Difference between revisions of "Convergence of a sequence"

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(Created page with "Like with continuity there are three forms for convergence of a Sequence Given a sequence <math>(a_n)^n_{n=1}</math> we may say that it converges to {{...")
 
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Like with [[Continuous map|continuity]] there are three forms for convergence of a [[Sequence]]
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#REDIRECT [[Limit (sequence)]]
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Given a sequence <math>(a_n)^n_{n=1}</math> we may say that it converges to {{M|a}} or <math>\lim_{n\rightarrow\infty}(a_n)=a</math> if and only if the following definition holds:
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==First form==
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Introductory form
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<math>\forall\epsilon>0\exists N\in\mathbb{N}:n> N\implies|a_n-a|<\epsilon</math>
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==Second form==
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[[Metric space]] form
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<math>\forall\epsilon>0\exists N\in\mathbb{N}:n> N\implies d(a_n-a)<\epsilon</math>
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==Third form==
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[[Topological space|Topological]] form
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<math>\forall N_a\exists N\in\mathbb{N}: n> N\implies a_n\in N_a</math> where <math>N_a</math> denotes a neighbourhood of <math>a</math>
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==Cauchy Criterion==
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Convergence can be shown without knowing what exactly the sequence converges to, see the [[Cauchy criterion for convergence]] page
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==Note on norms==
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Recall from [[Norm|norm]] that we can simply define <math>d_{\|\cdot\|}(x,y)=\|x-y\|</math>, thus we can also have a slight variation of the metric form:
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<math>\forall\epsilon>0\exists N\in\mathbb{N}:n> N\implies \|a_n-a\|<\epsilon</math>
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Is is worth noting because in [[Functional Analysis]] norms are considered and if we deal with a [[Metric space|metric space]] we are inside a branch of
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[[Topology|topology]]
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{{Todo|preserve "interesting" example}}
 
==Interesting examples==
 
==Interesting examples==
 
===<math>f_n(t)=t^n\rightarrow 0</math> in <math>\|\cdot\|_{L^1}</math>===
 
===<math>f_n(t)=t^n\rightarrow 0</math> in <math>\|\cdot\|_{L^1}</math>===
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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>
 
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>
  
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{{Definition|Real Analysis|Topology|Functional Analysis}}
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{{Definition|Real Analysis|Topology|Functional Analysis|Metric Space}}

Latest revision as of 13:30, 5 December 2015

Redirect to:

TODO: preserve "interesting" example

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 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 [ilmath]0[/ilmath], we think of this from the sequence [math](\|f_n-0\|_{L^1})^\infty_{n=1}\rightarrow 0\iff f_n\rightarrow 0[/math]

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