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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} = \sqrt{e_x^...


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 $1$.


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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 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 ...


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

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 ...


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


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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 ...


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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

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

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 1+x\log(1-...


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

Given that A is symmetric it can be diagonalised. Therefore we can write down the solution of the system of ODEs directly, and simply subtract the discreteized solution front the exact.


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

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

I think of neural networks as a construction kit for functions. The basic building block - called a "neuron" - is usually visualized like this: It gets a variable number of inputx $x_0, x_1, \dots, x_n$, they get multiplied with weights $w_0, w_1, \dots, w_n$, summed and a function $\phi$ is applied to it. The weights is what you want to "fine tune" to ...


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

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 A\|}{\|A\|}\|x+\...


1

Clearly the main task here is to reach a precise mathematical question. So, let us assume that the setting is that one starts from some vector $U_1$ in $\mathbb R^3$, that one first transforms $U_1$ into an intermediate vector $U_2$ in $\mathbb R^3$ using $T_1$ and some noise, thus $$U_2=T_1U_1+W_1$$ where $W_1$ has mean zero and covariance matrix $\Sigma_1$,...


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

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

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 \xi)=\dfrac{\xi}...


1

One usually uses first order differentials for this, so $e[3]\approx \frac d{db}(\frac ba) e[2]+\frac d{da}(\frac ba) e[1]=\frac 1ae[2]-\frac b{a^2}e[1]$ and you want to add, not subtract, the two terms (I didn't include absolute value signs). Added: sometimes it is easier to work in relative error instead of absolute. In this case we get $\frac {e[3]}{\...


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

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

The partial derivatives with respect to $x$ and $y$ play a similar role as the derivative does in the one-variable case. We have $$\frac{\partial f}{\partial x}=\frac{x}{\sqrt{x^2+y^2}} \quad\text{and}\quad \frac{\partial f}{\partial y}=\frac{y}{\sqrt{x^2+y^2}}.$$ Thus the error $\Delta f$ is given approximately by $$\frac{x}{\sqrt{x^2+y^2}} \Delta x +\frac{...


1

The error is typically approximated using the differential; i.e. the first-order term of the Taylor series: $$ \Delta f \approx df(\Delta x, \Delta y) = \frac{\partial f}{\partial x} \Delta x + \frac{\partial f}{\partial y} \Delta y.$$ For your particular $f$ we have $\partial f / \partial x = x/f, \partial f / \partial y = y/f$.


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 ...



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