There is a related question to what I'm asking, which has been closed as too broad here: What are some ways of thinking about exponentiation, other than repeated multiplication?

I would like to limit the scope of my question to natural numbers. I think my question (as presented at the end) is a bit more focused in what kind of answer I'm looking for as well.

Please comment if you think this needs to be refined further (or my tags could be improved...).

At some point in school, addition is taught, probably using something like an addition table:

    1 2 3 4 ... 
1 | 2 3 4 5 
2 | 3 4 6 6 
3 | 4 5 6 7 
4 | 5 6 7 8 ... 

And then it's shown that problems like

+ 789 

are composed of smaller related problems: $9+6$ and carry into the shifted (i.e., times 10) $5+8$ ...

Later on a similar process is used to teach multiplication. A multiplication table is memorized and then it is taught how to solve multi-digit multiplication problems by breaking apart the problem into smaller problems of multiplication and addition. For example

× 789 

would first be $9 × 456$ which is then itself broken into $9 × 6$ add that to "shifted" (i.e., times 10) $9 × 5$ ...

(This may not be an accurate representation of elementary school learning, but the details about the schooling process are not so important here)

But then this process stops. There is no "power table" (or higher!) learned in school (nor is "exponentiation" as a binary operator associative...).

Typically when encountering an integral exponential problem there isn't any "exponentiation" that occurs. For example, take $57^4$. This would be solved by squaring $57$ (multiplication) and then squaring (multiplication) the result again.

Now, I think there is a semantic objection that goes something like: "But you are doing exponentiating, because, by definition, it is simply repeated multiplication." But I would respond by pointing out how there's something rote about multiplication and addition that is simply lacking when applying exponents. So technically, yes, this is exponentiation, but the point of posting this question here is to draw this distinction between applying an algorithm (for exponentiation) and using more fundamental/rote components (for multiplication and addition).

So my question is something like: In the same way that it's possible to break down addition problems into smaller addition problems, and the same way it's possible to break down multiplication problems into smaller multiplication/addition problems, is it possible to break down exponentiation problems into smaller exponentiation/multiplication/addition problems?

Or alternatively, am I simply stuck on a semantic issue, and addition, multiplication, and exponentiation are all "the same" in some sense?

  • $\begingroup$ I honestly struggle to find what the actual question really is. $\exp(x)$ is a function, there are multiple and equivalent ways to define it. You may think to $\exp(37)$ in any way you like, as $e^{12}\cdot e^{5^2}$, as $\lim_{n\to +\infty}\left(1+\frac{37}{n}\right)^n$, as $\sum_{n\geq 0}\frac{37^n}{n!}$ etcetera. Mathematics, like art, is a wonderful thing exactly because there are multiple ways to represent the very same objects. $\endgroup$ – Jack D'Aurizio Oct 20 '17 at 20:39
  • $\begingroup$ @JackD'Aurizio sorry, raising a positive integer to the power of another positive integer, not exponentiation like $e^x$. $\endgroup$ – Burnsba Oct 20 '17 at 21:09
  • $\begingroup$ By exponentiation I mean the process of raising some real number $a> 0$ to some real number $b$. In such a context, $a^b$ can be defined as $\exp(b\log a)$, where $\log$ is the inverse function of $\exp$. This is a natural extension of the definition of $a^b$ where $a$ and $b$ are positive natural numbers (as $a\cdot a\cdot\ldots$ multiple times). So, for what it is worth, one may also define $a^b$ as $\exp(b\log a)$ in the last particular case. And think to $a^b$ as $$\sum_{n\geq 0}\frac{b^n\log^n(a)}{n!},$$ for instance. $\endgroup$ – Jack D'Aurizio Oct 20 '17 at 21:14
  • $\begingroup$ Of course you are correct (I swear I knew this at one point...). Anyways, I don't think that a limit or sum necessarily makes the problem "simpler" or "smaller" as I was trying to hint at in my question in a decomposable way. But this could be a problem of (my) intuition. For instance, in the same way multiplication decomposes based on the decimal system, I thought perhaps exponentiation could be broken into prime factors, but I think there are other problems with this approach. $\endgroup$ – Burnsba Oct 20 '17 at 21:32
  • $\begingroup$ It depends on the situation, of course. To prove $\frac{d}{dx} e^x = e^x$ through the series definition is trivial, to prove it by invoking that $e^x = \lim_{n\to +\infty} e^{a_n}$ for any sequence of rational numbers $\{a_n\}\to x$ is still possible but much more involved. The point of having multiple representations for the same objects is exactly this one: before getting a reasonable amount of experience in the field, one never really knowns which representation is best suited for the job. In many occasions, words like precursor or fundamental are just states of mind. $\endgroup$ – Jack D'Aurizio Oct 20 '17 at 21:38

The answer is yes, of course. The multiplication process you describe depends on using a radix representation to multiply a given number by single digits, shift them by appropriate number of places and then add the results together. You can do something very similar with exponentiation. For example, suppose you want to evaluate $x^{13}$. Express $13=1101_2$ in binary. Then perform the calculations $a_0=1,\; a_1=a_0^2\cdot x,\; a_2=a_1^2\cdot x,\; a_3=a_2^2,\; a_4=a_3^2\cdot x,\;$ reading the binary digits left to right, with final result $a_4=x^{13}.\;$ A similar process works for any other radix, but it gets more complicated. You can do something similar with Fibonacci numbers. For example, to compute $x^{13},$ do this, $\;a_1=a_2=x,\;a_3=a_2a_1,\;a_4=a_3a_2,\;a_5=a_4a_3,\;a_6=a_5a_4,\;a_7=a_6a_5=x^{13}.$

  • $\begingroup$ This is not different from performing a repeated multiplication. It is performed in a different (more efficient) way, but still is a repeated multiplication. $\endgroup$ – Jack D'Aurizio Oct 20 '17 at 21:24
  • $\begingroup$ @JackD'Aurizio Yes, just exactly as multiplication is repeated addition (just more efficient). $\endgroup$ – Somos Oct 20 '17 at 22:39

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