User:Tohline/Appendix/Mathematics/ToroidalConfusion - Revision history

Tohline: /* Proper Interpretation of DLMF Expression */

2018-06-13T04:43:48Z

Proper Interpretation of DLMF Expression

← Older revision		Revision as of 04:43, 13 June 2018
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	which matches expression #2, above. But it does not ''appear'' to match expressions #1 or #4.		which matches expression #2, above. But it does not ''appear'' to match expressions #1 or #4.

			<table border="1" cellpadding="5" align="center" width="80%"><tr><td align="left">
			The standard "Euler reflection formula for gamma functions" is usually presented in the form,
			{{ User:Tohline/Math/EQ_Gamma01 }}
			If we make the association,
			<div align="center">
			<math>~z \leftrightarrow (m - n + \tfrac{1}{2}) \, ,</math>
			</div>
			with <math>~m</math> and <math>~n</math> both being either zero or a positive integer, then, this Euler reflection formula becomes,
			<table border="0" cellpadding="5" align="center">

			<tr>
			<td align="right">
			<math>~\Gamma(m - n + \tfrac{1}{2}) ~ \Gamma(n - m + \tfrac{1}{2} )</math>
			</td>
			<td align="center">
			<math>~=</math>
			</td>
			<td align="left">
			<math>~\pi \biggl\{ \sin\biggl[ \pi(m - n + \tfrac{1}{2}) \biggr] \biggr\}^{-1}</math>
			</td>
			</tr>

			<tr>
			<td align="right">

			</td>
			<td align="center">
			<math>~=</math>
			</td>
			<td align="left">
			<math>~\pi (-1)^{m+n} \, .</math>
			</td>
			</tr>
			</table>

			</td></tr></table>

	However, in our situation the so-called "Euler reflection formula for gamma functions" gives the relation,		However, in our situation the so-called "Euler reflection formula for gamma functions" gives the relation,

Tohline: /* Published Expressions */

2018-06-13T02:04:16Z

Published Expressions

← Older revision		Revision as of 02:04, 13 June 2018
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	<span id="Gil">So far, this gives me three ''similar'' but not identical formulae for the same function mapping!</span> As per equation (8) in (yet another source!) [http://adsabs.harvard.edu/abs/2000JCoPh.161..204G A. Gil, J. Segura, & N. M. Temme (2000, JCP, 161, 204 - 217)], the relationship is:		<span id="Gil">So far, this gives me three ''similar'' but not identical formulae for the same function mapping!</span> As per equation (8) in (yet another source!) [http://adsabs.harvard.edu/abs/2000JCoPh.161..204G A. Gil, J. Segura, & N. M. Temme (2000, JCP, 161, 204 - 217)], the relationship is:
			{{ User:Tohline/Math/EQ_Toroidal02 }}

	~~<div align="center">~~		This expression from Gil et al. (2000) means, for example, that by identifying <math>~x</math> with <math>~\coth\eta</math>, we have <math>~\lambda = \cosh\eta</math>, and,
	~~<table border="0" cellpadding="5" align="center">~~

	~~<tr>~~
	~~<td align="right">~~
	~~<math>~Q_{n-1 / 2}^m (\lambda)</math>~~
	~~</td>~~
	~~<td align="center">~~
	~~<math>~=</math>~~
	~~</td>~~
	~~<td align="left">~~
	~~<math>~(-1)^n~~
	~~\frac{\pi^{3/2}}{\sqrt{2} \Gamma(n-m+1 / 2)} (x^2-1)^{1 / 4} P_{m-1 / 2}^n(x) \, ,~~
	~~</math>~~
	~~</td>~~
	~~</tr>~~
	~~</table>~~
	~~</div>~~
	~~where, <math>~\lambda \equiv x/\sqrt{x^2-1}</math>.~~ This expression from Gil et al. (2000) means, for example, that by identifying <math>~x</math> with <math>~\coth\eta</math>, we have <math>~\lambda = \cosh\eta</math>, and,

	<div align="center">		<div align="center">

Tohline: /* Published Expressions */

2018-06-11T01:28:07Z

Published Expressions

← Older revision		Revision as of 01:28, 11 June 2018
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	</table>		</table>

	So far, this gives me three ''similar'' but not identical formulae for the same function mapping! As per equation (8) in (yet another source!) [http://adsabs.harvard.edu/abs/2000JCoPh.161..204G A. Gil, J. Segura, & N. M. Temme (2000, JCP, 161, 204 - 217)], the relationship is:		<span id="Gil">So far, this gives me three ''similar'' but not identical formulae for the same function mapping!</span> As per equation (8) in (yet another source!) [http://adsabs.harvard.edu/abs/2000JCoPh.161..204G A. Gil, J. Segura, & N. M. Temme (2000, JCP, 161, 204 - 217)], the relationship is:

