From 62b397677d21590ddeec386ebbaee909938239c9 Mon Sep 17 00:00:00 2001 From: Bryan Weber Date: Sun, 16 Jun 2024 12:28:21 -0400 Subject: [PATCH] Update time-since-periapsis-and-keplers-equation/universal-variables.md --- .../universal-variables.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/time-since-periapsis-and-keplers-equation/universal-variables.md b/time-since-periapsis-and-keplers-equation/universal-variables.md index e2ebce1..c8ef22a 100644 --- a/time-since-periapsis-and-keplers-equation/universal-variables.md +++ b/time-since-periapsis-and-keplers-equation/universal-variables.md @@ -312,7 +312,7 @@ The Laguerre algorithm can be implemented as: :::{math} :label: -\chi_{i + 1} = \chi_{i} - \frac{n f(\chi_i)}{f'(\chi_i) \pm \left[\left(n - 1\right)^2 f'(\chi_i)^2 - n\left(n - 1\right) f(\chi_i)f''(\chi_i)\right]^{1/2}} +\chi_{i + 1} = \chi_{i} - \frac{n f(\chi_i)}{f'(\chi_i) \pm \left[\left(n - 1\right)^2 \left(f'(\chi_i)\right)^2 - n\left(n - 1\right) f(\chi_i)f''(\chi_i)\right]^{1/2}} ::: The sign ambiguity in the denominator is determined by taking the sign of the numerical value of $f'(\chi_i)$. In addition, the solution is relatively insensitive to the choice of the value of $n$, which is an integer constant. It seems as though $n = 5$ is a reasonable value. Choosing $n = 1$ gives the standard Newton's algorithm.