anonymous
  • anonymous
Determine the centre, radius, and interval of convergence of the power series
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
anonymous
  • anonymous
for the radius i keep getting 1
anonymous
  • anonymous
x<1

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anonymous
  • anonymous
and for the interval of convergence -3
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle \sum_{ n=1 }^{ \infty } ~ \frac{ (-1)^n(x+1)^n }{n^32^n}}\)
SolomonZelman
  • SolomonZelman
you mean that \(|x|\le1\) ?
anonymous
  • anonymous
the interval is right, but the solution says that the limit is 2
anonymous
  • anonymous
i mean the radius is 2
anonymous
  • anonymous
SolomonZelman
  • SolomonZelman
The series will converge for any x that satisfies: \(|x|\le1\) (Because even if x=1, the alternating p-series with p=3 will converge)
anonymous
  • anonymous
yea thats what u got using the ratio test ,but can u look at the solution i posted
SolomonZelman
  • SolomonZelman
Wrong, not n+1, but x+1, so you don't get what you got
anonymous
  • anonymous
why does it say 1/L=2
SolomonZelman
  • SolomonZelman
Look, let me do the ratio test for you, knowing that must be: (1) |r|<1 (and check for ±r=1) (2) r = \(a_{n+1}/a_n\)
anonymous
  • anonymous
yea my ratio test simplified to |x+1|/2<1
anonymous
  • anonymous
x<1
SolomonZelman
  • SolomonZelman
\(\large\color{slate}{\displaystyle\lim_{n \rightarrow ~\infty }\left|\frac{(-1)^{n+1}(x+1)^{n+1}}{(n+1)^32^{n+1}}\div \frac{(-1)^{n}(x+1)^{n}}{(n)^32^{n}}\right|}\)
SolomonZelman
  • SolomonZelman
\(\large\color{black}{\displaystyle\lim_{n \rightarrow ~\infty }\left|\frac{(-1)^{n+1}(x+1)^{n+1}}{(n+1)^32^{n+1}}\times \frac{(n)^32^{n}}{(-1)^{n}(x+1)^{n}}\right|}\) \(\large\color{black}{\displaystyle\lim_{n \rightarrow ~\infty }\left|\frac{(x+1)^{n+1}}{(n+1)^32^{n+1}}\times \frac{(n)^32^{n}}{(x+1)^{n}}\right|}\) -1 has gone away due to absolute value. And then algebra. \(\large\color{black}{\displaystyle\lim_{n \rightarrow ~\infty }\left|\frac{n^3(x+1)}{2(n+1)^3}\right|}\) You need this so called common ratio to be <1, so: \(\large\color{black}{\displaystyle\left|\frac{(x+1)}{2}\right|<1}\) \(\large\color{black}{\displaystyle\left|x+1\right|<2}\) \(\large\color{black}{\displaystyle-2
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle \sum_{ n=1 }^{ \infty } ~ \frac{ (-1)^n(\color{red}{-3}+1)^n }{n^32^n}}\) \(\large\color{black}{ \displaystyle \sum_{ n=1 }^{ \infty } ~ \frac{ (-1)^n(-2)^n }{n^32^n}}\) \(\large\color{black}{ \displaystyle \sum_{ n=1 }^{ \infty } ~ \frac{ 1 }{n^3}}\) ---> CONVERGES
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle \sum_{ n=1 }^{ \infty } ~ \frac{ (-1)^n(\color{blue}{1}+1)^n }{n^32^n}}\) \(\large\color{black}{ \displaystyle \sum_{ n=1 }^{ \infty } ~ \frac{ (-1)^n(2)^n }{n^32^n}}\) \(\large\color{black}{ \displaystyle \sum_{ n=1 }^{ \infty } ~ \frac{ (-1)^n }{n^3}}\) ---> CONVERGES
anonymous
  • anonymous
yeah but what about the radius? thats what I'm confused about
SolomonZelman
  • SolomonZelman
So interval of convergence is: x ∈ [-3,1] And the radius is always half the interval (whether interval is closed or open)
tanya123
  • tanya123
all hail SolomonZelman
SolomonZelman
  • SolomonZelman
Hello, tanya!
