For instance, when a=3, it equals


4x^3 + 18x^2 + 22x + 6
----------------------------------------------
x^4 + 6x^3 + 11x^2 + 6x

The numerator is the derivative of the denominator. I'm very confused here, /sci/.

>>7972770
It follows from [math]\displaystyle \frac{d}{dx} \log(f(x)) = \frac{f'(x)}{f(x)}
[/math] since [math]\displaystyle \sum_{n=0}^a \frac{1}{x+n} = \frac{d}{dx} \log \prod_{n=0}^a (x+n)
[/math].

>>7972817
Then why does the sum turn into a product function in the numerator if f(x) stays the same the whole way through? I get it logically in the denominator, but not the numerator

>>7972836
>>7972817
Okay wait nvm. I get why it's the derivative now. But why does [math]\displaystyle \sum_{n=0}^a \frac{1}{x+n} = \frac{d}{dx} \log \prod_{n=0}^a (x+n)
[/math]. ?
Where does the log come from? Is that just an identity I need to know?

>>7972856
Use rules of logs to convert log of product to sum of logs, then take the derivative.