- #1

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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

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- Thread starter EngWiPy
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- #1

- 1,367

- 61

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

- #2

jedishrfu

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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

jedishrfu

Mentor

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

- #5

- 1,367

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

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

- #6

- 16,692

- 8,597

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

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[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

mfb

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- #9

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

- #10

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$$\prod_{k=1}^K (\frac {x_k} {x_k + 1})= exp[log(\prod_{k=1}^K (\frac {x_k} {x_k + 1}))]=exp[\sum_{k=1}^K log(\frac {x_k} {x_k + 1})]$$

peace,

Fred

- #11

mfb

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That's an exponential.

Its argument has a sum.

Its argument has a sum.

- #12

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$$P(x)=\prod_{k=0}^K(x+z_k)$$

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 $K$.

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