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17:51:34 my laptop is the brokenness now. :(
17:51:39 "was fun while it lasted".
17:52:08 although in retrospect ~1550eur was a bit much for a machine that worked for six days.
18:42:44 uh
18:44:31 that's over 250eur/day, after all.
19:35:20 ZeroOne: btw, while cycling back home i got a new idea regarding that language where one task can be done in only one way
19:35:47 elaborate. clog's back.
19:41:10 so, if every program is presented as a mathematical function, say f(x), then the language could be defined so that only the form of f(x) that is the "most" simplified is legal
19:42:03 fizzie: at least with polynomials it should not be possible to get a function (read: program) that behaves in the same way, unless the functions are identical
19:42:42 so the core idea is that there needs to be some set of simplication rules
19:42:56 wellll, yes, if you limit your mathematical notation enough.
19:45:12 polynomials are safe in that way.
19:46:17 actually I think we proved in one of our maths courses that n-degree polynomials are.. ah, what's the word? n-ulotteisen avaruuden kanta, tavallaan. yksikäsitteiset koordinaatit ja näin.
19:46:28 :)
19:47:52 it's still pretty limited, though, but at least your 'programs' can now take a set of real numbers as input and return a set of real numbers as output.
19:49:33 hmm. interesting.
19:50:02 fizzie: "radix of n-dimensional space, sort of. unambiguous coordinates etc."
19:50:17 thanks.
19:50:27 it was the 'radix' word I wasn't so sure of.
19:52:42 complex functions make for some really weird plots.
19:53:13 if you happen to have mathematica and/or an equivalent piece of software (perhaps gnuplot would be enough), plot Plot3D[Arg[f2[x, y]], {x, -5, 5}, {y, -5, 5}, PlotPoints -> 100,
19:53:16 Mesh -> False]
19:53:20 erhm.
19:53:25 lindi-: so every program would be a polynomial of degree n, or something?
19:53:30 there's a piece of misplaced copypasting.
19:53:44 since it doesn't include the actual function.
19:54:36 f2[x_, y_] := x^3 + 3x y^2 - 3x + i(y^3 + 3x^2y - 3y), where 'i' is the imaginary unit. when you plot the magnitude of that as a surface, it's as clean and smooth as.. well, as any function you'd like to name, but the argument plot..
19:55:18 ok... unfortunately I don't have any math software installed... I guess I should soon get something.
19:55:34 ZeroOne: well, if you extend the simplification rules you can support many other types too
19:56:08 although I once made a program that was able to draw 2d-diagrams of polynomial functions. in qbasic, even. ;P=
19:56:19 i wrote one in ti86 basic :P
19:56:58 and you need to limit yourself to finite-degree polynomials, otherwise you could have an infinitely-long program which would do the same as 'sin x' (if you write some simplification rules to allow trig. funcs)
19:56:59 yeah. but TI features all math functions built-in, qbasic doesn't.
19:57:27 heh, 2d plots of polynomials seems to be a popular project.
19:57:29 fizzie: gotta love mr. Taylor.
19:57:48 I think it was an exercise-or-sort-of in my high school.
19:57:57 fizzie: but what if "sin x" is not allowed but only the 'sarjakehitelmä'
19:58:08 mooz at least wrote a.. pretty advanced one, in qbasic.
19:58:47 fizzie: that reminds me, where does mooz irc nowadays? /whois shows no channels but maybe they are secret/private
19:58:56 heh. :) we only had some really stupid piece of some software the teacher had made. copied that from paper and got disappointed when it did practically nothing: "oh, no, it shouldn't do anything, it's just the initialization function thingie!" was the teacher's response ;PP
19:59:59 who is mooz?
20:00:05 I'm not sure, really. I speak mostly in a query. he's been gradually moving to an apartment in kamppi, probably parted from #da during that time.
20:02:21 ZeroOne: apt-get install maxima and plot3d(x^3 + 3*x*y^2 - 3*x + %I*(y^3 + 3*x^2*y - 3*y), [x,-5,5], [y,-5,5]); if you want to see the function fizzie pasted
20:02:30
20:03:31 hmm, you can even rotate the plot
20:03:54 lindi-: apt-get doesn't work under winblows, you know.
20:04:57 how does maxima plot complex-valued two-variable functions?
20:05:08 ZeroOne: http://prdownloads.sourceforge.net/maxima/maxima-5.9.0.exe?download
20:05:35 with mathematica I need to use either Abs[] or Arg[] (well, or Re[] or Im[]) in front of it.
20:05:37 fine... :p
20:07:12 fizzie: iki.fi/lindi/maxima_plot3d.png
20:07:28 ah, but that's just the real part of it.
20:07:46 you want to use abs() or arg() (or corresponding maxima functions) if you want to see the strange-ness I mentioned.
20:08:39 and actually while the magnitude plot with abs() looks smooth, looking at it with the ranges [-2, 2] reveals otherwise.
20:12:23 fizzie: plot3d(cabs(x^3 + 3*x*y^2 - 3*x + %I*(y^3 + 3*x^2*y - 3*y)), [x,-5,5], [y,-5,5]); looks like iki.fi/lindi/maxima_plot3d_2.png
20:13:13 yes, doesn't it look pretty smooth?
