fraction in combination...please tell me the calculationby maverick6664 Tags: calculation, combinationplease, fraction 

#1
Dec2405, 07:56 AM

P: 80

Hi,
I'm reading the proof of Rodriguez recurrence formula [tex]P_l(x) = \frac{1}{2^l l!} \frac{d^l}{dx^l} (x^21)^l[/tex] This formula itself isn't a problem. But during the proof I got [tex](12xt+t^2)^{\frac{1}{2}} = \sum_n \left( \begin{array}{c} \frac{1}{2} \\ n \end{array} \right) (2xt)^n(1+t^2)^{(\frac{1}{2})n} [/tex] and wondering what the fraction in [tex]\left( \begin{array}{c} \frac{1}{2} \\ n \end{array} \right)[/tex] means (and that it's negative)... and I don't know the range of [tex]n[/tex] in this summation (maybe 0 to indefinate?). Actually if this fraction is allowed, this formula makes sense. Will anyone show me the definition of this kind of combination? Online reference will be good as well. Thanks in advance! and Merry Christmas! 



#2
Dec2405, 08:24 PM

Sci Advisor
P: 5,942

The sum (n) goes from 0 to infinity. The coefficient you are looking at is the generalization of the binomial coefficient.
The first few terms are 1, 1/2, (1/2)(3/2)/2!, (1/2)(3/2)(5/2)/3!. If you look at a binomial expansion of (a+b)^{c}, you have a=1+t^{2}, b=2xt and c=1/2. 



#3
Dec2405, 10:17 PM

P: 80

Thanks! It helps me a LOT. The keywords are what I needed
I will learn Pochhammer symbol. EDIT: oh...thinking of Taylor expansion, proof is easy, but I've never seen that form of binomial expantion 


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