Timeline for Is the value of $\sum\limits_{k=1}^{\infty}\frac1{(C_k)^n}$ known?
Current License: CC BY-SA 4.0
9 events
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Jun 3, 2018 at 15:50 | history | edited | T. Amdeberhan | CC BY-SA 4.0 |
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Jun 3, 2018 at 15:46 | comment | added | user64494 | You answer is completely unbased and unclear to me. Please, elaborate is. In particular, please explain why $a_2=f(1)$. Maple produces a complicated expression for the general solution of $z^2(z-16)f''(z)+5z^2f'(z)+4(z-1)f(z)+4=0$ in terms of integrals of the Legendre functions (see en.wikipedia.org/wiki/Legendre_function). | |
Jun 3, 2018 at 15:21 | comment | added | T. Amdeberhan | What do you? I don't understand. | |
Jun 3, 2018 at 14:53 | comment | added | user64494 | Can you base your modified answer? | |
Jun 3, 2018 at 13:43 | history | edited | T. Amdeberhan | CC BY-SA 4.0 |
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Jun 3, 2018 at 13:43 | comment | added | T. Amdeberhan | @user64494: I am so sorry, I missed the term $f(z)$ from $4(z-1)f(z)$. | |
Jun 3, 2018 at 7:13 | comment | added | user64494 | Also up to Maple (the code on demand), the right limit of $f'(z)$ at $z=0$ is $-\infty$. | |
Jun 3, 2018 at 6:56 | comment | added | user64494 | Can you elaborate your answer? According Maple, the general solution of $z^2(z-16)f''(z)+5z^2f'(z)+4(z-1)+4=0$ is $$f \left( z \right) =-4096\, \left( z-16 \right) ^{-3}-{\frac {409600}{ 3\, \left( z-16 \right) ^{4}}}+128\, \left( z-16 \right) ^{-2}-16\, \left( 3\,z-48 \right) ^{-1}-1/4\,{\frac {{\it \_C1}}{ \left( z-16 \right) ^{4}}}-{\frac {\ln \left( z \right) z \left( {z}^{3}-64\,{z} ^{2}+1536\,z-16384 \right) }{ \left( z-16 \right) ^{4}}}+{\it \_C2} . $$ The function is singular at $z=0$. | |
Jun 3, 2018 at 3:53 | history | answered | T. Amdeberhan | CC BY-SA 4.0 |