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I am studying this pdf file. in the file they say:

$2.1.1$ Errors in sum and difference. Let us consider the exact
numbers $X_1, X_2, . . . , X_n$ and their corresponding approximate
number be respectively $x_1, x_2, . . . , x_n$. Assumed that $\Delta > x_1, \Delta x_2, . . . , \Delta x_n$ be the absolute errors in $x_1, > x_2, . . . , x_n$. Therefore, $X_i = x_i ± \Delta x_i$ , $i = 1, 2, . > . . , n$. Let $X = X_1 + X_2 + · · · + X_n$ and $x = x_1 + x_2 > + · · · + x_n$. The total absolute error is: $|X − x| = |(X_1 − x_1) + (X_2 − x_2) + · · · + (X_n − x_n)| ≤ |X_1 − x_1| + |X_2 − > x_2| + · · · + |X_n − x_n|$ This shows that the total absolute error in the sum is $∆x = ∆x_1 + ∆x_2 + · · · + \Delta x_n$.

I don't understand how they conclude $∆x = ∆x_1 + ∆x_2 + · · · + \Delta x_n$. I mean the inequality $|(X_1 − x_1) + (X_2 − x_2) + · · · + (X_n − x_n)|≤ |X_1 − x_1| + |X_2 − x_2| + · · · + |X_n − x_n|$, shows $∆x \le ∆x_1 + ∆x_2 + · · · + \Delta x_n$. but why they put equal sign between those exression?

Etemon
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    You are right. They probably mean the best upper bound on the error. –  Nov 05 '20 at 13:03
  • @YvesDaoust It is strange because our teacher also said the same thing as $∆x = ∆x_1 + ∆x_2 + · · · + \Delta x_n$. – Etemon Nov 05 '20 at 13:04
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    They probably mean the best upper bound on the error. A majorant on the error. A worst-case error. Note that such a bound is often pessimistic, but this is another topic. –  Nov 05 '20 at 13:05
  • @YvesDaoust Yes, I got it. thank you. – Etemon Nov 05 '20 at 13:10

1 Answers1

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The wording in the text you are reading is a bit sloppy.

In fact, the problem is this: if you can measure quantities $X_1,\ldots X_n$ with absolute errors $\Delta x_1,\ldots,\Delta x_n$, so your measurements $x_1,\ldots,x_n$ are guaranteed to satisfy $|x_1-X_1|\le\Delta x_1,\ldots |x_n-X_n|\le\Delta x_n$, then what can you say about the sum $x_1+\ldots+x_n$ as an approximation of $X_1+\ldots+X_n$? The answer is:

$$|(x_1+\ldots+x_n)-(X_1+\ldots+X_n)|=|(x_1-X_1)+\ldots+(x_n-X_n)|\le|x_1-X_1|+\ldots+|x_n-X_n|\le\Delta x_1+\ldots+\Delta x_n$$

In other words, the error you make is at most $\Delta x_1+\ldots+\Delta x_n$.

On the other hand, you can easily see that you cannot assume you will make a smaller error than that, because it may happen that, for example $x_1=X_1+\Delta x_1,\ldots, x_n=X_n+\Delta x_n$ - i.e. you have "overshot" all the values by precisely the maximum you could, and in that case you will also have the equality:

$$|(x_1+\ldots+x_n)-(X_1+\ldots+X_n)|=|(x_1-X_1)+\ldots+(x_n-X_n)|=|\Delta x_1+\ldots+\Delta x_n|=\Delta x_1+\ldots+\Delta x_n$$

The same would happen if you "undershot" all the values by as much as their corresponding absolute error.

Thus, $\Delta x_1+\ldots+\Delta x_n$ is the smallest guaranteed absolute error, so it makes sense to call that number $\Delta x$. Your sum $x_1+\ldots+x_n$ is certainly at most that far from the correct sum $X_1+\ldots+X_n$, and this error may actually happen in the worst case(s).