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3

The first one is correct from the perspective of interval arithmetic. Meaning that if you are absolutely certain that each volume is within $0.05$ mL of the true value, then you are absolutely certain that the average volume is within $0.05$ mL of the true value as well. This works out for the reason you described. The second one is never correct. Dividing ...


3

It means rounding to $n$ decimal places. Regarding $\pi$, note that $\pi = 3.14159\dots$. Therefore you round the third decimal place to $2$ instead of $1$.


3

There are basically two seemingly reasonable possibilities. The $n$th decimal digit and all preceding digits are correct. Then the error is at most $10^{-n}$. But this is not sufficient. For instance, if the true number is $0.199$, then your error bound must be less than $0.001$ to ensure that the first decimal digit is correct. The $n$th decimal digit is ...


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

As $Harish has already shown, the maximum uncertainty is much smaller than what you have found. Given that you are using the general formula for error propagation (using partial derivatives, etc.) I'm guessing you are not looking for the maximum uncertainty but for the "one-sigma" uncertainty. So your answer should be a smaller than what he found. Also, ...


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

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 ...


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} ...


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 error of the denominator can be estimated using relative error formulas for floating point evaluations as $$ (1-(\cos(x)·(1+δ_1))^3·(1+δ_2))·(1+δ_3) \\ = (1-\cos(x)^3)·(1+δ_3)-\cos(x)^3·(3δ_1+δ_2)+O(δ^2) $$ where $|δ_k|<\mu=2^{-52}$ (the last for double). Thus the relative error of the denominator computation is $$ ...


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

Because, if you don't use radians the derivative of $\sin\theta$ is not $\cos\theta$, and so your formula $dy=\cos\theta\,d\theta$ doesn't hold (it needs a coefficient).


1

Although I realize that this is not a general solution to this problem, it was possible to approximate $c_t$ as function of $x$ using a first order Taylor expansion in this scenario: $c_t(x) = f(\bar{x},y_1...) + \frac{\partial f(\bar{x},y_1...)}{\partial x} (x - \bar{x})$ This can be written to return $x$: $x = (c_t(x) - ...


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

I don't like this problem because the accuracy in bearing given by the GPS depends on may factors and is calculated using also doppler effect on the received signals (so is usually very accurate, as long as the speed is large enough). Moreover, to calculate the variances on positions and bearings it is better to use tools as Kalman filters (which GPSs ...


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

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

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

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 ...



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