@heather There is hardly any intuition in a lot of pure math , unfortunately. I suggest, if you're interested, to get a topology text and start there since all it requires is basic set theory.
@heather If you want to learn what @0celo7 is talking about in chat, i suggest you learn topology the right way (not frantically searching wiki pages like I do sometimes when I'm bored). But, I suggest you don't try to learn what he's talking about until you finish your primary math education first.
@SirCumference, I'm going out of order. I dunno anything about conics, but I do know Algebra I and some of Algebra II; I'm doing Algebra II after trig.
@0celo7 Actually, I don't see why you can't learn abstract algebra before a basic algebra 2 course. (Not that I'm suggesting you do so) theoretically, it doesn't assume knowledge of algebra 2 because it's all built up axiomatically
@heather Then you can read the notes, which describe logic and basic set theory, followed by calc
@Obliv Yeah, but it won't play much use at those points
@heather The difficulty of learning calc depends on where you learn it from. Some teachers make it confusing as hell, others explain it in a really intuitive way
@SirCumference, yeah, trig is pretty short; I'm already more than 50% of the way through the Khan Academy course. Sounds good about the calc; that shouldn't be too bad.
So I guess, I have the question of what I should use to learn Algebra II/topology/abstract algebra
@heather It could also help to learn some precalc from Khan Academy. Despite its name, precalc is not actually useful for or related to calc, but it will enlighten you on some concepts you probably missed out on when learning trig and algebra 2
Just remember precalc paradoxically isn't related to calc. It's just useful math knowledge one should know before leaving high school or starting some higher mathematics.
As I said, calc can be easy and enjoyable, if you learn it from a good teacher
> However, here's another way of understanding why $e^{ix} = \cos{x} + i\sin{x}$. It too involves some calculus, but I can describe the calculus involved more easily.
@rob when maximizing a curve $f(x,y)$ by a constraint $g(x,y)=k$ they must both lie on the same tangent plane, right? Doesn't that mean either can be a scalar multiple of the other? So in a sense $\lambda \nabla f(x,y) = \nabla g(x,y) \equiv \nabla f(x,y) = \lambda \nabla g(x,y)$?
@SirCumference I hadn't done so yet ... and after the spate of \renewcommand{\foo}{I'm a doofus} that went on in the sidebar the other day I think I'll refrain.
@rob it wasn't a real question though. I was just wrapping my head around the two functions having to lie on the same tangent plane. only way those two would be equivalent would be $\lambda_2 = \frac{1}{\lambda_1}$
@Obliv I think I've parsed your statement, but I don't understand it straightaway. Why would your function to maximize and its constraint need to share a tangent plane?
like, imagine if g(x,y) = k passed the point marked on the picture then never touched f(x,y) = 11 or f(x,y) = 9 again. Would they still share a tangent plane at f(x,y) = 10?
