach is to use the language of Dedekind cuts.

For these reasons, the "repeated multiplication" definition fails for almost every interesting value of $k$. The goal is to find a definition that reduces to "repeated multiplication" in the special case that $k \in \mathbb{Z}^+$, but that works more generally for all real numbers when the base is positive.

To address your edit, which is not something I've actually encountered, but I will comment nonetheless:

If you define $n^{k+1} = n \cdot n^k$, this definition is satisfactory for all values of $k$, presuming you have defined exponentiation rigorously. This differs from the repeated multiplication definition in that you may start from an arbitrary $k$. This is actually one step in the process of defining exponentiation. More generally, we wish to declare that $n^{k_1}n^{k_2} = n^{k_1+k_2}$ regardless of the classes of real number to which $k_1$ and $k_2$ belong. The "recursive" definition you presented is a special case of this.

  • $\begingroup$ +1 The "should be easy to generalize" is a very good argument for choosing a notation over another. $\endgroup$ – Winther Sep 15 '14 at 16:46
  • $\begingroup$ I'm not sure what you mean when you say that defining $n^k$ for $k\in\Bbb R\setminus\Bbb Q$ is “more complicated than one might expect”. What I would expect is that if $q_1,q_2,\ldots$ is a sequence of rationals converging to $r$, then the sequence $n^{q_1},n^{q_2},\ldots$ also converges, and so we might try defining $n^r$ as the limit of that sequence. For this to make sense we would have to show that the value of $n^r$ obtained in this way is the same regardless of which sequence $q_1,q_2,\ldots$ we choose, as long as it converges to $r$. But this is the case. Dedekind cuts are not required. $\endgroup$ – MJD Sep 15 '14 at 17:50
  • $\begingroup$ @mjd Of course, it's not more complicated than you expect, but using convergent sequences is probably more complicated than expected for someone accustomed to defining exponentiation as repeated multiplication. It's not obvious at all until you learn those topics. And cuts are not necessary, which is why I suggested that they are but one way. I find them convenient as addition and subtraction of cuts is well defined, and so defining the product rule for exponentials is easier. $\endgroup$ – Emily Sep 15 '14 at 17:56
  • $\begingroup$ @MJD: That definition works, but it's extremely hard to work with. Suppose you want to differentiate the function $f(x) = y^x$. It would be very tedious to do it with approximations. On the other hand if you simply define $x^y$ to be $e^{y\log x}$ then it becomes a couple of routine lines. $\endgroup$ – Zavosh Sep 16 '14 at 9:19
  • $\begingroup$ Where $e^x$ would be defined as $1 + x + \frac{x^2}{2} + \frac{x^3}{3!} + ...$ $\endgroup$ – Zavosh Sep 16 '14 at 9:20

Because only multiplication of two numbers is defined at that point. Also, if it's not yet defined to be associative, that expression may be ambiguous.


Why do we write $ab$ instead of $$\underbrace{b+b+\cdots+b}_{a \textrm{ times}}~~~?$$

  • $\begingroup$ I did not ask, why we define the power. I want to know, why we define the power via recursion and not via the expression $\underbrace{n\cdot n\ldots n}_{k \text{ times}}$... $\endgroup$ – Stephan Kulla Sep 15 '14 at 16:28
  • $\begingroup$ I'm sorry, I had an ambiguous question before... $\endgroup$ – Stephan Kulla Sep 15 '14 at 16:34
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    $\begingroup$ @tampis Saying "$k$ times" is recursive. If I want to multiply $n$ by itself three times, first I have to do it twice, then once more. Tomato, tomato. $\endgroup$ – Slade Sep 15 '14 at 16:38

Although this is hardly the reason that cause confusion because people have common sense, "..." is not well defined if you want to be picky.

For example $ \{1,2,3,4,...,100\} $ could mean set of natural number less or equal to 100 or the image of that set under function $f(n)=(n-1)(n-2)(n-3)(n-4)(n-100)+n$

Define things using recursive remove the "ambiguity".


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