anonymous
  • anonymous
Using cauchy's criterion |a_n+p - a_n|
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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anonymous
  • anonymous
Using \[|a_{n+p}-a _{n}|<\] Show that \[a_{n}=\frac{ \cos x }{ 3 }+\frac{ \cos 2x }{ 3^{2} } + \frac{ \cos 3x }{ 3^{3} } + .... +\frac{ \cos nx }{ 3^{n} }\] is a cauchy sequence
anonymous
  • anonymous
\[|a_{n+p}-a _{n}|< \epsilon \]
anonymous
  • anonymous
since cosine is bounded above by 1 and below by -1 i believe this is the same as saying \[\frac{1}{3^{n+p}}-\frac{1}{3^n}|<\epsilon\]

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anonymous
  • anonymous
will that be enough ?:D
anonymous
  • anonymous
should be although maybe we need to be a little careful
anonymous
  • anonymous
the largest the absolute value can be is if one cosine is -1 and the other is 1 but the difference is bounded by 2
anonymous
  • anonymous
so you can factor a 2 out of the whole thing, makes no difference in the proof
anonymous
  • anonymous
by which i mean \[|\frac{\cos((n+p)x)}{a^{n+p}}-\frac{\cos(nx)}{a^n}\leq 2|\frac{1}{3^{n+p}}-\frac{1}{3^n}|\]
anonymous
  • anonymous
\(a\) should be \(3\)
anonymous
  • anonymous
|dw:1353710521064:dw|
anonymous
  • anonymous
Thank you both very much :D

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