(Free42) roundoff for complex SQRT

04032018, 01:28 PM
Post: #1




(Free42) roundoff for complex SQRT
Nitpicking, perhaps, but if the original 42S gets it right  then so should Free42 ;)
Try SQRT(a+ib) with a=0 and b/2 a perfect square eg. b=8 Free42 shows 2+i2, which SHOW reveals to be really 2+i(21E33). Almost any b you throw at it has a rounding error like that, in the real or complex part. The 12digit 42S gets the exact result  it must have special code for a=0, and return SQRT(ABS(b)/2)*(1+i*SIGN(b)) in that case  even when b/2 would underflow. Cheers, Werner 

« Next Oldest  Next Newest »

User(s) browsing this thread: 1 Guest(s)