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## 物理代写|理论力学代写theoretical mechanics代考|Error Propagation

We now start from $K$ physical quantities $x_{k}$ with average errors $\sigma_{k}$ and consider the derived quantity $y=f\left(x_{1}, \ldots, x_{K}\right)$. Here all the quantities $x_{k}$ will be independent of each other. What is then the average error in $y$ ?
To begin with, the error $\varepsilon$ in $f\left(x_{1}, \ldots, x_{K}\right)$ is to first order
$$\varepsilon=\sum_{k=1}^{K} \frac{\partial f}{\partial x_{k}} \varepsilon_{k}$$
and hence
\begin{aligned} \sigma^{2} &=\left\langle\varepsilon^{2}\right\rangle \ &=\int \cdots \int_{-\infty}^{\infty}\left(\sum_{k=1}^{K} \frac{\partial f}{\partial x_{k}} \varepsilon_{k}\right)^{2} \rho\left(\varepsilon_{1}, \ldots, \varepsilon_{K}\right) \mathrm{d} \varepsilon_{1} \cdots \mathrm{d} \varepsilon_{K} \ &=\left\langle\sum_{k=1}^{K} \frac{\partial f}{\partial x_{k}} \varepsilon_{k} \cdot \sum_{l=1}^{K} \frac{\partial f}{\partial x_{l}} \varepsilon_{l}\right\rangle=\sum_{k, l=1}^{K} \frac{\partial f}{\partial x_{k}} \frac{\partial f}{\partial x_{l}}\left\langle\varepsilon_{k} \varepsilon_{l}\right\rangle . \end{aligned}
Since the quantities $x_{k}$ and $x_{l}$ should not depend upon each other, they are not correlated to each other (the property $x_{l}$ does not care how large $x_{k}$ is – correlations will be investigated in more detail in Sect. 6.1.5). With $\left\langle\varepsilon_{k}^{2}\right\rangle=\sigma_{k}^{2}$ this leads to
$$\left\langle\varepsilon_{k} \varepsilon_{l}\right\rangle=\left{\begin{array}{cc} \left\langle\varepsilon_{k}\right\rangle\left\langle\varepsilon_{l}\right\rangle & \text { for } k \neq l, \ \sigma_{k}{ }^{2} & \text { for } k=l . \end{array}\right.$$
Here, $\left\langle\varepsilon_{k}\right\rangle=0$ holds for all $k$ (and $l$ ). Therefore, the law of error propagation follows:
$$\sigma^{2}=\sum_{k=1}^{K}\left(\frac{\partial f}{\partial x_{k}}\right)^{2} \sigma_{k}^{2} .$$
In the proof, no normally distributed errors were necessary -thus other distributions with the properties $\left\langle\varepsilon_{k}^{0}\right\rangle=1,\left\langle\varepsilon_{k}^{1}\right\rangle=0$, and $\left\langle\varepsilon_{k}^{2}\right\rangle=\sigma_{k}^{2}$ deliver the error propagation law and with it the basis for all further proofs in this section. In particular, we may invoke this law for repeated measurements of the same quantity, as we shall now do.

## 物理代写|理论力学代写theoretical mechanics代考|Finite Measurement Series and Their Average Error

If we consider the expression
$$\langle x\rangle \approx \bar{x} \equiv \frac{1}{N} \sum_{n=1}^{N} x_{n}$$
as $\bar{x}=f\left(x_{1}, \ldots, x_{N}\right)$, then we can use it in the law of error propagation and deduce that $\partial f / \partial x_{n}=N^{-1}$. Hence, all single measurements enter into the error estimate with the same weight-as already for the estimated value $x_{0}$.

In order to determine the error $\sigma_{n}$, we think of an average over several measurement series, each with $N$ measurements. In this way, we can introduce the average error of the single measurement and find that all single measurements have the same average error $\Delta x$. Therefore, the law of error propagation for $N$ equal terms $N^{-2}(\Delta x)^{2}$ delivers
$$(\Delta \bar{x})^{2}=N \cdot N^{-2}(\Delta x)^{2}=\frac{(\Delta x)^{2}}{N}$$
The average error $\Delta \bar{x}$ in the mean value of the measurement series is thus the $\sqrt{N}$ th part of the average error in a single measurement: the more often measurements are made, the more accurate is the determination of the mean value. However, because of the square root factors, the accuracy can be increased only rather slowly.

