Finding the Summation Equivalent of a Product

[EngWiPy]

Hello,

I have the following product, and I am looking for a summation equivalent

\prod_{k=1}^K\left(1-\frac{1}{x_k+1}\right)

Is this doable? I tried to use partial fraction but got nowhere!

Thanks in advance

 

[jedishrfu]

Have you tried to combine the fraction to see where that leads?

 

[EngWiPy]

Have you tried to combine the fraction to see where that leads?

I did. The fraction becomes $x_k/(x_k+1)$, but didn't know where to go from there!

 

[jedishrfu]

Next question is it $x_k$ or $x^k$ ?

 

[EngWiPy]

Next question is it $x_k$ or $x^k$ ?

It is $x_k$. I have $K$ different variables.

 

[PeroK]

It is $x_k$. I have $K$ different variables.

You could let $y_k = \frac{1}{x_k + 1}$ and then expand. The product is a polynomial with roots $y_k$. The polynomial is the sum.

 

[EngWiPy]

For $K=3$ I have

1-\frac{1}{x_1+1}-\frac{1}{x_2+1}-\frac{1}{x_3+1}+\frac{1}{(x_1+1)(x_3+1)}+\frac{1}{(x_2+1)(x_3+1)}-\frac{1}{(x_1+1)(x_2+1)(x_3+1)}

I cannot see how this can be written as a closed form summation in the form of $\sum_{k=1}^K(.)$ or something similar.

 

[mfb]

If you have no further information about x_k then there won't be a useful way to write it as sum. It is like asking to convert $$\prod_{k=1}^K z_k$$ to a sum. While technically possible you don't learn anything from it.

 

[EngWiPy]

I don't have any further information about $x_k$.

 

[Fred Wright]

If you want a sum here is a way:
$$\prod_{k=1}^K (\frac {x_k} {x_k + 1})= exp[log(\prod_{k=1}^K (\frac {x_k} {x_k + 1}))]=exp[\sum_{k=1}^K log(\frac {x_k} {x_k + 1})]$$

peace,
Fred

 

[mfb]

That's an exponential.
Its argument has a sum.

 

[NickP]

First, consider the polynomial
$$P(x)=\prod_{k=0}^K(x+z_k)$$
Define the nth elementry symmetric function $e_n(Z)$ for the set of roots $Z=\{z_k\}_{k=0}^K$ to be
$$e_n(Z)=\sum_{1\leq i_1<i_2<\cdots<i_n\leq K} z_{i_1}z_{i_2}\cdots z_{i_n}$$
(Think of this as a sum of all possible -product- combinations of $n$ distinct elements from the set $X$.) As a personal preference, I will write $e_n(Z)=e_Z^n$. Confirm that
$$P(x)=\prod_{k=0}^K(x+z_k)=\sum_{k=0}^K e_Z^{K-k}x^k$$
Thus, we find that your product is the subcase for the root set $z_k=\frac{-1}{x_k+1}$ evaluated at $P(1)$.\\

For the more general product: given sets $X$ and $Z$ as defined above,
$$\prod_{k=0}^K(x_k+y_k)=e_X^K\ast e_Z^K$$
where $\ast$ is the discrete convolution with respect to $K$.