So, what it is; I'm hoping for help from a mathematician, should one happen to read this. But first, a complaint, the discussion of which may well end up taking longer than the actual topic…

The commonly given definition of a mathematical exponent (rather than an exponent of mathematics, which would be an enthusiastic maths teacher; I'm here all week, folks!) is that it represents how many times a given number should be multiplied by itself. Even as a child—a rather pedantic child, I should admit, but I know for a fact that I wasn't the only one—when I was given this explanation by a maths teacher, I found it both misleading and ambiguous.

The ambiguity first. If I'm told to multiply, say, three by itself more than once, just what does "multiplied by itself" mean when I come to perform the second multiplication—nine times nine, or three times nine? It's not exactly clear what the "itself" in the definition is referring to, by this point in the proceedings. (A point of pedantry regarding sloppy use of English which I got so hung up on that for a month or so I had to remind myself that "two to the power of three is eight," and work backwards from that to "two times two times two" in order to remember what a given exponent was telling me to do.)

The misleading—indeed, plainly wrong—aspect of the definition lies in the "how many times" part. Let's write out 3^{4} as a series of multiplications:

3 × 3 × 3 × 3

I may not have been an experienced mathematician at the age of eleven (I'm still not any kind of mathematician, I should hasten to add), but I was well-enough experienced as a user of the English language to spot that the above is *not* "three multiplied by itself four times," as the given definition of that superscripted "4" implies, but (just count the number of operations!) rather it is three multiplied by itself *three* times. Obviously the purely mathematical description containing only three multiplication symbols was the correct one, or else my maths exercises, and those of several classmates, would have been returned covered in enough large red crosses to denude an entire ambulance service. Therefore (don't even *whisper* it!) an adult, and a teacher to boot, was not only capable but actually guilty of uttering an incorrect statement. Sacrebleu!

All of which problems with the definition, as given, might be put down to mere childish pedantry, except our growing conviction that maths teachers would make terrible teachers of English was confirmed when we were told, with no explanation given as to why, that *n* to the power of one equals *n*, and *n* to the power of nought equals one. Why did this confirm the "bad English" theory? Because, for instance, two multiplied by itself one time is quite obviously four, while not only does "two multiplied by itself zero times" seem ominously close to "the sound of one hand clapping," but the given definition supplies no reason why "two multiplied by itself zero times" should equal one, rather than the more obvious conclusion that if we start with "two" and zero operations are performed upon it, then we should still have "two" at the end of the process. You don't, after all, take a ton of iron ore, describe some things which might be done to it which you then proceed not to do, and end up with a Model-T Ford or even a 3⁄8 Whitworth bolt. You end up with a ton of iron ore. (For Sale. Model-T Ford. Some assembly required….)

In other words, if the exponent truly is the number of times the given number should be multiplied by itself, then:

2^{0} = "2 multiplied by itself 0 times" (2 with nothing done to it) = 2

2^{1} = "2 multiplied by itself 1 time" (2 × 2) = 4

2^{2} = "2 multiplied by itself 2 times" (2 × 2 × 2 ) = 8

And so on. Or more famously, E = mc^{1}

Clearly, either countless generations of mathematicians have been using exponents incorrectly, or the person who originally came up with the commonly given definition should be keelhauled, tarred, feathered, placed in the stocks for a month, and then locked in a cell and forced to listen to *One Direction's Greatest Hit.* On continuous loop. For life.

The sad thing, really, is that a better definition is available which removes the ambiguity completely and, for the more important part, "how many times," is just as pithy, just as easily understood and just as memorable, whilst having the distinct advantage of being true. Here, for instance, are the first few powers of three:

3^{0} : One multiplied by three, zero times, is 1

3^{1} : One multiplied by three, one time, is 1 × 3 = 3

3^{2} : One multiplied by three, two times, is 1 × 3 × 3 = 9

3^{3} : One multiplied by three, three times, is 1 × 3 × 3 × 3 = 27

3^{4} : One multiplied by three, four times, is 1 × 3 × 3 × 3 × 3 = 81

And suddenly things make sense. No number is being "multiplied by itself"; instead, we're multiplying by an unambiguously described constant. The exponent really does specify the number of operations. And "one," with no operations performed upon it, really does remain, as logic, plain sense and a bloody gert lump of iron ore labelled "Model-T kit car" suggest it should, "one."

The exponent of a number, then, if we're looking for an English-language definition, represents "how many times the number 'one' should be multiplied by that number": a definition which is both factual and more explanatory than the one usually given, and is just as easily understood by (let alone less possibly-confusing for) the pupils it's being given to.

