From: ullrich@math.okstate.edu (David C. Ullrich) Subject: Re: Band-limited wiggle limit? Date: Sun, 10 Sep 2000 14:57:31 GMT Newsgroups: sci.math Summary: [missing] On Sun, 10 Sep 2000 14:13:19 GMT, Ron Hardin wrote: >David C. Ullrich wrote: >> >Replace the FOURIER TRANSFORM of the original band-limited function by >> >a linear combination of its derivatives. The new Fourier transform >> >will have the same (or smaller) compact support. The effect on the >> >original function is to multiply it by a polynomial whose coeficients >> >are the same as those of the linear combination (except for factors of >> >i). Choose the polynomial to have lots of zeros on the given >> >interval. >> >> Why yes, that _is_ nice. > >That is what I didn't follow in the original functional notation. > >It multiplies the FT by a polynomial (in iy), not the bandlimited function, >and the zeros are in frequency space not time. No, the FT _is_ the band-limited function. Here in math land we don't really have a time side and a frequency side - the two are really symettric. Jargon: a function f on the line "has compact support" if it vanishes outside some bounded interval [a,b]; ie if f(x) = 0 except for a <= x <= b. Now the FT of a function with compact support is a band-limited function, and _also_ the FT of a band-limited function has compact support. >So either I'm missing something elegant, which is possible - that's the >sort of thing I miss a lot - or it's close to an idea that does work: >put a lot of band limited functions in play in the interval, approximate >powers of x by linear combinations of them, and build any old polynomial >you want to any accuracy you want over the interval, in particular having >as many zeros as you want. Kinda more or less sort of. Petry's original exposition was very precise, with formulas and variables and things. Kotlow's version put everything in words instead. If you didn't follow the English version you should look again at Petry's version. I don't see the point to restating it, I'm not going to say anything he didn't say. ok. I'll restate it all just as explicitly as I can: First choose a function F which is infinitely differentiable but has compact support (and is also not identically 0). Think of this F as living on the frequency side. Now pick the interval I where you want your band-limited function to have lots of wiggles (I is an interval on the time side.) Choose a polynomial P such that P has that many zero-crossings on the interval I (you can make a polynomial look like whatever you want on any given bounded interval like I - a slighlty more precise version of this is a theorem of Weierstrass.) Now define a function f by f(x) = F^(x) P(x). So f is the product of the polynomial P times the FT of the function F. This f does what you want: First, it has lots of zeroes on the interval I because P has lots of zeroes there. And it's band-limited because it is the FT of a function with compact support (in particular it is the FT of a certain linear combination of derivatives of F.) >There's no intuition that x^n can't be approximated by a band limited >function over some interval, at least in the way you might intuit that wiggles >can't be approximated by a band limited function. (It may be of course >that the other intuition is wrong...) >-- >Ron Hardin >rhhardin@mindspring.com > >On the internet, nobody knows you're a jerk. ============================================================================== From: David C. Ullrich Subject: Re: Band-limited wiggle limit? Date: Sun, 10 Sep 2000 18:59:53 GMT Newsgroups: sci.math In article <39BBB16D.3B1C@mindspring.com>, Ron Hardin wrote: > Ron Hardin wrote: > > Okay thanks. The problem was a culture clash - an engineer's frequency space > > fourier transforms are not differentiable for band limited functions, so > > I kept reading differentiation of the time function and nothing quite worked. > > Although, in this form, I now have a misgiving: namely, if the time domain > function is multiplied by a polynomial, it will blow up and the fourier > transform will not exist. A good thing to worry about but here it's no problem. It would be easier to explain if you'd quoted a bit - I don't remember what letters were used for what this morning, and I don't feel like going up two posts, looking it up and then coming back (more trouble than it sounds like from where I'm sitting this second.) Anyway, I think it was F that was the frequency-domain function. We started by saying F had compact support and was also infinitely differentiable. It follows that F^(x) -> 0 very fast as x -> infinity; in fact F^ -> 0 so fast that P(x) times F^(x) still tends to 0 even though P blows up. Amazing but true. > Usually I ignore those misgivings, but it seems to be related to the > original question, whether those wiggles can exist in a band-limited > function. > > With band limited functions, the time domain doesn't tail off very fast. Depends on where you got your band-limited