How do I prove that any second-order equation with three regular singular points can be transformed into a...
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Consider the hypergeometric equation $$z(1-z)u''+(c-(a+b+1)z)u'-abu=0.$$ I read that it should be possible to show that any second-order equation with three regular singular points can be transformed into a hypergeometric equation by a combination of Möbius maps $tilde{z}=(az+b)/(cz+d)$ (changing the variable), and multiplication operations $tilde{u}=L(z)u(z)$ for fixed functions $L$. But I could not find any explanation about how/why anywhere on the internet. How does one prove that?
ordinary-differential-equations hypergeometric-function
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Consider the hypergeometric equation $$z(1-z)u''+(c-(a+b+1)z)u'-abu=0.$$ I read that it should be possible to show that any second-order equation with three regular singular points can be transformed into a hypergeometric equation by a combination of Möbius maps $tilde{z}=(az+b)/(cz+d)$ (changing the variable), and multiplication operations $tilde{u}=L(z)u(z)$ for fixed functions $L$. But I could not find any explanation about how/why anywhere on the internet. How does one prove that?
ordinary-differential-equations hypergeometric-function
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add a comment |
$begingroup$
Consider the hypergeometric equation $$z(1-z)u''+(c-(a+b+1)z)u'-abu=0.$$ I read that it should be possible to show that any second-order equation with three regular singular points can be transformed into a hypergeometric equation by a combination of Möbius maps $tilde{z}=(az+b)/(cz+d)$ (changing the variable), and multiplication operations $tilde{u}=L(z)u(z)$ for fixed functions $L$. But I could not find any explanation about how/why anywhere on the internet. How does one prove that?
ordinary-differential-equations hypergeometric-function
$endgroup$
Consider the hypergeometric equation $$z(1-z)u''+(c-(a+b+1)z)u'-abu=0.$$ I read that it should be possible to show that any second-order equation with three regular singular points can be transformed into a hypergeometric equation by a combination of Möbius maps $tilde{z}=(az+b)/(cz+d)$ (changing the variable), and multiplication operations $tilde{u}=L(z)u(z)$ for fixed functions $L$. But I could not find any explanation about how/why anywhere on the internet. How does one prove that?
ordinary-differential-equations hypergeometric-function
ordinary-differential-equations hypergeometric-function
asked Jan 27 at 2:51
user638526user638526
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