1
$\begingroup$

I have two products I hope to compare symbolically and I am hoping to use Mathematica for this. They are as follows: $$ \prod_{1 \leq i < j \leq n} {(a_i + \cdots + a_{j-1}) + j - i \over j- i} \tag{1} $$ and $$ {\prod_{\ell = 1}^{n-1}\left(\prod_{j = \ell}^{n-1}\left[\sum_{i = 1}^j a_i + j - 1\right]\right) \over (n-1)!\cdot (n - 2)!\cdots 2!\cdot 1!} \tag{2} $$ However, I'm not very familiar with the Mathematica syntax required to do this. Regards

$\endgroup$
  • $\begingroup$ Ponder on Product[(Sum[a[k], {k, i, j - 1}] + j - i)/(j - i), {j, 1, n}, {i, 1, j - 1}], and you might be able to figure how to write what you need. $\endgroup$ – J. M. will be back soon Jun 18 '13 at 1:31
  • $\begingroup$ @0x4A4D I have come up with the code (Product[(Sum[a[k], {k, 1, j}] + j - 1), {l, 1, n - 1}, {j, l, n - 1}])/(Product[(n - i)!, {i, 1, n - 1}]) for the second product, but I am confused about how to implement the first product with the $1 \leq i < j \leq n$. I have the following so far: Product[(Sum[a[k], {k, i, j - 1}] + j - i)/(j - i), {1 <= i < j <= n}] $\endgroup$ – Moderat Jun 18 '13 at 2:54
  • $\begingroup$ Possibly using the Boole function? $\endgroup$ – Moderat Jun 18 '13 at 2:57
  • 1
    $\begingroup$ Well, for the first: did you already ponder upon how I built the snippet in my first comment? $\endgroup$ – J. M. will be back soon Jun 18 '13 at 3:00
  • $\begingroup$ Ah! I just saw how your answer was actually a response to the first product. Many thanks. Does Mathematica have an == expression to test equivalence for two products such as these? $\endgroup$ – Moderat Jun 18 '13 at 3:02
3
$\begingroup$

Your first product can be written as:

Product[(Sum[a[k], {k, i, j - 1}] + j - i)/(j - i), {j, 1, n}, {i, 1, 
  j - 1}]

and your second one:

Product[Sum[a[k], {k, 1, j}] + j - 1, {l, 1, n - 1}, {j, l, n - 1}]/
 Product[i!, {i, 1, n - 1}]
$\endgroup$

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.