-1
$\begingroup$

Say I want to create the list
$\{ \sum_{j=1}^{5} 2j, \sum_{j=1}^{6} \frac{5j}{2} \}$
I guessed that this can be done using the code
Sum[j/{1, 2}*{2, 5}, {j, 1, {5, 6}}]
However what mathematica gives is
$\Big{\{} \{ \sum_{j=1}^{5} 2j, \sum_{j=1}^{5} \frac{5j}{2} \}, \{ \sum_{j=1}^{6} 2j, \sum_{j=1}^{6} \frac{5j}{2} \} \Big{\}}$
I guess I could just somehow pick out the right element of the result, but is there a more efficient way to achieve this?

EDIT: Eventually I want to apply this to a case, where I have something like
Sum[f[list,j],{j,1,list}]

$\endgroup$
  • 1
    $\begingroup$ Maybe: MapThread[Sum[# j, {j, 1, #2}] &, {{2, 5/2}, {5, 6}}] $\endgroup$ – Coolwater Jun 4 at 15:03
  • $\begingroup$ How can I apply this to an arbitrary list and an arbitrary function of that list, e.g. something like Sum[f[list,j],{j,1,list}]? $\endgroup$ – lomby Jun 4 at 16:14
1
$\begingroup$

Your question is not very well-defined. It seems that you have a list of different functions, f1, f2, f3, etc., and you want to sum each of those functions to different maximum values. Is that correct? If so:

f1[j_]=2j;
f2[j_]=5j/2;
funclist={f1,f2};
maxsumlist={5,6};
result=MapThread[Sum[#1@j,{j,1,#2}]&,{funclist,maxsumlist}]
$\endgroup$
2
$\begingroup$

Perhaps

sum[a_, b_, c_] := Sum[a j, {j, b, c}]
SetAttributes[sum, Listable]

Examples:

sum[{2, 5}/{1, 2}, 1, {5, 6}]

{30, 105/2}

sum[{a, b}, 1, {n, m}]

{1/2 a n (1 + n), 1/2 b m (1 + m)}

sum[{foo[j], bar[j]}, q, {n, m}] // TeXForm

$\left\{\sum _{j=q}^n j \text{foo}(j),\sum _{j=q}^m j \text{bar}(j)\right\}$

$\endgroup$

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.