Copy wrong
- The other day I wanted to write f(x(1 + y)) on the board, but wrote something else instead.
- What did you write?
- f(x)(1 + f(y)).
- That is almost the same.
- Yes, I agree, and my students thought the slight change didn’t matter!
- Does it?
- What do you mean?
- Are there functions f for which the two give the same results for all x and y?
- I doubt it very much.
Problem source: Berkeley Math Circle Contest.
June 7th, 2010 at 12:44 am
The identity function: f(x)=x (which of course means that f(y)=y also). Both give x(1+y) as output. But that’s boring. I wonder if there’s anything more interesting.
June 7th, 2010 at 4:50 pm
For convenience in notation, write a for f(1). Where I use n later, let it be an integer.
Substituting x=1 and y=1 into equation 1:
f(2) = f(1)(1 + f(1)) = a + a²
Substituting x=1 and y=2 into equation 1:
f(3) = f(1)(1 + f(2))
= a(1 + a + a²)
= a + a² + a³
Substituting x=1 and y=3 into equation 1:
f(4) = f(1)(1 + f(3))
= a(1 + a + a² + a³)
= a + a² + a³ + a^4 ___ [2]
Substituting x=2 and y=1 into equation 1:
f(4) = f(2)(1 + f(1))
= (a + a²)(1 + a)
= a + 2a² + a³ ___ [3]
From equations 2 and 3,
a + a² + a³ + a^4 = a + 2a² + a³
a^4 = a²
so a=0 or a=1 or a=-1.
If f(1) = -1, then, substituting x=1 in equation 1,
f(1+y) = -(1+f(y)) ___ [4]
so
f(2+y) = -(1+f(1+y)) = f(y)
that is, f is periodic with period 2. In particular,
f(2) = 0
f(3) = -1 ___ [5]
Substituting x=2, y=1/2 in equation 1,
f(3) = 0
contradicting equation 5. Thus f(1) cannot equal -1.
If f(1) = 0, then, substituting x=1 in equation 1,
f(1+y) = 0
yielding f(x)=0 for all x.
If f(1) = 1, then, substituting x=1 in equation 1,
f(1+y) = 1 + f(y)
determining f(x) for all x once f(x) has been defined on the interval (0, 1):
f(x) = floor(x) + f(frac(x))
(a result that can be proved by induction on floor(x)), where floor(x) is the greatest integer less than or equal to x, and frac(x) = x – floor(x).
f(x) on the interval (0, 1) can’t be arbitrarily chosen, however. Substituting x=1/n, y=n-1 in equation 1:
1 = f(1) = f(1/n) * n
so f(1/n) = 1/n and f(x) = x for every x that is a multiple of 1/n. Thus f(x) = x for all rational x. If f must be continuous, f(x)=x for all x.
To summarise, if f must be continuous, the only possibilities are f(x)=0 for all x and f(x)=x for all x.
June 7th, 2010 at 4:51 pm
Where my equation 1 was Jan’s equation:
f(x(1 + y)) = f(x)(1 + f(y)) ___ [1]
June 7th, 2010 at 6:22 pm
The domain of f was not stated in the problem.
Richard has shown that if the domain is R and is continuous f is rather boring.
What if the domain is not the R? Can the domain have ten elements, one hundred, million and many?