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