# Product to sum

• I
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!!

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

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
Mentor
Next question is it ##x_k## or ##x^k## ?

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

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

PeroK
Homework Helper
Gold Member
2020 Award
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.

jedishrfu
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
Mentor
If you have no further information about xk 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.

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

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

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 $K$.