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graydarl

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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  1. graydarl
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    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
  2. graydarl
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    \[|a_{n+p}-a _{n}|< \epsilon \]

    • one year ago
  3. satellite73
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    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
  4. graydarl
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    will that be enough ?:D

    • one year ago
  5. satellite73
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    should be although maybe we need to be a little careful

    • one year ago
  6. satellite73
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    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
  7. satellite73
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    so you can factor a 2 out of the whole thing, makes no difference in the proof

    • one year ago
  8. satellite73
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    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
  9. satellite73
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    \(a\) should be \(3\)

    • one year ago
  10. mahmit2012
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    |dw:1353710521064:dw|

    • one year ago
  11. graydarl
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    Thank you both very much :D

    • one year ago
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