Hi Branden, On Wed, Nov 01, 2023 at 11:56:32AM -0500, G. Branden Robinson wrote: > Hi Alex, > > At 2023-11-01T16:00:16+0100, Alejandro Colomar wrote: > > On Wed, Nov 01, 2023 at 02:02:10PM +0000, Helge Kreutzmann wrote: > > > Without further ado, the following was found: > > > > > > Issue: [-pi,pi] means both -pi and pi are included, this does not > > > make sense, either one must be out of the interval? > > > > > > "The logarithm B<clog>() is the inverse function of the exponential > > > " "B<cexp>(3). Thus, if I<y\\ =\\ clog(z)>, then I<z\\ =\\ > > > cexp(y)>. The " "imaginary part of I<y> is chosen in the interval > > > [-pi,pi]." > > > > I don't know this function. Please suggest a fix, and CC glibc so > > that they can review the change. > > The complex logarithm is the inverse function of the complex > exponential, with which you may be familiar if you've taken a course in > ordinary differential equations. Yup, I have. This reminds me of some old TODO I had: take a course on multivariable calculus from a professor I like: <https://www2.math.upenn.edu/~ghrist/BLUE.html> The course on single-variable calculus from him was quite good. <https://www2.math.upenn.edu/~ghrist/calculus.html> > > Euler's formula famously relates the trigonometric and exponential > functions. > > exp(i*x) = cos x + i sin x > > While the exponential function is aperiodic, that is for y=exp(x), no > value of y ever occurs more than once for any real x, this is not true > of a complex x or (equivalently[1]), the complex exponential, or real x > multiplied by the imaginary unit i as seen in Euler's formula. > > (Recall that for any x, sin(x) and cos(x) take on values in the interval > [-1, 1], and i is a constant imaginary unit that we can interpret as a > y axis. Thus the complex exponential maps any real x to a point on the > unit circle.) e^(ix) is something I can visualize thanks to Euler, and e^(ix + y), well, not so much, but by combination of e^(ix) and e^y I can understand, but > > In other words, for us to have an inverse function for the complex > exponential, we must impose a restriction on its range, lest it give us > an infinite vector of solutions. Geometrically, the complex logarithm > asks, "given a point on the unit circle, which value of x to the complex > exponential corresponds to it?" But there is so single answer to that > question. It is still a useful one to ask, so we can apply a constraint > on the range of the solution which will make the complex logarithm > one-to-one. I can't yet visualize a complex or simple imaginary logarithm. I'm not so imaginative at the moment. :| > > We do a similar thing for the arc sine function. Given a value c in > [-1, 1], what angle theta has c as its sine? There is an unbounded > number of answers. If you plot y=sin(x), you will see that y takes on > every real value from -1 to 1,[2] repeatedly and unendingly. > > At any rate, I think this statement: > > > > Issue: [-pi,pi] means both -pi and pi are included, this does not > > > make sense, either one must be out of the interval? > > Needs more support. clog(1) = pi and clog(-1) = -pi. The limits of > machine representation are applicable here, so you'll only ever get > values "close to" ±pi anyway. > > Also, POSIX and the ISO C committee didn't seem troubled by this; > the same closed interval is issued in POSIX Issue 8 Draft 3, which says > (in so many words) that the function's definition comes from ISO C99. > > Oh, now that I'm about ready to send this, I see Jakub Wilk made the > same point far more concisely. Who's surprised? > > Regards, > Branden > > [1] ...by applying of properties of powers such as x^(ab) = x^a * x^b. > > [2] I, uh, don't actually have a proof of the claim "every value". And > in fact I am unlikely to ever have one. Per Niven's theorem, the > elementary trigonometric functions never take any rational values > _except_ 0, ±1, and ±1/2.[3] This may be a startling result to > insightful students of elementary calculus, as, if one is paying > attention, one should then wonder why we can say with certainty that > any of these functions are differentiable, since there are jump > discontinuities in them--in fact a countably infinite number of > such discontinuities.[4] You mean discontinuities in the representation, right? Not in the function, I presume. Students should have in mind that a representation cannot be perfect in some cases. You can ask the same student to measure the rope that circles a circumference of radius 1; it'll have a hard time with the ruler. :P Cheers, Alex > > Since our domain of discourse is computer arithmetic with finite > precision, it's an academic question, as with sufficient precision > you can have a sine function that will produce every representable > value within the function's range[5]. > > [3] https://en.wikipedia.org/wiki/Niven%27s_theorem > > [4] I think it's a set of measure zero but I am now exceeding the limits > of my training. I need more topology and theory of functions of a > real variable. And to learn Lebesgue integration. > > [5] I don't actually have a proof of that, either. It seems not > impossible to me that the way IEEE floating point is defined means > that functions with the properties that trig functions have might > skip some representable values due to the nonlinearity of that > representation (the magnitude of error in floating-point math is not > constant, which is a reason some applications prefer fixed-point). > An "obvious" possibility is the set of rational values that have a > precise floating point representation in base 2, so 1/4th, -1/8th, > and so. I need more numerical analysis, too.[6] Anyway, one > counterexample disproves my claim, so let's settle for "almost > every", where we are using "almost" in a hand-wavy manner, not the > way someone studying the Dirichlet indicator function might. > > [6] Obligatory citation of mandatory reading (Goldberg 1991): > > https://ece.uwaterloo.ca/~dwharder/NumericalAnalysis/02Numerics/Double/paper.pdf > > Press et al.'s books on numerical methods ("numerical recipes in > $LANG") are surprisingly controversial. -- <https://www.alejandro-colomar.es/>
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