Question

for the power series (2x)^n/n^4 find the radius and integral of convergence.

Answer 1

this is what i've done and am getting the wrong answer...

Answer 2

\[\frac{ 2x ^{n+1} }{ (n+1)^{4} }*\frac{ n ^{4} }{ (2x)^{n} }\]

\[\lim_{n \rightarrow \inf}\left| \frac{ 2xn ^{4} }{ (n+1)^{4} } \right|\]

\[\left| 2x \right|\lim_{n \rightarrow \inf}\frac{ n ^{4} }{ (n+1)^{4} }\]

\[\left| 2x \right|\]

so \[-1<2x <1\]

meaning R=1

solve for x...

integral is -1/2 to 1/2.

but Webwork says that's wrong.....

Answer 3

@Directrix do you know how to do this? >_<

Answer 4

Very funny, but I don't know how which is why I'm asking you haha

Answer 5

She's been on for an hour just waiting which is why I was asking, I never learned this and was googling things to learn it but I still don't know.

Answer 6

Careful reading brackets, second denominator is 2^n * x^n

Answer 7

what do you mean?

Answer 8

i wrote the top part wrong, but did it right. it's supposed to be

\[\frac{ (2x)^{(n+1)} }{ (n+1)^{4} }*\frac{ n ^{4} }{ (2x)^{n} }\]

Answer 9

so the n+1-n still cancel each other leaving 2x at the top.

Answer 10

then your radius should be\[\lim_{x \rightarrow \infty} \left| a_n \over a_{n+1} \right|\] if it exists.

Answer 11

http://tutorial.math.lamar.edu/Classes/CalcII/PowerSeries.aspx

this says the radius is \[\frac{ a_{n+1} }{ a _{n} }\]

Answer 12

so i'm confused. which is it?

Answer 13

Ups, my mistake, got a bit rusty, it's yours. :)

Answer 14

(n+1)/n

or

n/(n-1)

:)

Answer 15

ok, so since that's what i did, why is webwork saying i got the wrong answer?

Answer 16

how do you determine that your results are wrong?

Answer 17

can you take a screen shot ?

Answer 18

yeah, hang on

Answer 19

try a radius of .5, since you have unlimited attempts :)

Answer 20

that was correct....so why was my radius .5 and not 1?

Answer 21

\[|2x|<1\]
\[2|x|<1\]
\[|x|<1/2\]

Answer 22

Think the way a "circle" radius work, it goes from the center to the side, not that one side to the opposite :)

Answer 23

so the radius is gonna be the absolute value of the endpoint?

Answer 24

if your interval ... the diameter, is from -.5 to .5; the radius (half the diameter) is .5 in length

Answer 25

in general;

\[|f(x-a)|~\lim~c\]
\[R=\frac1c\]

Answer 26

you should only pull out the naked x parts, and leave everything else inside the limit. So pulling out the 2 alongside the x was a bit off putting i think

Answer 27

and as Paounn stated, you cant really have a radius that is larger than half the interval

Answer 28

ok. i think i understand. maybe. we'll see when i attempt the next problem...thanks amistre64 and paounn. you've been very helpful.

Answer 29

good luck ;)