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3

The first assertion assumes one takes mean squared errors, which in probabilistic terms translates into standard deviations. Now, probability says that the variance of two independent variables is the sum of the variances. Hence, if $z = x + y$ , $\sigma_z^2 = \sigma_x^2 + \sigma_y^2 $ and $$e_z = \sigma_z = \sqrt{\sigma_x^2 + \sigma_y^2} = ...


2

The arithmetic mean is just a scaled version of the sum, so you just need to know that the error scales as the quantity itself under scaling; thus the error in the arithmetic mean is $\sqrt{e_1^2+e_2^2}/2$. (You need to enclose the argument of the root in curly braces instead of parentheses to have it displayed under the square root.) For more general error ...


2

Because of relative in "relative accuracy". The relative error is the absolute error divided by the magnitude of the exact value. See here.


2

It represents random error. Random error is a bit of 'spreading out', but systematic error means the data is centered around the wrong spot. Put another way: Say your wife asks you your anniversary. If you guess the date slightly wrong, that's random error. You know roughly what's going on, but the data isn't exact. If you give the correct date of her ...


2

The simple thing to do is to say $(1-A)^x \approx 1-xA$ for $A \ll 1$. This leads to $f(x)=x$ as you say. You can test $A$ and change to $f(x)$ when it is small. The problem this leads to is you have a jump in $f(x)$ as you cross the transition. Maybe this is a problem, maybe not. You can use more terms of the Taylor series: $(1-A)^x \approx ...


2

If the random variables $X_1,\ldots,X_n$ are independent and normally distributed with mean $\mu_i$ and variance $\sigma_i^2$ (i.e., standard deviation $\sigma_i$), then the sum $$ S_n = X_1 + \ldots + X_n $$ is normally distributed with mean $\mu = \mu_1 + \ldots + \mu_n$ and variance $\sigma^2 := \sigma_1^2 + \ldots + \sigma_n^2$. Thus, the average $$ ...


2

Firstly, you should understand fractional uncertainty $= \frac{\Delta x}x$ and percentage uncertainty $= \frac{\Delta x}x \times 100$ Also be careful where you are given a length with its uncertainty and required to calculate, for example, density of a cube. The Volume is the cube of the length, and so the power rule applies to the uncertainty of the ...


2

I think you missed reading the well-known literatures such as "T. Soderstrom, G.W. Stewart, On the numerical properties of an iterative method for computing the Moore-Penrose generalized inverse, SIAM J. Numer. Anal. 11 (1974), 61-74." in this field. According to the above-mentiond paper and under a typical seed with machine precision, basically ALMOST ...


2

Here's a formula from Numerical Analysis by Burden and Faires (chapter 4.1). \begin{align*} f'(x_0) &= \frac{f(x_0 + h) - f(x_0 - h)}{2h} - \frac{h^2}{6} f^{(3)}(\xi_0). \end{align*} Notice that if the third derivative of $f$ is huge, the error might be huge. There are other formulas for numerically computing derivatives, and they have similar ...


2

The notation gets confusing, so allow me to set $e_x=\epsilon_x$ and $e_y=\epsilon_y$. The trick here is to observe that $x^y = e^{y \log{x}}$ and Taylor expand as follows: $$\begin{align}(x+\epsilon_x)^{y+\epsilon_y} &= e^{(y+\epsilon_y)\log{(x+\epsilon_x)}}\\ &=e^{y \log{(x+\epsilon_x)}} e^{\epsilon_y \log{(x+\epsilon_x)}} \\ &= e^{y \log{x} ...


2

Let's write your stuff in a cleaner way: $$n_\text{avg} = 2\sin(63°) = 1.7820130483767356$$ $$n = n_\text{avg} \pm^{u}_l \ .$$ Then $$u = 2\sin(63.5°) - 2\sin(63°)$$ $$l = 2\sin(63°) - 2\sin(62.5°)$$ The way your friend does it is via first order Taylor approximation: $$\Delta n \approx \left.\frac{dn}{d\theta}\right|_{\theta=\theta_\text{min}} \cdot ...


1

Use a trigonometry identity $$\sin(a\pm b)=\sin a\cos b\pm\sin b \cos a$$ So, you will get $$2\sin\theta=2(\sin63^{\circ}\cos0.5^{\circ}\pm\sin0.5^{\circ}\cos63^{\circ})\approx1.78195\pm0.00792353$$


1

First: You don't necessarily get an uncertainty of $\sin(x)\pm\sin(e)$. Remember that the sine function is periodic. So if you have that $x = 0$ and $e = 2\pi$. Then you would not get an uncertainty of $\sin(0) \pm \sin(2\pi) = 0 \pm 0$. In such a case you would get $0\pm 1$ since on the interval $[-2\pi, 2\pi]$ sine takes all values. Now for another ...


1

The main clock is running all the time so when the robot presses for the first split 0.244 seconds have elapsed and the stopwatch displays 0.24. The clock is still running, it has not stopped and been restarted so when the robot presses for the second split the elapsed time is 0.488 seconds so the clock displays 0.48 when the robot presses for the third ...


1

What you're doing wrong is assuming too much about the error distribution. There are two common ways to look at the error in a measurement: normally distributed and absolute bounds. Normal (or Gaussian) distributions have a bell shape, with a 66% chance of the actual value being within one standard deviation of your measured value and a 95% chance of the ...


