# Formatting Sandbox II: please test stuff here

Spoiler warning: Be aware that this page contains a lot of MathJax, so it will probably need quite a while to load completely.

User of the old sandbox will have noticed that with the amount of answers gathered there, it is very hard on your computer to get past the posts to the answer field. (Even deleting answers would not solve this problem for 10k users.)

Here you have a new playground... have fun.
(Use this for testing stuff instead of bumping random posts to the mainspace a million times.)

Before you delete a post here, please reduce it to one line without MathJax.

Old formatting sandboxes:

• This Sandbox II is extremely convenient and I thank those who came up with it! But I have noticed that some of the posts are getting upvoted. I though this was just a 'scratchpad' area. Does upvoting count toward rep or is it just a sincere pro forma sign of appreciation? Either way, I will gladly upvote posts here and in the first Sandbox: I never would have been able to format a definite integral without seeing how it was done by someone else! – Ed V Aug 7 at 2:16
• @EdV posts on meta have no influence on reputation, so even the voting is part of playing around. To crazy! – Martin - マーチン Aug 8 at 0:44

Wrong space around stretchy parentheses

\begin{align} y&=a\cdot b/(c\cdot d)\tag{ok}\\[6pt] y&=a\cdot b/\left(c\cdot d\right)\tag{?!}\\[6pt] y&=a\cdot b/{\left(c\cdot d\right)}\tag{ok}\\[6pt] y&=a\cdot b/\mathord{\left(c\cdot d\right)}\tag{ok} \end{align}

\begin{align} y&=\sum\limits_{i = 1}^n a_i\cdot b_i/(c_i^2\cdot d_i^2)\tag{ok}\\[6pt] y&=\sum\limits_{i = 1}^n a_i\cdot b_i/\left(c_i^2\cdot d_i^2\right)\tag{?!}\\[6pt] y&=\sum\limits_{i = 1}^n a_i\cdot b_i/{\left(c_i^2\cdot d_i^2\right)}\tag{ok}\\[6pt] y&=\sum\limits_{i = 1}^n a_i\cdot b_i/\mathord{\left(c_i^2\cdot d_i^2\right)}\tag{ok} \end{align}

\begin{align} f(x)&=x^2\tag{ok}\\[6pt] f\left(x\right)&=x^2\tag{?!}\\[6pt] f{\left(x\right)}&=x^2\tag{ok}\\[6pt] f\mathord{\left(x\right)}&=x^2\tag{ok} \end{align}

\begin{align} y&=\log(x)\tag{?}\\[6pt] y&=\log\left(x\right)\tag{?}\\[6pt] y&=\log{\left(x\right)}\tag{?}\\[6pt] y&=\log\mathord{\left(x\right)}\tag{?} \end{align}

• This is weird here too. But, your example does the same thing with LaTeX which mine doesn't. We probably need a \renewcommand that manipulates the padding. – pentavalentcarbon Oct 13 '16 at 14:22

# Decision threshold and detection limit

## General aspects

$$\textbf{Quantities and symbols}\\ \begin{array}{ll} \hline \text{Symbol}&\text{Name}\\ \hline y&\text{Estimate of the measurand}\\[-3pt] &\text{(e.g. measurement result of the measurand)}\\ u{\left(y\right)}&\text{Standard uncertainty of the measurand}\\[-3pt] &\text{(associated with the measurement result }y)\\ \tilde y&\text{True value of the measurand }\\ \tilde u{\left(\tilde y\right)}&\text{Standard uncertainty of an estimator of the measurand}\\[-3pt] &\text{(as a function of the true value of the measurand }\tilde y)\\ y^*&\text{Decision threshold}\\ y^\#&\text{Detection limit}\\ \hline x&\text{Estimate of the input quantity}\\ \hline \alpha&\text{Probability of the error of the first kind}\\ \beta&\text{Probability of the error of the second kind}\\ k_{1-\alpha}&\text{Quantile of the standardized normal distribution for the probability }\alpha\\ k_{1-\beta}&\text{Quantile of the standardized normal distribution for the probability }\beta\\ \hline \end{array}$$

In general, the result of a measurement $$y$$ is only an approximation or estimate of the value of the measurand and thus is complete only when accompanied by a statement of the uncertainty $$u{\left(y\right)}$$ of that estimate. Nevertheless, it is understood that the result of the measurement $$y$$ is the best estimate of the value of the measurand. The alternative best estimate $$\hat y$$, which explicitly takes into account the fact that the measurand is non-negative according to ISO 11929 and which differs from the measurement result $$y$$, is not used in the following examples.

For the provision and numerical calculation of the decision threshold $$y^*$$ and of the detection limit $$y^\#$$, the standard uncertainty $$\tilde u$$ of the measurand is needed as a function $$\tilde u{\left(\tilde y\right)}$$ of the true value $$\tilde y$$ of the measurand. This function is be determined in a way similar to $$u{\left(y\right)}$$ in accordance with ISO/IEC Guide 98-3. Usually, $$\tilde u{\left(\tilde y\right)}$$ can be explicitly specified, provided that $$\tilde u{\left(\tilde y\right)}$$ is given as a function of the primary measurement result $$x$$, which is taken as input quantity in the model of the evaluation $$y{\left(x\right)}$$ for the calculation of $$y$$. In this case, $$y$$ is be formally replaced by $$\tilde y$$ and the equation for $$y{\left(x\right)}$$ is solved for $$x$$. The result replaces $$x$$ in the equation of $$u{\left(y\right)}$$, which finally yields $$\tilde u{\left(\tilde y\right)}$$.

