The expression involving summation and factorials, e^{-(\lambda + \mu)}\sum_{k=0}^w \frac{\lambda^k \mu^{(w-k)}}{k!(w-k)!}, simplifies to e^{-(\lambda + \mu)}\frac{(\lambda + \mu)^w}{w!}. This simplification resembles a binomial expansion, confirming that the sum can be interpreted as the expansion of (λ + μ)^w. The final result effectively combines the exponential decay factor with the binomial coefficient.
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This is already pretty simplified. Do you mean that you want to expand it?Polymath89 said:I need to simplify this expression and I don't know how to deal with the factorials in the sum
e^{-(\lambda + \mu)}\sum_{k=0}^w \frac{\lambda^k \mu^{(w-k)}}{k!(w-k)!}
Can anybody give me a hint on how to sum over the factorials?
Thank you very much.mathman said:The sum looks almost like a binomial expansion.
net result e-(λ+μ)(λ+μ)w/w!.