# Special Functions

1. Jun 27, 2011

### Ted123

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. Jun 27, 2011

### ardie

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
3. Jun 27, 2011

### Ted123

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

4. Jun 27, 2011

### ardie

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

5. Jun 27, 2011

### Ted123

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

6. Jun 27, 2011

### ardie

yes c has to be positive and bigger than a+b, and z must equal 1

7. Jun 27, 2011

### ardie

but the thing is u already mention this, when u wrote Re(c-b-a)>0.
so i dont see what the question is

8. Jun 27, 2011

### Ted123

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
9. Jun 27, 2011

### ardie

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

10. Jun 27, 2011

### Ted123

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
11. Jun 27, 2011

### ardie

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

12. Jun 27, 2011

### Ted123

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
13. Jun 27, 2011

### ardie

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

14. Jun 27, 2011

### Ted123

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...

15. Jun 27, 2011

### ardie

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

16. Jun 27, 2011

### Ted123

So it looks like as long as $$c>a+b$$ it will be valid?

17. Jun 27, 2011

### ardie

yes... and that c is a positive integer

18. Jun 27, 2011

### Ted123

So for part (c), do I have to show that

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

If so, how?

The definition of the shifted factorial (Pochhammer symbol) is:
$$(a)_n = \frac{\Gamma(a+n)}{\Gamma(a)}$$

Last edited: Jun 27, 2011
19. Jun 27, 2011

### ardie

i believe if u do the summation in the power series you will find by induction that both represent the same rational number

20. Jun 27, 2011

### Ted123

Still struggling with evaluating either side...