# Strange pattern in exponents

1. Sep 21, 2004

### StatusX

I was looking at the different ways the operations +, *, and exponentiation can work on three natural numbers x, y, and z. I found a weird pattern when the second operation performed is exponentiation. These are the expressions:

$$(x+y)^z \ x^{(y+z)} \ (x \cdot y)^z \ x^{(y \cdot z)} \ (x^y)^z \ x^{(y^z)}$$

They are arranged in what I think is the most natural way: from "weakest" to "strongest", in the sense + < * < ^, and exponentiation is more powerful when the bigger number is the power, not the base. (This assumes x, y, z are relatively close in size). Here's the pattern:

$$(x+y)^z \ \ \ \ \ \ \ \ \ x^{(y+z)} \ \ \ \ \ \ \ \ \ (x \cdot y)^z \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x^{(y \cdot z)} \ \ \ \ \ \ \ \ \ (x^y)^z \ \ \ \ \ \ \ \ \ x^{(y^z)}$$

largest ......<- identical ->....... | .......<- indentical ->...... largest

written ........ written ............ | ............ value .............. value
formula ........ formula

I'm sorry if this doesn't format right, but I'll explain what it means. $$(x+y)^z$$ has the largest identity expression, in terms of the size of the written formula: the binomial theorem. $$x^{(y+z)}$$ and $$(x \cdot y)^z$$ are equal to $$x^y \cdot x^z$$ and $$x^z \cdot y^z$$ respectively, so the shape of their written formulas are identical. $$x^{(y \cdot z)}$$ is equal in value to $$(x^y)^z$$. And finally, $$x^{(y^z)}$$ has the largest value, for x,y,z>>1.

This seems like a very bizzare link between the "man-made" (sort of) written formulas and the "completely natural" values of these expressions. Is there anything to this, or is it just a coincidence? I'm really not a crackpot, I think there is something here that needs at least some basic explanation, but if someone can explain logically why I'm wrong, I'd be more than willing to accept it.

2. Sep 22, 2004

### matt grime

I think you're looking for far too much here, and you're ignoring other possibilities such as xyz or xy/z or x/yz which contradict your apparent "link"

What is true is that often when some operation that we use frequently has a large expression, then we find a shorter way of writing it, eg n! rather than 1.2.3....n.(n-1).

There is also the fact that if you were to write out what x^{y^z} meant in a similar way to your binomial expansion having written x as 1+1+1...+1 x times etc that the expression wouldn't be all that small.

3. Sep 22, 2004

### HallsofIvy

Staff Emeritus
You've posted this before. You seem to think there is something "natural" about the way you have ordered these when, in fact, it is an arbitrary choice.

4. Sep 22, 2004

### StatusX

Matt, thanks for your help, but like I said, I'm just talking about operations on three numbers, with one of the operations being an exponent. I already looked at operations like xyz and x(y+z), and they had simple patterns because of multiplication and additions associativity and commutativity, which exponents don't have. (and I'm ignoring inverse operations right now cause I want to stay in the natural numbers) And I don't get your point about writing x out as all 1's. I'm talking about normal algebra, and assuming the way were doing it is the simplest way it can be done(ie., you have to use the binomial expansion, but you would be wasting time writing out x as all 1s)

HallsofIvy, i know I already posted this, but I asked you then why you thought this was arbitrary, and in this post I tried to make clear the reason I put them this way (let me note that I put them this way and then found the pattern, not the other way around). Could you just please tell me why you think this is arbitrary? Did you read my reasoning?

5. Sep 22, 2004

### matt grime

Your posting an opinion about how you think mathematic symbols ought to be interpreted. Your post doesn't actually make any sense really: what on earth do *you* mean about size of expressions? And note yo'ure making a lot of completely arbitrary choices (why not division, why only integers?)

"largest identity expression"

that phrase in particular makes no sense to me as a mathematician.

Last edited: Sep 22, 2004
6. Sep 22, 2004

### StatusX

yea, ur right. I was having a hard time deciding what to call the first property, and I ended up going with "written expression," but now that I think of it, that's pretty meaningless. The only reason I thought there was something here has to do with the way I found it, so maybe if I explain that it will help you see where I'm coming from, and you can either tell me theres still nothing here or tell me how to reword my idea.

I arranged the 18 expressions involving x_y_z, where the blanks are filled by these three operations, and then the parantheses can go on either the first or second pair. There was an obvious way to pair them off for the ones where the second operation(outside the parantheses) was not exponentiation. For example, x+(yz) and (xy)+z, x(y+z) and (x+y)z, and (x+y)+z and x+(y+z). These are trivial because of commutativity and associativity. I noticed the pairs always had parantheses on opposite sides. Then I noticed you could pair off x^(yz) and (x^y)^z, cause they were equal, and x^(y+z) and (xy)^z seem to go together cause of the way they distribute. All that was left was (x+y)^z and x^(y^z). And I was pissed off cause I thought "damn, (x+y)^z won't fit with anything cause it's so big," as in its expansion. Then I realized x^(y^z) was big too, in a different way. That's all I found, and this chart might be stretching it a bit. So I can see now why this could just be a coincidence, but is it?

7. Sep 23, 2004

### matt grime

"damn, (x+y)^z won't fit with anything cause it's so big," as in its expansion.

but what on earth does that mean?

8. Sep 23, 2004

### StatusX

expand (x+y)^z. its a huge mess, someting like:

x^z + z*(x^(z-1))*y + ... + z*x*(y^(z-1)) + y^z

as opposed to, say, x^(y+z), which is just (x^y)*(x^z). thats what I meant, but now I see why theres no significant pattern. I didn't think through what I meant by my descriptions, like "big" and "written expression". thanks for the help.

9. Sep 26, 2004

### arildno

What's the huge mess about it?
Letting z be an integer, you have:
$$(x+y)^{z}=\sum_{i=0}^{i=z}\binom{z}{i}x^{i}y^{z-i}$$