anonymous
  • anonymous
but the way i learnt it using the ratio test, u get the radius by |x+1|/2<1
anonymous
  • anonymous
thats how it shows in my textbooks
SolomonZelman
  • SolomonZelman
Yes, and that is good, I suppose :O
anonymous
  • anonymous
it doesn't say anything about the radius being half of the interval
SolomonZelman
  • SolomonZelman
Just by a simple geometric series that you learn in high school. |r|<1, and that can be easily shown (what to hear?)
tanya123
  • tanya123
Hiiiii, you work soo hard lad, you must take a break and drink some water, all this math
SolomonZelman
  • SolomonZelman
Was one of my favorite topics in calc 2 - power series! Juicy
anonymous
  • anonymous
so using |x+1|/2<1 should |x|<1
SolomonZelman
  • SolomonZelman
No, |x+1|, not |x|
anonymous
  • anonymous
so |x+1|<2?
anonymous
  • anonymous
oh, right!
anonymous
  • anonymous
yes i get it now! thanks a lot!
SolomonZelman
  • SolomonZelman
Yes, one more recap to finish it up....
anonymous
  • anonymous
also is the centre that just the half way of the interval?
SolomonZelman
  • SolomonZelman
The idea comes from: \(\large\color{slate}{\displaystyle (1+r+r^2+r^3+r^4+r^{5})/(1-r)=1-r^5 }\) I will show why that is so: \(\Large \color{blue}{ \displaystyle ^{\color{red}{~~~~~~~~~~~~~~~~~~~~~~~~~{\rm r}^4~~~~~+~~~~{\rm r}^3~~~~+~~~~{\rm r}^2~~~~+~~~~{\rm r}~~~~+~~~1}}_{\Huge _\text{_______________________________}}}\) \(\large\color{blue}{ \displaystyle -{\rm r}+1{\huge|}~~-{\rm r}^5~~+~0{\rm r}^4~+~~0{\rm r}^3~~+~~0{\rm r}^2~+~~0{\rm r}~~+~1}\) \(\large\color{red}{ \displaystyle -{\rm r}^5~~+~~{\rm r}^4 }\) \(\large\color{blue}{ \displaystyle ^\text{____________} }\) \(\large\color{red}{ \displaystyle -{\rm r}^4~~+~~0{\rm r}^3 }\) \(\large\color{red}{ \displaystyle -{\rm r}^4~~+~~{\rm r}^3 }\) \(\large\color{blue}{ \displaystyle ^\text{_____________} }\) \(\large\color{red}{ \displaystyle -{\rm r}^3 ~~+~~0{\rm r}^2 }\) \(\large\color{red}{ \displaystyle -{\rm r}^3 ~~+~~~{\rm r}^2 }\) \(\large\color{blue}{ \displaystyle ^\text{_______________} }\) \(\large\color{red}{ \displaystyle -{\rm r}^2 ~~+~~0{\rm r} }\) \(\large\color{red}{ \displaystyle -{\rm r}^2 ~~+~~{\rm r} }\) \(\large\color{blue}{ \displaystyle ^\text{______________} }\) \(\large\color{red}{ \displaystyle -{\rm r}~+~1 }\) \(\large\color{red}{ \displaystyle -{\rm r}~+~1 }\) \(\large\color{blue}{ \displaystyle ^\text{___________} }\) \(\large\color{red}{ \displaystyle 0 }\) If you agree with (and understand) the above polynomial division, then you should get an intuitive understanding of why: \(\color{black}{ \displaystyle \color{blue}{(-{\rm r}^{\rm n}+1)}\div \color{red}{(-{\rm r}+1)} ~~= \color{green}{{\rm r}^{{\rm n}-1}~+~{\rm r}^{{\rm n}-2}~+~....~+~{\rm r}^3~+~{\rm r}^2~+~{\rm r}~+~1} }\) \((\)For all natural number n that are greater than 1 \()\) Thus we get: \(\large\color{black}{ \displaystyle \sum_{k=1}^{n}\left(r^{k-1}\right)=1+r+r^2+r^3+...