Difference between revisions of "(n+2lambda)C n^(lambda)(x)=2lambda(C n^(lambda+1)(x)-xC (n-1)^(lambda+1)(x))"

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(Created page with "==Theorem== The following formula holds: $$(n+2\lambda)C_n^{\lambda}(x) = 2\lambda(C_n^{\lambda+1}(x)-xC_{n-1}^{\lambda+1}(x)),$$ where $C_n^{\lambda}$ denotes [[Gegenbauer C]...")
 
 
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==Theorem==
 
==Theorem==
 
The following formula holds:
 
The following formula holds:
$$(n+2\lambda)C_n^{\lambda}(x) = 2\lambda(C_n^{\lambda+1}(x)-xC_{n-1}^{\lambda+1}(x)),$$
+
$$(n+2\lambda)C_n^{\lambda}(x) = 2\lambda \left(C_n^{\lambda+1}(x)-xC_{n-1}^{\lambda+1}(x) \right),$$
 
where $C_n^{\lambda}$ denotes [[Gegenbauer C]].
 
where $C_n^{\lambda}$ denotes [[Gegenbauer C]].
  

Latest revision as of 01:26, 20 December 2017

Theorem

The following formula holds: $$(n+2\lambda)C_n^{\lambda}(x) = 2\lambda \left(C_n^{\lambda+1}(x)-xC_{n-1}^{\lambda+1}(x) \right),$$ where $C_n^{\lambda}$ denotes Gegenbauer C.

Proof

References