can anyone help out with power series?

- anonymous

can anyone help out with power series?

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- schrodinger

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- anonymous

Yes

- anonymous

so waht is your question?

- 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!

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- 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

Do you mind giving me the last few lines of your ratio test?

- anonymous

well it boiled down to lim=|2x/n+1|

- anonymous

COOL i got that too

- anonymous

so limn->inifinity, 2x/(n+1) -> 0

- anonymous

yupp, exactly

- anonymous

that mean the interval is (-infin, infin) and radius of convergence is infinity

- anonymous

that means*

- anonymous

ok, just so i understand though what is the R value

- anonymous

oh is it infinity

- anonymous

yes

- anonymous

so how was i able to decipher that the R value is infinity just from knowing the series converged absolutely?

- anonymous

i guess im asking why does geting L=0 from the ratio test help me

- anonymous

u can interpret it this way lim n->infin |x|(2/(n+1))<1 , then |x|<1/ 0 |x|

- 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

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

YEP that is it!!

- anonymous

thankyou very much i appreciate it. and just for learning purposes, do i know this series is centered at 0?

- anonymous

yes

- anonymous

and that is bc the x-a isnt there?

- anonymous

Becasue it is the form of c0+c1(x)+c2(x)^2 +....cn(x)^n

- anonymous

yep!!

- anonymous

so its in the c0 form

- anonymous

i think i get this

- 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

and in my equation here the 2^2/n! is the Cn part

- anonymous

2^n/n!

- anonymous

u got it?

- anonymous

right right ok im dumb

- anonymous

LOL no i know that is you meant.

- anonymous

2^2 is just a typo~~

- anonymous

yup my fault

- 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

LOL my pleasure and honor :P

- 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

LOL alright if u have any further questions, just post here :) i am gonna go now
Bye :)

- 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

welp i think the ratio test gives you the limit to approach zero again, so i guess its infinity again?

- anonymous

Thanks lol i was arguing with my friend about the convergence of this series.

- anonymous

yeah like at the end of my ratio test i got x+3 ontop

- anonymous

but the numerator is like e^3n+1 time e^3n+2 times e^3m+3 so thats infinity

- anonymous

i just meant denominator* sorry so basicall that makes the denom infinity and the numerator x+3

- anonymous

yea haha Thanks!~

Looking for something else?

Not the answer you are looking for? Search for more explanations.