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Equation with factorials

  1. Apr 9, 2008 #1
    hello everyone,
    I usually post my questions on one small czech mathematical forum, but here is an equation noone knows how to "solve". I`ve came to it by accident, when I made an mistake in one combinatorics equation.


    [tex]x! + (x-3)! = 16x - 24[/tex]

    its fairly simple to solve in one way(x has to be greater than 2, and from some point left side is greater than right side, because (x-3)! is always greater then 0 and we don`t have to care about -24 on the right side, so we can check for which x is x! > 16x. this leaves us only very few possibilities for x to check).

    this is nice, but I`d like to know if its possible to get it in form x = something. I can`t think of any way how to do it.

    also, this equation doesn`t have any solution, its rather theoretical question.
     
  2. jcsd
  3. Apr 9, 2008 #2
    If x can be a real number (using Gamma function) you can find 4 solutions betweeen 1 and 4.5 (by plotting method or std numeric method).
    x1=1.1837
    x2=1.3134
    x3=2.1222
    x4=4.4099
    There are also 5 negative solution... None is integer.
     
  4. Apr 9, 2008 #3
    It is simple to check that there is no integer solution. Factorials increase far more rapidly than a linear function, more rapidly than exponentials even. We know x>2. For x=4, the LHS is smaller than the RHS, for x=5 it is larger, thus there is no integer x where they are equal.
     
  5. Apr 16, 2008 #4
    I have also tried this in several different ways, I got:

    x! = ( 8x (x-2) (x-1) (x-1.5) )/( x (x-1) ( x-2) +1 )

    which means no integer values......

    Is this really all????
     
  6. Apr 16, 2008 #5
    Plotting the function (remebering x!=Gamma(x+1))
    G(x)=Gamma(x + 1) + Gamma(x - 2) - 16x + 24
    you can see that for x>0 there are only four zeros.
    For x<0 there are infinite solutions near 0,-1,-2,-3...none exacly integer.
    There Gamma function is singular (has a pole) and change sign.
     
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