2
$\begingroup$

I have difficulties in writing some equations in summation/product notation.

I want to write this in summation notation. $$p_1p_2+p_1p_3+...+p_1p_n+p_2p_3+p_2p_4+...+p_2p_n+...+p_{n-1}p_n$$

is it $$\sum_{i=1}^{n-1}\sum_{j=i+1}^{n}p_ip_j \quad ?$$

or

$$\sum_{i=1}^{n-1}\sum_{j=1}^{n}p_ip_{i+j} \quad ?$$

or both wrong?

I want to write this in product notation. $$(1-p_1p_2)(1-p_1p_3)...(1-p_1p_n)(1-p_2p_3)(1-p_2p_4)...(1-p_2p_n)...(1-p_{n-1}p_n)$$

is it

$$\prod_{i=1}^{n-1}\prod_{j=i+1}^{n}(1-p_ip_j) \quad ?$$

or

$$\prod_{i=1}^{n-1}\prod_{j=1}^{n}(1-p_ip_{i+j}) \quad ?$$

or both wrong?

$\endgroup$
1
  • 1
    $\begingroup$ try with $\sum_{1\le i<j\le n}$ $\endgroup$ – Exodd Apr 2 '17 at 9:39
1
$\begingroup$

The first double sum $$\sum_{i=1}^{n-1}\sum_{j=i+1}^{n}p_ip_j$$ is quite ok.

It is useful to recall the validity of \begin{align*} \sum_{i=1}^{n-1}\sum_{j=i+1}^{n}p_ip_j=\sum_{1\leq i<j\leq n}p_ip_j=\sum_{j=2}^n\sum_{i=1}^{j-1}p_ip_j \end{align*}

The second double sum $$\sum_{i=1}^{n-1}\sum_{j=1}^{n}p_ip_{i+j}$$ is not correct. When looking at the term with index $i=n-1$ and $j=n$ we obtain $$p_{n-1}p_{2n-1}$$ which is not part of $$p_1p_2+p_1p_3+...+p_1p_n+p_2p_3+p_2p_4+...+p_2p_n+...+p_{n-1}p_n$$

We have quite the same situation when looking at the products. Again, it is useful to recall the validity of \begin{align*} \prod_{i=1}^{n-1}\prod_{j=i+1}^{n}(1-p_ip_j)=\prod_{1\leq i<j\leq n}(1-p_ip_j)=\prod_{j=2}^n\prod_{i=1}^{j-1}(1-p_ip_j) \end{align*}

$\endgroup$
2
$\begingroup$

For the sums: your first expression is correct, while your second expression should take the limit of the $j$-sum to be $n-i$.

Likewise the products: the limit of your second expression's $j$-product should be $n-i$.

As Exodd says, it's clearer just to use $$\sum_{1 \leq i < j \leq n}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.