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graydarl
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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
 one year ago
 one year ago
graydarl Group Title
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
 one year ago
 one year ago

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graydarl Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago

graydarl Group TitleBest ResponseYou've already chosen the best response.0
\[a_{n+p}a _{n}< \epsilon \]
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
since cosine is bounded above by 1 and below by 1 i believe this is the same as saying \[\frac{1}{3^{n+p}}\frac{1}{3^n}<\epsilon\]
 one year ago

graydarl Group TitleBest ResponseYou've already chosen the best response.0
will that be enough ?:D
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
should be although maybe we need to be a little careful
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
so you can factor a 2 out of the whole thing, makes no difference in the proof
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
by which i mean \[\frac{\cos((n+p)x)}{a^{n+p}}\frac{\cos(nx)}{a^n}\leq 2\frac{1}{3^{n+p}}\frac{1}{3^n}\]
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
\(a\) should be \(3\)
 one year ago

mahmit2012 Group TitleBest ResponseYou've already chosen the best response.1
dw:1353710521064:dw
 one year ago

graydarl Group TitleBest ResponseYou've already chosen the best response.0
Thank you both very much :D
 one year ago
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