Calculus of Piecewise FunctionsDate: 06/07/2003 at 21:09:28 From: Matt Subject: Calculus of the modulus function The modulus function (absolute value function) is a piecewise function that is customarily written as f(x)=|x|. Functions such as the floor and ceiling functions are also piecewise, but they are written concisely with such notations as g(x)=[x]. Most of the time, with some exceptions, the integrals and derivatives of piecewise functions are piecewise themselves. Take for example INT(|x|)dx. Its integral is piecewise, i.e. x^2/2 + C for x=>0 and -x^2/2 + C for x<0. This can be written |x^2/2| + C for all x. But what if the expression within the absolute value signs is more complicated than just x? What if the integral is INT(|u|)u'dx? Then the arbitrary C's are different for each case of the modulus, and there would seem to be no way to consolidate the two cases into one expression using case notation. I guess my real question is, can I take the integral or derivative of a piecewise function like the floor function [u] or the absolute value function |u| and still somehow notate it in concise form, such as |U| or [U]? Date: 06/07/2003 at 23:17:42 From: Doctor Peterson Subject: Re: Calculus of the modulus function Hi, Matt. First (and this will lead into what I want to say anyway), your answer for INT(|x| dx) is wrong. As you say, it is x^2/2 + C for x >= 0 -x^2/2 + C for x < 0 But that is not the same as |x^2| + C, since this is always positive, and the correct antiderivative is not! So let's introduce the function that helps us deal with this more effectively, the signum, or sign, function sgn(x) = 1 if x > 0 0 if x = 0 -1 if x < 0 Using this, we can write your antiderivative as sgn(x) * x^2/2 + C Further, what did you do with the two different C's in your antiderivative? It looks as if you ignored the distinction, but it worked out okay. What you have to do is to find the connection between them, which is based on the fact that the antiderivative has to be continuous. You find the C in one piece that makes it match up with the one next to it. In our example, we really have x^2/2 + C1 for x >= 0 -x^2/2 + C2 for x < 0 and to be continuous, we need 0^2/2 + C1 = -0^2/2 + C2 In this case, this means that C1=C2 and we can just call it C. In harder cases, you might find that, say, C2 = C1+1, and you would have to use C+1 in the second piece. Now I'll demonstrate a way to handle absolute values in integrals that can be a little risky sometimes if you aren't careful, but can get you more directly to an answer. Let's write the absolute value function in terms of signum: |x| = sgn(x) * x Do you see why this is true? And do you see that, since signum is constant within each piece of its definition, its derivative is zero and it commutes with the integral sign? Then INT(|x| dx) = INT(sgn(x) * x dx) = sgn(x) INT(x dx) = sgn(x) x^2/2 + C Note, by the way, that this approach also makes it obvious that d/dx |x| = sgn(x), if you ignore x=0. I can't guarantee that this will always work correctly; we're being a little loose with theorems about integration here, and the discontinuity of signum at x=0 in particular makes this potentially illegal. In particular, we are saved by the fact that x^2 is zero when sgn(x) is zero, which makes the result continuous. You have to check things like that before blindly accepting the result, so the best thing to do is to use this method to find a neat answer, use the piecewise approach to get the right answer, and then compare them. If you have any further questions, feel free to write back. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
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