Special Functions: Valid Gauss Formula Parameters

[Ted123]

Homework Statement

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

The series (Gauss Formula)
[PLAIN]http://img830.imageshack.us/img830/2365/gaussz.png
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)

The Attempt at a Solution

Anyone help?

 

[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

 

[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

 

[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

 

[Ted123]

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

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

 

[ardie]

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

 

[ardie]

but the thing is u already mention this, when u wrote Re(c-b-a)>0.
so i don't see what the question is

 

[Ted123]

but the thing is u already mention this, when u wrote Re(c-b-a)>0.
so i don't see what the question is

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

 

[ardie]

well b can start from 1 and a from 1 so c must start from 2. I am assuming that the gamma function cannot take negative arguements. last ones just numbercrunching i suppose

 

[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)?

 

[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

 

[Ted123]

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

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...

 

[ardie]

yes you can still have b=-1 with a=0 and c=0 and you will still not run into any trouble.
you can't have for example c=10 with b = 10 and a = 5
you also can't have c=-10 with b = 10 and a = 5

 

[Ted123]

yes you can still have b=-1 with a=0 and c=0 and you will still not run into any trouble.
you can't have for example c=10 with b = 10 and a = 5
you also can't have c=-10 with b = 10 and a = 5

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...

 

[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

 

[Ted123]

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

So it looks like as long as c&gt;a+b it will be valid?

 

[ardie]

yes... and that c is a positive integer

 

[Ted123]

yes... and that c is a positive integer

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)}

 

[ardie]

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

 

[Ted123]

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

Still struggling with evaluating either side...

 

[ardie]

calculate the LHS and confirm it is equal to the RHS?

 

[Ted123]

calculate the LHS and confirm it is equal to the RHS?

How do I calculate the LHS though?

 

[ardie]

ok I am not entirely sure about the notation used in your book, but in my reference books, the a,b,c's on that LHS with the subscripts are pretty hefty recurance relations which I cannot remember off by heart. So you basically just write the numbers in and keep putting new n's. since you have a n! in ur denomenator, youll find that the series converges pretty quickly, maybe after 4-5 terms, then you just calculate and they should be the same