Question

If the series divergence what is the radius of convergence? is it zero?

Answer 1

*diverges

Answer 2

radius of convergence is for a power series
if the radius of convergence is 0, then it only converges if \(x=0\) and not otherwise

Answer 3

what is the interval of convergence is (-infinity , infinity) whats the radius of convergence?

Answer 4

*if

Answer 5

unless the power series is expanded about \(a\) i.e. the terms look like \((x-a)^n\) in which case it only converges if \(x=a\) i.e. all the terms are zero

Answer 6

i think there might be some confusion here
can you say exactly what the question is? like a specific example
because i don't want to mislead you

Answer 7

hmm okay, hold on

Answer 8

\[\lim_{n \rightarrow \infty}\frac{ x }{ 2n+2}\] after i did the ratio test, this is equals to 0 right?

Answer 9

hmm hold on a sec, i don't think so

Answer 10

then the interal of convergence would be \[(-\infty,\infty)\]

Answer 11

no lets go slow

Answer 12

okay the orginal qn is

Answer 13

find the mclaurin series for f(x) = sin(x^4) and find its radius of convergence

Answer 14

ooh i am sorry, i thought that was the sum
nvm if that is what you have to take the limit of, then you are right , the limit is \(0\) for all \(x\)

Answer 15

\[\sum_{n=0}^{\infty}\frac{ (-1)^n(x ^{(8n+4)}) }{ (2n+1)! }\] i got it to be

Answer 16

sorry sorry you are right of course
i thought that was the question
you have
\[\lim_{n\to \infty}\frac{|x|}{2n+2}=0\] and so radius of converence is \(\infty\) and interval is \((-\infty, \infty)\)

Answer 17

haha okay then thanks! :D

Answer 18

sorry about the confusion,
you have it correct

Answer 19

ahh! so radius would be infinity? why so?

Answer 20

what else would it be? if the radius was \(2\) the interval would be \((-2,2)\) or maybe \([-2,2)\) or something like that
saying the radius is infinity is the same as saying it converges for all \(x\)

Answer 21

ohh.. but sin(x^4) according to wolfram doesnt converge so why would it converge for all x?

Answer 22

the word "radius" will make more sense when you talk about convergence of series of complex variables
then it actually is a "radius"

Answer 23

??

Answer 24

i mean i checked my answer, with wolframalpha but it says it diverges

Answer 25

can you show me?

Answer 26

http://www.wolframalpha.com/input/?i=radius+of+convergence+of+sin%28x%5E4%29

Answer 27

i'm not sure what to put in radius of convergence since it doesnt converge.. does it mean i put NIL or something? haha

Answer 28

http://www.wolframalpha.com/input/?i= \sum+_{k%3D0}^{\infty+}+\frac{%28-1%29^k+\left%28x^4\right%29^{2+k%2B1}}{%282+k%2B1%29!}

Answer 29

oops copy and paste one

Answer 30

oooooh you were trying to add!

Answer 31

\[\sum _{k=0}^{\infty } \frac{(-1)^k \left(x^4\right)^{2 k+1}}{(2 k+1)!}\] is not the same as
\[\sum\sin(x^4)\] at all
the first one is the maclaurin series, it converges for all \(x\) like you found
the second one is the sum of sines
not at all the same thing

Answer 32

\[\sin(x^4)=\sum _{k=0}^{\infty } \frac{(-1)^k \left(x^4\right)^{2 k+1}}{(2 k+1)!}\]
you were trying to sum a sum

Answer 33

clear i hope?

Answer 34

so that means the frm is incorrect?i tought we just need to slot in the x values

Answer 35

*form

Answer 36

you are correct
everything you did was correct
except the thing you typed in to wolfram to confuse yourself

Answer 37

ohhh.. so it means the function converges then? haha

Answer 38

you checked your answer by writing the wrong thing in to wolram, and then got confused

Answer 39

yes, you are right
right right right
everything right
except the silly thing you typed it to wolfram, which was wrong wrong wrong

Answer 40

i see hahah thanks! xD so the radius would be infinity then (: but what if the function diverges? what would be the radius?

Answer 41

before i answer that, suppose you had a polynomial, i.e. a power series with finite non zero terms
like for example \(p(x)=1+x+x^2\) a 3 term power series, clearly converges for all \(x\) i.e you can evaluate it at any number

what you did was try to check your answer in wolfram by typing in
\[\sum_{x=0}^{\infty}{1+x+x^2}\]

Answer 42

haha yeah its not the same because i add the summation xD my bad

Answer 43

now it is possible that a power series converges for all \(x\) , for some \(x\) or only for 1 \(x\)
l.e. the radius of convergence could be \(\infty\) converges for all \(x\)
it could be \(R\) converges for \(-R+a<x<R+a\) or
radius is \(0\) converges only if \(x=a\)

Answer 44

i see, so there wont be radius of convergence if it diverges then.. haha i think there wont be any silly question like that haha xD, thank yous so mucch! it's so clear now! :D

Answer 45

wait hold the phone
"it diverges" does not really make any sense without referring to \(x\)

Answer 46

for example even
\[\sum n!x^n\]converges if \(x=0\)

Answer 47

oh okay, so the x should be a point of reference then (: got it! thanks!

Answer 48

yw