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Integration of Modular Arithmetic Functions

  1. Jul 6, 2011 #1
    Hello, I have been searching and can't seem to find anything on the topic of integrating modular arithmetic functions. So far I have created an equation for a function in the form of f(x)=mod(x,a):
    [itex]\int mod (x,a) dx=\frac{a(x-mod (x,a))+mod (x,a)^2}{2}+c[/itex]
    But, now I am investigating how to integrate a modular arithmetic function of the reverse: mod(a,x). Does anybody have any ideas or information on the topic?
     
  2. jcsd
  3. Jul 6, 2011 #2
    How do you define mod for non-integer x?

    One reason there's not much literature on this (besides mod only being defined for integers) is that mod is not a binary operation in math, it's an equivalence relation. It's only in programming languages that we see 5 mod 3 = 2.

    In math we would say that 5 = 2 (mod 3), but we would also say that 5 = -4 (mod 3). So you can only make mod into a function if you say that n mod a is the least positive member of the set of integers congruent to n mod a. In math it's rare to use that formulation.
     
  4. Jul 15, 2011 #3
    Also, integrals only operate on continuous functions.

    mod usually operates on integers or natural numbers. f(n):natural numbers-> real numbers is by definition not a continuous function.

    There are a few ways to extend the modulo operation to the real numbers. The greatest integer function or floor function is one natural way.

    a newMod b =floor(a) ( mod floor(b) )
    But as the floor function is very much like a staircase, it's not particularly useful in this context.
     
  5. Jul 17, 2011 #4
    If you ignore negative a's and x's, then [itex]f(x) = \{^{a-x \; \mbox{if} \; x \le a}_{a \; \mbox{if} \; x > a}[/itex]. The easy way to integrate that, is to just integrate each piece of the piecewise.
     
  6. Dec 30, 2012 #5
    but modular arithmetic can be pseudo continuous if the mod is prime, because every unit has an inverse.
     
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