The Strongest Matthew For Whenever Options

[DustyHaru]

Going to see Jonny Craig tonight.

Yes.

[Rhiannon]

I wonder if my advisor will care if I suddenly do everything she told me not to do and not do what she told me to do.

[Elnendil]

One of my other professors didn't agree with me on the idea I posted on here earlier. I'll go over it again and see what I can pull out of it. I don't think I worded it well enough.

I should probably stop talking so much about math before this becomes Math General though, so I guess i'll keep it to PM from now on. Soon i'll be able to just do most of these problems with relative ease, can't wait.

This post has been edited by Elnendil: Oct 8 2010, 12:27 AM

[Lloyd Seegymont the Rasier]

You don't need to do that. I think it's pretty cool and shit. I'm actually learning something on the internet, whoda' thunk it?

[Frisk]

I agree with Rasler. This is the perfect topic to talk about this kind of stuff.

[chrishawke]

I don't think we really care, Elnendil. Besides, what else are we going to talk about.

OH, LOOK. TACOS.

[Rhiannon]

YO IT'S COOL. Nothing's off-topic here.

And I'm so lucky/happy/whatnot, my Linguistics teacher was cool and randomly reviewed the stuff I missed due to being tooill that one day. So now I don't really feel behind, though I'm pretty sure I didn't do all that great on the quiz.

[Lloyd Seegymont the Rasier]

SO YEAH VAMPIRE KNIGHT IS PRETTY GOOD SO FAR.

[chrishawke]

Why are you watching/reading that.

[Frisk]

That's not important. The important thing is why ISN'T he reading Tsubasa.

[Lloyd Seegymont the Rasier]

That's not important. The important thing is why ISN'T he reading Tsubasa.

I've never gotten around to buying/renting/loaning this.

Why are you watching/reading that.

I saw a topic called Shiki, so I put that into the Google search engine. Vampire Knight came up, so I wanted to watch it for shits and giggles. It seems ok. I haven't watched much anime in a while, so anything new is good :x

[Elnendil]

YO IT'S COOL. Nothing's off-topic here.

And I'm so lucky/happy/whatnot, my Linguistics teacher was cool and randomly reviewed the stuff I missed due to being tooill that one day. So now I don't really feel behind, though I'm pretty sure I didn't do all that great on the quiz.

Don't worry about it. I honestly didn't do very well on my first exams, but thats because I haven't worked with enough practice problems to flesh out my weaker spots. I still hang around after my math courses and just talk with the professors about ideas i've thought on etc. At the very least, despite shortcomings, you're showing that you're willing to learn.

This post has been edited by Elnendil: Oct 8 2010, 05:04 AM

[Lloyd Seegymont the Rasier]

So whats up with the channel?

[Raijinili]

Yep. If a|b, there has to be some integer q that, when multiplied by a, equals b.

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.

I've been throwing some other concept around recently, has to do with using arithmetic to find square roots. This isn't a theorem, just a plausible observation:

q=lim(x->1+) sqrt(b)/x

When solving, it makes sense in a way that well, if x=1, then q=sqrt(b). But the observation is that you can get an approximation if q is an integer, when sqrt(b)/x and x approaches 1 from the right. Like how sqrt(33)=5.744... So there is some sqrt(x) that you can multiply into 5 that will equal sqrt(33) [5*sqrt(x)=sqrt(33)]. Then I divided both sides by sqrt(x), and I noticed that every time I pulled out a new decimal place, sqrt(x) approached 1 from the right, hence the observation. I showed this to the professor and he mentioned some other theorem...Newtons...gah, I forget the name. Its in my calculus book, i'll look into it later; has to deal with constantly finding approximations to find roots. I'll also say from observation that 1<=x<2 and q is the greatest integer of the root. Anyways, what do you think Rai?

wut

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...

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?

I should probably stop talking so much about math before this becomes Math General though, so I guess i'll keep it to PM from now on. Soon i'll be able to just do most of these problems with relative ease, can't wait.

I don't think I can be PM'd. I'm banned.

[Dr Strum]

I wonder if my advisor will care if I suddenly do everything she told me not to do and not do what she told me to do.

Like what?