Question

radius of convergence question

Answer 1

\[\sum \frac{ 1 }{ 4^{n} }(x-2)^{n}\]

Answer 2

the an+1 is \[\frac{ 1 }{ 4^{n+1} }(x-2)^{2n+1}\] right?

Answer 3

I am getting my interval of convergence to be between abs(x-2)<4 but when I do my test for the ends neither is convergent so I am thinking I got this wrong because it would make more sense if it was abs((x-2)^2)<4

Answer 4

we have to apply the definition of the radius of convergence

Answer 5

?

Answer 6

no, sorry please wait....

Answer 7

you have it right

Answer 8

\[|x-2|<4\] is what you need to solve

Answer 9

you get
\[2<x<6\] pretty much in your head, but you need to check the endpoints

Answer 10

that's the thing for the end points I dont get convergent

Answer 11

OOPS !!

Answer 12

yes! you have to solve those two inequalities

Answer 13

\[|x-2|<4\\
-4<x-2<4\\

-2<x<6\] DOE

Answer 14

lol i said pretty much in your head, and did it wrong!!

Answer 15

now put \(x=-2\) and see what you get

Answer 16

then put \(x=6\) you have to check at the endpoints as well

Answer 17

at x=6, we got a divergen series

Answer 18

I got divergent for both ends

Answer 19

for sure at the right because \[\frac{4^n}{4^n}=1\]

Answer 20

excuse me I wrote the original equation wrong

Answer 21

\[\sum \frac{ 1 }{ 4^{n} }(x-2)^{2n}\]

Answer 22

but is my original an+1 still incorrect?

Answer 23

@misty1212

Answer 24

we have to request that:
\[\frac{{{{\left( {x - 2} \right)}^2}}}{4} < 1\]

Answer 25

so my original an+1 should be \[\sum \frac{ 1 }{ 4^{n+1} }(x-2)^{2(n+1)}\]

Answer 26

everything is the same as you had in the previous one, except the radius of convergence has changed as @Michele_Laino wrote

Answer 27

solve
\[(x-2)^2<4\]

Answer 28

the way I was taught was use ratio test and that should be less than one. Where I think my discrepancy is, is my an+1 when i do the ratio test so I want to know what the right an+1 is

Answer 29

we have:
\[\frac{{{a_{n + 1}}}}{{{a_n}}} = \sqrt {{a_n}} = \frac{{{{\left( {x - 2} \right)}^2}}}{4}\]

Answer 30

that's still not what I'm asking for help with I'm asking about my set up. If my set up was incorrect or not. what is an+1/an without being simplified

Answer 31

here, we can write:
\[\Large {a_{n + 1}} = {\left[ {\frac{{{{\left( {x - 2} \right)}^2}}}{4}} \right]^{n + 1}} = \frac{{{{\left( {x - 2} \right)}^{2\left( {n + 1} \right)}}}}{{{4^{n + 1}}}}\]

Answer 32

ok thank you that makes it much clearer and was not specified in my lectures. That is what I needed.

Answer 33

thank you!