# Analysis Function Question

1. Oct 15, 2013

### lepton123

1. The problem statement, all variables and given/known data
Suppose f is continous on [0,1] and f(0)=f(1). Let n be a natural number. Prove that there is some number x, such that f(x)=f(x+1/n).

2. Relevant equations
The hints says to consider g(x)=f(x)-f(x+1/n)

3. The attempt at a solution
I've tried to consider the function g(x), but I haven't gotten anything useful from it. When I've tried various values of n, like 1/2, 1/3...I've noticed that there are repeating terms and I can manipulate the terms a bit to get like g(0)=-g(1/2) for n=1/2 and the like, but I am not sure where to go with this

2. Oct 15, 2013

### dirk_mec1

What's wrong with x =0 and n=1?

3. Oct 15, 2013

### jbunniii

You don't get to choose $n$. It is chosen for you. In other words, you have to show there is a solution for ANY $n$.

4. Oct 15, 2013

### jbunniii

If $g(0) = 0$ then you can simply take $x = 0$. If $g(0) > 0$, then I claim there must be some other point $x$ such that $g(x) < 0$, and then you can apply the intermediate value theorem. Can you prove this claim? Hint: consider $g(0) + g(1/n) + \ldots + g((n-1)/n)$.