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Comments:
<0> what's something like a lagrange polynomial that will allow specifying slope too?
<1> spline
<1> s
<0> k
<1> actually it depends on the context what's best, but the lagrange interpolation polynomial doesn't use slopes just points
<0> i'm going to have a hard look at polynomial regression
<2> _death, there's a general way to solve for a polynomials with specified values, derivatives, second derivatives, and so on.
<0> i thought that there must be
<2> but do you want a polynomial at the end?
<2> That's the first question.
<0> no
<2> I didn't understand your notation for your data.
<0> a polynomial would be acceptable though
<2> Well... in that case, write down a general polynomial of "sufficient" degree
<0> suppose i have several samples of the form (dy/dx, X, y) where X is a vector, y is a real value, and dy/dx is a ... i don't know what it is called
<1> _death : regression is another thing though
<2> first derivative?
<0> yes, but i'm not sure what to call that sort of derivative
<2> what's y?
<1> _death : is the function R^n->R =
<1> ?
<0> yes
<2> OH!
<0> i'm just trying to come up with a polynomial to fit a loess curve
<0> (though i think a trig function would be better)
<1> so that would be n-di splines or n-dim interpolation
<3> _death: a cubic spline is a good starting point
<2> dy/dx is then the partial derivatives?
<1> if you want to avoid the partition you get with splines you can try bernstein polynomials too
<1> kercyr : the gradient maybe ?
<0> kercyr: yes
<0> partition with splines...
<0> i am making a log of this conversation so i can look everything up later
<4> What is the cardinality of the set of smooth, continuous real->real functions ?
<3> _death: you need a different spline for each three points with cubic splines
<1> pgm : he has a R^n->R function
<3> oh
<2> Olathe, powerset of the continuum?
<1> actually splines will work, but it might be really messy
<4> kercyr : That seems right, thanks.
<0> i think that fitting trig functions of the form b_0 + b_1 * sin(a_{10} + a_{11} * x_1 + a_{12} * x_2 ....) + b_2 * sin(a_{20} + a_{21} ...) ... is a good idea
<2> I don't think _death actually wants a surface that hits all the points.
<0> i'm not even completely sure why i am interested in this problem to be perfectly honest; i'd like to be able to capture underlying form without necessarily storing all of the detail
<0> thanks to everyone who helped; i'll do a bit of reading before i ask anything else on this topic
<4> _death : Are you looking for compression ?
<2> http://math.uprm.edu/~wrolke/esma6665/locreg.htm
<2> he's doing local regression, I think.
<4> Oh.
<0> Olathe: looking for capacity for extrapolation and faster evaluation
<0> also looking for something that i can take a higher-level view of
<0> got to go
<5> if 2^x + 2^x + 2^x + 2^x = 2^7
<5> What's x?
<5> and why?
<6> 8^x
<5> why?
<6> add the like terms, unless im missing something significant here.
<2> random, does that even make sense? x=8^x?
<6> x doesn't equal 8^x
<6> 2^x+2^x+2^x+2^x=8^x
<1> 2^x + 2^x + 2^x + 2^x = 2^2*2^x=2^7
<1> 2^(x+2)=2^7
<1> x=5
<5> thanks.
<1> random : make sense ?
<5> wait
<5> how did you get 2^2 * 2^x?
<2> random, think about it.
<1> 2^2=4
<7> sysfault: no, that's not right
<2> 1^x+1^x does not equal 2^x, for example.
<5> right
<5> it's 2*1^x
<5> so
<5> 2^x+2^x = 2*2^x
<5> then x3
<5> and x4
<1> yes
<5> and 2^2=4
<5> got it
<5> thanks
<5> :)
<7> You have 4 * 2^x = 2^7, so 2^2 * 2^x = 2^7, and so 2^(2+x) = 2^7. Taking the base 2 logarithm of both sides, we get 2 + x = 7, and so x = 5.
<1> Cale : actually comparing exponents is good enough
<7> Well, how do you prove that's good enough? :)
<5> 2^2*2^x=>2^(2+x)
<1> Cale : rule of thumb
<7> kmh_afk: You take the log :)
<1> Cale : na
<7> "rule of thumb" isn't a proof of anything :)
<1> Cale : nor does the nlog
<8> O_o?
<1> :)
<2> x^x = x^(x^2-1)
<7> In particular, log is an injective function, so it's okay to say that if log u = log v, then u = v.
<1> Cale : injective on R+
<7> yes
<7> or, for that matter, f(x) = 2^x is as well.
<1> but you don't need it anyway
<8> Hiding pieces of thought process like that when you're learning isn't to be recommended.
<1> comparing is not hiding
<8> it is if the purpose of the lesson is to learn how to use logs ;)
<1> it is just using less theory (not being needed here)
<8> But checking the backlog, that doesn't seem to be the case.
<8> So I drop it. :D
<1> merus : well if the purpose is to learn logs - ok, that's different
<8> Yeah, I didn't read the whole backlog. My bad. ^^;
<8> But there are some things they do in algebra cl***es that just makes me shake my head.
<1> i just looked and it as some early precalc highschool thing
<7> Let's use another example... 1^2 = 1^3, so, comparing exponents, 2 = 3 :)
<2> you learn logs in precalc.
<1> i.e. all he has to know are integer exponents
<2> (don't you?)
<8> The whole metaphor between algebra and "moving things about" really screwed me up >:(
<1> kercyr : there's no such thing as precalc here
<7> This kind of lack of reasoning can get you in trouble :)
<1> Cale : it doesn't
<8> I wish they wouldn't try to teach calculus in HS here :/
<7> Of course, if I actually was forced to use log there, it would be obvious that I'm dividing by 0 :)
<1> Cale : you would use the cancellation law with zero divisors either
<1> but that's no argument against using where it is appropiate
<9> this channel's cool
<8> B)
<1> merus : here = where ?
<8> US
<7> But I've just applied the exact same reasoning you gave to a new situation and got a contradiction
<9> lol, I like how you ask a question with an equation :D
<2> I wish they'd stop teaching the intergrate math stuff.
<8> I wish they'd just teach two years of algebra, followed by two years of geometry
<8> or maybe the other way around
<1> Cale : not really, we were looking at 2 not 1
<7> It's a silly thing, since yes, you are allowed to do what you did, but you didn't actually say why :)
<8> but certainly not interwoven; this Alg 1 -> Geometry -> Alg 2 -> Pre-Calc track is annoying beyond annoying.
<1> Cale : my point was logs are not required for that particukar problem
<7> Well, something more is needed :)
<1> Cale : and not required for a why either (in that context)
<7> Injectivity of 2^x will do
<1> Cale : no it isn't
<1> Cale : yes
<2> Really, you could have just told random to check if x=5.
<2> and had a wunderbar proof.
<7> kercyr: However, you wouldn't show that that's the only solution that way
<1> kercyr : well to ***ure x=5 is the only solution you need injectivity
<1> wunderbar sounds sonderbar :)
<2> merus, model theory?
<2> kmh_afk, is random aware that there are no other solutions?
<8> http://en.wikipedia.org/wiki/Model_theory
<2> merus, I know... I was wondering why were you looking at it.
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