A proof within a proof (S.O.S. DUE TOMMORROW)

[Jamin2112]

A proof within a proof (S.O.S.! URGENT! DUE tomorrow!)

Homework Statement

So I'm stuck in the middle of a problem about uniform convergence. Nevermind what the problem asks, but please help me show

0 $$\leq$$ x $$\leq$$ 1 -----> |xx^-1/2n-x| < 1/2ⁿ

Homework Equations

None

The Attempt at a Solution

Multiply each side by something, subtract both sides by something, etc. I've tried that time and time again.

 

[╔(σ_σ)╝]

I know you said we should not worry about the question but it seems like what you are trying to prove is flawed .

$$|xx^{\frac{-1}{2^n}} -x |$$What happens when x=0 ?

The term x^((-1/2)^n) is not really defined.Perhaps you should post the complete problem since there might be as better way of going about the proof.

 

[Dickfore]

I know you said we should not worry about the question but it seems like what you are trying to prove is flawed .

$$|xx^{\frac{-1}{2^n}} -x |$$


What happens when x=0 ?

Isn't it 0?

 

[╔(σ_σ)╝]

Isn't it 0?

What about the X^(-1/2^n) term ?

 

[Dickfore]

Isn't it zero as well?

 

[╔(σ_σ)╝]

X^(-1/2^n) = 1/ X^(1/2^n)

 

[Dickfore]

Oh, god. I completely overlooked the fact that the exponent is negative.

EDIT:

But the term is:

$$x x^{-1/2^{n}} = x^{1 - 1/2^{n}} = 0, \; x = 0, \mathrm{for} \; n > 0$$

 

[Jamin2112]

I'm supposed to prove that f_n(x )= x^a_n is uniformly convergent (a_n = 1/2 + 1/4 + ... + 1/2ⁿ).

The problem suggests I show that |f_n(x) - f(x)| < 1 - a_n if x is in [0,1]

I've been using the fact that a_n = (2ⁿ-1)/2ⁿ.

 

[Dickfore]

Have you found the limit function $f(x)$?

 

[Jamin2112]

Have you found the limit function $f(x)$?

f(x) = x

 

[Dickfore]

Assume $x = 1 + t$. Because $0 \le x \le 1$, this implies:

$$\left{\begin{array}{l}<br /> 1 + t \ge 0 \\<br /> <br /> 1 + t \le 1<br /> \end{array}\right. \Rightarrow -1 \le t \le 0$$

Now, you might find it useful to use the following inequality:

$$(\forall t > -1) (\forall n \in \mathbb{N}) (1 + t)^{n} > 1 + n t$$

which may be proven by mathematical induction.