Absolute Error of Sum is not Greater than Sum of Absolute Errors

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Theorem

Let $X_1, X_2, \ldots, X_n$ be approximations to a collection of (true) values $x_1, x_2, \ldots, x_n$ respectively.

Let $\Delta X_i$ be the absolute error of $X_i$ in $x_i$ for $i \in \set {1, 2, \ldots, n}$.


Then:

$\ds \map \Delta {\sum_{i \mathop = 1}^n X_i} \le \sum_{i \mathop = 1}^n \Delta X_i$

That is, the absolute error of a sum of approximations is not greater than the sum of the absolute errors of those approximations.


Proof

The proof proceeds by induction.

For all $n \in \Z_{\ge 2}$, let $\map P n$ be the proposition:

$\ds \map \Delta {\sum_{i \mathop = 1}^n X_i} \le \sum_{i \mathop = 1}^n \Delta X_i$

for appropriately defined $X_1, X_2, \ldots, X_n$ and $x_1, x_2, \ldots, x_n$.


Basis for the Induction

$\map P 2$ is the case:

$\map \Delta {X_1 + X_2} \le \Delta X_1 + \Delta X_2$

We have:

\(\ds \Delta X_1 + \Delta X_2\) \(=\) \(\ds \paren {X_1 - x_1} + \paren {X_2 - x_2}\)
\(\ds \) \(=\) \(\ds \paren {X_1 + X_2} - \paren {x_1 + x_2}\)


This needs considerable tedious hard slog to complete it.
In particular: Until it is truly specified what the "approximation" is, we can't make sense of this.
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Thus $\map P 2$ is seen to hold.


This is the basis for the induction.


Induction Hypothesis

Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.


So this is the induction hypothesis:

$\ds \map \Delta {\sum_{i \mathop = 1}^k X_i} \le \sum_{i \mathop = 1}^k \Delta X_i$


from which it is to be shown that:

$\ds \map \Delta {\sum_{i \mathop = 1}^{k + 1} X_i} \le \sum_{i \mathop = 1}^{k + 1} \Delta X_i$


Induction Step

This is the induction step:


This theorem requires a proof.
In particular: I understand this intuitively, but because of the loose way we have defined errors and approximations, I can't make a sensible proof.
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So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.


Therefore:

$\forall n \in \N: n \ge 2: \ds \map \Delta {\sum_{i \mathop = 1}^n X_i} \le \sum_{i \mathop = 1}^n \Delta X_i$


Sources

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