The default precision is double unless yue specify otherwise (float32 or long double (float128 or float96))
You can see this from: f(fsolve(f,1.01)) # 1.7763568394002505e-15 The last line should be: >>> fsolve(f,1.01) - float64(1.034324523462345) 8.8817841970012523e-16 Nadav -----הודעה מקורית----- מאת: numpy-discussion-boun...@scipy.org בשם Adam Ginsburg נשלח: ש 17-אוקטובר-09 20:08 אל: numpy-discussion@scipy.org נושא: [Numpy-discussion] double-precision sqrt? Hi folks, I'm trying to write a ray-tracing code for which high precision is required. I also "need" to use square roots. However, math.sqrt and numpy.sqrt seem to only use single-precision floats. Is there a simple way to make sqrt use higher precision? Alternately, am I simply being obtuse? Thanks, Adam Example code: from scipy.optimize.minpack import fsolve from numpy import sqrt sqrt(float64(1.034324523462345)) # 1.0170174646791199 f=lambda x: x**2-float64(1.034324523462345)**2 f(sqrt(float64(1.034324523462345))) # -0.03550269637326231 fsolve(f,1.01) # 1.0343245234623459 f(fsolve(f,1.01)) # 1.7763568394002505e-15 fsolve(f,1.01) - sqrt(float64(1.034324523462345)) # 0.017307058783226026 _______________________________________________ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
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