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Loser66

  • one year ago

Old problem confirm \[\sum_{n=1}^\infty \dfrac{z^{n^2}}{n}=\sum_{k=1}^\infty a_kz^k\] Then \(a_k =?? \) Please, help

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  1. Loser66
    • one year ago
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    @SithsAndGiggles

  2. anonymous
    • one year ago
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    If you're replacing \(n^2\) with \(k\), that means you have this pattern: \[\begin{array}{c|cccc} n&1&2&3&4&5&\cdots\\ \hline k&1&4&9&16&25&\cdots \end{array}\] So while \(a_n\) is defined for all positive integers \(n\), \(a_k\) is only defined for perfect squares. In other words, you can transform \(a_n\) to \[a_k=\begin{cases}a_{\sqrt n}&\text{ for }n\in\{1,4,9,16,25,\ldots\}\\0&\text{otherwise}\end{cases}\]

  3. Loser66
    • one year ago
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    |dw:1443666750718:dw|

  4. Loser66
    • one year ago
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    |dw:1443666782122:dw|

  5. Loser66
    • one year ago
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    Since I have to find\( lim sup \sqrt[k] a_k\), hence a_k must be clear.

  6. anonymous
    • one year ago
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    Yes, our definitions of \(a_k\) are identical.

  7. anonymous
    • one year ago
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    Notice \[a_n=\frac{1}{n}~~\implies~~a_{\sqrt n}=\frac{1}{\sqrt n}\] If \(n\) is a perfect square, i.e. \(n^2=k\), then \(a_k=a_{\sqrt n}=\dfrac{1}{\sqrt{n}}=\dfrac{1}{k}\).

  8. anonymous
    • one year ago
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    Sorry, swap \(n\) and \(k\) in the last set of equations :)

  9. Loser66
    • one year ago
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    I interpret it like this \(\sum_{n=1}^\infty \dfrac{z^{n^2}}{n}= z + \dfrac{z^4}{2}+\dfrac{z^9}{3}+...\) compare to \(\sum_{k=1}^\infty a_kz^k = a_1z +a_2z^2 +a_3z^3 +a_4z^4+....+a_9z^9+...\) hence the equivalent part is a_1 =1 a_4 = 1/2 a_9 = 1/3 all other a_j =0

  10. Loser66
    • one year ago
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    a_1 is k =1 when n=1 n^2 =1 a_4 = 1/2 when n =2, n^2 =4, hence k = 1/ sqrt n

  11. Loser66
    • one year ago
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    But I am not sure about the terminology I should put there.

  12. Loser66
    • one year ago
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    if \(a_k = \dfrac{1}{\sqrt k}\) , then \(limsup\sqrt[k]a_k= limsup\dfrac{1}{\sqrt[2k]k}\) and by squeeze theorem it is =1 still but if \(a_k = 1/k, \) we still have limsup =1 that is why I need confirm.

  13. imqwerty
    • one year ago
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    ok so i did this - \[\sum_{n=1}^{\infty}\frac{ z^\left( n^2 \right) }{ n }=\sum_{k=1}^{\infty}a _{k}z^k\] \[d \frac{ \sum_{n=1}^{\infty} z^\left( n^2 \right)n^\left( -1 \right)}{ dz}=d \frac{ \sum_{k=1}^{\infty}a _{k}z^k }{ dz}\] \[\sum_{n=1}^{\infty}\frac{ n^2z^\left( (n+1)(n-1) \right) }{ n }=\sum_{k=1}^{\infty}a _{k}\times k \times z^{k-1}\] \[\sum_{n=1}^{\infty}z ^{n+1}=\sum_{k=1}^{\infty}a_{k}\] ?

  14. Loser66
    • one year ago
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    And you know that when we turn it to k, everything must be calculated w.r.t k, not n anymore. n is there to show the relationship between the original one to the new one

  15. Loser66
    • one year ago
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    @imqwerty what are you doing?

  16. imqwerty
    • one year ago
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    i differentiated the summation on both sides of the equation

  17. Loser66
    • one year ago
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    why do we have to do that?

  18. imqwerty
    • one year ago
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    i did that so as to remove/factor out the n^2 from the z

  19. Loser66
    • one year ago
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    No need, but I think I got it, just turn the original one like this \[\dfrac{z^{n^2}}{\sqrt{n^2}}\] hence to k, it will be \(\dfrac{z^k}{\sqrt{k}}\)

  20. Loser66
    • one year ago
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    Am I right?

  21. Loser66
    • one year ago
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    Hence \( a_k = 1/\sqrt k\) if k = n^2

  22. imqwerty
    • one year ago
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    yes :)

  23. Loser66
    • one year ago
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    Thank you so much, both you @SithsAndGiggles and @imqwerty . I do appreciate:)

  24. imqwerty
    • one year ago
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    np :) i like ur questions but can we get the same result using my method? :/

  25. Loser66
    • one year ago
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    I don't think so, since a_k is a coefficient but in your method a_k is a series of variable z.

  26. imqwerty
    • one year ago
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    ok thanks :)

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