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butterflydreamer

  • one year ago

Find those values of x satisfying 0 <= x <= 2pi for which the geometrical series: 1 + 2cosx + 4cos^2x + 8cos^3x + ... has a limiting sum. I'm wondering how we'd approach this question? Maybe by sketching y= cos x for 0 <= x <= 2pi ?? :)

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  1. ganeshie8
    • one year ago
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    Start by finding the common ratio

  2. butterflydreamer
    • one year ago
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    common ratio = 2cosx

  3. ganeshie8
    • one year ago
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    whats the criterion for geometric series to converge ?

  4. ganeshie8
    • one year ago
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    common ratio must be between -1 and 1, yes ?

  5. butterflydreamer
    • one year ago
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    yesss

  6. ganeshie8
    • one year ago
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    -1 < 2cosx < 1 solve x

  7. butterflydreamer
    • one year ago
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    ohh okay so, \[\frac{ \pi }{ 3 } < x < \frac{ 2\pi}{ 3 } and \frac{ 4\pi }{ 3 } < x < \frac{ 5\pi }{ 3 }\]

  8. ganeshie8
    • one year ago
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    looks good http://www.wolframalpha.com/input/?i=solve+%7C2cos%28x%29%7C%3C1%2C+0%3Cx%3C2pi

  9. butterflydreamer
    • one year ago
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    thank you :) I forgot about the criterion for geometric series LOL.

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