Since we do not know the true value $x_{0}$ itself, but only its approximation $\bar{x}$, we still have to account for its uncertainty $\Delta \bar{x}$ in order to determine the average error of the single measurement:
$$(\Delta x)^{2}=\overline{\left(x-x_{0}\right)^{2}}=\overline{\left(x-\bar{x}+\bar{x}-x_{0}\right)^{2}}=\overline{(x-\bar{x})^{2}}+2 \overline{(x-\bar{x})}\left(\bar{x}-x_{0}\right)+\left(\bar{x}-x_{0}\right)^{2} .$$
Here, $\overline{x-\bar{x}}=\bar{x}-\bar{x}=0$ and thus $\overline{(x-\bar{x})^{2}}=\overline{x^{2}}-\bar{x}^{2}$ is rather easy to evaluate. For $\left(\bar{x}-x_{0}\right)^{2}$, we take $(\Delta \bar{x})^{2}=(\Delta x)^{2} / N$. Hence, because $1-N^{-1}=N^{-1}(N-1)$, the average error of the single measurement is
$$(\Delta x)^{2}=\frac{N}{N-1} \overline{(x-\bar{x})^{2}},$$
as claimed previously (see p. 46). And so we have the announced proof. For sufficiently large $N$, we may write $(\Delta x)^{2}=\overline{x^{2}}-\bar{x}^{2}$. The expression $\Delta x$ is referred to as the uncertainty of $x$ in quantum theory (see p. 275).

## 物理代写|理论力学代写theoretical mechanics代考|Error Analysis

How should we modify the result ohtained so far if the same quantity is measured in different ways: first as $x_{1} \pm \Delta x_{1}$, then as $x_{2} \pm \Delta x_{2}$, and so on? What is then the most probable value for $x_{0}$, and what average error does it have?

If the readings of the measurement were taken with the same instrument and equally carefully, the difference in the average errors stems from values $x_{n}$ from measurement series of different lengths. According to the last section, the average error of every single measurement in such a measurement series should be equal to $\Delta x_{n} \sqrt{N_{n}}$, and this independently of $n$ in each of the measurement series. Therefore, the mentioned values $x_{n}$ should contribute with the weight
$$\rho_{n}=\frac{N_{n}}{\sum_{k} N_{k}}=\frac{1}{\left(\Delta x_{n}\right)^{2}} / \sum_{k} \frac{1}{\left(\Delta x_{k}\right)^{2}},$$
whence $\bar{x}=\sum_{n} \rho_{n} x_{n}$ is the properly weighted mean value. The error propagation law delivers
\begin{aligned} \sigma^{2} &=\sum_{n} \rho_{n}^{2} \sigma_{n}^{2}=\frac{1}{\left(\sum_{k}\left(\Delta x_{k}\right)^{-2}\right)^{2}} \sum_{n} \frac{1}{\left(\Delta x_{n}\right)^{4}}\left(\Delta x_{n}\right)^{2}=\frac{\sum_{n}\left(\Delta x_{n}\right)^{-2}}{\left(\sum_{k}\left(\Delta x_{k}\right)^{-2}\right)^{2}} \ &=\frac{1}{\sum_{n}\left(\Delta x_{n}\right)^{-2}} \quad \Longrightarrow \quad \frac{1}{(\Delta x)^{2}}=\sum_{n} \frac{1}{\left(\Delta x_{n}\right)^{2}} \end{aligned}
The more detailed the readings of the measurement, the more important they are for the mean value and for the (un)certainty of the results. These considerations are only then valid without restriction, if the values are compatible with each other within their error limits. If they lie further apart from each other, then we have to take
$$(\Delta x)^{2}=\frac{1}{N-1} \frac{1}{\sum_{n}\left(\Delta x_{n}\right)^{-2}} \sum_{n} \frac{\left(x_{n}-\bar{x}\right)^{2}}{\left(\Delta x_{n}\right)^{2}}$$
Note that, if the values $x_{n}$ do not lie within the error limits, then systematic errors may be involved.

Thus, these two equations answer the questions raised in the general case, where measurements are taken with different instruments and different levels of care: to each value $x_{n}$, we must attach the relative weight $1 /\left(\Delta x_{n}\right)^{2}$.