function. Let's say again that F has compact support so that F^ is band-limited. If F was a very crude compact-support function, like maybe F(x) = x for |x| < 1 and F(x) = 0 for |x| > 1 then you're exactly right, F^ -> 0 quite slowly (in fact this F is not even continuous, and it follows that the integral of |F^| over the entire line must be infinite - this is one way of saying how slowly it -> 0, if it -> 0 fast then that integral is finite.) But this F is not even continuous while the F we're talking about is infinitely differentiable. Totally different story. Um. It's not that F is an infinitely differentiable function that we cut off, so it's no longer continuous at the endpoints of its support. F has compact support but it's nonetheless infinitely differentiable on the entire line. (The fact that there exists such a thing is not obvious. For example if F(x) = 0 for |x|>1 then all the derivatives of F vanish at x = 2, so the Taylor series for F centered at x_0 = 2 is identically 0. So F is one of those functions that doesn't equal the sum of its Taylor series. (So if we were talking to one of those engineers who thinks that every infinitely differentiable function is given by a Taylor series then there is no such F...) > -- > Ron Hardin > rhhardin@mindspring.com > > On the internet, nobody knows you're a jerk. > -- Oh, dejanews lets you add a sig - that's useful... Sent via Deja.com http://www.deja.com/ Before you buy. ============================================================================== From: Ron Hardin Subject: Re: Band-limited wiggle limit? Date: Sun, 10 Sep 2000 22:10:53 GMT Newsgroups: sci.math Okay, thanks. So the result is, you can have an interval of a band limited function when you get any number of zero crossings. I suppose you could have any function at all over an inverval as well? It's just a little more curve fitting with the polynomials, so almost any function anyway. In particular, you could have an interval of astounding high fidelity in the middle of band-limited audio mud? I have no intuition what would prevent this first. -- Ron Hardin rhhardin@mindspring.com On the internet, nobody knows you're a jerk. ============================================================================== From: Dan Kotlow Subject: Re: Band-limited wiggle limit? Date: Mon, 11 Sep 2000 06:12:50 GMT Newsgroups: sci.math On Sun, 10 Sep 2000 22:10:53 GMT, Ron Hardin wrote: >Okay, thanks. > >So the result is, you can have an interval of a band limited >function when you get any number of zero crossings. > >I suppose you could have any function at all over an inverval >as well? It's just a little more curve fitting with the polynomials, >so almost any function anyway. > >In particular, you could have an interval of astounding high >fidelity in the middle of band-limited audio mud? > >I have no intuition what would prevent this first. >-- >Ron Hardin >rhhardin@mindspring.com > >On the internet, nobody knows you're a jerk. Audio signals are usually not modeled by functions which have Fourier transforms -- these are transients. Audio signals are modelled by periodic or almost periodic functions. So let's ask instead how many zeros can a periodic or an almost periodic function have on an interval, given that it is band-limited. In the periodic case, f band-limited means that only harmonics up through the Nth are included in the Fourier series of f. The substitution z = exp[2 pi i t/T], where T = period, pulls the function back to the unit circle, where the (finite) Fourier series becomes z^-N * P(z), where P is a polynomial of degree 2N. Therefore, f can have no more than 2N zeros in any period. Can anybody get a result for almost periodic functions? Another question I've wondered about in the past is this: suppose a signal is a sinusoid plus bandlimited noise with known statistics. If a zero crossing is detected at time 0, what can be said about the statistics of the next zero crossing? Do whatever special case you can. -Dan ============================================================================== From: ullrich@math.okstate.edu (David C. Ullrich) Subject: Re: Band-limited wiggle limit? Date: Mon, 11 Sep 2000 13:23:44 GMT Newsgroups: sci.math On Mon, 11 Sep 2000 06:12:50 GMT, Dan Kotlow wrote: >On Sun, 10 Sep 2000 22:10:53 GMT, Ron Hardin >wrote: > >>Okay, thanks. >> >>So the result is, you can have an interval of a band limited >>function when you get any number of zero crossings. >> >>I suppose you could have any function at all over an inverval >>as well? It's just a little more curve fitting with the polynomials, >>so almost any function anyway. >> >>In particular, you could have an interval of astounding high >>fidelity in the middle of band-limited audio mud? >> >>I have no intuition what would prevent this first. >>-- >>Ron Hardin >>rhhardin@mindspring.com >> >>On the internet, nobody knows you're a jerk. > >Audio signals are usually not modeled by functions which have Fourier >transforms -- these