Is there a such function? (Math question)

Factual Questions
1

Let F be a function such that:

F is a one to one corresondence from the positive real numbers onto the positive real numbers, and

If x = 1.5y, then F(x) = 2F(y)

(If x equals one and a half times y, then F of x equals two times F of y.)

Intuitively, I thought there could be such a function.

But when I tried to work out a set of actual correspondences for such a function, it started to feel like the kind of construction that ends up being impossible.

Is there a proof that such a function can not exist? Alternatively, is there some formula that someone here can provide which, for some F, and for all positive real numbers r, takes r and yields the number which F maps them onto?

There is probably a very simple answer to this question,* and if that is the case, then my apologies in advance.

-FrL-

*I think maybe it has to do with logarithms, but its been over ten years since I have had to think about those, so my fuzzy intuition that they may be relevant is not anything I take very seriously.

2

Try f(x) = 4x/3.

3

f(12) = 16

f(18) = 24

18 = 1.5 * 12

but 24 /= 2 * 16.

(That was the first one that popped in my head, too, though.)

-FrL-

4

Yes, of course such a function exists. In fact there are a lot of them. The simplest is just
F(x) = 2[sup](log x)/(log 1.5)[/sup] ;
the exponent here can alternately be written as
(log x)/(log 1.5) = log[sub]1.5[/sub]x ,
so you see that multiplying x by 1.5 adds one to the value of the exponent and hence multiplies the result by 2.

Now you can multiply this function by any (positive) constant k; k F(x) is obviously a solution whenever F(x) is. And since you haven’t specified continuity, you can choose different values of k (not independently, if you want to make F bijective) over all real values in, say, [1,1.5) to get as ugly a function as you like.

But this probably doesn’t prove anything interesting about octaves or fifths.

5

I think you want to first take x, then take its log base 1.5, then raise 2 to that power. So you’d have 2^(log-base-1.5(x))

Log base 1.5 of x is the natural log of x divided by the natural log of 1.5.

So we want

2^(ln(x) / ln(1.5))
Let’s try some sample values:

f(1) = 1
f(1.5) = 2
f(2.25) = 4
Bingo!
I think there’s probably a way to clean that up further, but it’s also been a while since I’ve done logarithms…

Let’s see, some playing around gets us:
x^(ln(2) / ln(1.5)), so x^1.709…

6

Thanks!

I purposefully didn’t specify continuity, though at the same time, I was hoping there would be continuous functions of this type. If I am reading you correctly, that first example you give is continuous, correct?

Also, please remind me, does “is bijective” mean “is a one-to-one correspondence?” If not, what does it mean?

(Well, I guess I could look it up…)

Heheh, don’t draw attention. :slight_smile:

No, seriously, I’m not working on a proof of anything, I just started trying to really work out the ‘mechanics’ of one of the things I proposed in that thread.
-FrL-

7

It means that it is both one-to-one and onto.

8

Yes, the first example is continuous; it (and its constant multiples) are the smoothest solutions. But you can make other continuous solutions by choosing an arbitrary continuous bijective function f on [1,1.5] with f(1)=1, f(1.5)=2 as the generator, and reducing all values to this range.

(A bijection is one-to-one and onto. For a function with F(1.5x)=2F(x) over the positive reals, continuity already implies onto.)