A community for students.
Here's the question you clicked on:
 0 viewing
Loser66
 one year ago
What is the result?
(1)^(n^2/2) = 1if n is even
what is the result if n is odd?
Please, help
Loser66
 one year ago
What is the result? (1)^(n^2/2) = 1if n is even what is the result if n is odd? Please, help

This Question is Closed

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1This is the original problem: Find the radius convergence of \(\sum_{n=1}^\infty \dfrac{i^{n^2}}{n}\)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1We need an argument for i^n^2

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0To be fair the root of \(1\) can be \(+i\) or \(i\).

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.0n=even that thing is 1 n=odd then its i

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.0wait n=odd has got cases

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0Exactly, we can't tell if it's \(i\) or \(i\) right?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1\(i^{n^2}= (i^2)^{n^2/2}=(1)^{n^2/2}\)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1ok, then dw:1443285237007:dw, right?

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0i got afraid and deleted my answer.. hah

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0\[\pm i \]goddammit

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1So, what is the radius of convergence?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1@ganeshie8 how to argue \( (1)^{n^2/2}\) depend on n mod 2? the limit should be 1 and 1

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1Let me attach my prof's arguments but I don't quite understand it. :(

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1Just similar, not my problem but the argument is the same.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What do you mean by radius of convergence? That doesn't make sense to me. It's like asking the radius of convergence for \(\sum\limits_{n\ge1}\frac{1}{n}\). Are you talking about a power series at all?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1The original problem is Find the radius of convergence of \(\sum_{n=0}^\infty \dfrac{z^{n^2}}{n}\)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1I got it =1 by finding the \( lim sup\sqrt[k]a_k=1\)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1The second part is :"discuss the convergence of the power series if z =i "

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That's much better, thanks for clarifying.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1Since the question is about convergence, I think we need the radius of convergence also, right? clearly, when n is even, then sum is sum of 1/n which is convergent with radius R = infinitive, right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Like you said, the series converges with radius \(1\), which means for \(z<1\). What's \(z\) with \(z=i\)?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1oh, divergent!! right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I would say so. I'm pretty sure a power series cannot converge for all points within its convergence disk AND a finite number of points on the boundary of the disk, but maybe I'm wrong about that... (I've never seen it happen at any rate)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Hmm, looks like that might actually be possible: https://en.wikipedia.org/wiki/Radius_of_convergence#Convergence_on_the_boundary

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I don't think that's true. Consider \(\sum_{n\ge1}\frac{(1)^n}{n}\). You have two subsequences, \(\left\{1,\frac{1}{3},\frac{1}{5},\ldots\right\}\) and \(\left\{\frac{1}{2},\frac{1}{4},\frac{1}{6},\ldots\right\}\), both of which diverge, yet the original series converges.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1But both are real, on my problem, we have only one real subsequence, the leftover is imaginary sequence

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1The net is so bad at this end :(

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0My point in my last comment is that, in general, you can have two series, \(\sum a_n\) and \(\sum b_n\) that both diverge, yet, \(\sum (a_n\pm b_n)\) may still converge. In any case, you can split this series up casebycase, as you/your prof have already done: \[\begin{align*} \sum_{n\ge1}\frac{i^{n^2}}{n}&=\sum_{k\ge1}\left(\frac{i^{(2k)^2}}{2k}+\frac{i^{(2k1)^2}}{2k1}\right)\\[1ex] &=\sum_{k\ge1}\frac{i^{4k^2}}{2k}+\sum_{k\ge1}\frac{i^{4k^24k+1}}{2k1}\\[1ex] &=\sum_{k\ge1}\frac{(i^4)^{k^2}}{2k}+i\sum_{k\ge1}\frac{(i^4)^{k(k1)}}{2k1}\\[1ex] &=\sum_{k\ge1}\frac{1}{2k}+i\sum_{k\ge1}\frac{1}{2k1} \end{align*}\] For this to converge, both the real and imaginary parts must converge. They do not. So yes, your reasoning works *in this particular case*.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1Thanks for that. I would like to make the concept clear when I replace a_n by a_k

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1Like this \(\sum_{n =1}^\infty \dfrac{(z)^{n^2}}{n}=\sum_{k=1}^\infty a_k z^k\) where \(a_k = 1/k ~~if~~ k = n^2 \) and \(a_k =0 ~~if~~ k\neq n^2\)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1Honestly, I am on and off with this. It means, I sometimes see it clearly, sometimes, not. :( I interpret the original one as = z + 0*z^2 +0*z^3 +(z^4)/2)+ 0*z^5 +0*z^6 +...+ (z^9)/3 +...