	<div align="center">		<div align="center">

Tohline: /* Derivation From Scratch */

2018-06-11T00:58:38Z

Derivation From Scratch

← Older revision		Revision as of 00:58, 11 June 2018
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	</table>		</table>
	</div>		</div>
	Hence, we can write,		<span id="QPrelation">Hence, we can write,</span>
	<div align="center">		<div align="center">
	<table border="0" cellpadding="5" align="center">		<table border="0" cellpadding="5" align="center">

Tohline: /* Joel's Additional Manipulations */

2018-06-03T04:18:27Z

Joel's Additional Manipulations

← Older revision		Revision as of 04:18, 3 June 2018
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	<td align="right">		<td align="right">
	<math>~		<math>~
	\sum_{n=1}^{\infty} \epsilon_n		\sum_{n=0}^{\infty} \epsilon_n
	Q_{n-\frac{1}{2}}\left(\cosh\xi\right) \cos\left(n		Q_{n-\frac{1}{2}}\left(\cosh\xi\right) \cos\left(n
	\phi\right)		\phi\right)

Tohline: /* Joel's Additional Manipulations */

2018-05-31T20:52:36Z

Joel's Additional Manipulations

← Older revision		Revision as of 20:52, 31 May 2018
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	<td align="left">		<td align="left">
	<math>~ \biggl[		<math>~ \biggl[
	\dfrac{ ~~\Gamma\left( \tfrac{1}{2} \right) \left(\frac{1}{2}~~\pi\~~right)^{\frac{1}~~{2}} }{\left(\cosh\xi-\cos\phi\right)^{\frac{1}{2}} }\biggr]		\dfrac{ \pi/\sqrt{2} }{\left(\cosh\xi-\cos\phi\right)^{\frac{1}{2}} }\biggr]
	-		-
	2\sum_{n=1}^{\infty}		2\sum_{n=1}^{\infty}
	Q_{n-\frac{1}{2}}\left(\cosh\xi\right) \cos\left(n		Q_{n-\frac{1}{2}}\left(\cosh\xi\right) \cos\left(n
	\phi\right) \, .		\phi\right) \, .
			</math>
			</td>
			</tr>
			</table>
			</div>

			But this can be rewritten in the form,
			<div align="center">
			<table border="0" cellpadding="5" align="center">

			<tr>
			<td align="right">
			<math>~
			\sum_{n=1}^{\infty} \epsilon_n
			Q_{n-\frac{1}{2}}\left(\cosh\xi\right) \cos\left(n
			\phi\right)

			</math>
			</td>
			<td align="center">
			<math>~=</math>
			</td>
			<td align="left">
			<math>~ \biggl[
			\dfrac{ \pi/\sqrt{2} }{\left(\cosh\xi-\cos\phi\right)^{\frac{1}{2}} }\biggr]
	</math>		</math>
	</td>		</td>

Tohline: /* Joel's Additional Manipulations */

2018-05-31T20:46:30Z

Joel's Additional Manipulations

← Older revision		Revision as of 20:46, 31 May 2018
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	</div>		</div>

	~~In our case~~, we will set <math>~\mu = 0</math>, in which case the ''Sums'' expression becomes,		When dealing with Dyson-Wong tori, we will set <math>~\mu = 0</math>, in which case the ''Sums'' expression becomes,

	<div align="center">		<div align="center">

Tohline: /* Joel's Additional Manipulations */

2018-05-31T20:45:35Z

Joel's Additional Manipulations

← Older revision		Revision as of 20:45, 31 May 2018
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	</table>		</table>
	</div>		</div>
	In our case, we ~~only need~~
			In our case, we will set <math>~\mu = 0</math>, in which case the ''Sums'' expression becomes,

			<div align="center">
			<table border="0" cellpadding="5" align="center">

			<tr>
			<td align="right">
			<math>~Q_{-\frac{1}{2}}\left(\cosh\xi\right)</math>
			</td>
			<td align="center">
			<math>~=</math>
			</td>
			<td align="left">
			<math>~ \biggl[
			\dfrac{ \Gamma\left( \tfrac{1}{2} \right) \left(\frac{1}{2}\pi\right)^{\frac{1}{2}} }{\left(\cosh\xi-\cos\phi\right)^{\frac{1}{2}} }\biggr]
			-
			2\sum_{n=1}^{\infty}
			Q_{n-\frac{1}{2}}\left(\cosh\xi\right) \cos\left(n
			\phi\right) \, .
			</math>
			</td>
			</tr>
			</table>
			</div>

