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I need to convert n-dimensional vectors into hyperspherical coordinates, which involves doing calculating an arctan with an argument which is the square root of a sum of squares (i.e., the radius of a truncated vector) once for every non-radial dimension. The squares can all be precomputed and re-used, but the sum to be rooted is different every time.

Is there a faster way to compute the angles than just calculating a new square root followed by an arctan every time? Either a way to incrementally update the square root, or directly compute an arctan of squared inputs?

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  • $\begingroup$ Does this require atan() or atan2()? I suspect it is the latter, but it would be beneficial to clarify this in the question. $\endgroup$
    – njuffa
    Commented Aug 16 at 3:48
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    $\begingroup$ I am not convinced the arctangent computation has a material impact on overall performance. To get an idea of how much this computation would benefit from use of a fast atan2() implementation, you could try the fast, low-precision, algorithm from a previous answer of mine. $\endgroup$
    – njuffa
    Commented Aug 16 at 4:12
  • $\begingroup$ Indeed the summing of the $n$ components to get $x$ for $\arctan(\sqrt x)$ could be a dominating aspect for large enough $n$. The summation might be susceptible to optimization by "updating". There's not enough detail in the Question's body to discuss that idea. More notation would be welcome (use $\LaTeX$ for typeset mathematics). $\endgroup$
    – hardmath ♦
    Commented Aug 18 at 15:01
  • $\begingroup$ @hardmath The summation is indeed amenable to updating; there's only one square and one addition per coordinate. $\endgroup$
    – Logan R. Kearsley
    Commented Aug 18 at 16:57

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sqrt is fairly fast (about the cost of division), while arctan is more expensive. As such creating a function for x->arctan(sqrt(x)) is likely the way to go. The ideal implementation will depend a lot on the precision you're targeting.

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    $\begingroup$ I hope for the OP to supply more details about the precision required, in which case we could flesh out your idea of an ideal implementation of $\arctan(\sqrt x)$. $\endgroup$
    – hardmath ♦
    Commented Aug 18 at 15:18

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