From: rusin@vesuvius.math.niu.edu (Dave Rusin) Subject: Re: Riemann zeta function question Date: 23 Nov 1999 21:16:42 GMT Newsgroups: sci.math Keywords: images of zeta along vertical lines In article <7xeq8itimp33@forum.swarthmore.edu>, Leroy Quet wrote: >Let z_n= a +i b_n, where a is real and each b is a distinct real >=0. >Is there an example of where >zeta(z_1) =zeta(z_2) =zeta(z_3) not= 0 ? Yes. >What about the specific case of a=1/2? I would guess not. >If there is a case for 3 z's, is there a case for an infinite number >of z's? I would guess not; indeed I wouldn't even expect four z's. For a specific a, the function f(t)=zeta(a + i t) describes a map from the real line into the complex plane, i.e. a (smooth) curve. When a=1/2, this curve passes through the origin infinitely often; for other a it is conjectured never to pass through the origin. Here is a portion of (the image of) this curve when a=1/2 (for 0 < t < 20.5): + AA A A A A A A + AA A A A 1 + AA A A + A A A + A A +AA A 0.5 * A * AAAAAAAAAAAA AA * AAA AAA A * AA AA A -*-++-+-+-+-+-+-+-++-+-+-+-+-*-++-+-+-+**-+-+-++-+-+-+-+-**+-++-+-+-+-+-+*+ -1.5 -1 -0.5 0 *A 0.5 1 1.5 2 A +AA AA AA A A * AA AA AA A -0.5 * AAA AA A A +AA A AAA AAAA A +AAA AAA AAAAAAAA A A A A A A+ A A A A -1 + A A A + A A A A A + A A A A A A Note that there is a finite number of self-intersections. The map H(s,t) = f(s)-f(t) is a smooth map from the closed square to the complex plane, and so the pre-image of a point will normally be a finite set; in this case H^(-1)(0) consists of the four points roughly at (1.4, 13.4) (and (13.4, 1.4), of course) and (1.0, 20.4) (and (20.4, 1.0) ) corresponding respectively to the self-intersections of the curve near 0.3-.6i and .14-.7i . What's important here is the pairing of these four values of t=1.0, 1.4, 13.4, 20.4 : the sequence is, let's say, ABBA. Now, if you modify the parameter a just a little, you'll modify the curve, but you'll retain the basic shape (loosely speaking, it's a a little more than two complete loops encircling a region near 1 + 0 i) There are always two intersection points, corresponding to four real numbers t, paired off in the same order ABBA. On the other hand, when a = 2.5, the curve is still a little more than two loose coils around 1 + 0 i, but the points of self-intersection have shifted: the two points in H^(-1) (0) are near (2.5, 12.6), (12.3, 20.3) (and again H(t,s)=0 whenever H(s,t)=0 ) corresponding to points where the curve self-intersects near .89 -.15 i and .93 -.17 i respectively. But this time, the four relevant values of t = 2.5, 12.3, 12.6, 20.3 are paired off in the pattern ABAB. Well, with a little hand-waving, you can convince yourself that the four points in H^(-1)(0) vary continuously with a , and so with a mean-value argument you can convince yourself there must be a value of a (between a = 0.5 and a = 2.5, in fact) for which the two inner values of t coalesce, that is, there is a triple intersection of the curve f(t). Using Newton's method I find more precisely the values of t in the last case (a=2.5) to be t=2.485889757, 12.27813806, 12.59688499, 20.29518207. By modifying a gradually (using Newton's method for each a ) I find that around a = 1.8658816 we see the two inner values of t approximately agree (t = 12.71311045). So there appears to be a degenerate self-intersection in this particular curve. This gives the approximate solution zeta( a+ 2.254915348 i ) = zeta( a+ 12.71311045 i ) = zeta( a+ 20.44134167 i) = 0.813740069 - 0.242173699 i, when a = 1.8658816 . So we find three points on a vertical line where zeta takes on the same value. But note that small pertubations destroy this multiple intersection, that is, the property of having simple intersections is an open condition on the set of vertical lines. This leads me to expect that in general the curves have only simple intersection (and thus that the curve with a=1/2 has only simple intersections away from 0), and to imagine that in fact there are no quadruple intersections at all, save the multiple solutions of zeta( 1/2 + b i ) = 0. In particular, I would be very surprised to discover that there is a curve of this type with infinitely many arcs intersecting at a single point, as the poster conjectures. Of course, this is not a proof that such extraordinary intersections cannot occur. I cannot imagine such a proof is possible with today's tools. dave ============================================================================== From: rusin@vesuvius.math.niu.edu (Dave Rusin) Subject: Re: Riemann zeta function question Date: 24 Nov 1999 17:19:19 GMT Newsgroups: sci.math I wrote: > For a specific a, the function f(t)=zeta(a + i t) describes a map from > the real line into the complex plane, i.e. a (smooth) curve. When a=1/2, > this curve passes through the origin infinitely often; for other a it is > conjectured never to pass through the origin. > > Here is a portion of (the image of) this curve when a=1/2 (for 0 < t < 20.5): [image deleted] > Note that there is a finite number of self-intersections. In article <383B3F60.44DF076F@crypt0.demon.co.uk>, Hugo van der Sanden wrote: > >Even when a=1/2? Given the image you drew, and 'passes through the >origin infinitely often' it seems unlikely: are there infinitely many >loops, all passing through the origin, but none intersecting with each >other? I was definitely refering only to the _portion_ of the curve. Certainly I expect an infinite number of self-intersections for the whole curve. When a=1/2, the curve is known to pass through the origin infinitely often, so if there were only finitely many self-intersections then, restricting to the portion of the curve coming after the last intersection, we would have an infinite number of closed curves in the plane meeting only at the origin, perhaps resembling the Hawaiian Earrings or an infinite bouquet of circles. That would be an extremely strong geometric restriction on the behaviour of this curve, which from the data I have seen is instead a rather meandering path. Really I suspect that every single loop of the curve (i.e. the portion between successive zeros of zeta) is intersected infinitely often by other parts of the whole curve. I guess it's possible to _prove_ that any restriction of the curve to a bounded interval has only finitely many self-intersections, but that would require some analysis, not a mere appeal to compactness, as can be seen by the function f(t) = (1-t^2, t*(1-t^2)*sin(pi/(1-t^2))) which on the compact interval [-1,1] traces out a curve with self-intersections at each point ( 1/n, 0 ). In a separate post, Richard Carr reminds me that my statement > When a=1/2, this curve passes through the origin infinitely often; for > other a it is conjectured never to pass through the origin. is sloppy. The "trivial zeros" of zeta are the points -2, -4, -6, ... (but zeta(-n) is _not_ zero when n is an odd natural number). It follows from the functional equation that there are no other zeros of zeta outside the critical strip (0 <= a <= 1). By the way, I'm not claiming I have looked in detail at these curves except the special case a=1/2, so there may be interesting behaviour in some other cases which I have overlooked. dave