[R] summation equation whose numerator has subscript

David Winsemius dwinsemius at comcast.net
Mon Nov 30 23:27:46 CET 2015


> On Nov 30, 2015, at 1:20 PM, Sherouk Moawad via R-help <r-help at r-project.org> wrote:
> 
> Dear R experts 
> Please do you have any idea about how this summation can be written in R(the equation can be viewed in the following link): 
> http://s16.postimg.org/or2km30ph/equation.jpg 

Can you explain in natural language the goals of this expression. It makes little sense to me to start with an index of j_sub_l = 0 and to then iterate to up to j_sub_(l-1) -1 . How can there be a value for j_sub(l-1) with a starting point of zero. The notation saying to do something for l = 2:n is not helpful since values of “l” doesn’t really appear in the looped expression (noting that j_sub_l starts at 0, so it's not being determined by “l".

I believe the confused notation was the cause of this question being closed after it appeared last week on SO:

http://stackoverflow.com/questions/33882285/summation-equation-whose-numerator-has-subscript

And what intent is meant for the indices of the outer summation? The expression j_sub_1 = 0 seems to have no corresponding reference point inside the looped expression. So you would simply be summing the same value N times, but since N is not defined we cannot write any code.

> 
> I've tried out out this code but it gave me error for writing brackets in function of summation:

You should _always_, _always_, _always_ post the entire results of an error. We have no way of seeing your console. Error messages are usually informative.

> 
>>>> 
> x=matrix(c(6,2,1),3,1) 
> 
> for (l in 1:3){ 
> sum(sapply(1:3, function(j[l]){if(l>1){sum(sapply(1:j[l-1], function(j[l]){x[j[l]]*(j[l]<j[l-1])}))}}))}


You have three nested loops in the code above, but at least it appears you do understand that R is a 1-based language. But since the image-expression goes from 0 to some cryptic value (minus one) then the R version ought to go from one to "one more” than that expression.


> >>>Thank you
> 
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David Winsemius
Alameda, CA, USA



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