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

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

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.
 
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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.
 


╔(σ_σ)╝ said:
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?
 


Dickfore said:
Isn't it 0?

What about the X^(-1/2^n) term ?
 


Isn't it zero as well?
 


X^(-1/2^n) = 1/ X^(1/2^n)
 


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

EDIT:

But the term is:

<br /> x x^{-1/2^{n}} = x^{1 - 1/2^{n}} = 0, \; x = 0, \mathrm{for} \; n &gt; 0<br />
 


I'm supposed to prove that fn(x )= xan is uniformly convergent (an = 1/2 + 1/4 + ... + 1/2n).

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

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


Have you found the limit function f(x)?
 
  • #10


Dickfore said:
Have you found the limit function f(x)?

f(x) = x
 
  • #11


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

<br /> \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<br />

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

<br /> (\forall t &gt; -1) (\forall n \in \mathbb{N}) (1 + t)^{n} &gt; 1 + n t<br />

which may be proven by mathematical induction.
 
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