by **Joy Christian** » Mon Mar 30, 2020 9:16 am

This is still not quite right. A null vector is a vector without directon or magnitude, so the subsripts

a and

b on it seems wrong, or at least redundant. More seriously,

is identically equal to zero, so

is ambiguous.

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[quote="FrediFizzx"][quote="gill1109"]So you want

[tex]f(\lambda)=\,-1+\frac{2}{\sqrt{1+3(\frac{\lambda}{\pi})}}\,\, \text{with}\,\,\lambda \in [0, \pi].[/tex]

[tex]A({\bf a},\,{\bf s},\,\lambda) = \,-\, \text{sgn}({\bf a}\cdot{\bf s}),\,\text{if}\,(|({\bf a}\cdot{\bf s})|>f(\lambda)),\,\text{else}\,\text{sgn}({\bf s}\cdot{\bf n}),[/tex]

[tex]B({\bf b},\,{\bf s},\,\lambda) = \,+\, \text{sgn}({\bf b}\cdot{\bf s}),\,\text{if}\,(|({\bf b}\cdot{\bf s})|>f(\lambda)),\,\text{else}\,\text{sgn}({\bf s}\cdot{\bf n})[/tex]

where "n" is the "null vector?[/quote]

Thanks. Almost good.

[tex]A({\bf a},\,{\bf s},\,\lambda) = \,-\, \text{sgn}({\bf a}\cdot{\bf s}),\,\text{if}\,(|({\bf a}\cdot{\bf s})|>f(\lambda)),\,\text{else}\,\text{sgn}({\bf s}\cdot{\bf n_a}),[/tex]

[tex]B({\bf b},\,{\bf s},\,\lambda) = \,+\, \text{sgn}({\bf b}\cdot{\bf s}),\,\text{if}\,(|({\bf b}\cdot{\bf s})|>f(\lambda)),\,\text{else}\,\text{sgn}({\bf s}\cdot{\bf n_b})[/tex]

[/quote]

This is still not quite right. A null vector is a vector without directon or magnitude, so the subsripts [b]a[/b] and [b]b[/b] on it seems wrong, or at least redundant. More seriously, [tex]{\bf s}\cdot{\bf n_a}[/tex] is identically equal to zero, so [tex]\text{sgn}({\bf s}\cdot{\bf n_a})[/tex] is ambiguous.

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