The decision threshold $$y^*$$ is a value of the measurand that allows the conclusion that the physical effect of interest is present if the measurement result $$y$$ exceeds the decision threshold $$y^*$$; i.e. a determined measurement result $$y$$ is only significant for the true value of the measurand to differ from zero $$(\tilde y>0)$$ if it is larger than the decision threshold $$(y>y^*)$$. Otherwise, the result cannot be attributed to the physical effect; nevertheless, it cannot be concluded that the physical effect is absent.

If the physical effect is really absent $$(\tilde y=0)$$, the probability of taking the wrong decision that the effect is present $$(\tilde y>0)$$ shall not exceed the specified probability $$\alpha$$ (error of the first kind). The choice of the probability $$\alpha$$ of the error of the first kind depends on the application. A frequently cited choice is $$\alpha=5\ \%=0.05$$. The corresponding quantile of the standardized normal distribution for the probability $$1-\alpha=0.95$$ is $$k_{1-\alpha}\approx1.645$$.

According to ISO 11929, the equation for the decision threshold $$y^*$$ is given as

$$y^*=k_{1-\alpha}\cdot\tilde u{\left(0\right)}\tag1$$

where $$\tilde u{\left(0\right)}$$ is the standard uncertainty $$\tilde u{\left(\tilde y\right)}$$ of the measurand for a true value of the measurand of $$\tilde y=0$$.

The detection limit $$y^\#$$ is the smallest true value of the measurand, for which (by applying the decision rule according to the decision threshold $$y^*$$) the probability of the wrong assumption that the physical effect of interest is absent (error of the second kind) does not exceed the specified probability $$\beta$$. A frequently cited choice is $$\beta=5\ \%=0.05$$. The corresponding quantile of the standardized normal distribution for the probability $$1-\beta=0.95$$ is $$k_{1-\beta}\approx1.645$$.

According to ISO 11929, the equation for the detection limit $$y^\#$$ is given as

$$y^\#=y^*+k_{1-\beta}\cdot\tilde u{\left(y^\#\right)}\tag2$$

where $$\tilde u{\left(y^\#\right)}$$ is the standard uncertainty $$\tilde u{\left(\tilde y\right)}$$ of the measurand for a true value of the measurand of $$\tilde y=y^\#$$.

The comparison of the detection limit with a given guideline value allows a decision on whether or not the measurement procedure satisfies the requirements set forth by the guideline value and is therefore suitable for the intended measurement purpose. The measurement procedure satisfies the requirement if the detection limit is smaller than the guideline value.

The determined measurement result $$y$$ of the measurand is compared with the decision threshold $$y^*$$. If $$y\gt y^*$$, the physical effect quantified by the measurand is recognized as present, and the primary measurement result $$y$$ and its standard uncertainty $$u{\left(y\right)}$$ are reported. Otherwise, it is decided that the effect is absent. In this case the value of the detection limit $$y^\#$$ is reported as $$\lt y^\#$$.

## Example 1 – net count rate (without calibration factor)

$$\textbf{Quantities and symbols}\\ \begin{array}{ll} \hline \text{Symbol}&\text{Name}\\ \hline n_\mathrm g&\text{Number of counted pulses of the gross effect}\\ t_\mathrm g&\text{Measurement duration of the measurement of the gross effect}\\ r_\mathrm g&\text{Estimate of the gross count rate}\\ \hline n_0&\text{Number of counted pulses of the background effect}\\ t_0&\text{Measurement duration of the measurement of the background effect}\\ r_0&\text{Estimate of the background count rate}\\ \hline r_\mathrm n&\text{Estimate of the net count rate}\\ u{\left(r\right)}&\text{Standard uncertainty of the net count rate}\\[-3pt] &\text{(associated with the measurement result }r)\\ \tilde r_\mathrm n&\text{True value of the net count rate}\\ \tilde u{\left(\tilde r\right)}&\text{Standard uncertainty of an estimator of the net count rate}\\[-3pt] &\text{(as a function of the true value of the net count rate }\tilde r)\\ r_\mathrm n^*&\text{Decision threshold of the net count rate}\\ r_\mathrm n^\#&\text{Detection limit of the net count rate}\\ \hline \alpha&\text{Probability of the error of the first kind}\\ \beta&\text{Probability of the error of the second kind}\\ k_{1-\alpha}&\text{Quantile of the standardized normal distribution for the probability }\alpha\\ k_{1-\beta}&\text{Quantile of the standardized normal distribution for the probability }\beta\\ \hline \end{array}$$

This example relates to a sample of radioactive material. In particular, the measurand is the net count rate $$\rho_\mathrm n$$ of the sample. It is determined from counting the gross effect and the background effect with preselection of time.

