def rectangles(f, a, b, tol): sum = 0 n = 1 dx = (b - a)/n while abs(sum - original_value) > tol:

Thus, when you add squats to your workout routine, you will only make your daily life easier. You’ll be more accustomed to the sitting to standing motion, you’ll be more efficient when you reach for heavy objects, and you will be less likely to get injured in the process.,Or maybe it wouldn’t have been. Maybe the idea that Covid was something to be eliminated never took root in the West because it couldn’t. Because elites knew they couldn’t sell this idea to working and middle classes, who by now have grown too selfish and indolent and stupid to even understand the basic idea that you should try to crunch a pandemic’s cases right down to zero. Maybe Westerners are too intellectually and morally and socially weak now to have ever accepted such an idea, much less acted towards it. Maybe it would have seemed as absurd as the elites who never even tried broaching such a goal appear to have believed.,To achieve the given tolerance just below 10⁻⁶, we needed n≃145 in the most favourable case. Below, I show you how the numerical integral I(n) evolves with n, when estimated using the midpoint method for which the “effective m” is m=2.,It can be shown that the remainder of a numerical integration algorithm, defined as the difference between the approximated and true values of an integral, can be written as,The remainder tends to zero as 1/n² when n tends to infinity. Thus, if we plot I(n) as a function of 1/n², according to the above formula, we should expect I(n) to follow a linear behaviour. Indeed, here is how I(n) evolves as a function of 1/n²:,Of course, the apparent saving of a factor 145/24≃6 in computing time is, at least partially, lost because of the need to fit the data. However, a straight line fit is, in fact, straightforward, and can be achieved with a very simple algorithm, such that the time spent in doing it is negligible. In fact, while we needed 0.0066 s to compute the integral using the midpoint rule and n=145, the time needed to compute it with a precision better than an order of magnitude was, in fact, 0.0012 s.,Of course, the apparent saving of a factor 145/24≃6 in computing time is, at least partially, lost because of the need to fit the data. However, a straight line fit is, in fact, straightforward, and can be achieved with a very simple algorithm, such that the time spent in doing it is negligible. In fact, while we needed 0.0066 s to compute the integral using the midpoint rule and n=145, the time needed to compute it with a precision better than an order of magnitude was, in fact, 0.0012 s.,Indeed, the midpoint rule is a special case of the so-called Gaussian integration methods, which exploit the symmetries and other properties of the function to be integrated, to make a clever choice of the points in which f(x)=P(x,m), P(x,m) being a polynomial of degree m.,There are many answers to that question. Answers about cynical elites, selfish populations, failed public figures, decrepit attitudes, toxic values. None of them are good answers, though.,Thus, in order to achieve the same precision, it is enough to compute the same integral numerically for n=3, 5, 7 and 9, then fit the data against a straight line: its intercept coincides with the estimation of the integral for n=∞. The remainder, in this case, is 1.5⨉10⁻⁷, one order of magnitude better than the one obtained with n=145!,That, anyway, is what the East sees. Easterners look at the West today, like much of the rest of the world, and wonder, baffled: “why didn’t they try to eliminate Covid? Why don’t they still try? Why do they keep on playing nice with it, hoping to reduce cases by a few percent this week, here or there — instead of wiping them out? Why don’t they aim for zero cases, forever, like any sane society did, as we did?”,The integral will thus be given by p[1]. Its remainder can be statistically estimated as the square root of the corresponding element of the covariance matrix cov[1][1], as,In the end, evaluating f(x) just N=3 5 7 9=24 times, we obtained a better precision with respect to the case n=145, which requires to evaluate f(x) 145 times. Not bad, isn’t it?,The midpoint method consists of interpolating a curve in an interval [a,a h] using a constant corresponding to f(ξ), where ξ is the midpoint of the interval. It is equivalent to the trapezoidal method, yet its error is lower if the function is monotone in the interval [figure made by myself and published on “Scientific Programming”, ed. World Scientific],In fact, there is a trick. The trick consists of the fact that I used a special version of the rectangle rule, known as the midpoint rule. In this case, the height of the rectangle, used to approximate the function, is not f(a), nor f(b), but f((a b)/2). You will have no difficulty in showing that the area of a rectangle whose height is the average of the lengths of the bases of a trapezoid is equivalent to the area of the latter. The two methods are, in fact, completely equivalent. The difference in n is mainly due to the specific properties of f(x) (in particular, to the fact that f(x) is monotonically decreasing).