yay! i might have actually got this one right

[Sarahisme]

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....

[Dave]

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.

[Sarahisme]

ok yep sorry i goofed

 

yeah i get 3/2 aswell

 

thanks dave!

[DQW]

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] ?

[Sarahisme]

lol so the radius of convergence is 3/2 ??

 

i am getting a bit confuzzled here

[DQW]

Yes, it's 3/2

 

What's to be confuzzled about? When in doubt, recheck definitions.

[Sarahisme]

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)