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Hopefully this is appropriate for Math SE given that, ultimately, this questions stems from a math textbook (Zorich's Mathematical Analysis).

Zorich has shown using simple arguments with absolute values that if one has uncertainties for an approximation $\tilde{x}$ to some true value $x$ (defined as $\Delta_x(\tilde{x}) := (x-\tilde{x})$, and I emphasize that the $\Delta$ function is parametrized by the true value $x$ of interest) then the uncertainty in the approximation to the sum of two true values by the sum of their approximations is bounded above by the sum of the two uncertainties. Mathematically, $$\Delta_{x+y}(\tilde{x} + \tilde{y}) := |(x+y)-(\tilde{x} + \tilde{y})| \leq \Delta_x(\tilde{x}) + \Delta_y(\tilde{y}) $$ with the $\leq$ bit being the "theorem" part which follows immediately from the triangle inequality.

Now Zorich says the following:

For example, suppose your height has been measured twice by some device, and the precision of the measurement is ±0.5 cm. Suppose a sheet of paper was placed under your feet before the second measurement. It may nevertheless happen that the results of the measurement are as follows: H1 = (200 ± 0.5) cm and H2 = (199.8 ± 0.5) cm respectively.

It does not make sense to try to find the thickness of the paper in the form of the difference H2 − H1 , from which it would follow only that the thickness of the paper is not larger than 0.8 cm. That would of course be a crude reflection (if indeed one could even call it a “reflection”) of the true situation.

However, it is worthwhile to consider another more hopeful computational effect through which comparatively precise measurements can be carried out with crude devices. For example, if the device just used for measuring your height was used to measure the thickness of 1000 sheets of the same paper, and the result was (20 ± 0.5) cm, then the thickness of one sheet of paper is (0.02 ± 0.0005) cm, which is (0.2 ± 0.005) mm, as follows from formula (2.1).

Note that notation like $\tilde{a} \pm b$ was defined to mean that the given approximation's error (based on the measurement tool) has an upper bound of $b$, i.e. $\Delta_a(\tilde{a}) \leq b$.

In the first part, one simply defines the variable of interest (the width of the paper) by $P:=H_2 - H_1$, whence one arrives at $-0.2 \pm 1$ cm from a direct application of the theorem above.

It is the part beginning with "However, it is worthwhile..." that I cannot seem to obtain for myself. Here we are again interested in the variable $P$, and presumably one defined it as $P := S/1000$ where $S$ is the height of the stack, $\tilde{S} = 20$ cm. But I can't follow how one gets to Zorich's conclusion via the given theorem as he says.

We are interested I suppose in determining the error in taking the approximation $\tilde{P} := \tilde{S}/1000$ to $P$. Then $$\Delta_{P}(\tilde{P}) := |P - \tilde{P}| = |S/1000 - \tilde{S}/1000| = \Delta_S(\tilde{S})/1000 \leq 0.5/1000 \, \textrm{cm} = 0.0005 \, \textrm{cm}.$$ This gets me to the desired conclusion but I'm not sure if I'm missing something given that I have not used the theorem which he specifically mentions the result follows from. I wondered if I'm missing the crux of the argument and I somehow have to consider a "sum over the papers", but I couldn't make sense of that.


I suspect my issue is with defining $P := S/1000$ but I'm not really sure what else to do.

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$$\Delta_{x_1 + x_2 + \cdots + x_{1000}}(\tilde{x}_1 + \tilde{x}_2 + \cdots + \tilde{x}_{1000}) \\ \qquad \leq \Delta_{x_1}(\tilde{x}_1) + \Delta_{x_2} (\tilde{x}_2) + \cdots + \Delta_{x_{1000}} (\tilde{x}_{1000})$$

The left-hand side is the error in measuring the $1000$ pages together, which only induces one copy of the device's limited precision. On the right-hand side, we presume that the pages are all the same thickness, $x$, and that we have the same error for each of them, $\tilde{x}$. (That is, our single measurement's error results is a consistent and systematic bias in the estimated thickness of each page. We don't have independent errors on the right when we do this.)