And thus endeth the complaint.

So now that's out of the way, here, finally, is the problem I wanted help with. A friend's child had been introduced to the fable of the grains of rice on a chessboard. The solution to that is well-known:

The number of grains on the board is 2^{64} − 1

And the number of grains on the final square is 2^{63}

A more general solution:

Let S be the number of squares.

Let T_{1} be the number of grains on the board.

Let T_{2} be the number of grains on the final square.

T_{1} = 2^{S} − 1

T_{2} = 2^{S − 1}

But what happens, my young friend wanted to know, if the grains are tripled, quadrupled, etc, on each square. *Oh bugger,* thought I. *I don't know.*

Well, we painstakingly drew up tables for several multipliers, using addition and multiplication for each step, then took a look to see if a pattern could be spotted. Then we drew up loads more in order to check predictions we'd made using a formula we'd arrived at, to see if we could get it to fail. And so far, it hasn't, but I've been careful to explain that what we have, as far as we can tell, is a hypothesis, not a proven theorem. And that's where I'm stuck. Though I can *follow* the simpler of the two proofs of the original, doubling, version given at the Wikipedia page linked above, I don't trust my dimly remembered scraps of schoolboy maths enough to confidently try to extend that proof (if such be possible) beyond doubling.

Here's what we tentatively believe to be true:

To recap the definitions:

Let S be the number of squares.

Let T_{1} be the number of grains on the board.

Let T_{2} be the number of grains on the final square.

Let x be the multiplier.

T_{1} = (x^{S} − 1) ÷ (x − 1)

T_{2} = x^{S − 1}

The second part, the total number of grains on the final square, I'm confident of. It's a literal description of the operation being performed at each step (multiply the previous square's contents by x). The first part though; it *seems* to work, but my maths is rusty enough that I can't see *why* it works and so, while I'm confident enough that I'd use it myself should the need ever arise (and on my own head be it should I be wrong), I'm *not* confident enough to tell a schoolchild—and not even my own child—that what we have is a truth, rather than a little-tested rule of thumb or even a howler of a mistake. Or maybe we're right but there's a simpler way of stating it, that we haven't spotted.

Any takers? All help will be gratefully accepted.

—*Daz*

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on Tuesday, September 13, 2016 at 14:24Neil RickertMathematician here.

On reading your post, it is clear that you don’t need a mathematician. You need a psychologist.

Well, more accurately, you need a linguist. So here’s my favorite quote from a linguist. This was posted to usenet, by linguist Jacques Guy, on Nov 5, 2000 (hmm, that was Guy Fawkes day):

(If you are looking for it, that was posted to “sci.lang”).

Personally, I think Jacques was absolutely correct. We can have fun pointing out illogicalities in language, but that’s just the way that language is.

Back to exponents. I like to think of starting with one, and then multiplying that by the base number. So using 3 with an exponent of 4 is 1 x 3 x 3 x 3 x 3 and the “3” does occur 4 times, as does the multiply symbol. This way of describing also works for an exponent of 0 or 1. But you can still get into arguments over fractional exponents.

on Tuesday, September 13, 2016 at 16:35remigiusDaz, not a mathematician here – maybe I can help.

The first part though; it seems to work, but my maths is rusty enough that I can’t see why it works…In plain English whatever the multiplier is the total of each subsequent square will be the sum of all previous squares times the multiplier minus 1, plus 1.

i.e if the multiplier is 3 then the total on a subsequent square is double the total of the previous squares plus 1.

if the multiplier is 4 then it is triple the total plus 1.

if the multiplier is 5 then it is quadruple the total plus 1.

if the multiplier is 17 then it is 16 times the total plus 1.

And so on… ad infinitum

on Sunday, September 18, 2016 at 15:29tonyeDaz,

Yes Kraftwerk even cover maths.

Enjoy

on Saturday, October 1, 2016 at 07:50donotwashSo on the 1st square you have 1 grain. ON the 2nd you have x grains. On the third you have x times x grains, and on the nth square you have x^n grains.

Thus T_1 = 1+ x + x^2 + x^3 + … + x^S, the total number of squares.

Multiply T_1 by x:

xT_1 = x + x^2 + … + x^S + x^{s+1)

Thus xT_1 – T_1 = x^(S+1) – 1

and so T_1 = (x^(S+1) – 1) / (x – 1)

where ^ symbolises the exponential as you’ve probly guessed.

Find more at:

https://proofwiki.org/wiki/Sum_of_Geometric_Progression

Sorry, would have answered this earlier but it didn’t come into my email feed