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Comments:
<0> you would use the difference of two squares to factor it, right?
<0> which gives (x^3 - 1) (x^3 + 1) = 0
<0> is this correct so far?
<1> yes
<0> then how do you produce the solutions from this?
<1> if a*b = 0 then a = 0 or b = 0
<0> How do we use an index with sqrt over irc?
<0> rather, when we wish to take the root of something
<1> n-th root of x is x^(1/n)...
<0> right
<0> I'll use that method then
<0> given (x^3 - 1)(x^3 + 1) = 0
<0> then x = 1^(1/3)
<0> correct?
<1> thats 1 solution
<2> what is 1^(1/3)
<1> read what i wrote
<0> can't be right
<0> My book lists the following solutions
<1> why not
<1> what's 1^(1/3) = ?
<2> what is the cube root of 1^1
<0> cube root of 1 is 1
<2> so then, simplify that expression
<0> my book lists the following solutions: (-1 +/- (i*sqrt(3)))/2
<0> (1 +/- (i*sqrt(3)))/2 and +/- 1
<0> how did it derive the first two?
<2> There are also imaginary roots, the fundamental theorem of algebra says that for an nth degree polynomial there are n solution
<0> how do we use that to derive the first two solutions from (x^3 - 1)(x^3 + 1) = 0?
<0> further, if that is true, then how can we have 2nd degree polynomials with one solution?
<0> if we have a quadratic equation where the discriminate yields a value equal to zero, then it is said to have one solution
<2> that is called a 'double root'
<0> s/discriminate/discriminant/
<2> because technically the solutions are (b+0)/2a and (b-0)/2a
<2> -b rather
<0> aperson: I'm sorry, but I don't see how that follows from (x^3 - 1) (x^3 + 1) = 0
<2> the double root discussion is tangential because you brought it up
<2> I haven't solved polynomials with imaginary roots in awhile though
<3> aperson: are you aperson?
<2> I'm just letting you know that since you have x^3 * x^3 you automatically know you are looking for six solutions
<3> i mean aPerson
<0> right
<2> Sturgeon: My name wasn't originally derived from the programming function but I did stumble across it later :P
<0> I'm mostly wondering how they derived (-1 +/- (i*sqrt(3)))/2 from (x^3 - 1)
<3> I don't know what you're talking about!
<0> it is puzzling
<0> all I am seeing is x = 1^(1/3)
<3> I'm just asking if you are aPerson from undernet :)
<2> oh, no I'm not
<0> in fact, I don't see how they derived +/- 1 as a solution
<0> after thinking about it, that would have to be +1 only from x - 1 = 0
<0> because -1^3 yield -1
<0> which is an imaginary number and would yield i
<2> hmm?
<2> Look at (x^3 + 1)
<2> Insert -1 into it
<0> anyne have any idea as to how those solutions were derived?
<4> (New Paste) Nick: Jas-Nix (70.98.110.50) :: (Desc.: compiz "make" errors) :: Addy: http://nanoBin.nanoSouffle.net/1496
<5> hello
<5> does x^(-1) have an anti derivative?
<6> yes, it's log(x) + K.
<5> thanks
<7> are there any linux (or any OS I guess) programs around where I could draw a line, and it would do a linear/quad/exp regression on it?
<8> hmmmm
<9> draw with you hand?
<7> well, a mouse, but yea
<8> you mean you would have no equation of that line ?
<7> no, I want to anti-graph it more or less. :P
<8> but for a regression you have a set of points not a line
<8> the result of the regression might a line
<9> With help of your mind you can use OpeOffice.org Calc for it
<8> JoKa :?
<8> spreadsheets do regression, but again you need to enter a set of points not the line (being the result) to begin with
<8> narg : ?
<7> yea, thats what I thought. I doubt there is anything like that around, but it was worth asking. I'm trying to find a good equation for something.
<8> narg : no the problem your problem makes no sense that way
<7> How so?
<8> doing a (linear) regression on a line will get you as result exactly that line itsself (per definition)
<8> regression is to fit a curve into a given set of points
<7> ok, I want the equation for a line. What would you call that?
<8> input : set of points output: a curve (a line for instance)
<8> oh
<7> I was thinking it would just sample the line for points, and use them.
