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anonymous
 3 years ago
Using cauchy's criterion a_n+p  a_n<epsilon prove that a_n= cos(x)/3 + cos(2*x)/3^2 + ..... + cos(n*x)/3^n is a cauchy sequence
anonymous
 3 years ago
Using cauchy's criterion a_n+p  a_n<epsilon prove that a_n= cos(x)/3 + cos(2*x)/3^2 + ..... + cos(n*x)/3^n is a cauchy sequence

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Using \[a_{n+p}a _{n}<\] 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
 3 years ago
Best ResponseYou've already chosen the best response.0\[a_{n+p}a _{n}< \epsilon \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0since cosine is bounded above by 1 and below by 1 i believe this is the same as saying \[\frac{1}{3^{n+p}}\frac{1}{3^n}<\epsilon\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0will that be enough ?:D

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0should be although maybe we need to be a little careful

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the 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
 3 years ago
Best ResponseYou've already chosen the best response.0so you can factor a 2 out of the whole thing, makes no difference in the proof

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0by 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
 3 years ago
Best ResponseYou've already chosen the best response.0\(a\) should be \(3\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1353710521064:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Thank you both very much :D
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