Search All of the Math Forum:
Views expressed in these public forums are not endorsed by
Drexel University or The Math Forum.
|
|
|
|
[ap-calculus] An everywhere unbounded function
Posted:
Oct 4, 2006 1:02 PM
|
|
I had a little free time, so I thought I'd
discuss a function, related to the ruler function
<http://mathforum.org/kb/thread.jspa?messageID=4594375>,
whose behavior illustrates a possibility that some
might consider impossible for a function to have.
We define f from the reals to the reals as follows.
f(x) = 0 if x is irrational or x=0.
f(p/q) = q if x is rational and x = p/q in lowest
terms and x has the same sign as p (i.e. q > 0).
Then f has the property of being unbounded in
every interval. To see this, consider an arbitrary
interval J. Then, for each integer n > 0 (no matter
how large), there exist infinitely many reduced-form
fractions p/q in J such that p > n. This is easy
to see if you consider "grids" on the number line
formed by marking the points
..., -2(1/q), -1(1/q), 0, 1(1/q), 2(1/q), ...,
first for q = M+1, then for q = M+2, then for
q = M+3, and so on. If the interval J is very
short, the first few of these grids might not
produce any points in J, but once q gets large
enough, the grids will start producing points
in J. Of course, only those p/q's where p and q
are relatively prime will matter, but this is
easily taken care of by just considering those
integers q > M such that q is a prime number.
Note that the value of f is bounded at each point.
[All I'm saying is that for each real number r,
|f(r)| < oo, which is a no-brainer.] However, for
each point r, the limiting behavior of f at x=r
is "unbounded". By this, I mean that f is unbounded
in every neighborhood of x=r.
Incidentally, f is unbounded in only one way,
namely from above. However, if we replace
"f(p/q) = q" with "f(p/q) = [(-1)^p] * q",
then I believe we'll get a function whose graph
is unbounded, both from above and from below,
in every interval.
As badly behaved as this function is (it's worse
than being discontinuous at each point, because
a function can be discontinuous at each point and
still be bounded on the entire real line), its
graph is relatively sparse in the plane. For
example, the graph of this function is certainly
a subset of the union of all horizontal lines
having an integer y-intercept, and even this
larger set misses a lot of regions with positive
area in the plane (e.g. the interior of the circle
with center (1, 1/2) of radius 1/2).
As a challenge, see if you can define a function g,
much worse than this, so that the graph of g is
dense in the plane. This means that the interior
of *every* circle in the plane will contain points
belonging to the graph of g. This might be something
a particularly bright student would be interested
in working on, especially someone who had taken
BC calculus before their last year of high school
and is now working on more advanced topics in math
(self-study or at a nearby college).
By the way, I was going define such a function,
but then I decided it might be better to let
others have fun working on. Also, I've now used
up the "a little free time" that I had!
Dave L. Renfro
====
Course related websites:
http://apcentral.collegeboard.com/calculusab
http://apcentral.collegeboard.com/calculusbc
---
ap-calculus is an Electronic Discussion Group (EDG) of The College Board
TO CHANGE YOUR EMAIL ADDRESS, PASSWORD OR SETTINGS, go to
http://lyris.collegeboard.com/read/my_account/edit
To UNSUBSCRIBE click the unsubscribe button on your Forums page:
http://lyris.collegeboard.com/read/my_forums/
Send questions about the list to owner-ap-calculus@lyris.collegeboard.com
|
|
|
|