The estimate of the gross count rate is given by

$$r_\mathrm g=\frac{n_\mathrm g}{t_\mathrm g}\tag3$$

and the estimate of the background count rate is given by

$$r_0=\frac{n_0}{t_0}\tag4$$

Assuming Poisson statistics for the gross counts

\begin{align} u{\left(n_\mathrm g\right)}&=\sqrt{n_\mathrm g}\tag5\\[6pt] u^2{\left(n_\mathrm g\right)}&=n_\mathrm g\tag6\\[6pt] u{\left(r_\mathrm g\right)}&=\frac{u{\left(n_\mathrm g\right)}}{t_\mathrm g}\tag7\\[6pt] &=\frac{\sqrt{n_\mathrm g}}{t_\mathrm g}\tag8\\[6pt] u^2{\left(r_\mathrm g\right)}&=\frac{u^2{\left(n_\mathrm g\right)}}{t_\mathrm g^2}\tag9\\[6pt] &=\frac{n_\mathrm g}{t_\mathrm g^2}=\frac{r_\mathrm g}{t_\mathrm g}\tag{10} \end{align}

as well as for the background counts

\begin{align} u{\left(n_0\right)}&=\sqrt{n_0}\tag{11}\\[6pt] u^2{\left(n_0\right)}&=n_0\tag{12}\\[6pt] u{\left(r_0\right)}&=\frac{u{\left(n_0\right)}}{t_\mathrm g}\tag{13}\\[6pt] &=\frac{\sqrt{n_0}}{t_0}\tag{14}\\[6pt] u^2{\left(r_0\right)}&=\frac{u^2{\left(n_0\right)}}{t_0^2}\tag{15}\\[6pt] &=\frac{n_0}{t_0^2}=\frac{r_0}{t_0}\tag{16} \end{align}

The standard uncertainties $$u{\left(t_\mathrm g\right)}$$ and $$u{\left(t_0\right)}$$ of the measurement durations $$t_\mathrm g$$ and $$t_0$$ are neglected since the measurement duration can be measured far more exactly than all the other quantities involved and can thus be taken as a constant.

The model of evaluation for the net count rate $$r_\mathrm n$$ is

\begin{align} r_\mathrm n&=r_\mathrm g-r_0\tag{17}\\[6pt] &=\frac{n_\mathrm g}{t_\mathrm g}-\frac{n_0}{t_0}\tag{18}\\[6pt] \end{align}

The corresponding uncertainty is

\begin{align} u^2{\left(r_\mathrm n\right)}&=u^2{\left(r_\mathrm g\right)}+u^2{\left(r_0\right)}\tag{19}\\[6pt] &=\frac{n_\mathrm g}{t_\mathrm g^2}+\frac{n_0}{t_0^2}=\frac{r_\mathrm g}{t_\mathrm g}+\frac{r_0}{t_0}\tag{20}\\[6pt] u{\left(r_\mathrm n\right)}&=\sqrt{\frac{n_\mathrm g}{t_\mathrm g^2}+\frac{n_0}{t_0^2}}=\sqrt{\frac{r_\mathrm g}{t_\mathrm g}+\frac{r_0}{t_0}}\tag{21} \end{align}

Standard uncertainty of the estimator as a function of the true value of the measurand

According to $$\text(18)$$, the equation for the true value $$\tilde r_\mathrm n$$ of the net count rate is expected as

\begin{alignat}{2} &&\tilde r_\mathrm n&=\frac{n_\mathrm g}{t_\mathrm g}-\frac{n_0}{t_0}\tag{22}\\[6pt] &\Leftrightarrow\quad&n_\mathrm g&=\left(\tilde r_\mathrm n+\frac{n_0}{t_0}\right)\cdot t_\mathrm g\tag{23} \end{alignat}

and according to $$\text(19)$$ and $$\text(20)$$, the corresponding equation for the standard uncertainty $$\tilde u$$ of the the true value $$\tilde r_\mathrm n$$ of the net count rate is expected as

\begin{align} \tilde u^2{\left(\tilde r_\mathrm n\right)}&=u^2{\left(r_\mathrm g\right)}+u^2{\left(r_0\right)}\tag{24}\\[6pt] &=\frac{n_\mathrm g}{t_\mathrm g^2}+\frac{n_0}{t_0^2}\tag{25}\\[6pt] \end{align}

Inserting $$\text{(23)}$$ into $$\text{(25)}$$ yields:

\begin{align} \tilde u^2{\left(\tilde r_\mathrm n\right)}&=\frac{\left(\tilde r_\mathrm n+\frac{n_0}{t_0}\right)\cdot t_\mathrm g}{t_\mathrm g^2}+\frac{n_0}{t_0^2}\tag{26}\\[6pt] &=\frac{\tilde r_\mathrm n}{t_\mathrm g}+\frac{n_0}{t_0}\cdot\left(\frac1{t_\mathrm g}+\frac1{t_0}\right)\tag{27}\\[6pt] \tilde u{\left(\tilde r_\mathrm n\right)}&=\sqrt{\frac{\tilde r_\mathrm n}{t_\mathrm g}+\frac{n_0}{t_0}\cdot\left(\frac1{t_\mathrm g}+\frac1{t_0}\right)}\tag{28} \end{align}

Decision threshold

According to $$\text{(1)}$$, the equation for the decision threshold of the net count rate is:

$$r_\mathrm n^*=k_{1-\alpha}\cdot\tilde u{\left(0\right)}\tag{29}$$

Inserting $$\text{(28)}$$ with $$\tilde r_\mathrm n=0$$ yields:

\begin{align} r_\mathrm n^*&=k_{1-\alpha}\cdot\sqrt{\frac0{t_\mathrm g}+\frac{n_0}{t_0}\cdot\left(\frac1{t_\mathrm g}+\frac1{t_0}\right)}\tag{30}\\[6pt] &=k_{1-\alpha}\cdot\sqrt{\frac{n_0}{t_0}\cdot\left(\frac1{t_\mathrm g}+\frac1{t_0}\right)}\tag{31} \end{align}

