Question

Radius of Convergence...

Answer 1

@thomaster please help..

I don't know if it's right but my guess is that the Radius of Convergence of the sum of the two series is at least as large as r... Am I right? If it is I don't know how to prove it. just my guess..

Answer 2

@phi , @ash2326

Answer 3

I could be wrong but the way I see it is this:

sum(a_n x^n) converges if -r<x<r
sum(b_n x^n) converges if -s<x<s

sum[(a_n+b_n)x^n] can be written as
sum(a_n x^n) + sum(b_n x^n)

it seems intuitive to me that this will converge only if both series converge, and since s>r, then if -s<x<s both series should converge, which implies their sums converge, so the radius of convergence should be the larger of the two radii; s

Answer 4

what @TuringTest said
lowest will work

Answer 5

oh you said largest, i would say smallest

Answer 6

after thinking about it for a moment I realize my logic suggests the opposite, the radius should be the smaller so that they both converge

Answer 7

yeah just realized it lol

Answer 8

so the radius of convergence should be the smallest of the two?

Answer 9

think about it, you are with me up until

sum[(a_n+b_n)x^n] can be written as
sum(a_n x^n) + sum(b_n x^n)

right?

Answer 10

I got confused at the convergence part..sorry..

Answer 11

you mean you don;t understand convergence in general?

Answer 12

nope, i don't mean that. I mean the logic in which the radius of convergence of the sum of the given power series should be the smallest. because the only part that I am sure that it is convergent is when -r<x<r. I am not sure if the sum of the other part of the power series will converge at the part x E (r, s).

Answer 13

you mean you are not sure whether
sum[(a_n+b_n)x^n]
will converge for
r<x<s and -s<x<-r
?

Answer 14

that is, x between r and s

Answer 15

yeah, kind of like that... sorry _ _

Answer 16

no worries
do you see that we can rewrite
sum[(a_n + b_n) x^n] = sum(a_n x^n) + sum(b_n x^n)
?

Answer 17

yep, we can distribute the summation...

Answer 18

ok, and the right will only converge if both series converge
i.e. if one of the two series on the right diverges, the whole thing diverges, agreed?

Answer 19

agree..

Answer 20

so, if r<x<s, then we have that x>r, which means that the first summation, sum(a_n x^n), diverges

Answer 21

same for -s<x<-r, if x<-r then the first summation diverges, so the largest radius at which they BOTH converge is the smaller of the two, r