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1Hence, if k = 1 = 1^2 , \(a_k = 1/k =1\) if k = 2, \(a_k = 1/k =1/2\) and this k must be a complete square of a number n =2, right? Now, confused part. if k =9, then \(a_k = 1/9 \) and n =3, but as above, \(a_k = 1/3\) not 1/9. What am I missing?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It should be \(a_kz^k=\dfrac{1}{\sqrt k}z^k=\dfrac{1}{n}z^{n^2}=a_nz^{n^2}\).

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1\(a_k = 1/ \sqrt k ~~ if k = n^2\) , right?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1and the sum becomes \(\sum_{k=1}^\infty a_k z^k\) and we can apply lim sup a_k, right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes, \(\sqrt k\) can only be an integer if \(k\) is a perfect square, i.e. if \(k=n^2\) for \(n\in\mathbb{N}\).

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1But then, if z =1, should I use the original one or turn it to a_k?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1When z =1, the sum is \(\sum_{n=1}^\infty \dfrac{(1)^{n^2}}{n}\)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1And I have to argue for its convergence also.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You can use the same even/odd split to arrive at the same conclusion. You would end up with \[\sum_{n\ge1}\frac{(1)^{n^2}}{n}=\sum_{k\ge1}\left(\frac{1}{2k}+\frac{1}{2k1}\right)=\sum_{k\ge1}\frac{4k1}{2k(2k1)}\sim\sum_{k\ge1}\frac{1}{k}\] To use your \(n\to k\) replacement approach, you can still make an even/odd argument: \[\begin{align*}\sum_{\substack{k\ge1\\k\in S}}\frac{(1)^k}{\sqrt k}&=\sum_{\substack{p\ge1\\p\in S}}\left(\frac{(1)^{2p}}{\sqrt {2p}}+\frac{(1)^{2p1}}{\sqrt {2p1}}\right)\\[1ex] &=\sum_{\substack{p\ge1\\p\in S}}\left(\frac{1}{\sqrt {2p}}\frac{1}{\sqrt {2p1}}\right)\\[1ex] &=\sum_{\substack{p\ge1\\p\in S}}\frac{\sqrt{2p1}\sqrt{2p}}{\sqrt {4p^22p}}\\[1ex] &\sim \sum_{\substack{p\ge1\\p\in S}}\frac{1}{\sqrt p}\\[1ex] &=\sum_{\substack{q\ge1\\q\in\mathbb{N}}}\frac{1}{q} \end{align*}\]where \(S\) denotes the set of perfect squares.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1oh, I tried to put it into something totally different from yours , hihihi... but let me digest your solution first. My attempt: sum = 1 + (1/2) (1/3) +(1/4) (1/5) ....... and group them as \(a_n  a_{n+1}\) where \(a_n = 1/n\)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1and I still stuck on the first 1.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1oh, I cannot do it since it is infinite series, right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well there's nothing preventing you from doing that sort of pairing of the terms, but I'm not sure what you're trying to accomplish.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1One more question, how can you go from n to k at the first line of the argument?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1\(\sum_{n\ge1}\frac{(1)^{n^2}}{n}=\sum_{k\ge1}\left(\frac{1}{2k}+\frac{1}{2k1}\right)=\sum_{k\ge1}\frac{4k1}{2k(2k1)}\sim\sum_{k\ge1}\frac{1}{k}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That's the evenodd split.\[\begin{array}{ccc} k&2k&2k1\\ \hline 1&2&1\\ 2&4&3\\ 3&6&5\\\vdots&\vdots&\vdots \end{array}\]

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1so, you split it into n is even = 2k and if n is odd = 2k +1, right?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1I got it. Thank you so much. Thanks for being patient to me. :)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1But then, the odd series must be  , right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Hmm, I might have made a mistake somewhere above... If it were \(\dfrac{1}{2k}\dfrac{1}{2k1}\), then the series would converge...

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1Yes, I got \(\sum_{k=1} \dfrac{1}{4k^2 +2k}\) then, it converges since it is < sum (1/k^2) which is convergent. Thanks again.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.