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[Vector]

Integration won't really help with anything.

Yeah, I know... I was just proud I remembered that, even though I last did calculus more than five years ago ^_^;;

[MagmaMcFry]

I mentioned that to Helgoland, seeing as he thought it could be useful. I thought you were just answering his question.

[Skyrunner]

It's hard to get that on probability density graphs, the probability that X becomes a certain number rather than an interval is 0. The equation for getting the probability obviously says that that happens (integral from a to b), but it's hard to imagine that there is 0 chance for a single value o.o

[MagmaMcFry]

It's crazy, right? You even have probability 0 of getting a rational number, or even an algebraic number, or even a number that you can describe in a finite amount of words.

[Descan]

Can someone explain how you go from
(x-2)³ + 12(x²-x) + 8
to
x³ +6x² ?

One of (read: the only) the steps the book says is x³ -6x² +12x - 8 + 12x² -12x +8

And I have no idea where the (-6x² +12x) comes from. I get the x³ and I get the - 8 + 12x² - 12x + 8, but no idea where the fuck that little piece of shit came from :I

[Vector]

Can someone explain how you go from
(x-2)³ + 12(x²-x) + 8
to
x³ +6x²

(x-2)^3 + 12(x^2-x) + 8 = (x^2 -4x + 4)(x-2) + 12x^2 -12x + 8
                                   = (x^3 -2x^2 -4x^2 +8x +4x -8 + 12x^2 -12x + 8
                                   = x^3 -6x^2 +12x -8 +12x^2 -12x + 8
                                   = x^3 + 12x^2 -6x^2 + 12x -12x +8 -8
                                   = x^3 +6x^2

[Descan]

Ah, I see, that was my issue. I just went from (x-2)³ to x³ - 2³, and didn't unpack it.

[Vector]

>_______________>

[Descan]

Yay! There's a word for it!

[Powder Miner]

2^x - 2^-x = 12
I need to solve for x.
How do I simplify so I can make a reasonable logarithm?

[MagmaMcFry]

Substitute x=log2(u), then solve for u, then use x=log2(u) to find x.

Spoiler: Complete solution (click to show/hide)

2^x - 2^-x = 12
u - 1/u = 12
u^2 - 1 = 12*u
u^2-12*u-1=0
u=6+-sqrt(37)
x = log2(6+-sqrt(37))

[Descan]

u?

[Powder Miner]

Hmm. Is there another way to solve it?

[Jim Groovester]

Not a correct way, no.

[MagmaMcFry]

Hmm. Is there another way to solve it?

There are probably oodles of approaches to this. For example you could just skip the substitution and do the quadratic reduction manually:

2^x-2^-x = 12
4^x-1=12*2^x
(2^x)^2-12*2^x-1=0
(2^x)^2-12*2^x+36-36-1=0
(2^x-6)^2-37=0
(2^x-6)^2=37
2^x-6=+-sqrt(37)
2^x=6+-sqrt(37)
x=log2(6+-sqrt(37))