You are currently browsing the tag archive for the ‘summation’ tag.

I am caught with a little trouble to figure out an easier way to set up and solve a  summation of a function of several variables.  I just realized that Mathematica doesnt allow me to add a constraint in the summation operation.  My problem  is that, the summation should be performed over  the ordered partition set of a number

An example, will illustrate the problem definition a lot better.  Suppose I want to compute the sum of a function of 3 variables over all variables such that, the sum of the index variables always equal to a constant n

\displaystyle \sum_{\underset{i+j+k=n}{i,j,k}}f(i,j,k)

Clearly, the indices here runs over all the ordered partitions of the number 3.  Since 3 is not that big a number, we can simply write these partitions by hand and somehow get the job done.  Let us say denote the ordered partitions and unordered partitions of the number n by P(n) and Q(n) respectively. 

Q(3) =( (3), (2, 1),(1, 1, 1)). The ordered partition will have the permutations of each of these on the three positions! So (0,0,3) and (0,3,0) are different sets of P(3) whereas, they are considered as one (3) in Q(3).  The summation needs to be carried over the ordered partition list P(n).  Since both Q(n)  and P(n) grow pretty big with even modest n, the job of manual summation is not that appealing. I really hoped mathematica to aid me here. Unfortunately, I don’t see a way out here, other than the painful individual partition sum.  Anyway, for the curious reader, the sum I attempt are of the following types:

\displaystyle \sum_{n_1+n_2+\ldots+n_k=n}{(n_1+1)a_1\frac{n!}{n_1!n_2!\ldots n_k!}\frac{a_{1}^{n_1} a_{2}^{n_2} \ldots a_{k}^{n_k}}{\left(\sum_{i=1}^{k}{a_i}\right)^{n+1}} }

\displaystyle \sum_{n_1+n_2+\ldots+n_k=n}{(n_1+1-n_2)a_1\frac{n!}{n_1!n_2!\ldots n_k!}\frac{a_{1}^{n_1} a_{2}^{n_2} \ldots a_{k}^{n_k}}{\left(\sum_{i=1}^{k}{a_i}\right)^{n+1}} }

typical values for n is 20 and k is around 10. If any enthused reader finds a way/trick, I will be happy to sponsor a coffee 🙂 (Sorry at this recession time, nothing more 😦 sadly…).

In fact, the first one is easy (Interestingly, I just found a way, while typesetting the blog).  I can do a differential with respect to a_1 and then scale it to get the requisite sum.  It can be written as follows:

\displaystyle \sum_{n_1+n_2+\ldots+n_k=n}{(n_1+1)a_1\frac{n!}{n_1!n_2!\ldots n_k!}\frac{a_{1}^{n_1} a_{2}^{n_2} \ldots a_{k}^{n_k}}{\left(\sum_{i=1}^{k}{a_i}\right)^{n+1}} }

=\frac{a_1}{{\left(\sum_{i=1}^{k}{a_i}\right)^{n+1}}} \frac{\partial}{\partial a_1}\left[a_1 \left(a_1+a_2+\ldots+a_k\right)^{n}\right]

=\frac{a_1}{{\left(\sum_{i=1}^{k}{a_i}\right)^{n+1}}} \left[n a_1 \left(a_1+a_2+\ldots+a_k\right)^{n-1}+\left(a_1+a_2+\ldots+a_k\right)^{n}\right]

Second one is the trouble maker 😦

Oh man! power of a coffee.  As I am typing this on a moody Lausanne swiss weather, here comes the trick.  I just made a coffee and that seemed to have worked.  I guess I can apply a similar trick to the second one too.  Basically, we can split them into two expressions and then write each as differential versions of a multinomial sum. Here it is:

\displaystyle \sum_{n_1+n_2+\ldots+n_k=n}{(n_1+1-n_2)a_1\frac{n!}{n_1!n_2!\ldots n_k!}\frac{a_{1}^{n_1} a_{2}^{n_2} \ldots a_{k}^{n_k}}{\left(\sum_{i=1}^{k}{a_i}\right)^{n+1}} }

=\displaystyle \sum_{n_1+n_2+\ldots+n_k=n}{(n_1+1)a_1\frac{n!}{n_1!n_2!\ldots n_k!}\frac{a_{1}^{n_1} a_{2}^{n_2} \ldots a_{k}^{n_k}}{\left(\sum_{i=1}^{k}{a_i}\right)^{n+1}} }

-\displaystyle \sum_{n_1+n_2+\ldots+n_k=n}{(n_2)a_1\frac{n!}{n_1!n_2!\ldots n_k!}\frac{a_{1}^{n_1} a_{2}^{n_2} \ldots a_{k}^{n_k}}{\left(\sum_{i=1}^{k}{a_i}\right)^{n+1}} }

=\frac{a_1}{{\left(\sum_{i=1}^{k}{a_i}\right)^{n+1}}} \frac{\partial}{\partial a_1}\left[a_1 \left(a_1+a_2+\ldots+a_k\right)^{n}\right]

-\frac{a_1 a_2}{{\left(\sum_{i=1}^{k}{a_i}\right)^{n+1}}} \frac{\partial}{\partial a_2}\left[\left(a_1+a_2+\ldots+a_k\right)^{n}\right]

=\frac{a_1}{{\left(\sum_{i=1}^{k}{a_i}\right)^{n+1}}} \left[n a_1 \left(a_1+a_2+\ldots+a_k\right)^{n-1}+\left(a_1+a_2+\ldots+a_k\right)^{n}\right]

-\frac{a_1 a_2}{{\left(\sum_{i=1}^{k}{a_i}\right)^{n+1}}} \left[n \left(a_1+a_2+\ldots+a_k\right)^{n-1}\right]

Amazingly, we can simplify both to get a simple looking expression. I am glad! Here is what I finally got:

\displaystyle \sum_{\sum_{j}{ n_j}=n}{\frac{n!(n_1+1)a_1}{n_1!n_2!\ldots n_k!}\frac{a_{1}^{n_1} a_{2}^{n_2} \ldots a_{k}^{n_k}}{\left(\sum_{i=1}^{k}{a_i}\right)^{n+1}} }=\frac{a_1\left(na_1+\sum_{i}{a_i}\right)}{{\left(\sum_{i=1}^{k}{a_i}\right)^{2}}}

\displaystyle \sum_{\sum_{j}{ n_j}=n}{\frac{n!(n_1+1-n_2)a_1}{n_1!n_2!\ldots n_k!}\frac{a_{1}^{n_1} a_{2}^{n_2} \ldots a_{k}^{n_k}}{\left(\sum_{i=1}^{k}{a_i}\right)^{n+1}} }=\frac{a_1\left(n(a_1-a_2) +\sum_{i}{a_i}\right)}{{\left(\sum_{i=1}^{k}{a_i}\right)^{2}}}

\displaystyle \sum_{\sum_{j}{ n_j}=n}{\frac{n!(n_l)a_1}{n_1!n_2!\ldots n_k!}\frac{a_{1}^{n_1} a_{2}^{n_2} \ldots a_{k}^{n_k}}{\left(\sum_{i=1}^{k}{a_i}\right)^{n+1}} }=\frac{a_1 a_l n}{{\left(\sum_{i=1}^{k}{a_i}\right)^{2}}}

Pages

October 2017
M T W T F S S
« Mar    
 1
2345678
9101112131415
16171819202122
23242526272829
3031  

Enter your email address to follow this blog and receive notifications of new posts by email.

Join 86 other followers

Like

%d bloggers like this: