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Product Identity

  1. Nov 14, 2008 #1
    Challenge to find a identity for the product abc, wherein each term contains a triangular variable as a factor and a, b, c are each separately used as the sole variable of the argument in at least one of the variable triangular numbers.
    My solution is

    [Tex]abc = T_{(c+ab)} - aT_{b} - T_{c} - b^{2}T_{(a-1)} [/Tex]

    Can anyone do better?

    Anyone want to try one for the product "abcd"?

    FYI T_(x) = x(x+1)/2;
     
  2. jcsd
  3. Nov 14, 2008 #2
    Well, since T(n+1) - T(n) = n+1, you can always try

    abc = (T(a+1) - T(a) - 1) (T(b+1) - T(b) - 1) (T(c+1) - T(c) - 1)

    and a similar thing for abcd.

    -------- edit:
    Sorry, that was really stupid. Just use T(n) - T(n-1) = n to produce

    abc = (T(a) - T(a-1)) (T(b) - T(b-1)) (T(c) - T(c-1))
     
  4. Nov 15, 2008 #3
    Silly that I didn't think to prevent the writeup of your solution.
    What if I required a triangular number with all variables in the argument as part of the solution in addition to the original requirements?
     
  5. Nov 16, 2008 #4
    So, if I understood you correctly, you used the identity

    T(x+y) = x*y + T(x) + T(y)

    and applied it with x=a*b, y=c. This should give you an idea on how to extend it to abcd.

    Another possibility is to multiply the identity on T(a+b) by c, to obtain

    abc = c T(a+b) - c T(a) - c T(b)
    ---- Edit: Oh, sorry, this one has all variables in one term, but not all variables in the argument. Let be think a bit more about that.
     
    Last edited: Nov 16, 2008
  6. Nov 16, 2008 #5
    How I got this is another issue altogether, but you are on the right track even though I got this by another means. Yes, you can solve for "abcd" using T(ab+cd) also, but with a little more work.
     
  7. Nov 17, 2008 #6
    Here's another funny one, that reminds me of how the rows of the Pascal Triangle add up to a power of 2.

    Take three integers, a, b, c. (The same can be done with two, or four.) Let n be an integer of 3 bits, that is, between 0 and 2^3-1.

    Now define the function f(a,b,c,n) in the following way: each bit position in 'n' represents one of the integers a,b,c. The function f() will return the product of three triangular numbers: if the bits in 'n' are 111, the function returns T(a)*T(b)*T(c), but if the bits in 'n' are 000, f() returns T(a-1)*T(b-1)*T(c-1). And all the intermediate combinations: for example, if the bits in 'n' are 101, f() will return T(a)*T(b-1)*T(c). As you see, f() is defined such that the argument of each triangular number is (a) or (a-1), depending on the 100 bit of 'n'; the next argument is (b) or (b-1), depending on the 010 bit of 'n', and the next argument is (c) or (c-1), depending on the 001 bit of 'n'.

    Now the magic trick: with these definition, the sum of f(a,b,c,n), for three fixed a,b,c integers, and n going from 0 to 2^3-1 (all possible combinations of 3 bits), ... the sum of these f() values is = (a*b*c)^2.

    So the square of any number can be partitioned by products of triangular numbers (just apply the former to the prime factors of the number).
     
    Last edited: Nov 17, 2008
  8. Nov 17, 2008 #7
    But if p is prime then you have just T(p) + T(p-1) = p^2 which if deemed to qualify as a partition into the sum of products of triangular numbers would mean that any square could be partition into the sum of two such products.

    Honestly though, I am very much impressed with this trick

    So now we have two different ways to represent p^2 as a sum of triangular products: say my identity is called G(a,b,c) and your identity is called F(a,b.c) (the n part of your identity is not required since you take the sum of all possible combinations as n goes from 0 to 7).

    Now G(1,p,p) = G(p,1,p) = G(p,p,1) = G(1,1,p^2) = F(1,1,p) = p^2

    Is there a good cite for an explanation of your identity.
     
    Last edited: Nov 17, 2008
  9. Nov 17, 2008 #8
    I don't know - I just made it up during the weekend. But the "minimal case" T(n) + T(n-1) = n^2 is either in Wikipedia or in Mathworld. If you find some, let me know.
     
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