CSCI 2951 - G : Computational Protein Folding Fall 2012 Lectures 3 and 4 — 09 / 12 / 2012 and 9 / 17 / 2012

We will start with a review of some calculus. We start with one dimension to give intuition for higher dimensions. Given a function from one variable to one variable, for example, f(x) = x, we can approximate it via its derivatives. The most trivial approximation is without derivatives: we can approximate f(x) near c by just the constant f(c). But if f is rapidly changing (has large derivative) then this estimate will become rapidly bad. Approximating f(x) = x near c = 1 yields the approximation x ≈ 1 near 1. How bad is this approximation? If we evaluate the function at 1.1 then 1.1 = 1.21, and our trivial approximation is off by 0.21. Why does this number make sense? Well, our constant approximation is ignoring the first derivative – the derivative of x is 2x, which is 2 when evaluated at our center, c = 1. Thus if we move 0.1 away from our center and ignore a derivative of 2, we would expect our estimate to be off by roughly 0.2. And indeed, 0.21 is pretty close to 0.2. This suggests a more sophisticated approximation: the constant approximation at 1, plus our difference from 1 multiplied by the derivative at 1. This is the first-order approximation around c: f(x) ≈ f(c) + f ′(c) · (x− c). In our case, for f(x) = x, the approximation around 1 is x ≈ 1 + 2 · (x− 1) = 2x− 1. How accurate is this approximation? Well, for x = 1.1 the approximation yields 1.2 as compared to x = 1.21; namely, we are off by 0.01, which is much better than before. Where does this 0.01 come from? The second derivative. Analogously with what we just saw, where ignoring the 1st derivative gives an error proportional to the first derivative and proportional to the distance from the center c, ignoring the second derivative will yield error that is proportional to the second derivative and proportional to the distance from the center squared. The second-order approximation near c, analogously, will be f(x) ≈ f(c)+f ′(c)·(x−c)+ 1 2f ′′(c)·(x−c)2. For the case of f(x) = x the second derivative is 2, and thus the approximation around 1 is 1+2(x−1)+(x−1) which is exactly x. This makes sense since any errors in this approximation would come from the 3rd derivative, which is 0 everywhere for f(x) = x. (If you were wondering why we scale f ′′ by a half, this equality justifies why it cannot be any other way; in general, the nth derivative is scaled by n!, because the nth derivative corresponds to a term involving (x− c), and its nth derivative is n!, which we need to cancel out.) Some of the approximations of this form which are particularly useful are that x near 1 is roughly 1 + a(x− 1), e near 0 is roughly 1 + x (and log x near 1 is roughly x− 1), sin(x) near 0 is roughly x, and cos(x) near 0 is roughly 1− x/2. If any of the above is unclear, draw diagrams with Matlab.