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Special Functions

  1. Jun 27, 2011 #1
    1. The problem statement, all variables and given/known data

    It is known that Euler's integral representation
    [PLAIN]http://img12.imageshack.us/img12/5578/euler.png [Broken]
    is valid for Re(c)>Re(b)>0 and |z|<1.

    The series (Gauss Formula)
    [PLAIN]http://img830.imageshack.us/img830/2365/gaussz.png [Broken]
    on the other hand converges for Re(c-b-a)>0.

    For what values of the parameters a, b and c is the Gauss Formula valid? (Think carefully)

    3. The attempt at a solution

    Anyone help?
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Jun 27, 2011 #2
    c must be bigger than a+b ? sorry i mean x must equal 1 for the integral to become the second gamma expression on the nomenator
     
    Last edited: Jun 27, 2011
  4. Jun 27, 2011 #3
    Where's an x? Do you mean z? I'm not asked to derive the formula here - just to explain what values of a,b,c it is valid for. I was thinking c cannot be a non-negative integer
     
  5. Jun 27, 2011 #4
    im sorry ,z must equal 1. the integral is a beta function and is equal to the gamma (the second in the nomenator) of the second function when z = 1
     
  6. Jun 27, 2011 #5
    I'm not asked to derive the formula here - just to explain what values of a,b,c it is valid for. I was thinking c cannot be a non-negative integer
     
  7. Jun 27, 2011 #6
    yes c has to be positive and bigger than a+b, and z must equal 1
     
  8. Jun 27, 2011 #7
    but the thing is u already mention this, when u wrote Re(c-b-a)>0.
    so i dont see what the question is
     
  9. Jun 27, 2011 #8
    The full question is this: (I've done part (a) so far which is deriving it...)
    [PLAIN]http://img7.imageshack.us/img7/7766/fullqs.png [Broken]
     
    Last edited by a moderator: May 5, 2017
  10. Jun 27, 2011 #9
    well b can start from 1 and a from 1 so c must start from 2. im assuming that the gamma function cannot take negative arguements. last ones just numbercrunching i suppose
     
  11. Jun 27, 2011 #10
    Do a,b and c have to be integers? And why does a+b have to be less than c?

    How do I compute the sum of the power series with b=-1 for part (c)?
     
    Last edited: Jun 27, 2011
  12. Jun 27, 2011 #11
    the gamma functions are evaluated using the gauss integral method such that:
    gamme (n) = factorial (n-1)
    you want the factorial to have integer values as input otherwise it becomes really sad and starts to cry
    just plug in the numbers to get those values.
    factorial (-1) is a mathematical nonsense, hence why c-b-a must be positive for the function to be welldefined
     
  13. Jun 27, 2011 #12
    Part (c) says verify it for b=-1 but you've said that b must start from 1 for it to be valid in part (b)? a,b and c can be complex numbers...
     
    Last edited: Jun 27, 2011
  14. Jun 27, 2011 #13
    yes you can still have b=-1 with a=0 and c=0 and you will still not run into any trouble.
    you cant have for example c=10 with b = 10 and a = 5
    you also cant have c=-10 with b = 10 and a = 5
     
  15. Jun 27, 2011 #14
    But I thought we established for part (b) that the formula was only valid for a and b starting from 1 and c>2? This shows it can be valid for other values too...
     
  16. Jun 27, 2011 #15
    yes i think i made a mistake there again, but trying to minimise c, you can still have c=0 and then move upwards fitting all possible a and b values in
     
  17. Jun 27, 2011 #16
    So it looks like as long as [tex]c>a+b[/tex] it will be valid?
     
  18. Jun 27, 2011 #17
    yes... and that c is a positive integer
     
  19. Jun 27, 2011 #18
    So for part (c), do I have to show that

    [tex]\sum_{n=0}^{\infty} \frac{(a)_n(-1)_n}{(c)_n n!} = \frac{\Gamma(c)\Gamma(c-a+1)}{\Gamma(c-a)\Gamma(c+1)}\;?[/tex]

    If so, how?

    The definition of the shifted factorial (Pochhammer symbol) is:
    [tex](a)_n = \frac{\Gamma(a+n)}{\Gamma(a)}[/tex]
     
    Last edited: Jun 27, 2011
  20. Jun 27, 2011 #19
    i believe if u do the summation in the power series you will find by induction that both represent the same rational number
     
  21. Jun 27, 2011 #20
    Still struggling with evaluating either side...
     
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