Skip to main content
10 events
when toggle format what by license comment
May 31, 2023 at 17:16 comment added Giorgio Metafune It shoud follow by writing $f=\sum_k c_k u_k$ and computing $Lf$ by the recursion formulas. Then $Lf=\lambda f$ gives $c_{k+2}\sqrt{(k+2)(k+1)}=(\lambda+k)c_k$ and, after choosing $c_0$ or $c_1$ one imposes that the $(c_k)$ are square summable.
May 31, 2023 at 14:20 history edited mathamphetamine CC BY-SA 4.0
added 67 characters in body
May 31, 2023 at 14:19 comment added mathamphetamine @GiorgioMetafune Thanks, I see how we can derive the condition on $\lambda$ from the series formula. Do you any thoughts on the main claim ?
May 31, 2023 at 10:12 comment added Giorgio Metafune Writing $\sum_n a_n$, I checked (for the first series) that $a_{n+1}/a_n=(1+(\lambda-3/2)/n+O(n^{-2}))$. Taking logarithms and summing you get $a_n \equiv Cn^{\lambda-3/2}$ which gives the result.
S May 31, 2023 at 3:36 history edited LSpice CC BY-SA 4.0
Tidying
May 31, 2023 at 3:35 review Suggested edits
S May 31, 2023 at 3:36
May 30, 2023 at 21:06 history edited mathamphetamine CC BY-SA 4.0
deleted 38 characters in body
May 30, 2023 at 20:57 history edited mathamphetamine CC BY-SA 4.0
deleted 3 characters in body
May 30, 2023 at 20:51 history edited mathamphetamine CC BY-SA 4.0
added 64 characters in body
May 30, 2023 at 20:11 history asked mathamphetamine CC BY-SA 4.0