	=See Also=		=See Also=

Tohline: /* Joel's Additional Manipulations */

2018-05-31T20:34:40Z

Joel's Additional Manipulations

← Older revision		Revision as of 20:34, 31 May 2018
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	<div align="center">		<div align="center">
	<math>		<math>
	\boldsymbol{Q}^{\mu}_{n-\frac{1}{2}}\left(x\right)=e^{-\mu\pi i}\frac{Q^{\mu}_{n-\frac{1}{2}}\left(x\right)}{\Gamma\left(\mu+n + \tfrac{1}{2} \right)} \, ,		\boldsymbol{Q}^{\mu}_{n-\frac{1}{2}}\left(x\right)
			=
			e^{-\mu\pi i}\frac{Q^{\mu}_{n-\frac{1}{2}}\left(x\right)}{\Gamma\left(\mu+n + \tfrac{1}{2} \right)} \, ,
	</math>		</math>
	</div>		</div>
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	<tr>		<tr>
	<td align="right">		<td align="right">
	<math>~\~~boldsymbol~~{Q}^{\mu}_{-\frac{1}{2}}\left(\cosh\xi\right)</math>		<math>~e^{-\mu\pi i}\frac{Q^{\mu}_{-\frac{1}{2}}\left(\cosh\xi\right)}{\Gamma\left(\mu+ \tfrac{1}{2} \right)}</math>
	</td>		</td>
	<td align="center">		<td align="center">
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	2\sum_{n=1}^{\infty}		2\sum_{n=1}^{\infty}
	\frac{\Gamma\left(\mu+n+\tfrac{1}{2}\right)}{\Gamma\left(\mu+\tfrac{1}{2}		\frac{\Gamma\left(\mu+n+\tfrac{1}{2}\right)}{\Gamma\left(\mu+\tfrac{1}{2}
	\right)}\~~boldsymbol~~{Q}^{\mu}_{n-\frac{1}{2}}\left(\cosh\xi\right)\cos\left(n		\right)}
			\biggl[ e^{-\mu\pi i}\frac{Q^{\mu}_{n-\frac{1}{2}}\left(\cosh\xi\right)}{\Gamma\left(\mu+n + \tfrac{1}{2} \right)} \biggr] \cos\left(n
	\phi\right)		\phi\right)
			</math>
			</td>
			</tr>

			<tr>
			<td align="right">
			<math>~\Rightarrow ~~~Q^{\mu}_{-\frac{1}{2}}\left(\cosh\xi\right)</math>
			</td>
			<td align="center">
			<math>~=</math>
			</td>
			<td align="left">
			<math>~ e^{\mu\pi i} \Gamma\left(\mu+ \tfrac{1}{2} \right) \biggl[
			\dfrac{\left(\frac{1}{2}\pi\right)^{1/2}\left(\sinh\xi\right)^{\mu
			}}{\left(\cosh\xi-\cos\phi\right)^{\mu+(1/2)}}\biggr]
			-
			2\sum_{n=1}^{\infty}
			Q^{\mu}_{n-\frac{1}{2}}\left(\cosh\xi\right) \cos\left(n
			\phi\right) \, .
	</math>		</math>
	</td>		</td>
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	</table>		</table>
	</div>		</div>
			In our case, we only need

	=See Also=		=See Also=

Tohline: /* Joel's Additional Manipulations */

2018-05-31T20:20:19Z

Joel's Additional Manipulations

← Older revision		Revision as of 20:20, 31 May 2018
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	<div align="center">		<div align="center">
	<math>		<math>
	\boldsymbol{Q}^{\mu}_{n-\frac{1}{2}}\left(x\right)=e^{-\mu\pi i}\frac{Q^{\mu}_{n-\frac{1}{2}}\left(x\right)}{\Gamma\left(\mu+n - \tfrac{1}{2} \right)}.		\boldsymbol{Q}^{\mu}_{n-\frac{1}{2}}\left(x\right)=e^{-\mu\pi i}\frac{Q^{\mu}_{n-\frac{1}{2}}\left(x\right)}{\Gamma\left(\mu+n + \tfrac{1}{2} \right)} \, ,
	</math>		</math>
	</div>		</div>