Re: adding order matters for accuracy ?

"Matt " <xys@xxxxxxxxxxxx> wrote in message <h8u0tn$g7e$1@xxxxxxxxxxxxxxxxxx>...
"Bruno Luong" <b.luong@xxxxxxxxxxxxxxxxxxxx> wrote in message <h8tuq7$qpr$1@xxxxxxxxxxxxxxxxxx>...

There's something I still don't understand here. According to my numerical analysis books, the relative error does not accumulate when you add up numbers of the same sign.

Not sure where this wrong statement you read from.

From "Introduction to Numerical Analysis" by Stoer and Bulirsch, but the idea is very simple. If the true values being summed are x(i) and have a relative error of u(i), then the computed sum S is given by,

S=sum( x.*(1+u) )

Since the true value of the sum is

S_true=sum(x)

this produces a relative error in S is

RelativeError= ( S- sum(x) )/sum(x)

= sum u.* x/sum(x)

But since all the x have the same sign the coefficients x(i)/sum(x) are a partition of unity, leading to

abs(RelativeError)<=max(abs(u))

The above is about finite precision error due to truncation (and their propagation), not error due when making the sum between two numbers that have widely different magnitudes.

The later can accumulate as various examples have showed.

Bruno
.

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