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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 ?? :)
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 ?? :)

This Question is Closed

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Start by finding the common ratio

butterflydreamer
 one year ago
Best ResponseYou've already chosen the best response.1common ratio = 2cosx

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1whats the criterion for geometric series to converge ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1common ratio must be between 1 and 1, yes ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.11 < 2cosx < 1 solve x

butterflydreamer
 one year ago
Best ResponseYou've already chosen the best response.1ohh okay so, \[\frac{ \pi }{ 3 } < x < \frac{ 2\pi}{ 3 } and \frac{ 4\pi }{ 3 } < x < \frac{ 5\pi }{ 3 }\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1looks good http://www.wolframalpha.com/input/?i=solve+%7C2cos%28x%29%7C%3C1%2C+0%3Cx%3C2pi

butterflydreamer
 one year ago
Best ResponseYou've already chosen the best response.1thank you :) I forgot about the criterion for geometric series LOL.
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