"I STILL DON'T KNOW WHAT YOU'RE SAYING
Are you explaining divisibility to me? I was clarifying that it's not something you have to prove, because it's basically the definition." Yeah but it doesn't hurt to clarify.
"Uh, Newton's Method is to use the derivative to approximate the zeroes of a function. It applies to only functions satisfying certain properties on an open interval of values (i.e. every open interval for which the function has the right properties lets you use Newton's method on it). Off the top of my head: x_{i+1} = x_i - f(x_i)/f'(x_i). In other words, there's a sequence x_0, x_1, x_2, ... that converges to x where f(x) = 0. I still have no idea what you're talking about. Yes, when you divide a real number by a smaller integer, you get closer to 1. What?
If you want to solve 5*sqrt(x) = sqrt(33), then x = 33/25. I... don't think that's what you're saying..." I'll look it over again when I have the time.
"Uh... Define [x] as the greatest integer less than or equal to x (floor function). In other words, [x] is an integer such that [x]<= x < [x]+1. Then are you saying that 1 <= x / [x] < 2? Yes, that's true, for x >= 1. The first inequality is true by virtue of x >= [x] by definition, therefore x/[x] >= [x]/[x] = 1. Assume the second is false. Then x/[x] >=2. Then x >= 2*[x] = [x] + [x]. But since x >= 1, [x] >= 1. So x >= [x] + [x] >= [x] + 1, which contradicts the definition of [x]. Is that what you want?" We just hit this concept, so I sort of understand what you're saying, but i'll look it over some more after I get more studying and textbook examples down.
Also did the quote system break or something?
This post has been edited by Elnendil: Oct 8 2010, 02:55 PM
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QUOTE Kenji: Where else would I could get beaten with a phone that would make me unable to remember it? The ass?
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