Question

someone please help!!!! "Find the radius of convergence of summation of 3(x-2)^n from n=0 to infinity

Answer 1

this book comes directly from the textbook example. I am trying to understand how to do this. and here is my problem.......they use the ratio test and apply the limit as n approaches infinity. The catch is at the end they say "By the ratio test the series converges if the absolute value of (x-2) is less than one and diverges if the absolute value of (x-2) is greater than one. Therefore the radius of convergence is R=1. This statement doesnt make sense because if you plug one in for x in the absolute vqlue of x-2 then the statement says 1>1. I don't get it

Answer 2

\[\lim_{n \rightarrow \infty}\left| \frac{ 3(x-2)^{n+1} }{ 3(x-2)^{n} } \right|=\lim_{n \rightarrow \infty}\left| x-2\right|\]

Answer 3

this I understand

Answer 4

the radius is half the length, but not the interval itself

Answer 5

the 3 here is a red herring, it is just a number

Answer 6

but then when you evaluate the limit you get \[\left| x-2 \right|\] if that is stictly less than one it converges by the ratio test. So:
\[\left| x-2 \right| \lt 1\]

Answer 7

so if you evaluate that you would see that x has to be stirctly less than three

Answer 8

the limit is in \(n\) not in \(x\)

Answer 9

how is it then that the radius is one?

Answer 10

forget the \(x\), the limit is in \(n\)

Answer 11

ok I am confused I thought that you could simplify and all the "n" s were gone.

Answer 12

\(n\) is going to infinity, not \(x\)
the possible values of \(x\) are the ones for which \(|x-2|<1\)

Answer 13

yeah, the \(n\)'s are gone you are right

Answer 14

so you have to make sure that that limit, which is \(|x-2|\) is less than one

Answer 15

which is another way of saying that \(|x-2|<1\)

Answer 16

I agree but when you evaluate that doesn't x have to be striclty less than three? not equal?

Answer 17

if you solve \(|x-2|<1\) you get \(1<x<3\) an interval of length 2,

Answer 18

agreed

Answer 19

there is really no difference between
\[\sum x^n\] and \[\sum(x-2)^n\] except you are shifted right two units

Answer 20

but how is the radius then equal to one? wouldn't the radius of convergence have to be less than one?

Answer 21

first one converges for \(|x|<1\) and the second one converges for \(|x-2|<1\)

Answer 22

"radius" is half the length of the interval

Answer 23

the word "radius' will make more sense in complex valued functions, because it is literally a "radius" but for real valued functions it is an interval on the real line

Answer 24

I agree, I know it is a fine point, but it says that 1<x<3 which does not inlude 1 or 3, correct?

Answer 25

may include 1 or 3 or may not
but that does not change the length of the interval

Answer 26

in your case it does include 1 and does not include 3, as you can check
but the inclusion or exclusion of the number at the endpoint of the interval does not change the length of the interval

Answer 27

ok I understand the general idea, I just get hung up on the little points that seem trivial. Ok so we look for the interval and then check the endpoints?

Answer 28

this is clear right? \((1,3)\) has the same length as \([1,3]\)

Answer 29

Like two separate Ideas?

Answer 30

just saying that the lengths are the same
one point does not contribute anything to the length

Answer 31

Oh yeah!!! that makes sense...... the interval notation made it click.... you should be a math professor, I would sign up for your calculus classes lol

Answer 32

yeah they are separate ideas

Answer 33

But then you probably would stop helping so much lol

Answer 34

length of interval is one thing
inclusion of the endpoints, one or the other or both, is separate

Answer 35

thanks for your help.

Answer 36

yw