"Maass forms"의 두 판 사이의 차이

2010년 3월 5일 (금) 18:33 판

z = x + iy in the upper half-plane
Re(s) > 1
definition
\(E(z,s) ={1\over 2}\sum_{(m,n)=1}{y^s\over|mz+n|^{2s}}\)
Maass form
\(\DeltaE(z,s)=-y^2\left(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}\right)E(z,s) = s(1-s)E(z,s)\)
functional equation
\(E^{*}(z,s) = \pi^{-s}\Gamma(s)\zeta(2s)E(z,s) \)
\(E^{*}(z,s)=E^{*}(z,1-s)\)
a unique pole of residue 3/π at s = 1

@@ 11번째 줄: / 11번째 줄: @@
 * z = x + iy in the upper half-plane
 * Re(s) > 1
-* definition
+*  definition<br><math>E(z,s) ={1\over 2}\sum_{(m,n)=1}{y^s\over|mz+n|^{2s}}</math><br>
-* <math>E(z,s) ={1\over 2}\sum_{(m,n)=1}{y^s\over|mz+n|^{2s}}</math>
+*  Maass form<br><math>\DeltaE(z,s)=-y^2\left(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}\right)E(z,s) = s(1-s)E(z,s)</math><br>
+*  functional equation<br><math>E^{*}(z,s) = \pi^{-s}\Gamma(s)\zeta(2s)E(z,s) </math><br><math>E^{*}(z,s)=E^{*}(z,1-s)</math><br>
-<math>\DeltaE(z,s)=-y^2\left(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}\right)E(z,s) = s(1-s)E(z,s)</math>
+* a unique pole of residue 3/π at s = 1