Inverse fma

question

#1

Is there an inverse of fma, in the sense of

y = fma(x, A, B)
x2 = ???(y, A, B)  # this is what I am looking for
x ≈ x2             # "reasonably" close

I found that sometimes either (y - B)/A or fma(y, inv(A), -B/A) is more accurate, but it really depends on the magnitude of B vs A.


#2

You should not define “accuracy” based on the difference between x and x2. If x*A is smaller than B, then there will be many values of x that map to the same y and your inverse function cannot possibly know which one is your original x. Better to compare fma(x, A, B) to fma(x2, A, B).


#3

Good point, but I would still like a (y - B)/A that is as accurate as it can be, given B and A and recognizing that it can’t be a bijection.

Of course I can always go to BigFloat and back, but that is rather slow.


#4
function invfma(y, A, B) 
       x2 = (y-B)/A
       y2 = fma(x2, A, B)
       x2 + (y-y2)/A
end

#5

Another alternative to BigFloat is Base.TwicePrecision.