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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
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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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If you're replacing \(n^2\) with \(k\), that means you have this pattern: \[\begin{array}{ccccc} n&1&2&3&4&5&\cdots\\ \hline k&1&4&9&16&25&\cdots \end{array}\] 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 \[a_k=\begin{cases}a_{\sqrt n}&\text{ for }n\in\{1,4,9,16,25,\ldots\}\\0&\text{otherwise}\end{cases}\]

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Since I have to find\( lim sup \sqrt[k] a_k\), hence a_k must be clear.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes, our definitions of \(a_k\) are identical.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Notice \[a_n=\frac{1}{n}~~\implies~~a_{\sqrt n}=\frac{1}{\sqrt n}\] If \(n\) is a perfect square, i.e. \(n^2=k\), then \(a_k=a_{\sqrt n}=\dfrac{1}{\sqrt{n}}=\dfrac{1}{k}\).

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Sorry, swap \(n\) and \(k\) in the last set of equations :)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0I 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

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0a_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

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0But I am not sure about the terminology I should put there.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0if \(a_k = \dfrac{1}{\sqrt k}\) , 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.

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.0ok so i did this  \[\sum_{n=1}^{\infty}\frac{ z^\left( n^2 \right) }{ n }=\sum_{k=1}^{\infty}a _{k}z^k\] \[d \frac{ \sum_{n=1}^{\infty} z^\left( n^2 \right)n^\left( 1 \right)}{ dz}=d \frac{ \sum_{k=1}^{\infty}a _{k}z^k }{ dz}\] \[\sum_{n=1}^{\infty}\frac{ n^2z^\left( (n+1)(n1) \right) }{ n }=\sum_{k=1}^{\infty}a _{k}\times k \times z^{k1}\] \[\sum_{n=1}^{\infty}z ^{n+1}=\sum_{k=1}^{\infty}a_{k}\] ?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0And 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

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0@imqwerty what are you doing?

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.0i differentiated the summation on both sides of the equation

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0why do we have to do that?

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.0i did that so as to remove/factor out the n^2 from the z

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0No 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 \(\dfrac{z^k}{\sqrt{k}}\)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Hence \( a_k = 1/\sqrt k\) if k = n^2

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Thank you so much, both you @SithsAndGiggles and @imqwerty . I do appreciate:)

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.0np :) i like ur questions but can we get the same result using my method? :/

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0I don't think so, since a_k is a coefficient but in your method a_k is a series of variable z.
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