# Is there a function such that

1. Nov 16, 2004

### yayMath

... f(0)=f(1) but for all x in [0,3/5] f(x) does not equal f(x+2/5) ???

2. Nov 16, 2004

### Hurkyl

Staff Emeritus
I would say probably yes. Try splitting up the interval into (lots of) groups of points that differ by 2/5 and see if you can assign values to each point.

Now, have you stated the whole question? Is f supposed to be, say, continuous?

3. Nov 16, 2004

### Zurtex

Surely there are an infinite amount of functions that satisfy those conditions?

4. Nov 16, 2004

### yayMath

Function is continuous

Yes, unfortunately, the function is continuous.

5. Nov 16, 2004

### Hurkyl

Staff Emeritus
Well, you want to make sure that f(x) does not equal f(x+2/5)...

A common trick is to rewrite this constraint so that a part of it is constant: you want to make sure that f(x) - f(x+2/5) does not equal 0.

6. Nov 17, 2004

### yayMath

so, i've gotten to the point at which g(x) = f(x) - f(x+2/5) does nto equal zero for any x in [0, 3/5].... finding a form for g(x) is not difficult, but the constraint f(0) = f(1), i have no idea how to include it.

7. Nov 26, 2004

### sinkdeep

If f(x) is continuous, I'd say there's no such func. Here's the reason:

1.The simplest case, that f(x) have no zero points in (0,1), so the curve is like bridge. For instance, let f(x)=sin(4*pi*x), and g(x)=f(x)-f(x+2/5), then
g(0)<0, g(3/5)>0, and since g(x) is continuous, there should be at least one zero point in (0,3/5).

2.Then if f(x) have zero points in (0,1), however, the result is the same. Suppose f(x)=sin(2*pi*x), and in part of (0,1)--in this case, (0,1/2),we can find zero points in (0,1/2-2/5), the reason is the same as in case 1.

Last edited: Nov 27, 2004