... f(0)=f(1) but for all x in [0,3/5] f(x) does not equal f(x+2/5) ???
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?
Surely there are an infinite amount of functions that satisfy those conditions?
Function is continuous
Yes, unfortunately, the function is continuous.
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.
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.
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.
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