## 物理代写|理论力学代写theoretical mechanics代考|Error Propagation

$$\varepsilon=\sum_{k=1}^{K} \frac{\partial f}{\partial x_{k}} \varepsilon_{k}$$

$$\sigma^{2}=\left\langle\varepsilon^{2}\right\rangle \quad=\cdots \int_{-\infty}^{\infty}\left(\sum_{k=1}^{K} \frac{\partial f}{\partial x_{k}} \varepsilon_{k}\right)^{2} \rho\left(\varepsilon_{1}, \ldots, \varepsilon_{K}\right) \mathrm{d} \varepsilon_{1} \cdots \mathrm{d} \varepsilon_{K}=\left\langle\sum_{k=1}^{K} \frac{\partial f}{\partial x_{k}} \varepsilon_{k} \cdot \sum_{l=1}^{K} \frac{\partial f}{\partial x_{l}} \varepsilon_{l}\right\rangle=\sum_{k, l=1}^{K} \frac{\partial f}{\partial x_{k}} \frac{\partial f}{\partial x_{l}}\left\langle\varepsilon_{k} \varepsilon_{l}\right\rangle$$

$\$ \$$\mathrm{~ V e f t ~ \ l a n g l e \ v a r e p s i l o n _ { k } ~}$$
\left\langle\varepsilon_{k}\right\rangle\left\langle\varepsilon_{l}\right\rangle \quad \text { for } k \neq l, \sigma_{k}^{2} \quad \text { for } k=l .
$$正确的。$$
\text { Here, } \$\left\langle\varepsilon_{k}\right\rangle=0 \$ \text { holds forall } \$k \$(\text { and } \$1 \$) \text {. There fore, thelawo ferrorpropagation follows : }
$$\backslash sigma^ {2}=\backslash sum_{ {k=1}^{\wedge}{K} \backslash eft \left(\backslash\right. frac {\backslash partial f } \wedge partial \left.x_{-}{k}\right} \backslash right )^{\wedge}{2} \backslash sigma_{k}^{{2} 。 \ \$$

## 物理代写|理论力学代写theoretical mechanics代考|Finite Measurement Series and Their Average Error

$$\langle x\rangle \approx \bar{x} \equiv \frac{1}{N} \sum_{n=1}^{N} x_{n}$$

$$(\Delta \bar{x})^{2}=N \cdot N^{-2}(\Delta x)^{2}=\frac{(\Delta x)^{2}}{N}$$

$$(\Delta x)^{2}=\overline{\left(x-x_{0}\right)^{2}}=\overline{\left(x-\bar{x}+\bar{x}-x_{0}\right)^{2}}=\overline{(x-\bar{x})^{2}}+2 \overline{(x-\bar{x})}\left(\bar{x}-x_{0}\right)+\left(\bar{x}-x_{0}\right)^{2} .$$

$$(\Delta x)^{2}=\frac{N}{N-1} \overline{(x-\bar{x})^{2}}$$

## 物理代写|理论力学代写theoretical mechanics代考|Error Analysis

$$\rho_{n}=\frac{N_{n}}{\sum_{k} N_{k}}=\frac{1}{\left(\Delta x_{n}\right)^{2}} / \sum_{k} \frac{1}{\left(\Delta x_{k}\right)^{2}},$$

$$\sigma^{2}=\sum_{n} \rho_{n}^{2} \sigma_{n}^{2}=\frac{1}{\left(\sum_{k}\left(\Delta x_{k}\right)^{-2}\right)^{2}} \sum_{n} \frac{1}{\left(\Delta x_{n}\right)^{4}}\left(\Delta x_{n}\right)^{2}=\frac{\sum_{n}\left(\Delta x_{n}\right)^{-2}}{\left(\sum_{k}\left(\Delta x_{k}\right)^{-2}\right)^{2}} \quad=\frac{1}{\sum_{n}\left(\Delta x_{n}\right)^{-2}} \quad \Longrightarrow \quad \frac{1}{(\Delta x)^{2}}=\sum_{n} \frac{1}{\left(\Delta x_{n}\right)^{2}}$$

$$(\Delta x)^{2}=\frac{1}{N-1} \frac{1}{\sum_{n}\left(\Delta x_{n}\right)^{-2}} \sum_{n} \frac{\left(x_{n}-\bar{x}\right)^{2}}{\left(\Delta x_{n}\right)^{2}}$$

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