Loser66
  • Loser66
What is the result? (-1)^(n^2/2) = 1if n is even what is the result if n is odd? Please, help
Mathematics
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SOLVED
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chestercat
  • chestercat
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Loser66
  • Loser66
@dan815
DanJS
  • DanJS
try it
Loser66
  • Loser66
Can you?

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Loser66
  • Loser66
This is the original problem: Find the radius convergence of \(\sum_{n=1}^\infty \dfrac{i^{n^2}}{n}\)
Loser66
  • Loser66
We need an argument for i^n^2
ParthKohli
  • ParthKohli
To be fair the root of \(-1\) can be \(+i\) or \(-i\).
imqwerty
  • imqwerty
n=even that thing is 1 n=odd then its i
imqwerty
  • imqwerty
wait n=odd has got cases
ParthKohli
  • ParthKohli
Exactly, we can't tell if it's \(i\) or \(-i\) right?
Loser66
  • Loser66
\(i^{n^2}= (i^2)^{n^2/2}=(-1)^{n^2/2}\)
imqwerty
  • imqwerty
yea tru
Loser66
  • Loser66
ok, then |dw:1443285237007:dw|, right?
DanJS
  • DanJS
i got afraid and deleted my answer.. hah
ParthKohli
  • ParthKohli
\[\pm i \]goddammit
DanJS
  • DanJS
conjugate/
Loser66
  • Loser66
So, what is the radius of convergence?
Loser66
  • Loser66
@ganeshie8 how to argue \( (-1)^{n^2/2}\) depend on n mod 2? the limit should be 1 and -1
Loser66
  • Loser66
Let me attach my prof's arguments but I don't quite understand it. :(
Loser66
  • Loser66
Loser66
  • Loser66
Just similar, not my problem but the argument is the same.
anonymous
  • anonymous
What 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
  • Loser66
The original problem is Find the radius of convergence of \(\sum_{n=0}^\infty \dfrac{z^{n^2}}{n}\)
Loser66
  • Loser66
I got it =1 by finding the \( lim sup\sqrt[k]a_k=1\)
Loser66
  • Loser66
The second part is :"discuss the convergence of the power series if z =i "
anonymous
  • anonymous
That's much better, thanks for clarifying.
Loser66
  • Loser66
Since 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
  • anonymous
Like you said, the series converges with radius \(1\), which means for \(|z|<1\). What's \(|z|\) with \(z=i\)?
Loser66
  • Loser66
1
Loser66
  • Loser66
oh, divergent!! right?
anonymous
  • anonymous
I 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
  • anonymous
Hmm, looks like that might actually be possible: https://en.wikipedia.org/wiki/Radius_of_convergence#Convergence_on_the_boundary
Loser66
  • Loser66
|dw:1443289362976:dw|
anonymous
  • anonymous
I 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
  • Loser66
But both are real, on my problem, we have only one real subsequence, the leftover is imaginary sequence
Loser66
  • Loser66
The net is so bad at this end :(
anonymous
  • anonymous
My 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 case-by-case, as you/your prof have already done: \[\begin{align*} \sum_{n\ge1}\frac{i^{n^2}}{n}&=\sum_{k\ge1}\left(\frac{i^{(2k)^2}}{2k}+\frac{i^{(2k-1)^2}}{2k-1}\right)\\[1ex] &=\sum_{k\ge1}\frac{i^{4k^2}}{2k}+\sum_{k\ge1}\frac{i^{4k^2-4k+1}}{2k-1}\\[1ex] &=\sum_{k\ge1}\frac{(i^4)^{k^2}}{2k}+i\sum_{k\ge1}\frac{(i^4)^{k(k-1)}}{2k-1}\\[1ex] &=\sum_{k\ge1}\frac{1}{2k}+i\sum_{k\ge1}\frac{1}{2k-1} \end{align*}\] 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
  • Loser66
Thanks for that. I would like to make the concept clear when I replace a_n by a_k
Loser66
  • Loser66
Like 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
  • Loser66
Honestly, 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
  • Loser66
Hence, 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
  • anonymous
It should be \(a_kz^k=\dfrac{1}{\sqrt k}z^k=\dfrac{1}{n}z^{n^2}=a_nz^{n^2}\).
