anonymous
  • anonymous
can anyone help out with power series?
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
Yes
anonymous
  • anonymous
so waht is your question?
anonymous
  • anonymous
Find the radius of convergence and the interval of convergence of the power series. the sum from n=0 to infinity of 2^n x^n / n!

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anonymous
  • anonymous
so i used ratio test and found it converges absolutely, but im not sure how to find the Radius on convergence or interval
anonymous
  • anonymous
Do you mind giving me the last few lines of your ratio test?
anonymous
  • anonymous
well it boiled down to lim=|2x/n+1|
anonymous
  • anonymous
COOL i got that too
anonymous
  • anonymous
so limn->inifinity, 2x/(n+1) -> 0
anonymous
  • anonymous
yupp, exactly
anonymous
  • anonymous
that mean the interval is (-infin, infin) and radius of convergence is infinity
anonymous
  • anonymous
that means*
anonymous
  • anonymous
ok, just so i understand though what is the R value
anonymous
  • anonymous
oh is it infinity
anonymous
  • anonymous
yes
anonymous
  • anonymous
so how was i able to decipher that the R value is infinity just from knowing the series converged absolutely?
anonymous
  • anonymous
i guess im asking why does geting L=0 from the ratio test help me
anonymous
  • anonymous
u can interpret it this way lim n->infin |x|(2/(n+1))<1 , then |x|<1/ 0 |x|
anonymous
  • anonymous
X is a constant. So no mater what X is, the limit will go to 0, and the ratio test states that if the limit is <1, it converges. So X can be any real numbers
anonymous
  • anonymous
ah ok, so becasue this series converges no matter what x-value is plugged in: my x value can range from -infinity to +infinity bc it will always converge. and then the radius for that interval is infinity also
anonymous
  • anonymous
YEP that is it!!
anonymous
  • anonymous
thankyou very much i appreciate it. and just for learning purposes, do i know this series is centered at 0?
anonymous
  • anonymous
yes
anonymous
  • anonymous
and that is bc the x-a isnt there?
anonymous
  • anonymous
Becasue it is the form of c0+c1(x)+c2(x)^2 +....cn(x)^n
anonymous
  • anonymous
yep!!
anonymous
  • anonymous
so its in the c0 form
anonymous
  • anonymous
i think i get this
anonymous
  • anonymous
yea cn(x)^n then it is centered at 0, if it is cn(x-1)^n then it would be centered at x=1
anonymous
  • anonymous
and in my equation here the 2^2/n! is the Cn part
anonymous
  • anonymous
2^n/n!
anonymous
  • anonymous
u got it?
anonymous
  • anonymous
right right ok im dumb
anonymous
  • anonymous
LOL no i know that is you meant.
anonymous
  • anonymous
2^2 is just a typo~~
anonymous
  • anonymous
yup my fault
anonymous
  • anonymous
well i really do appreciate this. It may not seem it but you just helped someone who will most likely be getting early admission to UB med school ha ha Ive been getting pounded with organic and inorganic so this has been on the back burner
anonymous
  • anonymous
LOL my pleasure and honor :P
anonymous
  • anonymous
keep up the good work you are really patient, and a good listener. Guys like you are doing it for the love of the game if you will.
anonymous
  • anonymous
LOL alright if u have any further questions, just post here :) i am gonna go now Bye :)
anonymous
  • anonymous
oh can you find the radius of convergence for this power series : sum n = 0 to n =infinity ((x+3)^n)/(e^3n!)
anonymous
  • anonymous
welp i think the ratio test gives you the limit to approach zero again, so i guess its infinity again?
anonymous
  • anonymous
Thanks lol i was arguing with my friend about the convergence of this series.
anonymous
  • anonymous
yeah like at the end of my ratio test i got x+3 ontop
anonymous
  • anonymous
but the numerator is like e^3n+1 time e^3n+2 times e^3m+3 so thats infinity
anonymous
  • anonymous
i just meant denominator* sorry so basicall that makes the denom infinity and the numerator x+3
anonymous
  • anonymous
yea haha Thanks!~

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