are transients. Audio signals are modelled by >periodic or almost periodic functions. So let's ask instead how many >zeros can a periodic or an almost periodic function have on an >interval, given that it is band-limited. > >In the periodic case, f band-limited means that only harmonics up >through the Nth are included in the Fourier series of f. The >substitution z = exp[2 pi i t/T], where T = period, pulls the >function back to the unit circle, where the (finite) Fourier series >becomes z^-N * P(z), where P is a polynomial of degree 2N. >Therefore, f can have no more than 2N zeros in any period. > >Can anybody get a result for almost periodic functions? There are lots of definitions of "almost periodic". I think the most common is "uniform limit of (aperiodic) trigonometric polynomials". In the periodic case specifying the period says something about how many frequencies there are in a given interval on the frequency side; the bound on the number of zeroes depends on this. If f is almost periodic and band-limited then it can gave only finitely many zeroes in a bounded interval, but to get a bound on how many I think you're going to need to specify something analogous to specifying the period in the periodic case. (The spectrum of an almost-periodic function is certainly not discrete; there can be infinitely many frequencies in an interval...) >Another question I've wondered about in the past is this: suppose a >signal is a sinusoid plus bandlimited noise with known statistics. If >a zero crossing is detected at time 0, what can be said about the >statistics of the next zero crossing? Do whatever special case you >can. > >-Dan > ============================================================================== From: Dan Kotlow Subject: Re: Band-limited wiggle limit? Date: Tue, 12 Sep 2000 06:15:13 GMT Newsgroups: sci.math On Mon, 11 Sep 2000 13:23:44 GMT, ullrich@math.okstate.edu (David C. Ullrich) wrote: >On Mon, 11 Sep 2000 06:12:50 GMT, Dan Kotlow >wrote: > >>On Sun, 10 Sep 2000 22:10:53 GMT, Ron Hardin >>wrote: >> >>>Okay, thanks. >>> >>>So the result is, you can have an interval of a band limited >>>function when you get any number of zero crossings. >>> >>>I suppose you could have any function at all over an inverval >>>as well? It's just a little more curve fitting with the polynomials, >>>so almost any function anyway. >>> >>>In particular, you could have an interval of astounding high >>>fidelity in the middle of band-limited audio mud? >>> >>>I have no intuition what would prevent this first. >>>-- >>>Ron Hardin >>>rhhardin@mindspring.com >>> >>>On the internet, nobody knows you're a jerk. >> >>Audio signals are usually not modeled by functions which have Fourier >>transforms -- these are transients. Audio signals are modelled by >>periodic or almost periodic functions. So let's ask instead how many >>zeros can a periodic or an almost periodic function have on an >>interval, given that it is band-limited. >> >>In the periodic case, f band-limited means that only harmonics up >>through the Nth are included in the Fourier series of f. The >>substitution z = exp[2 pi i t/T], where T = period, pulls the >>function back to the unit circle, where the (finite) Fourier series >>becomes z^-N * P(z), where P is a polynomial of degree 2N. >>Therefore, f can have no more than 2N zeros in any period. >> >>Can anybody get a result for almost periodic functions? > > There are lots of definitions of "almost periodic". >I think the most common is "uniform limit of (aperiodic) >trigonometric polynomials". I think the most common one is "the set of its translates has compact closure in C_0". That's the one that gets generalized to LCA groups. Your characterization, IIRC, comes near the "end of the chapter". The original definition is: f is almost periodic if for every eps there exists L such that in every interval of length L there exists a tau such that abs[f(t+tau) - f(x)] < eps for all t. We see why that's not the most common one:-> Nevertheless, it does show why they're called "almost periodic". The "Fourier" coefficients of such an f can be defined as a_lambda = lim (1/2T) Int over [t,T] [f(t) exp[-i lambda t]dt (T->oo) These satisfy Parsival: sum abs[a_lambda]^2 <= Integral abs(f)^2 dt, which implies that only a countable number of the a_lambda can be non-zero. These constitute the spectrum of f. As far as I know, the spectrum can be any countable set, it can even be dense. The approximating trigonometric sums are sum a_lambda exp[i lambda t] over some finite subsets of the spectrum, but this is not a series because you don't know which lambdas need to be included to approximate f to a given accuracy. > > In the periodic case specifying the period says >something about how many frequencies there are in a >given interval on the frequency side; the bound on the >number of zeroes depends on this. If f is almost periodic >and band-limited then it can gave only finitely many >zeroes in a bounded interval, How do you get