Uniform Convergence?

Question

I need to show that fn(x)=cos(x/n) converges uniformly to f(x)=1 for x in [0, pi]. I need help finding an n from the inequality
|cos(x/n) - 1| < eps

Thanks...

Answer 1

|cos (x/n) -1| = 1 - cos(x/n), because cos(alpha) < 1

You must find n so that

cos(x/n) > 1- eps

or

1 - sin^2(x/n) > (1-eps)^2

or

sin^2(x/n) < 1 -(1-eps)^2 = 1-1+2 eps- eps^2< 2 eps

If
n > x/sqrt(2 eps)
then
sin^2(x/n) < (x/n)^2 < 2 eps

and you have the n that you needed.

Answer 2

The cosine function is continuous and cos(0) =1. So, for every eps > 0 there exists d >0 such that |cos(u) - 1| < eps for every real u with |u| < d Just choose d= eps..

For every x in [0, pi], 0 <= x <= pi. So, for every positive integer n > pi/d, we have 0 <= x/n <= pi * d/pi = d => |x/n| < d for x i [0, pi]. Hence, for every n > pi/d, we have

|cos(x/n) -1 | < eps for every x in [0, pi].

This shows f_n converges uniformly on [0,1] to the constant function f(x) = 1.

This is consequence of the fact that x/n converges uniformly to 0 on [0, pi] and cos is continuous at 0.

A general result: If f_n converges uniformly on an interval I to a constant function f(x) = a and g is continuous at a, then g(f(x_n)) converges uniformly on I to the constant function h(x) = g(f(a))