"슈바르츠 미분(Schwarzian derivative)"의 두 판 사이의 차이

2012년 7월 24일 (화) 20:18 판

\((Sf)(z) = \left({f''(z) \over f'(z)}\right)' - {1\over 2}\left({f''(z)\over f'(z)}\right)^2\)

\( = {f'''(z) \over f'(z)}-{3\over 2}\left({f''(z)\over f'(z)}\right)^2\)

\((Sf)(z)={f,z}\)

@@ 6번째 줄: / 6번째 줄: @@
 <h5>개요</h5>
+<math>(Sf)(z) = \left({f''(z) \over f'(z)}\right)' - {1\over 2}\left({f''(z)\over f'(z)}\right)^2</math>
+<math> = {f'''(z) \over f'(z)}-{3\over 2}\left({f''(z)\over f'(z)}\right)^2</math>
+<math>(Sf)(z)={f,z}</math>
@@ 74번째 줄: / 80번째 줄: @@
 * http://ko.wikipedia.org/wiki/
 * http://en.wikipedia.org/wiki/Schwarzian_derivative
-* [http://eom.springer.de/default.htm http://www.encyclopediaofmath.org/index.php/Schwarzian_derivative]
+* http://www.encyclopediaofmath.org/index.php/Schwarzian_derivative
 * [http://dlmf.nist.gov NIST Digital Library of Mathematical Functions]
 * [http://eqworld.ipmnet.ru/ The World of Mathematical Equations]