even this, may I ask? > but to get a bound on >how many I think you're going to need to specify something >analogous to specifying the period in the periodic case. >(The spectrum of an almost-periodic function is certainly >not discrete; there can be infinitely many frequencies >in an interval...) > >>Another question I've wondered about in the past is this: suppose a >>signal is a sinusoid plus bandlimited noise with known statistics. If >>a zero crossing is detected at time 0, what can be said about the >>statistics of the next zero crossing? Do whatever special case you >>can. >> >>-Dan >> > ============================================================================== From: ullrich@math.okstate.edu (David C. Ullrich) Subject: Re: Band-limited wiggle limit? Date: Tue, 12 Sep 2000 13:50:36 GMT Newsgroups: sci.math On Tue, 12 Sep 2000 06:15:13 GMT, Dan Kotlow wrote: >On Mon, 11 Sep 2000 13:23:44 GMT, ullrich@math.okstate.edu (David C. >Ullrich) wrote: > >>On Mon, 11 Sep 2000 06:12:50 GMT, Dan Kotlow >>wrote: [...] >>> >>>Audio signals are usually not modeled by functions which have Fourier >>>transforms -- these are transients. Audio signals are modelled by >>>periodic or almost periodic functions. So let's ask instead how many >>>zeros can a periodic or an almost periodic function have on an >>>interval, given that it is band-limited. >>> >>>In the periodic case, f band-limited means that only harmonics up >>>through the Nth are included in the Fourier series of f. The >>>substitution z = exp[2 pi i t/T], where T = period, pulls the >>>function back to the unit circle, where the (finite) Fourier series >>>becomes z^-N * P(z), where P is a polynomial of degree 2N. >>>Therefore, f can have no more than 2N zeros in any period. >>> >>>Can anybody get a result for almost periodic functions? >> >> There are lots of definitions of "almost periodic". >>I think the most common is "uniform limit of (aperiodic) >>trigonometric polynomials". > >I think the most common one is "the set of its translates has compact >closure in C_0". That's the one that gets generalized to LCA groups. >Your characterization, IIRC, comes near the "end of the chapter". > >The original definition is: f is almost periodic if for every eps >there exists L such that in every interval of length L there exists a >tau such that abs[f(t+tau) - f(x)] < eps for all t. We see why that's >not the most common one:-> Nevertheless, it does show why they're >called "almost periodic". Sorry. I didn't mean to say that there were lots of different equivalent formulations of this notion, although of course there are. What I meant was that there are lots of different notions of "almost periodic", the most common(?) of which is the one we give several characterizations of above. >The "Fourier" coefficients of such an f can be defined as > > a_lambda = lim (1/2T) Int over [t,T] [f(t) exp[-i lambda t]dt (T->oo) > > >These satisfy Parsival: sum abs[a_lambda]^2 <= Integral abs(f)^2 dt, And now for example the completion of these gizmos in that norm gives a somewhat different class of functions, which nonetheless have some right to be called "almost periodic". (Um, I _think_ the completion has a natural realization as a space of functions... still too early in the morning to be certain.) >which implies that only a countable number of the a_lambda can be >non-zero. These constitute the spectrum of f. As far as I know, the >spectrum can be any countable set, it can even be dense. Sure - like given a countable dense set you could take summable positive coefficients defined on that set - the sum of those coefficients times the exponentials is almost periodic (and I _think_ it's clear that the spectrum of that sum is the original set.) > The >approximating trigonometric sums are sum a_lambda exp[i lambda t] over >some finite subsets of the spectrum, but this is not a series because >you don't know which lambdas need to be included to approximate f to a >given accuracy. Actually I don't think that you _can_ necessarily approximate an AP function uniformly by such sums! Even in the periodic case: A continuous periodic function is AP, but the partial sums of the Fourier series need not even converge at every point, much less uniformly. (That's not _quite_ a proof of what I said, but I just got up - if it's possible to approximate a continuous periodic function unifomly by sums of "finite subsets of the Fourier series" I will be very very surprised.) You need summability methods to recover a continuous periodic function from the Fourier series. >> In the periodic case specifying the period says >>something about how many frequencies there are in a >>given interval on the frequency side; the bound on the >>number of zeroes depends on this. If f is almost periodic >>and band-limited then it can gave only finitely many >>zeroes in a bounded interval, > >How do you get even this, may I ask? If it's "band-limited" then it's the restriction to the line of an entire function. Um, that's