Difference between revisions of "C n^(lambda)'(x)=2lambda C (n+1)^(lambda+1)(x)"
From specialfunctionswiki
(Created page with "==Theorem== The following formula holds: $$\dfrac{\mathrm{d}}{\mathrm{d}x} C_n^{\lambda}(x)=2\lambda C_{n+1}^{\lambda+1}(x),$$ where $C_n^{\lambda}$ denotes Gegenbauer C....") |
(No difference)
|
Latest revision as of 01:29, 20 December 2017
Theorem
The following formula holds: $$\dfrac{\mathrm{d}}{\mathrm{d}x} C_n^{\lambda}(x)=2\lambda C_{n+1}^{\lambda+1}(x),$$ where $C_n^{\lambda}$ denotes Gegenbauer C.