Detection limit

According to $$\text{(2)}$$, the equation for the detection limit of the net count rate is:

$$r_\mathrm n^\#=r_\mathrm n^*+k_{1-\beta}\cdot\tilde u{\left(r_\mathrm n^\#\right)}\tag{32}$$

Inserting $$\text{(28)}$$ with $$\tilde r_\mathrm n=r_\mathrm n^\#$$ yields:

$$r_\mathrm n^\#=r_\mathrm n^*+k_{1-\beta}\cdot\sqrt{\frac{r_\mathrm n^\#}{t_\mathrm g}+\frac{n_0}{t_0}\cdot\left(\frac1{t_\mathrm g}+\frac1{t_0}\right)}\tag{33}$$

Since, according to $$\text{(32)}$$, $$r_\mathrm n^\#$$ depends on $$\tilde u{\left(r_\mathrm n^\#\right)}$$ and, according to $$\text{(28)}$$, $$\tilde u{\left(r_\mathrm n^\#\right)}$$ depends on $$r_\mathrm n^\#$$, the detection limit $$r_\mathrm n^\#$$ cannot be directly calculated using $$\text{(33)}$$. In principle, such equations can be solved for the detection limit; however, the procedure can be elaborate and the result can be unwieldy. In this case, solving $$\text{(33)}$$ for $$r_\mathrm n^\#$$ yields:

$$r_\mathrm n^\#=r_\mathrm n^*+\frac{k_{1-\beta}\cdot\left(k_{1-\beta}\cdot t_0+\sqrt{4\cdot r_\mathrm n^*\cdot t_0^2\cdot t_\mathrm g+k_{1-\beta}^2\cdot t_0^2+4\cdot n_0\cdot t_0\cdot t_\mathrm g+4\cdot n_0\cdot t_\mathrm g^2}\right)}{2\cdot t_\mathrm g\cdot t_0}\tag{34}$$

In practice, however, this step is usually unnecessary since typical spreadsheet software can automatically calculate formulas with circular references (i.e. when a formula refers back to its own cell) such as $$\text{(33)}$$ by using iteration (i.e. the repeated recalculation of a worksheet until a specific numeric condition is met). In Microsoft Office Excel, for example, iterative calculations are turned off by default and have to be enabled in the calculation options section. If necessary, $$r_\mathrm n^\#\approx2\cdot r_\mathrm n^*$$ may be used as initial approximation for the iteration.

$$\textbf{Numerical examples}\\ \begin{array}{llll|llll|l} \hline n_\mathrm g&t_\mathrm g&n_0&t_0&r_\mathrm n&u{\left(r_\mathrm n\right)}&r_\mathrm n^*&r_\mathrm n^\#&\text{Reported}\ r_\mathrm n\\ &\text{in}\ \mathrm s&&\text{in}\ \mathrm s&\text{in}\ \mathrm s^{-1}&\text{in}\ \mathrm s^{-1}&\text{in}\ \mathrm s^{-1}&\text{in}\ \mathrm s^{-1}&\text{in}\ \mathrm s^{-1}\\ \hline \hphantom{0}150&\hphantom{0}60&\hphantom{0}100&\hphantom{00}60&0.833&0.264&0.388&0.820&0.83\pm0.26\\ \hphantom{0}140&\hphantom{0}60&\hphantom{0}100&\hphantom{00}60&0.667&0.258&0.388&0.820&0.67\pm0.26\\ \hphantom{0}130&\hphantom{0}60&\hphantom{0}100&\hphantom{00}60&0.500&0.253&0.388&0.820&0.50\pm0.25\\ \hphantom{0}120&\hphantom{0}60&\hphantom{0}100&\hphantom{00}60&0.333&0.247&0.388&0.820&\lt0.82\\ \hphantom{0}110&\hphantom{0}60&\hphantom{0}100&\hphantom{00}60&0.167&0.242&0.388&0.820&\lt0.82\\ \hline \hphantom{0}150&\hphantom{0}60&6000&3600&0.833&0.205&0.276&0.598&0.83\pm0.21\\ \hphantom{0}140&\hphantom{0}60&6000&3600&0.667&0.198&0.276&0.598&0.67\pm0.20\\ \hphantom{0}130&\hphantom{0}60&6000&3600&0.500&0.191&0.276&0.598&0.50\pm0.19\\ \hphantom{0}120&\hphantom{0}60&6000&3600&0.333&0.184&0.276&0.598&0.33\pm0.18\\ \hphantom{0}110&\hphantom{0}60&6000&3600&0.167&0.176&0.276&0.598&\lt0.60\\ \hline 1500&600&\hphantom{0}100&\hphantom{00}60&0.833&0.179&0.288&0.580&0.83\pm0.18\\ 1400&600&\hphantom{0}100&\hphantom{00}60&0.667&0.178&0.288&0.580&0.67\pm0.18\\ 1300&600&\hphantom{0}100&\hphantom{00}60&0.500&0.177&0.288&0.580&0.50\pm0.18\\ 1200&600&\hphantom{0}100&\hphantom{00}60&0.333&0.176&0.288&0.580&0.33\pm0.18\\ 1100&600&\hphantom{0}100&\hphantom{00}60&0.167&0.176&0.288&0.580&\lt0.58\\ \hline \end{array}$$

