| T.R | Title | User | Personal
Name | Date | Lines |
|---|
| 1463.1 | some of the "monsters" | ALLVAX::JROTH | I know he moves along the piers | Wed Jun 26 1991 06:30 | 45 |
| There are standard examples that go back to Weierstrass and Riemann
that are constructed in terms of Fourier series; the function has
a Fourier series but its derivative has infinite energy in the spectral
components.
Here's an example:
f(t) = SUM(n > 0) sin(2*pi*n^2*t)/n^2
You can also construct them by fractals, say by subdividing a rectangle
recursively into 9 parts with a polyline connecting diagonal corners of
chosen sub-rectangles like this:
+-----------------+--------D
| | |
| | |
| | |
| | |
| | |
+--------B--------+ |
| | | |
| | | |
| | | |
| | | |
| | | |
| | | |
| +--------C--------+
| | |
| | |
| | |
| | |
| | |
| | |
A--------+-----------------+
Initially there is a diagonal line AD. Then the rectangle is divided.
Next draw a polyline ABCD and then recursively subdivide each rectangle
and redraw the diagonals in the obvious way.
Books on fractals and real analysis talk about these kinds of functions.
Now, how about some nowhere integrable functions :-)
- Jim
|
| 1463.2 | | GUESS::DERAMO | duly noted | Wed Jun 26 1991 06:31 | 16 |
| >>Are there functions that are continuous but not differentiable anywhere?
Yes.
These are easy to "construct" by taking limits of
a sequence of uniformly continuous functions.
sin(2^n x)
For example, I think f(x) = sum(n=0,...,oo) -----------
2^n
will work, or the same with sin(t) replaced by T(t) given
by T(t) = 2t for t in [0,1/2], T(t) = 2-2t for t in [1/2,1],
and T(t+1) = T(t) for all t (T periodic of period 1).
Dan
|
| 1463.3 | | GUESS::DERAMO | duly noted | Wed Jun 26 1991 06:32 | 1 |
| notes collision :-)
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| 1463.4 | note collision at 5:30 AM !!! | SMAUG::ABBASI | | Wed Jun 26 1991 12:01 | 6 |
| So Now I know the trick to be a good mathematician, work at
5:30 in the morning :-)
Thanks all for answers,
/nasser
|
| 1463.5 | continuous but not integrable in complex domain | SMAUG::ABBASI | | Sat Jun 29 1991 16:41 | 22 |
| ref .1 (JROTH)
> Now, how about some nowhere integrable functions :-)
there is one complex function i just came to know of that is
continuous but has no indefinite integral
f(z) = x-iy
this is because there not exist a complex function w=u(x,y)+i v(x,y)
whose derivative u + i v or -iu + v is equal to x-iy
x x y y
where u means partial u(x,y) w.r.t x etc..
x
but with real functions of real real variable, every SINGLE valued
valued continouse fucntion has an indefinite integral, so i guess
if a real contiuouse function will not have an indefinite integral
it must not be Single valued ?
/naser
|
| 1463.6 | | GUESS::DERAMO | duly noted | Sat Jun 29 1991 17:02 | 5 |
| A bounded function f:[a,b] -> R is Riemann integrable iff
it is continuous "almost everywhere" (i.e. except on a
set of measure zero).
Dan
|
| 1463.7 | | VINO::XIA | In my beginning is my end. | Sat Jun 29 1991 17:49 | 3 |
| Yea, it all depends on what one means by integration.
Eugene
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| 1463.8 | | GUESS::DERAMO | duly noted | Sat Jun 29 1991 21:36 | 25 |
| Let f be complex conjugation, f(x+iy) = x-iy. You can
still define the path integral over a path p in the
complex plane. For example, if p is a continuous,
piece-wise differentiable function from [0,1] into the
complex plane, p(t) = x(t) + i y(t), then the path
integral of f(z) dz over p will certainly exist. What
happens is that the path integral of f(z) dz over a path
from A to B depends on the path [for this particular f].
Two interesting results: the real part of the path
integral is path independent, and is always equal to
2 2
( |B| - |A| ) / 2
The path integral of f over a circle of radius r is
independent of the center and is equal to
2
2 pi i r
(by the first interesting result, you already knew the
second would be a pure imaginary)
Dan
|
| 1463.9 | complex functions talk | SMAUG::ABBASI | | Sun Jun 30 1991 05:43 | 43 |
| Dan,
The term "Riemann integrable" I did not understand, and cant find
any reference to, closest thing i found is the Riemann integral (which
i kind'a understand) which is a special case of Stieltjes integral
(which i dont undestand (yet..)). is this what you meant ?
but to go back to w=x-iy, this is not even an analytic function,(it
fails the cauchy-riemann rules) so as you pointed out, the path integral
is dependent on the path, so which path should one pick (as there are
infinite number of them) ? and hence infinite number of values of the
integral.
your result for real part looks intersting, but iam not sure how
you came up with right now. i'll have to study it more.
your result on the circle, is neat, this is how i proofed it:
take center to be origin, and f(z)= x-iy then
i$
z= re
i$
dz= ir e
$= 2Pi
/ / i$
| x-iy dz = | r cos($) - i r sin($) ir e d$
/ /
$=0
$=2Pi $=2Pi
/ i$ / i$
= ir^2 | cos($) e d$ + r^2 | sin($) e d$
/ /
$=0 $=0
= ir^2 (Pi) + r^2 (iPi) = i 2 r^2 Pi .
since origin is arbitrary point, result follows for any center, QED.
