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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

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  1. Loser66
    • one year ago
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    @dan815

  2. DanJS
    • one year ago
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    try it

  3. Loser66
    • one year ago
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    Can you?

  4. Loser66
    • one year ago
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    This is the original problem: Find the radius convergence of \(\sum_{n=1}^\infty \dfrac{i^{n^2}}{n}\)

  5. Loser66
    • one year ago
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    We need an argument for i^n^2

  6. ParthKohli
    • one year ago
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    To be fair the root of \(-1\) can be \(+i\) or \(-i\).

  7. imqwerty
    • one year ago
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    n=even that thing is 1 n=odd then its i

  8. imqwerty
    • one year ago
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    wait n=odd has got cases

  9. ParthKohli
    • one year ago
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    Exactly, we can't tell if it's \(i\) or \(-i\) right?

  10. Loser66
    • one year ago
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    \(i^{n^2}= (i^2)^{n^2/2}=(-1)^{n^2/2}\)

  11. imqwerty
    • one year ago
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    yea tru

  12. Loser66
    • one year ago
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    ok, then |dw:1443285237007:dw|, right?

  13. DanJS
    • one year ago
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    i got afraid and deleted my answer.. hah

  14. ParthKohli
    • one year ago
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    \[\pm i \]goddammit

  15. DanJS
    • one year ago
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    conjugate/

  16. Loser66
    • one year ago
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    So, what is the radius of convergence?

  17. Loser66
    • one year ago
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    @ganeshie8 how to argue \( (-1)^{n^2/2}\) depend on n mod 2? the limit should be 1 and -1

  18. Loser66
    • one year ago
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    Let me attach my prof's arguments but I don't quite understand it. :(

  19. Loser66
    • one year ago
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  20. Loser66
    • one year ago
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    Just similar, not my problem but the argument is the same.

  21. anonymous
    • one year ago
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    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?

  22. Loser66
    • one year ago
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    The original problem is Find the radius of convergence of \(\sum_{n=0}^\infty \dfrac{z^{n^2}}{n}\)

  23. Loser66
    • one year ago
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    I got it =1 by finding the \( lim sup\sqrt[k]a_k=1\)

  24. Loser66
    • one year ago
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    The second part is :"discuss the convergence of the power series if z =i "

  25. anonymous
    • one year ago
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    That's much better, thanks for clarifying.

  26. Loser66
    • one year ago
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    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?

  27. anonymous
    • one year ago
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    Like you said, the series converges with radius \(1\), which means for \(|z|<1\). What's \(|z|\) with \(z=i\)?

  28. Loser66
    • one year ago
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    1

  29. Loser66
    • one year ago
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    oh, divergent!! right?

  30. anonymous
    • one year ago
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    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)

  31. anonymous
    • one year ago
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    Hmm, looks like that might actually be possible: https://en.wikipedia.org/wiki/Radius_of_convergence#Convergence_on_the_boundary

  32. Loser66
    • one year ago
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    |dw:1443289362976:dw|

  33. anonymous
    • one year ago
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    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.

  34. Loser66
    • one year ago
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    But both are real, on my problem, we have only one real subsequence, the leftover is imaginary sequence

  35. Loser66
    • one year ago
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    The net is so bad at this end :(

  36. anonymous
    • one year ago
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    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*.

  37. Loser66
    • one year ago
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    Thanks for that. I would like to make the concept clear when I replace a_n by a_k

  38. Loser66
    • one year ago
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    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 \(a_k =0 ~~if~~ k\neq n^2\)

  39. Loser66
    • one year ago
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    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 +...

  40. Loser66
    • one year ago
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    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?

  41. anonymous
    • one year ago
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    It should be \(a_kz^k=\dfrac{1}{\sqrt k}z^k=\dfrac{1}{n}z^{n^2}=a_nz^{n^2}\).

  42. Loser66
    • one year ago
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    \(a_k = 1/ \sqrt k ~~ if k = n^2\) , right?

  43. Loser66
    • one year ago
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    and the sum becomes \(\sum_{k=1}^\infty a_k z^k\) and we can apply lim sup a_k, right?

  44. anonymous
    • one year ago
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    Yes, \(\sqrt k\) can only be an integer if \(k\) is a perfect square, i.e. if \(k=n^2\) for \(n\in\mathbb{N}\).

  45. Loser66
    • one year ago
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    But then, if z =-1, should I use the original one or turn it to a_k?

  46. Loser66
    • one year ago
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    When z =-1, the sum is \(\sum_{n=1}^\infty \dfrac{(-1)^{n^2}}{n}\)

  47. Loser66
    • one year ago
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    And I have to argue for its convergence also.

  48. anonymous
    • one year ago
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    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.

  49. Loser66
    • one year ago
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    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 \(a_n = 1/n\)

  50. Loser66
    • one year ago
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    and I still stuck on the first -1.

  51. Loser66
    • one year ago
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    oh, I cannot do it since it is infinite series, right?

  52. anonymous
    • one year ago
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    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.

  53. Loser66
    • one year ago
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    One more question, how can you go from n to k at the first line of the argument?

  54. Loser66
    • one year ago
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    \(\sum_{n\ge1}\frac{(-1)^{n^2}}{n}=\sum_{k\ge1}\left(\frac{1}{2k}+\frac{1}{2k-1}\right)=\sum_{k\ge1}\frac{4k-1}{2k(2k-1)}\sim\sum_{k\ge1}\frac{1}{k}\)

  55. anonymous
    • one year ago
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    That's the even-odd split.\[\begin{array}{c|c|c} k&2k&2k-1\\ \hline 1&2&1\\ 2&4&3\\ 3&6&5\\\vdots&\vdots&\vdots \end{array}\]

  56. Loser66
    • one year ago
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    so, you split it into n is even = 2k and if n is odd = 2k +1, right?

  57. Loser66
    • one year ago
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    I got it. Thank you so much. Thanks for being patient to me. :)

  58. Loser66
    • one year ago
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    But then, the odd series must be - , right?

  59. anonymous
    • one year ago
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    Hmm, I might have made a mistake somewhere above... If it were \(\dfrac{1}{2k}-\dfrac{1}{2k-1}\), then the series would converge...

  60. Loser66
    • one year ago
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    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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