Loser66
  • Loser66
\(a_k = 1/ \sqrt k ~~ if k = n^2\) , right?
Loser66
  • Loser66
and the sum becomes \(\sum_{k=1}^\infty a_k z^k\) and we can apply lim sup a_k, right?
anonymous
  • anonymous
Yes, \(\sqrt k\) can only be an integer if \(k\) is a perfect square, i.e. if \(k=n^2\) for \(n\in\mathbb{N}\).
Loser66
  • Loser66
But then, if z =-1, should I use the original one or turn it to a_k?
Loser66
  • Loser66
When z =-1, the sum is \(\sum_{n=1}^\infty \dfrac{(-1)^{n^2}}{n}\)
Loser66
  • Loser66
And I have to argue for its convergence also.
anonymous
  • anonymous
You can use the same even/odd split to arrive at the same conclusion. You would end up with \[\sum_{n\ge1}\frac{(-1)^{n^2}}{n}=\sum_{k\ge1}\left(\frac{1}{2k}+\frac{1}{2k-1}\right)=\sum_{k\ge1}\frac{4k-1}{2k(2k-1)}\sim\sum_{k\ge1}\frac{1}{k}\] To use your \(n\to k\) replacement approach, you can still make an even/odd argument: \[\begin{align*}\sum_{\substack{k\ge1\\k\in S}}\frac{(-1)^k}{\sqrt k}&=\sum_{\substack{p\ge1\\p\in S}}\left(\frac{(-1)^{2p}}{\sqrt {2p}}+\frac{(-1)^{2p-1}}{\sqrt {2p-1}}\right)\\[1ex] &=\sum_{\substack{p\ge1\\p\in S}}\left(\frac{1}{\sqrt {2p}}-\frac{1}{\sqrt {2p-1}}\right)\\[1ex] &=\sum_{\substack{p\ge1\\p\in S}}\frac{\sqrt{2p-1}-\sqrt{2p}}{\sqrt {4p^2-2p}}\\[1ex] &\sim \sum_{\substack{p\ge1\\p\in S}}\frac{1}{\sqrt p}\\[1ex] &=\sum_{\substack{q\ge1\\q\in\mathbb{N}}}\frac{1}{q} \end{align*}\]where \(S\) denotes the set of perfect squares.
Loser66
  • Loser66
oh, 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
  • Loser66
and I still stuck on the first -1.
Loser66
  • Loser66
oh, I cannot do it since it is infinite series, right?
anonymous
  • anonymous
Well 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
  • Loser66
One more question, how can you go from n to k at the first line of the argument?
Loser66
  • Loser66
\(\sum_{n\ge1}\frac{(-1)^{n^2}}{n}=\sum_{k\ge1}\left(\frac{1}{2k}+\frac{1}{2k-1}\right)=\sum_{k\ge1}\frac{4k-1}{2k(2k-1)}\sim\sum_{k\ge1}\frac{1}{k}\)
anonymous
  • anonymous
That's the even-odd split.\[\begin{array}{c|c|c} k&2k&2k-1\\ \hline 1&2&1\\ 2&4&3\\ 3&6&5\\\vdots&\vdots&\vdots \end{array}\]
Loser66
  • Loser66
so, you split it into n is even = 2k and if n is odd = 2k +1, right?
Loser66
  • Loser66
I got it. Thank you so much. Thanks for being patient to me. :)
Loser66
  • Loser66
But then, the odd series must be - , right?
anonymous
  • anonymous
Hmm, I might have made a mistake somewhere above... If it were \(\dfrac{1}{2k}-\dfrac{1}{2k-1}\), then the series would converge...
Loser66
  • Loser66
Yes, 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.

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