[cfp-interest 4004] Re: Rounding error in fma(r, s, t) - Not for inclusion in our Meetings

[Paul Zimmermann]

       Hi Damian,

please can you be more specific?

You have u = r*s+t+eps1, then v := fma(r,s,u) = r*s+u+eps2,
and w := o(v-t) = v-t+eps3, where eps1, eps2, eps3 are the rounding errors.

This yields w = 2*r*s+eps1+eps2+eps3.

What is your exact statement in terms of w?
Should it be v	= fma(r,s,-u) instead?
Do you consider rounding to nearest only?

Paul

> Date: Wed, 9 Sep 2026 09:50:03 +1000 (AEST)
> From: Damian McGuckin <damianm at esi.com.au>
> 
> 
> Thought I would tap into collective intelligence.
> 
> Give a computation with p = precision (24 for float, 53 for double)
> 
>  	u = fma(r, s, t)
> 
> where
>  	0.5 <= r <= 1 or strictly [0.5, 0.617..., 0.731..., 1]
> 
> and
>  	2**(1-p) < s < 1 (I can limit the upper bound to 0.25)
> and
>  	2**(-p-2) < |t| < 2**(-p-1)
> 
> why does
> 
>  	fma(r, s, u) - t
> 
> seem to yield the rounding error even when |t| > |r * s|
> 
> If you look at the requirements under which Kahan's Fast Two Sum works, 
> that formula will work for
> 
>  	|t| <= |r * s|
> 
> But it seems to work for my data when that inequality does not hold and I 
> am looking for a way to prove it (I have tested even 32-bit floating point 
> value in the range is 's' and it seems to but that doesn't prove it for 
> all possible double values).
> 
> Not urgent - Thanks - Damian
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