not quite as clear to me as it was yesterday: Yesterday I was still thinking about Fourier transforms. If our AP function has summable coefficients and is band-limited then it is in fact the FT of a certain measure with compact support, and is hence an entire function. For a general AP function that's not quite a proof. But it certainly seems like a band-limited AP function _should_ be an entire function... For the record I'm not nearly as certain of that as I am of a few other things above. >> but to get a bound on >>how many I think you're going to need to specify something >>analogous to specifying the period in the periodic case. >>(The spectrum of an almost-periodic function is certainly >>not discrete; there can be infinitely many frequencies >>in an interval...) >> >>>Another question I've wondered about in the past is this: suppose a >>>signal is a sinusoid plus bandlimited noise with known statistics. If >>>a zero crossing is detected at time 0, what can be said about the >>>statistics of the next zero crossing? Do whatever special case you >>>can. >>> >>>-Dan >>> >> > ============================================================================== From: ullrich@math.okstate.edu (David C. Ullrich) Subject: Re: Band-limited wiggle limit? Date: Wed, 13 Sep 2000 12:32:23 GMT Newsgroups: sci.math On Tue, 12 Sep 2000 13:50:36 GMT, ullrich@math.okstate.edu (David C. Ullrich) wrote: >On Tue, 12 Sep 2000 06:15:13 GMT, Dan Kotlow >wrote: > >>On Mon, 11 Sep 2000 13:23:44 GMT, ullrich@math.okstate.edu (David C. >>Ullrich) wrote: [...] > >>The "Fourier" coefficients of such an f can be defined as >> >> a_lambda = lim (1/2T) Int over [t,T] [f(t) exp[-i lambda t]dt (T->oo) >> >> >>These satisfy Parsival: sum abs[a_lambda]^2 <= Integral abs(f)^2 dt, I didn't read this carefully yesterday. What's actually true (if f is AP) is that sum abs[a_lambda]^2 = lim[T->infinity] 1/(2T) Integral[-T,T] abs(f)^2 dt (Pf: It's true if f is a trig polynomial by direct calculation. The right-hand side (with f*g in place of abs(f^2)) defines an inner product with respect to which the exponentials are orthonormal; it's easy to see that uniform convergence on R implies convergence in the corresponding norm.) > And now for example the completion of these gizmos in that >norm gives a somewhat different class of functions, which nonetheless >have some right to be called "almost periodic". > (Um, I _think_ the completion has a natural realization as a >space of functions... still too early in the morning to be certain.) Forget I said that. On reflection I believe it's nonsense - what the completion _is_ is L^2(bR), where bR is the Bohr "compactification" of R. Surely(?) R has measure 0 in bR, and hence these things are not functions on R at all. But what I said about a band-limited AP function being entire is correct - we have a proof: [...] > >>> In the periodic case specifying the period says >>>something about how many frequencies there are in a >>>given interval on the frequency side; the bound on the >>>number of zeroes depends on this. If f is almost periodic >>>and band-limited then it can gave only finitely many >>>zeroes in a bounded interval, >> >>How do you get even this, may I ask? > > If it's "band-limited" then it's the restriction to >the line of an entire function. > > Um, that's not quite as clear to me as it >was yesterday: Yesterday I was still thinking about >Fourier transforms. If our AP function has summable >coefficients and is band-limited then it is in fact the >FT of a certain measure with compact support, >and is hence an entire function. For a general >AP function that's not quite a proof. But it certainly >seems like a band-limited AP function _should_ >be an entire function... For the record I'm not >nearly as certain of that as I am of a few other >things above. That was yesterday. Today the proof goes like so: First we need a fact about Fourier transforms of measures with compact support. Suppose that mu is a measure with support contained in [-1,1], and let f be the Fourier transform of mu. Then f is certainly an entire function, for example f(i) is the integral of e^t d mu(t). Choose a smooth function phi with compact support, such that phi(t) = e^t for |t| < 1. Then f(i) = the integral of phi(t) d mu(t). It follows (assuming -1 = 1; not gonna get the minus signs and Pi's in the right place) that f(i) = integral over R of the product f * phi^ . Now the hypotheses on phi show that phi^ is in L^1, and hence there exists a constant c such that: If f = mu^ where mu has support in [-1,1] then |f(i)| <= c*||f||_R . Here ||f||_A is the sup of |f| on the set A, and R is the line. The same thing works with any other complex number in place of i, and the bounds are explicit, hence bounded on compact sets: the argument shows If K is a compact subset of C then there exists a constant c = c(K) such that ||f||_K <= c ||f||_R whenever f = mu^ for some complex measure mu with support in [-1,1]. Now for AP functions: Take