Product to sum

  • I
  • Thread starter EngWiPy
  • Start date
  • #1
1,367
61

Main Question or Discussion Point

Hello,

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

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

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

Thanks in advance
 

Answers and Replies

  • #2
11,485
5,010
Have you tried to combine the fraction to see where that leads?
 
  • #3
1,367
61
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!!
 
  • #4
11,485
5,010
Next question is it ##x_k## or ##x^k## ?
 
  • #5
1,367
61
Next question is it ##x_k## or ##x^k## ?
It is ##x_k##. I have ##K## different variables.
 
  • #6
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
12,319
5,120
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.
 
  • #7
1,367
61
For ##K=3## I have

[tex]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)}[/tex]

I cannot see how this can be written as a closed form summation in the form of ##\sum_{k=1}^K(.)## or something similar.
 
  • #8
34,043
9,891
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.
 
  • #9
1,367
61
I don't have any further information about ##x_k##.
 
  • #10
203
83
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
 
  • #11
34,043
9,891
That's an exponential.
Its argument has a sum.
 
  • #12
2
0
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 $K$.
 

Related Threads on Product to sum

  • Last Post
Replies
3
Views
19K
Replies
2
Views
2K
Replies
3
Views
1K
Replies
4
Views
1K
Replies
16
Views
3K
  • Last Post
Replies
1
Views
2K
  • Last Post
2
Replies
36
Views
14K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
2
Views
911
Top