overly geeky.

[katster]

This is what I ponder when I'm going to sleep. Well, sometimes.

Last night, I came up with the following (warning, math ahead):

xⁿ - x^n-1 = x^n-1(x-1)

Or in practice, for those of you who don't read math notation:

10³ - 10² = 10²(10-1) or:
1000 - 100 = 100 * 9

I haven't tested for some of the more esoteric cases, but thus far, it works with both negative x and negative n, so I'm fairly confident it will hold for all integer values of x and n. Whether it holds for fractional exponents is yet to be determined, as with non-integer values of x. (It appears it doesn't work with fractional exponents, so it's something to do with the properties of integers, I think.) [Note: I goofed the math, it worked with n=3/2 and x=2, so...maybe it does work with fractional exponents.]

But now, the next question is, why does it work? I haven't managed to figure out that one yet, but I'm sure one of the math geeks on my friends list can manage to give me an explanation.

I'll be out at court all afternoon (whee blah kill me now), so have fun with this.

(And yes, I'm a geek. And maybe even a nerd.)

Comments

[xiphias.livejournal.com]

Multiply it through.

x^(n-1)*(x-1) =
(x-1) * x^(n-1) =

(x * x^(n-1)) - (x^(n-1)) =
(x^n) - (x^(n-1))

It's just algebra.

[nvdaydreamer.livejournal.com]

Seconded. You've just factored out an x^(n-1) from both terms.

[underpope.livejournal.com]

Beats me. I'm still trying to figure out why 1 + 1 = 2 is a true axiom.

[xiphias.livejournal.com]

It's definitional. I mean, it's an axiom, so you can't prove it, so you CAN'T figure out why.

You are simply defining the integers as a set in which you have a starting value, an operation of addition, and a basic integral value, which is "1", and defining the set of positive integers as "things that start with this thing called '1', and then you do this thing called '+' with a '1' and another '1', and you get another element of the set, and then you can do the thing called '+' and the thing called '1' to THAT element, and you get ANOTHER element of the set."

So there isn't really a reason why, except that, y'know, that's what you call it and all.

[underpope.livejournal.com]

But isn't that all just an intellectual dodge? By saying that something is "axiomatic" or "definitional", you're really just equivocating, and, in fact, engaging in a circular argument. "It's definitional," you say, and I reply, "Yes, but WHY is it definitional?"

Watch out, man. I've got a philosophy degree and I'm not afraid to use it.

[xiphias.livejournal.com]

Of COURSE it's a circular argument. You need a starting point, and your starting point has no reason. There comes a point in any system where you're at the irreducable factors. You don't have reasons for them -- they just are. Philosophy and physics both work to take the things which were previously thought to be irreducable, and break them down further, but, even so, there MUST be a point beyond which you cannot go -- in both philosophy and physics, we're nowhere near that point yet, but it must exist.

In mathematics, those points are well-known and clearly delineated.

[inflection.livejournal.com]

If we really want to describe all of this in terms of the axioms commonly accepted in the study of mathematical logic, then we axiomatically define "1" as the cardinality of the set {{}}, and "2" as the cardinality of the set {{}, {{}}}.

Given x, then S(x) = x+1 is to be called the successor function. If x is the cardinality of a set A, then S(x) is the cardinality of the set A U {A}. This much so far is axiom.

We now claim that 1+1=2. By the definition of the successor function above, our claim is that S(1)=2, and by checking the cardinalities of the sets referenced by the definitions of S(1) and 2, we find that there is a bijective correspondence, and thus the finite quantities involved are equal.

Is any of this important to professional mathematicians? Only if you're a student of the underlying theory of mathematical logic, mostly. It's important for the integrity of the subject that the axioms be known and agreed, and that the theorems one is using be known to be derivable from those axioms. A scientist doesn't bother to confirm that a new bacterium he's studying has genes, he just assumes they do; I've been reliably informed that integer arithmetic works and don't worry myself about the underlying details.

Axioms are stated and accepted without proof. If you accept the axioms, you must accept their consequences; some people reject an axiom called the Axiom of Choice, and therefore are free to reject certain consequences of it such as the Banach-Tarski theorem. The choice of your axiom set is entirely dependent on the theorems you want to prove.

Interestingly, there are increasing suggestions that certain mathematical axioms got that way because they are genetically hardwired into our brain -- comparative order and successor functions seem to make sense because we have dedicated neurological circuits that understand things like "A is bigger than B," or "this is the next thing in this sequence."

[xiphias.livejournal.com]

Yep, that looks familiar. See, I've never studied any mathematics (where "mathematics" is defined as "stuff that doesn't have to do with numbers, because numbers are boring." Which, of course, isn't a REAL definition, because some mathemeticians will deal with 1, 0, e, pi, and/or i), but I've got friends who are mathemeticians, and they've explained some basic stuff like that to me.

[nekoneko.livejournal.com]

You broke my widdle brain.

[macklinr.livejournal.com]

xn - xn-1

xn can be stated as (xn-1 * x)

Thus:

(xn-1 * x) - (xn-1)

You have a common term clearly showing now, xn-1. Grouping by that term, you get:

xn-1 (	xn-1 * x	-	xn-1	)
	xn-1		xn-1

Or, to simplify: xn-1 (x - 1)

I hope this helps some. I did make a couple leaps in the explanation, but if needed I'd be happy to fill in any gaps.

[johnpalmer.livejournal.com]

Distributive law. a(b+c) = ab + ac.