# Tag Info

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 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 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 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 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} ...
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: $(1-A)^x \approx ... 1 Construct a polynomial (or rational) approximation of your function that gives good accuracy over the entire interval$[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 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 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 ... 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 I'm not sure if this will work or not. If you treat the two known vectors, I call$a$and$b$, as sides of a triangle, then the third side is$b-a$. Once you have found vectors for the three sides of the triangle, you can find their magnitude to get the lengths of the edges and then use the cosine rule to work out the desired angle. 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 time duration you're looking for appears to be the same as the time duration that the 4th signal is LOW. You can use the cursor to see how much time it takes to go from HIGH to LOW on the 4th channel to find your error bounds. So if your cursor moves in .02ms increments, and you find that the signal is HIGH at time$t$and LOW at time$t+.04$, then the ... 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 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 Your notation is a complete mess (I made a correction, but it's still all wrong). You cannot start to trying proving something if you cannot make sense of that something. Try, for example, to work out a numeric example, to get some consistent notation. I'll try. I define$e_X = (x_T- x_A)/x_T$, so$x_A = x_T (1-e_X)$($x_T$is the true value,$x_A$the ... 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 ...