Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Finding if an integer is odd through Riemann or some function?

  1. Sep 17, 2012 #1
    I'm trying to find if a number is odd or not, basically if X % 2 = 1.

    Can this be expressed through some function? Like the sum of 1 + 2 .. +n is n(n+1) /2

    Or as a Riemann sum?

    I'm trying to add only the odd numbers from a random set of N integers to a sum.
  2. jcsd
  3. Sep 17, 2012 #2
    Do you mean in some programming language? If '/' is integer division, then x is even if and only if

    x = 2 * (x/2). [using '=' as equality, not assignment]

    But if you already have a mod operator like % then you can just use that.
  4. Sep 17, 2012 #3
    i'm not entirely sure what you're getting at either, but
    2*n+1 is always an odd number, for any value of n.
  5. Sep 17, 2012 #4
    ohh okay i think i see what you're saying, you have a set of random numbers, and you only want to add the odd ones.
    yes the modulus function works well.
    if(x%2 == 1){
    alternatively, you can test the last bit to see if it's 1.
    if(x&1 == 1){
    this is slightly faster.
  6. Sep 18, 2012 #5
    Not what I meant but I did word this problem poorly. It's actually probably a bad and weird question anyway. I appreciate the help though

    For example
    Let's say the set of integers is from some function( not random anymore)
    [itex]\sum \limits_{i=1}^n A (A\mod{2})[/itex] is only adding odd terms but the equation is not reducible, so I was hoping the % could be expressed differently so I could solve or reduce the problem. If finding the number was odd or mod or % 2 was rewritable as a function or riemann sum, then I could reduce or solve the problem.
  7. Sep 18, 2012 #6


    User Avatar
    Science Advisor

    I fail to see how Riemann sums would be relevant in what appears to be a discrete math question.
  8. Sep 18, 2012 #7


    User Avatar
    Gold Member

    I think he wants do something like the trick with minus signs alternating by using n for an exponent on -1. Just a guess here.
  9. Sep 18, 2012 #8


    User Avatar
    Science Advisor

    So something like
    ##\frac{1 - (-1)^n}{2}##
  10. Sep 18, 2012 #9


    User Avatar
    Gold Member

    adding all the odds would look like this:

    [itex]\sum n(\frac{1 - (-1)^n}{2})[/itex]

    where n is a positive integer
  11. Sep 18, 2012 #10
    Ya that works, thanks guys. Sorry about the poor wording. I used that equation daily in my fourier series class, I'm surprised I forgot about that after one summer.
  12. Sep 18, 2012 #11

    I don't understand the problem here. All odd numbers equal 1 mod 2, so adding n odd numbers will give you the result in mod 2 the following:

    [itex]\sum \limits_{i=1}^n A\mod{2}) = n [/itex] so n odd numbers will give an even number if n is even and an odd number if n is odd. There is no need to use more complex functions.
  13. Sep 18, 2012 #12


    User Avatar
    Science Advisor

    He doesn't want to count the number of odd A, but to sum only the odds. So if
    ##A = (1,2,3,5,7,8,10)##, he wants an expression which evaluates 1+3+5+7.
  14. Sep 18, 2012 #13

    Oh I see. Well then how about

    x = [itex]\displaystyle \sum_{k=1}^n a_ksin^2(a_k\pi) [/itex]

    If ak is even, sin(ak[itex]\pi[/itex]) is 0; and if it's odd, sin(ak[itex]\pi[/itex]) is +/- 1.
  15. Sep 25, 2012 #14
    Hmm, this looks neat, so you are going to add all the odd terms in f(x)? So you need, as a function, [itex]\displaystyle \sum_{x=a}^n f(x)\frac{1-(-1)^{f(x)}}{2}[/itex]. That looks like a really cool and useful function! I have literally filled notebooks finding sums like that, so I have a huge collection, but I've never used a power of f(x) for anything other than f(x)=c or f(x)=x. Do you mind if I play with this? The most that I can simplify this too, without any extra exploration is:
    [itex]\displaystyle \sum_{x=a}^n f(x)\frac{1-(-1)^{f(x)}}{2} = \sum_{x=a}^n \frac{f(x)}{2}-\sum_{x=a}^n f(x)\frac{(-1)^{f(x)}}{2}[/itex]

    For my own purposes, I will want to look at x=0, specifically. If I want to look at specific parts of (f(x), I will use f(x-c).

    EDIT: I realised after working through this a bit that I forgot to multiply by f(x). The formulas above are corrected (I hope). However, I do have a list of solutions that relate to the correct version (where f(x)=xa for some non-negative integer a). Typically, the first part of that sum would be calculated with the Hurwitz Zeta function and the Riemann Zeta function (though I have a different way that I prefer). The second part, I believe, can be similarly computed. For cases where f(x) is a polynomial, this should not be difficult to compute, especially as it can be expressed as a linear combination of cnxn
    Last edited: Sep 25, 2012
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook