# Calculating absolute error. Teacher distributes the abs value signs.

Ill illustrate my confusion with an example:

It can be shown, assuming $E_xE_y=0$ that the error in an arithmetical multiplication will be:

$E_{xy}=xE_y+yE_x+\mu$

Where $\mu$ is the so called 'round-off error'.

Now if I'm told to calculate the absolute error in the multiplication I would find it by calculating:

$\mid xE_y+yE_x+\mu\mid= \mid E_{xy} \mid$

But my teacher, just calculates absolute errors of all kind like

$\mid xE_y\mid+\mid yE_x\mid+\mid\mu\mid= \mid E_{xy} \mid$

And for some reason, this works fine. Namely, it gives the correct result of

$\mid(xy)_{real}-(xy)_{given by machine}\mid$

Which is what were interested in calculating.

Any ideas why this works? I don't think you can distribute abs value signs that way.

• I don't quite know what's going on. In general, it is not true that if $a=b+c+d$ then $|a|=|b|+|c|+|d|$. But one can certainly say that $|a|\le |b|+|c|+|d|$ (Triangle Inequality). – André Nicolas Jun 9 '15 at 16:50
• I've found an example where my teachers method is wrong. I wanted to be sure that the way my teacher does it is wrong. Would you agree with me that that isn't the correct expression for abs value of the error? – DLV Jun 9 '15 at 17:29
• We almost never know the absolute value of the error, if we did we would know the exact answer. The best we can usually achieve is an upper bound for the error. The inequality $|a|\le |b|+|c|+|d|$ gives that. Your teacher may very well have said that the expression was an upper bound. Without full detail of exact wording and context, I cannot say the teacher is wrong. – André Nicolas Jun 9 '15 at 17:33
• But wouldn't knowing each of the terms (b,c, and d) in your example, imply that you can also know the absolute value of the error (and hence the exact answer too)? – DLV Jun 10 '15 at 3:26
• The roundoff error, for instance, is not known. At best we have an upper estimate for it. – André Nicolas Jun 10 '15 at 3:39