Convergence of sin(x/n) in A[0,pi]

1. Nov 27, 2007

desaila

1. The problem statement, all variables and given/known data

Show that the sequence real-valued functions fn(x) = sin(x/n) converges uniformly to f(x) = 0 on A=[0,pi]

2. Relevant equations

I don't believe there are any

3. The attempt at a solution

Well, my first thought was that if x/n = pi then sin(x/n) is obviously going to be zero. I really can't get past that, I mean it seems like stating that and that if x/n is any multiple of pi then sin(x/n) will be zero. I guess I can't see past the obvious and into how this problem might actually be challenging.

2. Nov 27, 2007

Avodyne

You are supposed to show that fn(x) goes to zero as n goes to infinity for *any* value of x between 0 and pi.

3. Nov 27, 2007

desaila

Oh. Well then. You can't use euler's because the function is real valued, correct? Or, can you expand it and just leave off the imaginary part?

I can't think of another way to show that this function is going to zero.

4. Nov 27, 2007

Office_Shredder

Staff Emeritus
Well, as n goes to infinity, x/n goes to zero for all x on [0,pi] And sin(0)=0

5. Nov 27, 2007

ZioX

If the sup norm ( sup |fn(x)-f(x)|, over x in X) goes to zero then f_n converges uniformly to f on X. This is biconditional.

I suspect you can also show this from the definition of uniform convergence.