1

The author does not really mean that, for example, the errors in measuring the two masses are the same. (But admittedly, that is exactly what the author writes!) What is meant is that we have a bound $\Delta m$ on the absolute value of the error in measuring a mass, and similarly a bound on the absolute value of the error in the density. The errors could ...


1

The relation $$ (A+\delta A)(x+\delta x)=b, $$ implies that $$ Ax+\delta A (x+\delta x)+A\delta x=b, $$ and since $Ax=b$, the above becomes $$ \delta A (x+\delta x)+A\delta x=0, $$ and hence $$ \delta x= -A^{-1}\delta A (x+\delta x), $$ which implies that $$ \|\delta x\|\le \|A^{-1}\|\|\delta A\|\|x+\delta x\|=\kappa(A)\cdot\frac{\|\delta ...


1

Here are the general ideas: 1) Treat errors coming from different sources as independent with mean 0. Moral: When two errors add together, the covariance matrices merely add up. The least square procedure in your case merely returns the average of 4 observations, so the final matrix $A=\frac 1{16}\sum_i A_i$ and each $A_i$ is the sum of the covariance ...


1

A suggestion: when dealing with error propagation, try always to reduce the problem as much as possible. It's terribly easy to make a mistake when dealing with those little numbers. Now, note that for positive $\cos t$ you have: $$\sin t= \tan t \cos t=\frac{\tan t}{\sqrt{1+\tan ^2t}}.$$ Substituting $t=\arctan \xi$, you get:$$ \sin (\arctan ...


1

I would use the distance between vertices as a metric, ie, AC+DE. It will be zero only if the two are coincident.


1

There are several methods to approach the problem of adjusting estimated regression coefficients for rounding errors in the data. Most of these techniques perform an adjustment to the main diagonal of the sample covariance matrix of the variables. One of the most commonly used method is the so-called Sheppard's correction. This is based on a Taylor expansion ...


1

1) This is an EXACT formula, so be careful with truncating infinite sums! It does not make sense to study it for "small" $T$ values. Keep in mind that Riemann proved it working basically with Fourier analysis (read the first chapter of the Edward's book), so what you want to do is something similar to stop a Fourier series of some stepwise function to the ...


1

If you have just one error, $\dfrac{\partial f}{\partial x_{i}}\sigma_{x_{i}}$ gives the error in $f$. It essentially comes from the Taylor series, linearizing the value of $f$ around the correct value. If you have a number of errors, we think of them as uncorrelated, so it is a random walk. The root sum square is the expected distance of a random walk.


1

As you say, this is essentially truncating Taylor's series after the linear term(s). By Lagrange's form of the remainder in one variable (see Wikipedia or your favorite calculus textbook): $$ f(x) = f(x_0) + f'(x_0) (x - x_0) + \frac{f''(\xi)}{2!}(x - x_0)^2 $$ where $x_0 \le \xi \le x$. The approximation you cite is valid as long as the second-order term ...


1

Construct a polynomial (or rational) approximation of your function that gives good accuracy over the entire interval $[0,1]$, and evaluate this approximation instead of evaluating your original function. Constructing a good approximation is a fair bit of work, so it may not be worthwhile for you. The best approach is Chebyshev approximation. You should be ...


1

$$a = \frac{T_\text{prop}}{T_\text{frame}}=\dfrac{20 \times 10^{-3}}{\tfrac{L}{4 \times 10^3}} = \dfrac{20 \times 10^{-3} \; \times \; 4 \times 10^3}{L} = \dfrac{80}{L}$$ so $$U \ge 0.5 \iff a = \frac{1}{2U} - \frac{1}{2} \le 0.5 \iff L=\frac{80}{a} \ge 160.$$


1

Look at the simplest functions $x(t)$ and compute the exact expressions $v(h,t) = \frac{x(t+h)-x(t)}{h}$. For $x(t) = at$ you have $v(h,t) = \frac{x(t+h)-x(t)}{h} = a$ and therefore the error term $O(h)$ is zero. For $x(t) = at^2$ you have $v(h,t) = 2 at+ah$ and $O(h)=ah.\;$ Thus the error is constant in time, it only depends on $a,h.$ For $x(t) = at^3$ ...


1

More generally, if $x$ is twice continuously differentiable, there exists a $\xi\in [0,1]$ for which $\frac{x(t+h)-x(t)}{h}=\frac{h}{2!}x''(t+\xi h)$, so that you need to estimate the second derivative: $v_1 = 10 \pm \frac{0.2}{2} v''_{max}$


1

$V(r,l)=\pi r^2l$, so $$\begin{align} V(r+\Delta r,l+\Delta l)&\approx V(r,l)+\frac{\partial V}{\partial r}\Delta r+\frac{\partial V}{\partial l}\Delta l\\ &\approx V(r,l)+2\pi r l\Delta r+\pi r^2\Delta l \end{align}$$ So error in $V$ is $2\pi r l\Delta r+\pi r^2\Delta l$. Note this does not factor in statistics/probability. If the given ...


1

By definition relative error is given by $\delta f / f$ so f here is sin30 and the numerator is the difference you have written in your question. To calculate percentage error just multiply relative error by 100.



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