Calculating (sqrt(n+1)-sqrt(n))*1e9

[kster59]

I need to calculate (sqrt(n+1)-sqrt(n))*1e9 on the XMOS with n=[1,100e3] rounded off to the nearest int?

Previously I stored an unsigne short lookup table * 1e5 for values n=[1,2048] but now I ran out of memory and need the memory for other stuff. I'd also like a bigger table so want to calculate it on the fly.

Any suggestions? On a processor with doubles I could use an iterative method but am new to a non floating point.

[kster59]

Nevermind I was able to use the C implement of sqrt with an Xc wrapper which was fast enough.

[Folknology]

You also have 8K of OTP ROM to play around with which is useful for things like tables, worth remembering in these situations

regards
Al

[segher]

I need to calculate (sqrt(n+1)-sqrt(n))*1e9 on the XMOS with n=[1,100e3] rounded off to the nearest int?

sqrt(n+1) and sqrt(n) are very close together, for bigger n, so you should try not to
calculate these separately. M*(sqrt(n+1)-sqrt(n)) = M/(sqrt(n+1)+sqrt(n)), which
behaves much better.

M/(sqrt(n+1)+sqrt(n)) is very close to M/sqrt(4n+2), maybe that is good enough
for your purposes? They differ by about M / (64*n*n*sqrt(n)), so no difference
at all for n>300 or so.

Either way, the best is usually to do a small lookup table and then refinement
(using Newton-Raphson for example).