chris@umcp-cs.UUCP (Chris Torek) (01/21/85)
[I've moved this from net.physics]
Speaking of discontinous functions ...
One of my favorite functions is
{ 1/q, x rational and expressed as p/q in lowest terms
f(x) = {
{ 0, x irrational
This thing is continuous nowhere, yet differentiable everywhere.
(f'(x) = 0 for all x.)
Does anyone else have a favorite weird function that is also simple to
define?
--
(This line accidently left nonblank.)
In-Real-Life: Chris Torek, Univ of MD Comp Sci Dept (+1 301 454 7690)
UUCP: {seismo,allegra,brl-bmd}!umcp-cs!chris
CSNet: chris@umcp-cs ARPA: chris@marylandosman@sprite.DEC (Eric Osman, dtn 283-7484) (01/28/85)
The function f(x) = 1/q or 0 for p/q and irrational respectively is indeed quite discontinuous. But please explain to us all what makes it differentiable ? I thought only *smoothies* could be differentiated, since the derivative is a nice tangent line. Such a line can't be unique on bumpy or discontinuous functions, can it ?
ken@turtlevax.UUCP (Ken Turkowski) (02/05/85)
> Does anyone else have a favorite weird function that is also simple to > define? My favorite is the delta function, defined as: delta(x) = 0, x != 0 infinity integral delta(x) = 1 -infinity -- Ken Turkowski @ CADLINC, Menlo Park, CA UUCP: {amd,decwrl,nsc,seismo,spar}!turtlevax!ken ARPA: turtlevax!ken@DECWRL.ARPA
ken@turtlevax.UUCP (Ken Turkowski) (02/13/85)
In article <3188@umcp-cs.UUCP> chris@umcp-cs.UUCP (Chris Torek) writes: >If this Dirac delta thing (I don't want to call it a function) is defined as > > f(x) = 0, x <> 0 > > integral from -infinity to infinity f(x) = 1 > >then in order to make any sense at all, its integral must also be one over >any interval containing the point zero, and zero for all others. *Most* >peculiar... Great deduction, Sherlock. >Whatever does one do with this beast, anyway? (I have a vague idea that >Dirac integrals are useful in quantum mechanics or something like that.) I thought that everyone knew what the Dirac delta function was. I guess not everyone on the net is an electrical engineer... The Dirac delta distribution (formally not a function) is the formal derivative of the unit step function (er, distribution), u(x), which is 0 for x less than zero, and 1 for x greater than zero. Still not satisfied because we haven't found a function? Well, integrate it one more time, and you have the unit ramp function, defined as: ramp(x) = 0, x <= 0 x, x >= 0 Take the formal derivative of the unit ramp, and you have a unit step. One more derivative gives a unit impulse or delta, and another derivative gives a unit doublet. The delta function is a unit under the operation of convolution: infinity f(x)*g(x) = integral f(t)g(x-t)dt -infinity where * denotes convolution, not multiplication. We have f(x)*delta(x) = f(x) A linear differential system can be described by a convolution, where f(x) is called the impulse response, and g(x) is the driving function or input. Such differential equations can be solved simply by purely algebraic techniques. Many of the netlanders are probably familiar with the concept of frequency. The delta function is related to frequency through the Fourier transform: -infinity F(t) = integral f(x)exp{-itx}dx infinity When f(x) is a delta function, F(t) = exp{-itx} = cos(xt) - i sin(xt), a sine wave of frequency x, so that a sine wave in time corresponds to a delta function in frequency. -- Ken Turkowski @ CADLINC, Menlo Park, CA UUCP: {amd,decwrl,nsc,seismo,spar}!turtlevax!ken ARPA: turtlevax!ken@DECWRL.ARPA
jfh@browngr.UUCP (John "Spike" Hughes) (02/14/85)
If you are willing to consider non-standard analysis, the delta function
can be defined as being infinity at 0 and 0 elsewhere, with an integral
that is in the monad of 1 (i.e. differs from 1 by an infinitesimal).
This makes perfect sense, in an odd sort of way. See E. Nelson on
Non-Standard Anaylsis in the Bulletin of the Amer. Math. Soc., about 2-4
years ago (sorry about the vague reference). Write to me for further
info.
-jfh