20:15:55 yep
20:16:14 well, what about either a) the range [-2, 2] or b) the argument of the result?
20:16:28 argument?
20:16:34 also called 'angle'.
20:16:39 I don't know which maxima function it is.
20:18:35 extra credit for proving in which points of C that function is a) complex differentiable b) analytic. (the b part was relatively easy, but I'm not sure how I'm supposed to do that a part.)
20:21:01 fizzie: now it does not look that smooth either indeed, iki.fi/lindi/maxima_plot3d_3.png
20:22:09 uh-huh.
20:28:57 eh, scratch that, now I'm not quite sure of the b part either. :p
20:29:13 no problem, i'm even more unsure :)
20:29:27 fizzie: were you in L1 or C1 math btw?
20:29:52 L1.
20:31:13 I thought the fact the function satisfies the cauchy-riemann equations only in the points of the imaginary and real axis, and therefore not in any real region of C, would be enough in stating it's not analytical. it still _might_ be, but..
20:32:48 the canonical definition would be that a complex function is analytical in point z0 if a derivative f' exists for all points z in a neighbourhood of z0, but in order to use that would require me solving the a) question.
20:33:02 hrm
20:34:24 some days I'm not quite sure why I'm in the L* series of maths courses anyway.
20:36:04 ah, right. in order for f to be analytical in a region G it would have the partial derivatives of its component functions satisfying cauchy-riemann, and since that isn't the case it can't be analytical. (I think.)
20:37:40 I still need some way to solve that 'a' part. I'm not sure how I'm supposed to prove that lim h->0 (f(z+h)-f(z))/h exists when h is a complex number and can approach 0 from any direction in the complex plane.
20:37:54 (and this for any z.)
20:38:35 don't know much about handling those (yet)
20:40:02 neither do I, but I'm sort-of supposed to. there could be a better way to do this though.
20:42:38 but at least i've learned something this week. when i cycled back home today i could also do some partical fraction decompositions in my head :P
20:42:52 s/partical/partial/
20:45:33 I freely admit I don't remember ~anything about how to do a LU-decomposition of a matrix.
21:03:06 anyway, need to sleep so that i can be awake on the math lecture
21:03:11 s/on/during/ :P
21:03:19 hnu.
21:03:23 g'night.
21:26:23 Oh, yippee... Complex analysis.
21:29:08 loads of fun.
21:30:09 Actually it's not too bad. I'm taking a course on it this semester.
21:37:48 oh? then you can tell me what's the easiest way to show where the derivative f'(z) for a complex function f(z) exists.
21:38:40 looking at the definition of f' (that is, f'(z) = lim h->0 (f(z+h)-f(z)/h) is awfully untrivial since h can approach 0 from any direction and I need to look at all points z.
21:41:07 Well... "Easiest" depends on the situation. But the Cauchy-Riemann equations are a safe bet.
21:45:48 but are CR enough? I mean, they definitely are enough to prove whether the function is analytical anywhere, but f'(z) could still exist in some isolated points without it being analytical there.
21:49:38 A function f=u+iv from an open subset of the complex plane to the complex plane itself is holomorphic if u and v are differentiable, and if the partial derivatives satisfy the relations du/dx = dv/dy and du/dy = -dv/dx (where the ds are "soft ds". I.e. partial differentiation.)
21:50:11 s/holomorphic if/holomorphic if and only if/
21:51:36 It can be applied to single points of the function as well. To prove differentiability in, say, (x_0,y_0).
21:54:12 I'm not sure what you mean by f'(z) existing in isolated points without being analytic in said points.
22:01:15 my definition of an analytic function says a function is analytic at point z0 if f'(z) exists for all z in a neighbourhood of z0, and it has a side remark saying "existance of f'(z) in a single point is not enough"
22:01:20 this in my lecture notes.
22:04:33 Hm. My notes don't mention analyticity in a single point. In gives the usual limit definition of differentiability and says a function is holomorphic if it is differentiable in all points. I suppose it makes sense that there needs to be a neighbourhood around z_0. You need more than a single point for an open subset.
22:06:25 If you can read Danish, you can take a look yourself. They're online at http://math.ku.dk/noter/ The course is called 2KF.
22:23:04 well, maybe my notes try to say that it needs to analytic in the open subset, but simply having f'(z) (that is, the limit existing) at a single point doesn't yet make it analytical.
22:23:15 s/al.$/./
22:23:31 who knows, I was probably at least half-asleep when I wrote that.
22:24:39 but our exercise question gives us a function and asks "in what subset of C the function f: z -> ... is (a) differentiable (b) analytic".
22:25:35 and I can easily say it's analytic nowhere in C, because it satisfies CR only at z=0, but I'm not quite sure about the differentiability thing.
22:26:19 Well... It's differentiable in the points where the limit exists... I guess you'll have to do it the hard way.
22:28:05 agh. I tried writing open "f(z+h)-f(z)" (with z=x+iy and h=hx+ihy) and ran out of paper width. granted, it's only a c5 envelope I'm writing on, since all proper paper is several meters right and I can't get myself off the chair
22:28:32 What's the function?
22:29:37 f: z=x+iy -> x^3 + 3xy^2 - 3x + i(y^3 + 3x^2y - 3y). I'm tempted to abuse the component functions u(z) and v(z) somehow here, since they are more humane.