/nasser
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| 1463.10 | correction | SMAUG::ABBASI | | Sun Jun 30 1991 05:48 | 2 |
| i just thought, that was not a "proof" since i used a special case of
f(z), but it is easy offourse to show same for general f(z) same way.
|
| 1463.11 | | GUESS::DERAMO | duly noted | Sun Jun 30 1991 10:36 | 5 |
| By "Riemann integrable" I meant a function the "Riemann
integral" of which exists (over the specified class of
paths).
Dan
|
| 1463.12 | | GUESS::DERAMO | duly noted | Sun Jun 30 1991 10:44 | 21 |
| For the result that the real part of the path integral is
path independent, I broke the path into segments going
from A = z to z , from z to z , ..., from z to z = B
0 1 1 2 n-1 n
Then I estimated the path integral as the sum from i=0 to n-1
of f(w ) delta z where delta z = z - z and for w
i i i i+1 i i
I took the midpoint of z and z (not necessarily on the path!)
i i+1
Writing each z as x + iy (that i is sqrt(-1), not a
j j j
summation index like the previous one) and computing the
sum, I found that it the real part telescoped into
(1/2)(|B|^2 - |A|^2)
I suppose this is subject to taking w not on the path
being valid. i
Dan
|
| 1463.13 | some stuff on the topic | SMAUG::ABBASI | | Sun Jun 30 1991 16:59 | 27 |
| ref .10 (me)
>i just thought, that was not a "proof" since i used a special case of
>f(z), but it is easy offourse to show same for general f(z) same way.
this is wrong.
if f is *analytic* on the contuor and at every point inside the closed
region then /
| f(z) dz = 0 for any shape closed contuor
/
anti clockwise
this is offcourse the cauchy theorm.
an anlytic function at a point is one that has a derivative at that
point, and the derivative is contiuouse. (which means the function has
a derivative at the point and all points around the point), it also
means the function has a convergent taylor series around the point and
the radius of convergence is the distance to the first point where the
function do not have a derivative or singular.
offcourse if the integral=0 that does not mean that the function is
analytic in the region.
so Dan's result of (i2 Pi r^2) applies (only?) to the
non-analytic function f(z)= x-iy .
/nasser
|
| 1463.14 | and on a totally, completely, unrelated note ... >:-) | GUESS::DERAMO | duly noted | Sun Jun 30 1991 17:45 | 8 |
| There is a function z -> f(z) which is not analytic
anywhere in the complex plane, but such that the
iterated function z -> f(f(z)) [iterated two times]
is analytic in an open set containing the origin,
and such that z -> f(f(f(f(f(f(f(f(z)))))))) [iterated
eight times] is analytic over the entire complex plane.
Dan
|
| 1463.15 | now, how did you come up with this one ? | SMAUG::ABBASI | | Sun Jun 30 1991 23:33 | 1 |
| let me see, you want us to guess this one, or actualy derive it :-)
|
| 1463.16 | | GUESS::DERAMO | duly noted | Sun Jun 30 1991 23:41 | 4 |
| Actually, it should be well known to the math conference
crowd. >:-)
Dan
|
| 1463.17 | | GUESS::DERAMO | duly noted | Mon Jul 01 1991 13:45 | 4 |
| In fact, it should be well known to the readers of this
topic! >:-)
Dan
|
| 1463.18 | A set that is not Lebesque measurable | DECWET::BISHOP | two bee ore naught too b | Mon Jul 01 1991 17:31 | 22 |
| Given x,y in (0,1), define the equivalence relation ~ by
x~y iff x-y=r is rational.
Now let S be a set containing exactly one element from each
equivalence class arising from ~ (by the axiom of choice, such
a set exists). Then S is not Lebesque measurable.
Proof: Exercise for the bored programmer.
Now define f() by
/ 1 if x is in E
f(x) = I (x) = <
S \ 0 otherwise.
Then f is not Lebesque integrable.
(I can't claim the credit for this -- it's from Rudin "Real and
Comlex Analysis," p. 55)
Avery
|
| 1463.19 | Hey Dan, it's July 1, not April 1! | GUESS::DERAMO | duly noted | Mon Jul 01 1991 17:48 | 19 |
| re .14,
> There is a function z -> f(z) which is not analytic
> anywhere in the complex plane, but such that the
> iterated function z -> f(f(z)) [iterated two times]
> is analytic in an open set containing the origin,
> and such that z -> f(f(f(f(f(f(f(f(z)))))))) [iterated
> eight times] is analytic over the entire complex plane.
Let f(z) map z = x + iy into its complex conjugate x - iy.
Any even number of iterations gives the identity function.
The clause "an open set containing the origin" does
after all describe the entire complex plane.
0:-)
Dan
|