| T.R | Title | User | Personal
Name | Date | Lines |
|---|
| 731.1 | not a very nice f | VINO::JMUNZER | | Tue Jul 14 1987 15:10 | 19 |
| Idea: consider the reals in pairs (x,y), and define f s.t.
f(x) = y f(y) = -x f(-x) = -y f(-y) = x
One way seems to be to use pairs (x,y), where
2 * k < x <= 2 * k + 1; y = x + 1
-2 * k > x >= -2 * k - 1; y = x - 1
(k = 0,1,2,3,...)
So, for instance,
ff(2.2) = f(3.2) = -2.2
ff(8) = f(-7) = -8
Also need f(0) = 0.
John
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| 731.2 | | KIRK::KOLKER | Conan the Librarian | Tue Jul 14 1987 17:44 | 26 |
| re .1
Remarkably close. I will place a solution after the vertical form
feed.
Define P(k) = {x in real | k < x <= k + 1}
Q(k) = {x in real | -k - 1 <= x < -k}
where k is a positive integer
Define f as follows:
f (0) = 0
f : P(k) -> P(k+1) for k even
f : P(k+1) -> Q (k) for k even
f : Q(k) -> Q ( k + 1) for k even
f : Q (k+1) -> P(k) for k even
notice that P(k) = -1 * Q (k) (applied pointwise)
Q(k) = -1 * P (k)
real = {0} U union (over all integers k) P(k) U union (over all
integers k) Q(k)
|
| 731.3 | | CLT::GILBERT | Builder | Tue Jul 14 1987 19:51 | 11 |
| A 'problem' with .1 and .2 is that they require that the reals be divided
into an infinite number of ranges. Can a finite number of ranges suffice?
Consider, for example,
1/x if x < -1
f(x) = -1/x if -1 < x < 0
-1/x if 0 < x < 1
1/x if 1 < x
This satisfies the problem, except that f(0), f(-1) and f(1) are undefined.
What is the minimum number of ranges that yield a solution over all reals?
|
| 731.4 | | BEING::POSTPISCHIL | Always mount a scratch monkey. | Wed Jul 15 1987 14:39 | 13 |
| Re .3:
Intuitively, consider dividing the reals with n points in addition to
0. This provides n+2 ranges. For any non-zero x, f(x) = y, f(y) = -x,
f(-x) = -y, so you need to map one range onto another onto a third onto
a fourth. You need a multiple of four ranges, so n+2 should be a
multiple of four. But you need to handle the n points, whether they
are endpoints in the ranges or handled separately. n and n+2 cannot
both be multiples of four, so I do not think a finite number of ranges
will suffice.
-- edp
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| 731.5 | | CLT::GILBERT | Builder | Wed Jul 15 1987 18:21 | 12 |
| We can show that f(0) must be 0.
Suppose that f(0) = y. Then f(f(0)) = 0 is equivalent to f(y) = 0.
But f(f(y)) = -y is equivalent to f(0) = -y. Thus f(0) = y = -y,
so y = 0, and f(0) = 0.
Suppose we are trying to solve f(f(x)) = g(x) for f. Then
g(f(x)) = f(f(f(x))) = f(g(x)). If z is a fixed-point of g
(that is g(z) = z), then f(z) is also a fixed-point of g, since
g(f(z)) = f(g(z)) = f(z).
P.S. (re .4) Right! It was important to recognize the 4-cycles.
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