<8> well drawing a line by by hand and getting the equation of that line you can use geometric software like geogebra but that would only work for lines
<7> Well, I'm actually going for an inverted exponential growth, so that wouldn't really work. I'll just go back to guessing and graphing. Thanks :)
<8> but if you draw any curve and wnat to sample its points to compute a (differenT) regression curve through it
<8> then you might to have to split that into parts:
<8> 1.) draw and receive the (sample) point set
<8> 2.) p*** the point set to a regression procedure
<8> there are plenty of programs doing 2.) (like spreadsheets mentioned by joka or most CAS)
<7> right, the first part is what I'm after
<8> 1.) might be more tricky and i haven't seen anything doing 1.) + 2.) in 1 step
<7> My ti calc can do the second.
<8> narg : yes many calcs do that too, and TI89 and up actually have a CAS (Derive) implemented
<8> for 1.) you probably can use drawing programs, which allow access to their drawing data (internally the curve is usually a set of points to them)
<7> Go manually read inkscape's file format perhaps?
<8> maybe (no idea)
<7> well, thanks for the ideas.
<8> actually if don't mind to write a program, you can write such thing yourself
<8> relatively easy, since you contril/capture the mouse coordinates and in addition to draw them you simply dump a copy into a file
<7> true
<7> Might be an interesting project to do...
<8> which also means some tool capturing mouse coordinates might do the job too
<8> narg : i think you can doit it fairly quick with java/delpi/C#
<7> pity I use python and C++ then; Its possible with Qt though.
<8> they will workk to
<8> you just need an (easy) interface exposing the mouse data and draw in a window
<8> because if any of them needs to be done yourself through lowlevel stuff then it is more work and less of a quickie
<7> I'm going to check how transparent SVGs are, and then I'll look into that
<8> good luck
<10> what about using a bunch of taylor series to approximate the curve you want.... like, from 0 to .5, use equation a, from .5 to 1, use equation b. get a bunch of points from those two tangent lines
<10> i mean, it's kindof a hack, but it's a far cry from implementing some crazy software package
<8> tjcoder : the problem is to retrieve the points of the curve not the regression itsself
<10> _kmh_ right..... but if you had some points and some approximate slopes you could generate as many points as you wanted
<10> as long as each individual tangent you drew was fairly short
<10> or, rather, the part of each slope you used was fairly short
<10> blah... stupid people touring in chicago
<10> and not seattle
<9> narg: There are programs that recover data points from plots.
<7> any examples come to mind?
<9> http://getdata-graph-digitizer.com/
<9> http://www.datatrendsoftware.com/
<11> mbot, dance.
<11> Pfft. :P
<12> Hi, I need a little help with basic mathematical notation.
<12> That is, I don't know how to write one thing in text, such as plain text here in a terminal, and how to name it.
<12> It's how do differentiate between different kind of log().
<12> Forinstance this:
<12> log(1000) = 3 <- If I translate from what I remember from Swedish school it would be called "the ten logarithm of 1000 is three.
<12> Should I write that as log_10(1000) = or log^10(1000) = 3 or what?
<13> log_10 will work
<12> Ah, ok. Is log^10 the same thing or something else?
<13> log^10 might confuse some people. anotherone is just log10()
<12> I just read a paper where it seems they meant log^x was something else.
<13> which is how programs do it
<12> Oh!
<9> log_10^2(1000) = (log_10(1000))^2 as for sin^2()
<12> So this is the best notation then: log10(1000)=3 and log2(8)=3
<13> log10 is a function, take the log base 10 of the number, log2(), etc
<12> Right.
<13> Mole2: shrug, log_10 is just as well
<12> JoKa: Ah, I see what you mean withy log^x then, thanks.
<12> So what I was looking for was log_10() or log10().
<12> Good.
<12> Thanks a lot guys.
<12> Just if anyone is curios. The reason I was needing this was to write stuff about how many lookup steps different kinds of p2p DHTs take. (DHT = Distributed Hash Table, the kind of distributed databases we use in advanced p2p systems.)
<12> Oh, one more thing, when people write just log() what base do most people ***ume that means? 2 or 10 or e or something else?
<12> I know p2p people and programmer usually ***umes it means 2...
<12> *programmers
<9> Mathematica default log base is e. C language ("math.h") log function gives natural logarithm (base e). Calc log gives log_10, ln gives log_e.
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