2
$\begingroup$

My question refers to the following equations:

enter image description here

(The summation over $\gamma$ is described here: Summation over a product of binomial coefficients)

When I perform the substitution $\beta=\alpha-\lambda+\mu+\sigma$, write everything as factorials and collect the terms again I arrive at eq. (3.13), except that my summation starts from $\beta=\sigma+\mu-\lambda$.

How do they end up with the sum starting from $\beta=0$?

(Remarks: all variables are integers, $\sigma,\lambda,\mu\geq 0$, $\lambda\geq\mu$)

$\endgroup$
2
$\begingroup$

The factor $$\binom{\lambda + \beta - \mu}{\sigma}$$ is $0$ for $\beta < \sigma + \mu - \lambda$, so the sum is the same, whether you start at $0$ or at $\sigma + \mu - \lambda$. Taking $0$ leads to simpler typography.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.