+r^{n-1} = \frac{-r^n+1}{-r+1}}\) This is where: \(\large\color{black}{ \displaystyle \sum_{k=1}^{n}\left(r^{k-1}\right)= \frac{1-r^n}{1-r}}\) and \(\large\color{black}{ \displaystyle \sum_{k=1}^{n}\color{orangered}{a_1}\left(r^{k-1}\right)= \frac{\color{orangered}{a_1}\left(1-r^n\right)}{1-r}}\) come from. ---------------------------------- Now, convergence of an infinite geometric series will be therefore determined by the convergence of the sequence of (1-r\(^n\))/(1-r) \(\large\color{slate}{\displaystyle\lim_{n \rightarrow ~\infty}(1+r+r^2+r^3+...+r^{n-1})=\lim_{n \rightarrow ~\infty}\left(\frac{1-r^n}{1-r}\right)}\) after applying limit properties, we get: \(\large\color{slate}{\displaystyle\lim_{n \rightarrow ~\infty}\left(\frac{1-r^n}{1-r}\right)=\left(\frac{1}{1-r}\right)\lim_{n \rightarrow ~\infty}\left(1-r^n\right) \\[1.9 em] \large \displaystyle =\left(\frac{1}{1-r}\right)\left(1-\lim_{n \rightarrow ~\infty}r^n\right)}\) \({\large \displaystyle =\left(\frac{1}{1-r}\right)-\left(\frac{1}{1-r}\right)\lim_{n \rightarrow ~\infty}r^n}\) \(\scriptsize\color{ slate }{\scriptsize{\bbox[5pt, royalblue ,border:2px solid royalblue ]{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ }}}\) So, \(r\ne1\) (because when r=1 we get an indetermine sum for the series) And when r>1 the limit will go into infinity. So 0>r>1 is so far verfied. \(\large\color{slate}{\displaystyle\lim_{n \rightarrow ~\infty}(r^n)}\) for -1
SolomonZelman
  • SolomonZelman
Sorry for a long reply, just a geometric convergence proof, for why is it so that |r|<1. And as you have probably heard or figured, the same idea of |r|<1 is applied here, in Ratio test (and in root test)... EXCEPT that for r=1 the test is inconclusive which can be explained due to the fact that the limit in ratio test is not quite the common ratio.
anonymous
  • anonymous
wow that was a long reply, lol thanks for the effort
SolomonZelman
  • SolomonZelman
Anytime !
anonymous
  • anonymous
also u said that
anonymous
  • anonymous
shouldn't it be 2/n^3
anonymous
  • anonymous
i mean isnt (-2)^n(-1)^n = (2)^2n
SolomonZelman
  • SolomonZelman
No
SolomonZelman
  • SolomonZelman
Missapplication of exponent rules: \(\LARGE a^n\cdot b^n=(a \cdot b)^n\)
SolomonZelman
  • SolomonZelman
\(\LARGE a^n\cdot b^n=(a \cdot b)^n\) YES \(\LARGE a^n\cdot b^n=(a \cdot b)^{2n}\) NO
anonymous
  • anonymous
i thought u add powers in multiplication?
SolomonZelman
  • SolomonZelman
when you have the same base, you add powers: \(\LARGE x^n\cdot x^m=x^{n+m} \) YES
anonymous
  • anonymous
oh, sorry my maths is really rusty
anonymous
  • anonymous
thanks for clearing it up
SolomonZelman
  • SolomonZelman
You just made some confusion about the rules.... anytime :0

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