Hello,

I came across this question the other day:

Find as many real-valued functions f and g as you can so that

i) f(f(x)) = -x for all real x

ii) g(g(x)) = 1/x for all real x (not equal to 0)

OK, that's interesting. For i), first of all if we weren't restricted to real functions then f(x) = ix would do the trick.

Anyway, the two questions are actually identical. If f(x) satisfies i) then g(x) = e^{f(ln x)} satisfies ii), and all solutions are of this form.

So to concentrate on i) briefly... it is clear that f(x) is a one-one function.

[If f(r) = f(s) then f(f(r)) = f(f(s)), so -r = -s so r = s. Therefore if r and s are distinct then f(r) =/= f(s). Also for every c there exists x such that f(x) = c. In fact x = f(-c).]

Now suppose f(a) = b.

Now f(f(a)) = f(b)

So f(b) = -a
f(f(b)) = f(-a)

So: f(-a) = -b

In other words the function f is odd: f(-a) = -f(a). Therefore f(0) = 0.

I'll see if I can get any further than this...

Michael

Let's try this one:

Let [x] = highest integer so that [x] < = x, and let x = [x] + {x}.

So [5.7] = 5 and {5.7} = 0.7 while [4] = 4 and {4} = 0. Basically [ ] is the integer part and { } is the decimal part.

Now we define our function f(x).

If x > 0 and {x} = 0 then let f(x) = x -0.5
If x > 0 and {x} = 0.5 then let f(x) = -(x + 0.5)
If x > 0 and 0 < {x} < 0.5 then let f(x) = x + 1 -2{x}
If x > 0 and 0.5 < {x} < 1 then let f(x) = -(x + 1 -2{x})

For negative x, let f(x) = -f(-x) and also let f(0) = 0. This defines f(x) for all x and it will satisfy f(f(x)) = -x.

Michael

Another one.

For all positive x:

If [x] is odd then let f(x) = x-1
If [x] is even then let f(x) = -(x+1)

Again for negative x, f(x) = -f(-x) and also f(0) = 0.