Rewrite of function into closed-form

[Big-Daddy]

Attached is a document with a function P(a) which contains several "if" sections, i.e. a different function applies depending on which value the dummy variables take.

Can anyone help me by rewriting this is as a straight, closed-form (sigma operators are fine) function in a? Or explaining why in this particular case it is impossible, if it is? I'd like something I can use to compute values of P(a) from a. In the current form, it cannot be used to compute values directly as far as I know.

My instinct is that you'd just need to separate out the operators for each different section. But really I have no idea. If you could help that would be great.

 

[mfb]

This is a very unusual way to use the "forall" operator.

You can split the second sum in a sum from i₂=1 to i_i-1 and consider i₂=i₁ as separate sum (with just 1 entry, so you can drop the sum sign and set i₂=i₁). This allows to get rid of the cases in the first bracket. The same can be done for the second part.

As all the sum indices do not depend on free parameters, you could just feed it to a computer algebra system and look at the result.

 

[Big-Daddy]

This is a very unusual way to use the "forall" operator.

Apologies for that, I wrote the function and so far I don't have too much experience with mathematical notation. What would be a better operator to use? (Not that it matters much - this thread is about removing the need!)

You can split the second sum in a sum from i₂=1 to i_i-1 and consider i₂=i₁ as separate sum (with just 1 entry, so you can drop the sum sign and set i₂=i₁). This allows to get rid of the cases in the first bracket. The same can be done for the second part.

As all the sum indices do not depend on free parameters, you could just feed it to a computer algebra system and look at the result.

Could you demonstrate, for the cases in the first bracket? I'll try and apply the same to the second bracket without asking back.

 

[mfb]

With my own example, as I don't want to open Word again (pdf or LaTeX would be better):
$$\sum_{i_1=1}^{6} \sum_{i_2=1}^{i_1} \begin{cases} 1\, \mathrm{for}\, i_1=i_2 \\ 2\, \mathrm{for}\, i_1>i_2 \end{cases}\\
= \sum_{i_1=1}^{6} \sum_{i_2=1}^{i_1-1} 2
+ \sum_{i_1=1}^{6} \sum_{i_2=i_1}^{i_1} 1
= \sum_{i_1=1}^{6} \sum_{i_2=1}^{i_1-1} 2
+ \sum_{i_1=1}^{6} 1$$
As the first summand is pointless for i₁=1, this case can be dropped
$$= \sum_{i_1=2}{6} \sum_{i_2=1}^{i_1-1} 2
+ \sum_{i_1=1}^{6} 1$$

 

[Big-Daddy]

Thanks. I think I can see how to split one sigma function, but here we've got both cases and one multiplying the other, all within the summation operators. What's the rewriting then?

I've reattached as a PDF. The "forall" operator is still there as I'm not sure what to replace it with, but that is a secondary discussion.

 

[mfb]

I would replace it with "for" or |. You can do the same splitting in both separate sums again.

 

[Big-Daddy]

Please check my solution (attached) to tell me if anything is wrong.

If there are any obvious simplifications that always hold, I'd love to hear, but I don't mind it being this long. Main thing is accuracy and functionality.