## Example 2 – activity (with calibration factor)

$$\textbf{Quantities and symbols}\\ \begin{array}{ll} \hline \text{Symbol}&\text{Name}\\ \hline n_\mathrm g&\text{Number of counted pulses of the gross effect}\\ t_\mathrm g&\text{Measurement duration of the measurement of the gross effect}\\ r_\mathrm g&\text{Estimate of the gross count rate}\\ \hline n_0&\text{Number of counted pulses of the background effect}\\ t_0&\text{Measurement duration of the measurement of the background effect}\\ r_0&\text{Estimate of the background count rate}\\ \hline r_\mathrm n&\text{Estimate of the net count rate}\\ \tilde r_\mathrm n&\text{True value of the net count rate}\\ r_\mathrm n^*&\text{Decision threshold of the net count rate}\\ r_\mathrm n^\#&\text{Detection limit of the net count rate}\\ \hline \varphi&\text{Calibration factor}\\ u{\left(\varphi\right)}&\text{Standard uncertainty of the calibration factor}\\ \hline A&\text{Estimate of the activity}\\ u{\left(A\right)}&\text{Standard uncertainty of the activity}\\[-3pt] &\text{(associated with the measurement result }A)\\ \tilde A_\mathrm n&\text{True value of the activity}\\ \tilde u{\left(\tilde y\right)}&\text{Standard uncertainty of an estimator of the activity}\\[-3pt] &\text{(as a function of the true value of the activity }\tilde A)\\ A_\mathrm n^*&\text{Decision threshold of the activity}\\ A_\mathrm n^\#&\text{Detection limit of the activity}\\ \hline \alpha&\text{Probability of the error of the first kind}\\ \beta&\text{Probability of the error of the second kind}\\ k_{1-\alpha}&\text{Quantile of the standardized normal distribution for the probability }\alpha\\ k_{1-\beta}&\text{Quantile of the standardized normal distribution for the probability }\beta\\ \hline \end{array}$$

This example is an expansion of the first example (see above). Again, it relates to a sample of radioactive material. In this case, the measurand is the activity $$A$$ (in $$\mathrm{Bq}$$) of the sample. It is determined from the net count rate $$r_\mathrm n$$ by multiplication by a calibration factor $$\varphi$$. This calibration factor may include various calibration, correction or influence quantities, or conversion factors that apply to the sample preparation and the actual measurement procedure. The value of the calibration factor may also be obtained from the measurement of the net count rate for a calibration source with a known activity.

$$A=\varphi\cdot r_\mathrm n\tag{35}$$

It is usually convenient to calculate the corresponding standard uncertainty $$u{\left(A\right)}$$ based on the propagation of the relative standard uncertainties:

\begin{align} \left(\frac{u{\left(A\right)}}{A}\right)^2&=\left(\frac{u{\left(\varphi\right)}}{\varphi}\right)^2+\left(\frac{u{\left(r_\mathrm n\right)}}{r_\mathrm n}\right)^2\tag{36}\\[6pt] \frac{u{\left(A\right)}}{A}&=\sqrt{\left(\frac{u{\left(\varphi\right)}}{\varphi}\right)^2+\left(\frac{u{\left(r_\mathrm n\right)}}{r_\mathrm n}\right)^2}\tag{37}\\[6pt] u{\left(A\right)}&=\sqrt{\left(\frac{u{\left(\varphi\right)}}{\varphi}\right)^2+\left(\frac{u{\left(r_\mathrm n\right)}}{r_\mathrm n}\right)^2}\cdot A\tag{38}\\[6pt] \end{align}

Inserting $$\text{(35)}$$ yields:

\begin{align} u{\left(A\right)}&=\sqrt{\left(\frac{u{\left(\varphi\right)}}{\varphi}\right)^2+\left(\frac{u{\left(r_\mathrm n\right)}}{r_\mathrm n}\right)^2}\cdot\varphi\cdot r_\mathrm n\tag{39}\\[6pt] &=\sqrt{r_\mathrm n^2\cdot u^2{\left(\varphi\right)}+\varphi^2\cdot u^2{\left(r_\mathrm n\right)}}\tag{40}\\[6pt] u^2{\left(A\right)}&=r_\mathrm n^2\cdot u^2{\left(\varphi\right)}+\varphi^2\cdot u^2{\left(r_\mathrm n\right)}\tag{41} \end{align}

Note that the same result may be obtained using the general equation for the law of propagation of uncertainty as described in ISO/IEC Guide 98-3:

\begin{align} u^2{\left(A\right)}&=\left(\frac{\partial A}{\partial\varphi}\right)^2 u^2{\left(\varphi\right)}+\left(\frac{\partial A}{\partial r_\mathrm n}\right)^2 u^2{\left(r_\mathrm n\right)}\tag{42}\\[6pt] &=r_\mathrm n^2\cdot u^2{\left(\varphi\right)}+\varphi^2\cdot u^2{\left(r_\mathrm n\right)}\tag{43} \end{align}

The general equation may be useful in case of complicated expressions for the standard uncertainty. Note that, when the nonlinearity of the considered function is significant, it might be necessary to include higher-order terms in the Taylor series expansion in the expression for its uncertainty.

