Determaning a function from a progression?

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christian0710
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Hi,
If i have a progression that looks like thisn | progression | Mathmatical formula
--------------------------------------
1 | x(x-1) .....| x(x-n)
2 | x(x-2)(x-1) .....| x(x-n)(x-n-1)
3 | x(x-3)(x-2)(x-1)...| x(x-n)(x-n-1)(x-n-2))

Every time n increases by one a new term (x-n) is added to the previous terms. How do I turn this into a function of n? I know that as n increases then we get (x-(n-1))*(x-(n-2)) ... (x-(n-(n-1)))) n times but how do you express this as a mathematical function? A function that adds keeps adding n terms (x-n)(x-n-1)(x-n-2)etc... all the way down to the last term (x-n-n-1)

Can anyone help me or give me a hint about how to construct a function of n, such that the above progression holds true for any n we put into the function?

I know that x! = x*x-1*x-2*--*x-x+1

but saying f(n)= (x-n)! does not make sense since we have two variales (x and n) , I'm stuck and would really appreciate a hint :)
 
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Wait I think i might have igured it out - or am i doing the math wrong?

I know that
∏k from 1 to n is defined as ∏k = 1*2*3*4...n

Therefore it must be true that

k(x/k - 1) going from 1 to n is defined as

∏k(x/k-1) =( (1*x)/1 -1) * ((2*x)/2-2) * ((3*x)/3-3) = (x -1)(x-2)(x-3) etc.. p to n :) Is this corret usage of the mathematical multiplication operator ∏?

So the function for this progression is
x*∏k(x/k-1)
 
See the Pochammer symbol for this function and the related closed form thanks to the Gamma function :
x(x-1)(x-2)...(x-n)=Gamma (x+1) / Gamma(x+1-n)