(Consider how we would do this in the other direction: If we measure you $1000$ times, we can estimate the height of a stack of $1000$ of you, but accumulating $1000$ copies of the error, we only know the height of that stack to within $\pm 500$ cm. When we accumulate repeated error, we have the right-hand side and so the knowledge of the left-hand side is very imprecise. When we measure the entire stack once, we have the left-hand side, so one copy of the error is distributed over many measurements on the right.)

\begin{align*} \Delta&_{x_1 + x_2 + \cdots + x_{1000}}(\tilde{x}_1 + \tilde{x}_2 + \cdots + \tilde{x}_{1000}) \\ &\leq \Delta_{x}(\tilde{x}) + \Delta_{x} (\tilde{x}) + \cdots + \Delta_{x} (\tilde{x}) \\ &= 1000 \Delta_x(\tilde{x}) \end{align*}

Getting $\Delta_x(\tilde{x})$ alone on the right, i.e., distributing the total error on the left among the thousand errors on the right, is just division by $1000$.

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  • $\begingroup$ I am not sure I totally follow the argument. We are looking to conclude that $\Delta_x(\tilde{x}) = \Delta_S(\tilde{S})/1000$ are we not? I don't see how your last line conveys that but, like I said, I'm not sure I understand the entirety of the argument. $\endgroup$
    – EE18
    Commented Mar 1 at 23:59
  • $\begingroup$ @EE18 : $\Delta_x(\tilde{x}) = \frac{1}{1000} \Delta_{x_1 + x_2 + \cdots + x_{1000}}(\tilde{x}_1 + \tilde{x}_2 + \cdots + \tilde{x}_{1000}) = \frac{1}{1000}\Delta_S(\tilde{S}) \text{.}$ $\endgroup$ Commented Mar 2 at 6:21
  • $\begingroup$ Ah, sorry for perseverating on this but I just want to be sure I understand precisely what assumptions we're making when we do the above before I accept. I've reread your answer and think I understand now but want to confirm: we have $\Delta_{x_1 + x_2 + \cdots + x_{1000}}(\tilde{x}_1 + \tilde{x}_2 + \cdots + \tilde{x}_{1000}) \leq \Delta_{x_1}(\tilde{x}_1) + \Delta_{x_2} (\tilde{x}_2) + \cdots + \Delta_{x_{1000}} (\tilde{x}_{1000})$ and then we argue that each term on the RHS here is bounded above by our uncertainty (call it $\Delta$) on the paper width, so that... $\endgroup$
    – EE18
    Commented Mar 2 at 17:44
  • $\begingroup$ ...$\Delta_{x_1 + x_2 + \cdots + x_{1000}}(\tilde{x}_1 + \tilde{x}_2 + \cdots + \tilde{x}_{1000}) \leq \Delta_{x_1}(\tilde{x}_1) + \Delta_{x_2} (\tilde{x}_2) + \cdots + \Delta_{x_{1000}} (\tilde{x}_{1000}) \leq 1000 \Delta$, but this still leaves $\Delta \geq \Delta_{x_1 + x_2 + \cdots + x_{1000}}(\tilde{x}_1 + \tilde{x}_2 + \cdots + \tilde{x}_{1000})/1000$ as far as I can tell? I'm also not totally clear on the status of $\tilde{x}_1 + \tilde{x}_2 + \cdots + \tilde{x}_{1000}$ as an approximation, since we don't really observe those values but rather just the value $S$ of the stack, so... $\endgroup$
    – EE18
    Commented Mar 2 at 17:46
  • $\begingroup$ ...it's not clear to me why we can break things out term-wise like we have. $\endgroup$
    – EE18
    Commented Mar 2 at 17:46

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