Inserting $$\text{(17)}$$ or $$\text{(18)}$$ into $$\text{(35)}$$ yields the complete model of evaluation:

\begin{align} A&=\varphi\cdot\left(r_\mathrm g-r_0\right)\tag{44}\\[6pt] &=\varphi\cdot\left(\frac{n_\mathrm g}{t_\mathrm g}-\frac{n_0}{t_0}\right)\tag{45} \end{align}

Inserting $$\text{(18)}$$ and $$\text{(20)}$$ in $$\text{(41)}$$ or $$\text{(43)}$$ yields the corresponding uncertainty:

\begin{align} u^2{\left(A\right)}&=\left(\frac{n_\mathrm g}{t_\mathrm g}-\frac{n_0}{t_0}\right)^2\cdot u^2{\left(\varphi\right)}+\varphi^2\cdot\left(\frac{n_\mathrm g}{t_\mathrm g^2}+\frac{n_0}{t_0^2}\right)\tag{46}\\[6pt] u{\left(A\right)}&=\sqrt{\left(\frac{n_\mathrm g}{t_\mathrm g}-\frac{n_0}{t_0}\right)^2\cdot u^2{\left(\varphi\right)}+\varphi^2\cdot\left(\frac{n_\mathrm g}{t_\mathrm g^2}+\frac{n_0}{t_0^2}\right)}\tag{47} \end{align}

Standard uncertainty of the estimator as a function of the true value of the measurand

According to $$\text{(45)}$$, the equation for the true value $$\tilde A$$ of the activity is expected as

\begin{alignat}{2} &&\tilde A&=\varphi\cdot\left(\frac{n_\mathrm g}{t_\mathrm g}-\frac{n_0}{t_0}\right)\tag{48}\\[6pt] &\Leftrightarrow\quad&n_\mathrm g&=\left(\frac{\tilde A}{\varphi}+\frac{n_0}{t_0}\right)\cdot t_\mathrm g\tag{49} \end{alignat}

and according to $$\text{(46)}$$, the corresponding equation for the standard uncertainty $$\tilde u$$ of the true value $$\tilde A$$ of the activity is expected as

$$\tilde u^2{\left(\tilde A\right)}=\left(\frac{n_\mathrm g}{t_\mathrm g}-\frac{n_0}{t_0}\right)^2\cdot u^2{\left(\varphi\right)}+\varphi^2\cdot\left(\frac{n_\mathrm g}{t_\mathrm g^2}+\frac{n_0}{t_0^2}\right)\tag{50}\\[6pt]$$

Inserting $$\text{(49)}$$ into $$\text{(50)}$$ yields:

\begin{align} \tilde u^2{\left(\tilde A\right)}&=\left(\frac{\left(\frac{\tilde A}{\varphi}+\frac{n_0}{t_0}\right)\cdot t_\mathrm g}{t_\mathrm g}-\frac{n_0}{t_0}\right)^2\cdot u^2{\left(\varphi\right)}+\varphi^2\cdot\left(\frac{\left(\frac{\tilde A}{\varphi}+\frac{n_0}{t_0}\right)\cdot t_\mathrm g}{t_\mathrm g^2}+\frac{n_0}{t_0^2}\right)\tag{51}\\[6pt] &=\tilde A^2\cdot\left(\frac{u{\left(\varphi\right)}}{\varphi}\right)^2+\varphi^2\cdot\left(\frac{\tilde A}{\varphi\cdot t_\mathrm g}+\frac{n_0}{t_0}\cdot\left(\frac1{t_\mathrm g}+\frac1{t_0}\right)\right)\tag{52}\\[6pt] \tilde u{\left(\tilde A\right)}&=\sqrt{\tilde A^2\cdot\left(\frac{u{\left(\varphi\right)}}{\varphi}\right)^2+\varphi^2\cdot\left(\frac{\tilde A}{\varphi\cdot t_\mathrm g}+\frac{n_0}{t_0}\cdot\left(\frac1{t_\mathrm g}+\frac1{t_0}\right)\right)}\tag{53} \end{align}

Decision threshold

According to $$\text{(1)}$$, the equation for the decision threshold of the activity is:

$$A^*=k_{1-\alpha}\cdot\tilde u{\left(0\right)}\tag{54}$$

Inserting $$\text{(53)}$$ with $$\tilde A=0$$ yields:

\begin{align} A^*&=k_{1-\alpha}\cdot\sqrt{0^2\cdot\left(\frac{u{\left(\varphi\right)}}{\varphi}\right)^2+\varphi^2\cdot\left(\frac0{\varphi\cdot t_\mathrm g}+\frac{n_0}{t_0}\cdot\left(\frac1{t_\mathrm g}+\frac1{t_0}\right)\right)}\tag{55}\\[6pt] &=k_{1-\alpha}\cdot\varphi\cdot\sqrt{\frac{n_0}{t_0}\cdot\left(\frac1{t_\mathrm g}+\frac1{t_0}\right)}\tag{56} \end{align}

Note that the decision threshold $$A^*$$ of the activity $$A$$ is not affected by the uncertainty $$u{\left(\varphi\right)}$$ of the calibration factor $$\varphi$$.

