Sarahisme Posted July 30, 2005 Share Posted July 30, 2005 yay! i might have actually got this one right hopefully, how'd i go... i reckon R = 2 , where R is the radius of convergence of the power series.... Link to comment Share on other sites More sharing options...
Dave Posted July 30, 2005 Share Posted July 30, 2005 I got the radius of convergence coming out as [imath]\frac{3}{2}[/imath]. I got it by working out: [math]R = \lim_{k\to\infty} \left| \frac{a_k}{a_{k+1}} \right|[/math] Because of that 3n in the denominator, you should definately be left with a 3 hanging around somewhere. Link to comment Share on other sites More sharing options...
Sarahisme Posted July 30, 2005 Author Share Posted July 30, 2005 ok yep sorry i goofed yeah i get 3/2 aswell thanks dave! Link to comment Share on other sites More sharing options...
DQW Posted July 31, 2005 Share Posted July 31, 2005 I think Dave meant to say that he got 3/2 by working out : [math] \lim_{k\to\infty} \left| \frac{a_k}{a_{k+1}} \right| < 1[/math] ? Link to comment Share on other sites More sharing options...
Sarahisme Posted July 31, 2005 Author Share Posted July 31, 2005 lol so the radius of convergence is 3/2 ?? i am getting a bit confuzzled here Link to comment Share on other sites More sharing options...
DQW Posted July 31, 2005 Share Posted July 31, 2005 Yes, it's 3/2 What's to be confuzzled about? When in doubt, recheck definitions. Link to comment Share on other sites More sharing options...
Sarahisme Posted August 7, 2005 Author Share Posted August 7, 2005 ok thanks guys oh, one other thing, the centre of convergence is at x = 0, isnt it? because that where the orginally summation is 0 (if you put x = 0 into it) Link to comment Share on other sites More sharing options...
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