Difference between revisions of "Cauchy criterion for convergence"

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(Created page with "If a sequence converges, it is the same as saying it matches the Cauchy criterion for convergence. ==Cauchy Sequence== A sequence <math>(a_n)^\infty_{n=1}</math>...")
 
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{{Todo|proof, easy stuff}}
 
{{Todo|proof, easy stuff}}
  
==Interesting examples==
 
===<math>f_n(t)=t^n\rightarrow 0</math> in <math>\|\cdot\|_{L^1}</math>===
 
Using the <math>\|\cdot\|_{L^1}</math> [[Norm|norm]] stated here for convenience: <math>\|f\|_{L^p}=\left(\int^1_0|f(x)|^pdx\right)^\frac{1}{p}</math> so <math>\|f\|_{L^1}=\int^1_0|f(x)|dx</math>
 
  
We see that <math>\|f_n\|_{L^1}=\int^1_0x^ndx=\left[\frac{1}{n+1}x^{n+1}\right]^1_0=\frac{1}{n+1}</math>
 
 
This clearly <math>\rightarrow 0</math> - this is <math>0:[0,1]\rightarrow\mathbb{R}</math> which of course has [[Norm|norm]] {{M|0}}, we think of this from the sequence <math>(\|f_n-0\|_{L^1})^\infty_{n=1}\rightarrow 0\iff f_n\rightarrow 0</math>
 
  
 
{{Definition|Real Analysis|Functional Analysis}}
 
{{Definition|Real Analysis|Functional Analysis}}
 
{{Theorem|Real Analysis|Functional Analysis}}
 
{{Theorem|Real Analysis|Functional Analysis}}

Revision as of 17:10, 8 March 2015

If a sequence converges, it is the same as saying it matches the Cauchy criterion for convergence.

Cauchy Sequence

A sequence [math](a_n)^\infty_{n=1}[/math] is Cauchy if:

[math]\forall\epsilon>0\exists N\in\mathbb{N}:n> m> N\implies d(a_m,a_n)<\epsilon[/math]

Theorem

A sequence converges if and only if it is Cauchy


TODO: proof, easy stuff


Real Analysis

Functional Analysis