A comparison of $$\text{(31)}$$ and $$\text{(56)}$$ shows that the decision threshold $$A^*$$ of the activity $$A$$ is equal to the decision threshold $$r_\mathrm n^*$$ of the net count rate $$r_\mathrm n$$ multiplied by the calibration factor $$\varphi$$:

$$A^*=\varphi\cdot r_\mathrm n^*\tag{57}$$

which is similar to the initial model that is expressed as $$\text{(35)}$$. However, the analogous simple relationship does not apply to the following detection limit.

Detection limit

According to $$\text{(2)}$$, the equation for the detection limit of the activity is:

$$A^\#=A^*+k_{1-\beta}\cdot\tilde u{\left(A^\#\right)}\tag{58}$$

Inserting $$\text{(53)}$$ with $$\tilde A=A^\#$$ yields:

$$A^\#=A^*+k_{1-\beta}\cdot\sqrt{{A^\#}^2\cdot\left(\frac{u{\left(\varphi\right)}}{\varphi}\right)^2+\varphi^2\cdot\left(\frac{A^\#}{\varphi\cdot t_\mathrm g}+\frac{n_0}{t_0}\cdot\left(\frac1{t_\mathrm g}+\frac1{t_0}\right)\right)}\tag{59}$$

Similar to the case of $$\text{(33)}$$, according to $$\text{(58)}$$, $$A^\#$$ depends on $$u{\left(A^\#\right)}$$ and, according to $$\text{(53)}$$, $$u{\left(A^\#\right)}$$ depends on $$A^\#$$. Therefore, $$A^\#$$ cannot be directly calculated using $$\text{(59)}$$. Again, it is usually unnecessary to solve $$\text{(59)}$$ for $$A^\#$$ since typical spreadsheet software can automaticall calculate formulas with circular references by using iteration.

$$\textbf{Numerical examples}\\ \begin{array}{llllll|llll|l} \hline n_\mathrm g&t_\mathrm g&n_0&t_0&\varphi&u{\left(\varphi\right)}&A&u{\left(A\right)}&A^*&A^\#&\text{Reported}\ A\\ &\text{in}\ \mathrm s&&\text{in}\ \mathrm s&\text{in}\ \mathrm{Bq\ s}&\text{in}\ \mathrm{Bq\ s}&\text{in}\ \mathrm{Bq}&\text{in}\ \mathrm{Bq}&\text{in}\ \mathrm{Bq}&\text{in}\ \mathrm{Bq}&\text{in}\ \mathrm{Bq}\\ \hline \hphantom{0}150&\hphantom{0}60&\hphantom{0}100&\hphantom{00}60&4.0&0.2&3.333&1.067&1.551&3.304&3.3\pm1.1\\ \hphantom{0}140&\hphantom{0}60&\hphantom{0}100&\hphantom{00}60&4.0&0.2&2.667&1.041&1.551&3.304&2.7\pm1.0\\ \hphantom{0}130&\hphantom{0}60&\hphantom{0}100&\hphantom{00}60&4.0&0.2&2.000&1.016&1.551&3.304&2.0\pm1.0\\ \hphantom{0}120&\hphantom{0}60&\hphantom{0}100&\hphantom{00}60&4.0&0.2&1.333&0.991&1.551&3.304&\lt3.3\\ \hphantom{0}110&\hphantom{0}60&\hphantom{0}100&\hphantom{00}60&4.0&0.2&0.667&0.967&1.551&3.304&\lt3.3\\ \hline \hphantom{0}150&\hphantom{0}60&6000&3600&4.0&0.2&3.333&0.838&1.106&2.408&3.3\pm0.8\\ \hphantom{0}140&\hphantom{0}60&6000&3600&4.0&0.2&2.667&0.805&1.106&2.408&2.7\pm0.8\\ \hphantom{0}130&\hphantom{0}60&6000&3600&4.0&0.2&2.000&0.771&1.106&2.408&2.0\pm0.8\\ \hphantom{0}120&\hphantom{0}60&6000&3600&4.0&0.2&1.333&0.738&1.106&2.408&1.3\pm0.8\\ \hphantom{0}110&\hphantom{0}60&6000&3600&4.0&0.2&0.667&0.705&1.106&2.408&\lt2.4\\ \hline 1500&600&\hphantom{0}100&\hphantom{00}60&4.0&0.2&3.333&0.734&1.150&2.334&3.3\pm0.7\\ 1400&600&\hphantom{0}100&\hphantom{00}60&4.0&0.2&2.667&0.724&1.150&2.334&2.7\pm0.7\\ 1300&600&\hphantom{0}100&\hphantom{00}60&4.0&0.2&2.000&0.716&1.150&2.334&2.0\pm0.7\\ 1200&600&\hphantom{0}100&\hphantom{00}60&4.0&0.2&1.333&0.707&1.150&2.334&1.3\pm0.7\\ 1100&600&\hphantom{0}100&\hphantom{00}60&4.0&0.2&0.667&0.703&1.150&2.334&\lt2.3\\ \hline \end{array}$$

## References

• I have looked into it, but could not find a valid way how to break these long names in the first column. I believe there is no handling of \parboxes implemented. Right now it just wouldn't fit into the margins and the units are swallowed by the right menu... – Martin - マーチン Aug 3 '16 at 4:31
• Loong, are you bored? – M.A.R. Aug 7 '16 at 20:14

# Image size test for small mobile devices

(test results)

## Resolution

Normal resolution (6.25 KB):

2 subpixels per pixel (10.74 KB):

2 or 3 subpixels per pixel (34.56 KB):

(test results)

Comparison of font size

(use of MathJax font for function graphs and similar images)

Image (png):

MathJax: $$0\quad1\quad2\quad3\quad4\quad5\quad6\quad7\quad8\quad9\quad10$$ $$x\ \text{axis title}$$

\left< \right> outputs the same thing as \left\langle \right\rangle and is much easier to type!