the definition "uniform limit of trig polynomials". Suppose that f is AP and band-limited - say the spectrum of f is a subset of [-1/2,1/2]. Now f is the uniform limit (on R) of a sequence of trig polynomials P_n. A few technical details show that we can in fact take the trig polynomials to have frequencies contained in [-1,1]. So they are in fact Fourier transforms of measures with support in [-1,1]. Now P_n -> f uniformly on R; the inequality above shows that hence P_n is uniformly Cauchy on compact subsets of the plane, and hence P_n converges to an entire function, the restriction of which to the line is f. QED. I haven't checked, but it seems very likely that the constant c(K) above satisfies c(K) = O(e^Y) if |Im(z)| <= Y for all z in K. If so then it follows that an AP function with bounded spectrum is actually an entire function of exponential type; that would be the easy half of the Paley-Wiener theorem for AP functions, raising the question of whether the converse holds. A brief outline of the "tecnical details" alluded to above: Say f is AP and the spectrum of f is contained in [-1/2, 1/2]. Now by definition there exist trig polynomials Q_n which converge to f uniformly on R. We don't know that the frequencies in Q_n lie in [-1/2, 1/2], but: Let chi be a smooth cut-off function: chi is infinitely differentiable with compact support, the support of chi is contained in [-1,1], and chi=1 on [-1/2, 1/2]. Let P_n be the trig polynomial obtained by multiplying the coefficients of Q_n by chi. Um, what I just said makes sense _if_ we're indexing the coefficients by the frequencies: If Q is a trig polynomial then Q(t) = sum a(x)*e^(ixt), where the sum ranges over all x in R, but all but finitely many of the a(x) vanish. Assuming that Q_n(t) = sum a_n(x)*e^(ixt), I'm setting P_n(t) = sum chi(x)*a_n(x)*e^(ixt). It follows that P_n is a trig polynomial with frequencies contained in [-1,1]. Also (as above) in fact if Chi is the FT of chi then P_n = Chi ** Q_n, where ** denotes convolution (because I've been using * for multiplication). Since Q_n ->f uniformly on R and Chi is in L^1(R) it follows that P_n-> Chi ** f uniformly on R. So we're done if Chi**f = f. But it's clear that Chi**f = f, because Chi**f and f have the same "Fourier coefficients" in the sense you defined above. (It must be a well-known fact that an AP function with all Fourier coefficients = 0 must be 0. It's true in any case: The AP function "is" a continuous function on bR, and the "Fourier coefficients" are the ordinary Fourier transform on bR; now abstract mumbling about locally compact abelian groups shows f = 0.) [deletia --djr] ============================================================================== From: sparge@globalnet.co.uk (Andy Spragg) Subject: Re: Band-limited wiggle limit? Date: Tue, 12 Sep 2000 00:19:35 GMT Newsgroups: sci.math On Sun, 10 Sep 2000 22:10:53 GMT, Ron Hardin wrote: ^ Okay, thanks. ^ ^ So the result is, you can have an interval of a band limited ^ function when you get any number of zero crossings. ^ ^ I suppose you could have any function at all over an inverval ^ as well? It's just a little more curve fitting with the polynomials, ^ so almost any function anyway. ^ ^ In particular, you could have an interval of astounding high ^ fidelity in the middle of band-limited audio mud? ^ ^ I have no intuition what would prevent this first. Hmm, I have been keeping my head down in this thread for lack of "shop floor experience" but the penultimate sentence prompted me to dig back into my archives and relocate a (Web-published) paper called "Black holes, bandwidth and Beethoven" which I have been keeping while I try and figure out what it means. No idea where I came across it, but this ASCIIised abstract will convey the flavour: It is usually believed that a function F(t) whose Fourier spectrum is bounded can vary at most as fast as its highest frequency component fmax . This is in fact not the case, as Aharonov, Berry and others drastically demonstrated with explicit counter examples, so­called superoscillations. The claim is that even the recording of an entire Beethoven symphony can occur as part of a signal with 1Hz bandwidth. Superoscillations have been suggested to account e.g. for transplanckian frequencies of black hole radiation. Here, we give an exact proof for generic superoscillations. Namely, we show that for every fixed bandwidth there exist functions which pass through any finite number of arbitrarily prespecified points. Further, we show that the behavior of bandlimited functions can be reliably characterized through an uncertainty relation: The standard deviation DT of samples F(tn ) taken at the Nyquist rate obeys DT >= 1/(4.fmax). This uncertainty relation generalizes to time­varying bandwidths. I have sent you the (postscript) original via email. HTH, Andy Andy -- sparge at globalnet point co point uk Look after the sins of write-commission, and the sins of read-omission will take care of themselves.