overly geeky.
Nov. 28th, 2006 10:55 amThis is what I ponder when I'm going to sleep. Well, sometimes.
Last night, I came up with the following (warning, math ahead):
xn - xn-1 = xn-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.)
Last night, I came up with the following (warning, math ahead):
xn - xn-1 = xn-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.)
no subject
Date: 2006-11-28 07:04 pm (UTC)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.
no subject
Date: 2006-11-28 07:18 pm (UTC)no subject
Date: 2006-11-28 07:27 pm (UTC)no subject
Date: 2006-11-28 07:52 pm (UTC)no subject
Date: 2006-11-28 07:55 pm (UTC)xn - xn-1xncan 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-1xn-1Or, 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.
no subject
Date: 2006-11-28 07:59 pm (UTC)no subject
Date: 2006-11-28 08:54 pm (UTC)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.
no subject
Date: 2006-11-28 09:54 pm (UTC)Watch out, man. I've got a philosophy degree and I'm not afraid to use it.
no subject
Date: 2006-11-28 10:11 pm (UTC)In mathematics, those points are well-known and clearly delineated.
no subject
Date: 2006-11-29 12:59 am (UTC)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."
no subject
Date: 2006-11-29 01:24 am (UTC)