\left< test^2 \middle| test^3 \right>

$\left< test^2 \middle| test^3 \right>$

If $\text{MathJax} \implies \LaTeX$ this wouldn't be such a problem...

\left\langle \psi \middle| \hat{A} \middle| \psi \right\rangle

$$\Large \left\langle \psi \middle| \hat{A} \middle| \psi \right\rangle$$

\left\langle \chi_{\mu} | \mathbf{rr}^{T} | \chi_{\nu} \right\rangle

$$\Large \left\langle \chi_{\mu} | \mathbf{rr}^{T} | \chi_{\nu} \right\rangle$$

\left\langle \chi_{\mu} \middle| \mathbf{rr}^{T} \middle| \chi_{\nu} \right\rangle

$$\Large \left\langle \chi_{\mu} \middle| \mathbf{rr}^{T} \middle| \chi_{\nu} \right\rangle$$

\left\langle \chi_{\mu} \left| \mathbf{rr}^{T} \right| \chi_{\nu} \right\rangle

$$\Large \left\langle \chi_{\mu} \left| \mathbf{rr}^{T} \right| \chi_{\nu} \right\rangle$$

\left\langle \, \chi_{\mu} \middle| \left. \! \mathbf{rr}^{T} \right| \chi_{\nu} \! \right\rangle

$$\Large \left\langle \, \chi_{\mu} \middle| \left. \! \mathbf{rr}^{T} \right| \chi_{\nu} \! \right\rangle$$

ಠ_ಠ

• $$\Huge \color{\red}{\text{ಠ_ಠ}}$$ – orthocresol Nov 4 '16 at 14:05

### Nested environments for the vertical alignment inside a table

\begin{array}{clr} \hline \text{Reaction} & E^\circ/\pu{V} \\ \hline \begin{align} \ce{Fe^3+ + e- &-> Fe^2+} \\ \ce{Cu^2+ + 2e- &-> Cu} \\ \ce{Fe^3+ + 3e- &-> Fe} \\ \ce{Fe^2+ + 2e- &-> Fe} \end{align} & \begin{array}{r} +0.77 \\ +0.34 \\ -0.04 \\ -0.41 \end{array} \\ \hline \end{array}

### Polymer bond through bracket

$$\require{enclose}\ce{\enclose{horizontalstrike}{\;(}HNCH2CONHCH2CH2NHCO\enclose{horizontalstrike}{)_n}}$$

Smallcaps experiments

• D-Glucose and L-rhamnose.
(plain text; incorrect normal size capital D/L)
• D-Glucose and L-rhamnose.
(<small>D/L</small> supported already or not?)
• ᴅ-Glucose and ʟ-rhamnose.
(U+1D05/U+029F LATIN LETTER SMALL CAPITAL D/L; potentially problematic)
• d-Glucose and l-rhamnose.
(<span style="font-variant: small-caps">d/l</span> unsupported (ever?), problematic – plaintext/non-functional rendering has lowercase d/l)
• $$\ce{\small\text{D}\normalsize\text{-Glucose}}$$ and $$\ce{\small\text{L}\normalsize\text{-rhamnose}}$$.
($$\ce{\small\text{D/L}\normalsize}$$)

Comment test sandbox ↓

• Keyboard <kbd> test: <kbd>Ctrl</kbd>+<kbd>C</kbd> <kbd>Ctrl</kbd>+<kbd>V</kbd> – mykhal Oct 8 '18 at 17:30
• [ tour ] teplate test: tour – mykhal Nov 29 '18 at 9:42
• italic test (1*R*)-, (1<i>R</i>)-, (1_R_)-, :( – mykhal Jan 4 at 17:40
• italic ................ (1⁠_R_⁠) – mykhal Jan 13 at 14:28
• (2​*R*​​)- ............. Fooboo Foo_boo_ (1*R*,2*S*)- – mykhal Jan 14 at 17:28
• (4aR,8aR)-decahydroquinoline – mykhal Jan 15 at 13:07

To find the the effective mass we have to find total energy $$E=\int_m\frac12u^2\,\mathrm dm$$ where $$E$$ is total energy and $$\mathrm dm=\left(\frac{\mathrm dy}{L}\right)m$$ which gives us

$$E=\int_0^L\frac12u^2\left(\frac{\mathrm dy}L\right)m$$ We assume that the velocity in a spring varies linearly with $$y$$ giving us $$u=\frac{vy}L$$ which on sustituting gives us

$$E=\frac12\frac mL\int_0^L\left(\frac{vy}L\right)^2\,\mathrm dy$$ which on solving gives

$$E=\frac12\frac m3v^2$$

HBr, and HCl are strong acids.

• This answer actually was "HI, HBr, and HCl are strong acids." However, salutations (e.g. "Hi") are automatically removed, which leaves "HBr, and HCl are strong acids." – Loong Jun 16 at 7:41

“Your question is taken.” Robert Mueller, 2019.