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Anonymous Wed 14 Apr 10:23:52 2021 No.12973221 View Reply Original Report

Quoted By: >>12973255 >>12973865

First time I saw this function, What the fuck is this? Confluent hypergeometric function of the 2nd kind?

Anonymous

Anonymous Wed 14 Apr 2021 10:36:02 No.12973255 Report

Quoted By: >>12973284

>>12973221
https://mathworld.wolfram.com/ConfluentHypergeometricFunctionoftheSecondKind.html
The Confluent Hypergeometric Function of the Second Kind has the Confluent Hypergeometric Function of the First Kind and the Confluent Hypergeometric Function as terms, which makes it a complicated topic. Do you need a book suggested?

Anonymous

Anonymous Wed 14 Apr 2021 10:48:03 No.12973284 Report

Quoted By: >>12973621

>>12973255
yeah, name some good textbooks for these confluent hypergeometric functions

Anonymous

Anonymous Wed 14 Apr 2021 12:32:58 No.12973621 Report

Quoted By:

>>12973284
Whittaker and Watson has a whole chapter on them.

Anonymous

Anonymous Wed 14 Apr 2021 13:06:46 No.12973755 Report

Quoted By: >>12973779

You'll be please to know that this has no solutions. The expanded form includes $\Gamma(0)$ and $\Gamma(-1)$ as terms.

Anonymous

Anonymous Wed 14 Apr 2021 13:11:25 No.12973779 Report

Quoted By: >>12973807

>>12973755
Proof?

Anonymous

Anonymous Wed 14 Apr 2021 13:20:22 No.12973807 Report

Quoted By:

>>12973779
$U(a,b,c)$ contains $\Gamma(2-b)$ . $L^{k}_{n}(x)$ contains $U(n,k+1,x)$ , which contains $\Gamma(1-k)$ .
In your case, $b$ is 3 and $k$ is 2, which makes the whole thing impossible to deal with.

Anonymous

Anonymous Wed 14 Apr 2021 13:42:38 No.12973865 Report

Quoted By:

>>12973221
https://youtu.be/dzuUpEFKV3E

Anonymous

Anonymous Wed 14 Apr 2021 15:50:27 No.12974257 Report

Quoted By: >>12974262 >>12974295

I expanded it for you, OP.

$y(x)=y'(x)-y''(x)+y'''(x)$
$\therefore$
$\begin{matrix} y(x)= & c_{1}x & + \\ & c_{2}x \int_{1}^{x} \frac {e^{-0.5(1+\sqrt{5})\xi} \pi} {sin \left( 3\pi \frac{ \sum\limits_{k=0}^{\infty} \frac {2\Gamma \left( \frac{3\sqrt{5}-1} {2\sqrt{5}+k \right) \sqrt{5}^{k}} {\Gamma(3+k) \Gamma \left( \frac {3\sqrt{5} -1} {2\sqrt{5} \right) k!}} {\Gamma \left( \frac {3\sqrt{5}-1}{2\sqrt{5}}-2 \right)} - <br />\frac {\sum\limits_{k=0}^{\infty} \frac {\Gamma \left( \frac {3\sqrt{5}-1} {2\sqrt{5}} -2+k \right) \Gamma{-1}\sqrt{5}^k} {\Gamma(k-1) \Gamma \left( \frac {3\sqrt {5}-1} {2\sqrt{5}} -2 \right) k!} {5\Gamma \left(\frac {3\sqrt{5}-1}{2\sqrt{5}} \right)} \right) } d\xi & + \\ & c_{3}x \int_{1}^{x} \frac {(-1)^{ \frac{-15\sqrt{5}}{10}} 5\sqrt{5} \zeta^{3} \pi} {sin \left( 2\pi \frac {\sum\limits_{k=0}^{\infty} \frac {\Gamma \left( \frac{-15\sqrt{5}}{10} +k \right) (\sqrt{5}\zeta)^{k}}{ \Gamma(2+k) \Gamma \left( \frac{-15\sqrt{5}}{10} \right) k!} {\Gamma \left( \frac{-15\sqrt{5}}{10} -1 \right) } - (\sqrt{5}\zeta)^{-1} {\frac {\sum\limits_{k=0}^{\infty} \frac {\Gamma \left( \frac{-15\sqrt{5}}{10} -1+k \right) \Gamma (0) (\sqrt{5}\zeta)^{k}} {\Gamma (k) \Gamma \left( \frac{-15\sqrt{5}}{10} -1\right) k! }}{\Gamma \left( \frac{-15\sqrt{5}}{10} \right) } \right) } d\zeta & \end{matrix}$

Anonymous

Anonymous Wed 14 Apr 2021 15:51:28 No.12974262 Report

Quoted By:

>>12974257
aw fuck

Anonymous

Anonymous Wed 14 Apr 2021 15:58:09 No.12974295 Report

Quoted By: >>12974310

>>12974257
Hope this works

$y(x)=y'(x)-y''(x)+y'''(x)$
$\therefore$
$\begin{matrix} y(x)= & c_{1}x & + \\ & c_{2}x \int_{1}^{x} \frac {e^{-0.5(1+\sqrt{5})\xi} \pi} {sin \left( 3\pi \frac{ \sum\limits_{k=0}^{\infty} \frac {2\Gamma \left( \frac{3\sqrt{5}-1} {2\sqrt{5}+k \right) \sqrt{5}^{k}} {\Gamma(3+k) \Gamma \left( \frac {3\sqrt{5} -1} {2\sqrt{5} \right) k!}} {\Gamma \left( \frac {3\sqrt{5}-1}{2\sqrt{5}}-2 \right)} - <br />\frac {\sum\limits_{k=0}^{\infty} \frac {\Gamma \left( \frac {3\sqrt{5}-1} {2\sqrt{5}} -2+k \right) \Gamma{-1}\sqrt{5}^k} {\Gamma(k-1) \Gamma \left( \frac {3\sqrt {5}-1} {2\sqrt{5}} -2 \right) k!} {5\Gamma \left(\frac {3\sqrt{5}-1}{2\sqrt{5}} \right)} \right) }} d\xi & + \\ & c_{3}x \int_{1}^{x} \frac {(-1)^{ \frac{-15\sqrt{5}}{10}} 5\sqrt{5} \zeta^{3} \pi} {sin \left( 2\pi \frac {\sum\limits_{k=0}^{\infty} \frac {\Gamma \left( \frac{-15\sqrt{5}}{10} +k \right) (\sqrt{5}\zeta)^{k}}{ \Gamma(2+k) \Gamma \left( \frac{-15\sqrt{5}}{10} \right) k!} {\Gamma \left( \frac{-15\sqrt{5}}{10} -1 \right) } - (\sqrt{5}\zeta)^{-1} {\frac {\sum\limits_{k=0}^{\infty} \frac {\Gamma \left( \frac{-15\sqrt{5}}{10} -1+k \right) \Gamma (0) (\sqrt{5}\zeta)^{k}} {\Gamma (k) \Gamma \left( \frac{-15\sqrt{5}}{10} -1\right) k! }}{\Gamma \left( \frac{-15\sqrt{5}}{10} \right) } \right) } }d\zeta & \end{matrix}$

Anonymous

Anonymous Wed 14 Apr 2021 16:00:42 No.12974310 Report

Quoted By: >>12975168

>>12974295
$\begin{matrix} y(x)= & c_{1}x & + \\ & c_{2}x \int_{1}^{x} \frac {e^{-0.5(1+ \sqrt{5})\xi} \pi} {sin \left( 3\pi \frac{ \sum\limits _{k=0}^{\infty} \frac {2\Gamma \left( \frac{3\sqrt{5} -1} {2\sqrt{5}+k \right) \sqrt{5}^{k}} {\Gamma(3+k) \Gamma \left( \frac {3\sqrt{5} -1} {2\sqrt{5} \right) k!}} {\Gamma \left( \frac {3\sqrt{5}-1}{2\sqrt{5}}-2 \right)} - <br />\frac {\sum\limits_{k=0}^{\infty} \frac {\Gamma \left( \frac {3\sqrt{5}-1} {2\sqrt{5}} -2+k \right) \Gamma{-1}\sqrt{5}^k} {\Gamma(k-1) \Gamma \left( \frac {3\sqrt {5}-1} {2\sqrt{5}} -2 \right) k!} {5\Gamma \left(\frac {3\sqrt{5}-1} {2\sqrt{5}} \right)} \right) }} d\xi & + \\ & c_{3}x \int_{1}^{x} \frac {(-1)^{ \frac{-15\sqrt{5}}{10}} 5\sqrt{5} \zeta^{3} \pi} {sin \left( 2\pi \frac {\sum\limits_{k=0}^{\infty} \frac {\Gamma \left( \frac{-15\sqrt{5}}{10} +k \right) (\sqrt{5}\zeta)^{k}}{ \Gamma(2+k) \Gamma \left( \frac{-15\sqrt{5}} {10} \right) k!} {\Gamma \left( \frac{-15\sqrt{5}} {10} -1 \right) } - (\sqrt{5}\zeta)^{-1} {\frac {\sum\limits_{k=0}^{\infty} \frac {\Gamma \left( \frac{-15\sqrt{5}} {10} -1+k \right) \Gamma (0) (\sqrt{5}\zeta)^{k}} {\Gamma (k) \Gamma \left( \frac{-15\sqrt{5}} {10} -1\right) k! }}{\Gamma \left( \frac{-15\sqrt{5}} {10} \right) } \right) } }d\zeta & \end{matrix}$

Anonymous

Anonymous Wed 14 Apr 2021 18:44:19 No.12975168 Report

Quoted By:

>>12974310
$y(x)= c_{1}x + c_{2}x \int_{1}^{x} \frac {e^{-0.5(1+sqrt{5}\xi}\pi} {sin \left( \3pi \frac{ \sum\limits_{k=0}^{\infty} \frac{2\Gamma(A+k) \sqrt{5}^{k} }{\Gamma(3+k)\Gamma(A)k!} } {\Gamma(A-1)} - \frac{ \sum\limits_{k=0}^{\infty} \frac{\Gamma(A-2k) \Gamma(-1) \sqrt{5}^{k} }{\Gamma(k-1) \Gamma(A-2) k!} } {5\Gamma(A)} \right)} d\xi + c_{3}x \int_{1}^{x} \frac {\pi(-1)^{B}(\sqrt{5}\zeta)^{3}} {sin \left( \2pi \frac{ \sum\limits_{k=0}^{\infty} \frac{\Gamma(B+k) (\sqrt{5}\zeta)^{k} }{\Gamma(B) \Gamma(2+k) k!} } {\Gamma(B-1)} - (\sqrt{5}\zeta)^{-1} \frac{ \sum\limits_{k=0}^{\infty} \frac{\Gamma(B-1+k) \Gamma{0} (sqrt{5}\zeta)^{-1} }{\Gamma(k) \Gamma(B-1) k!} } {\Gamma(B)} \right) } d\zeta$
$A=\frac{3\sqrt{5}-1}{2\sqrt{5}, B={-15+\sqrt{5}}{10}$

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