Floor and Mod Functions Discontinuous
Date: 10/10/2001 at 08:52:30 From: Alkhaled Subject: The derivative of --> a MOD b Dear Dr. Math, I'll be more than greatful, having an answer to the following: What is the derivative of ---> a MOD b ? To my understanding --> (a MOD b) = a - [floor(a/b) * b] even in this case, i have no idea what is the derivative of: [floor(a/b)] Can [a MOD b] be represented in basic maths? i.e. using (+,-,/,*) or something that can be differentiated? Many thanks.
Date: 10/10/2001 at 12:56:05 From: Doctor Peterson Subject: Re: The derivative of --> a MOD b Hi, Alkhaled. You haven't said whether you want the derivative with respect to a or b; I'll assume you mean f(x) = x mod b for some fixed b. It would be possible to treat both a and b as variables and determine the partial derivatives of f(x,y) = x mod y but I'll leave that for later if you need that. The mod function, like any function whose definition involves conversion from real numbers to integers (the floor function, in this case), is discontinuous. If you graph it, it will be a sawtooth, rising from 0 to b, then dropping back to zero (instantaneously) and rising again. If you picture it right, you will immediately see what its derivative has to be: 1 everywhere except at multiples of b, where it is undefined. The floor function is a step function, rising vertically at each integer and then remaining constant until the next. Therefore, its derivative is zero, but undefined at integers. If you then apply the chain rule, you see that d/dx[x mod b] = d/dx[x - b floor(x/b)] = d/dx[x] - b d/dx[floor(x/b)] = 1 - b * 0 = 1 for values of x except multiples of b. The fact that floor and mod are discontinuous is the reason there is no simple expression involving the (continuous) basic operations that can yield either of them. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/
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