7
$\begingroup$

In general, is $\sum_i \sum_j f(i,j) = \sum_j\sum_i f(i,j)$ ?

With $f(i,j)$ I mean some expression that depends on $i$ and $j$.

If yes, how could I prove that?

$\endgroup$
9
  • 1
    $\begingroup$ This holds for FINITE sums. As for the proof: isn't it obvious? Addition is commutative. $\endgroup$
    – Crostul
    Jun 4 '15 at 10:53
  • $\begingroup$ @Crostul sure. What are the infinite cases? $\endgroup$
    – Simon S
    Jun 4 '15 at 10:54
  • $\begingroup$ Maybe the OP should be more specific, since its summations are dependent on some unspecified set of indices. $\endgroup$
    – Crostul
    Jun 4 '15 at 10:55
  • $\begingroup$ @SimonS Infinite cases are those that add up to infinity; these series are typically found in taylor series and such $\endgroup$ Jun 4 '15 at 10:55
  • $\begingroup$ By infinite cases I mean when the index sets for $i$ and $j$ are (countably) infinite. What's interesting then is when the sums do commute. I'm hesitating in giving an answer because I think I know a very strong sufficient condition, but I'd rather have someone chime in who works with these sorts of expressions more than I do. $\endgroup$
    – Simon S
    Jun 4 '15 at 11:02
5
$\begingroup$

$$\sum_i\sum_jf(i,j) = \sum_i( f(i,j_1) + f(i,j_2)+...+f(i,j_n)) =$$$$ \sum_i f(i,j_1)+\sum_if(i,j_2)+...+\sum_i f(i,j_n)=$$ $$[f(i_1,j_1)+f(i_1,j_2)+...+f(i_1,j_n)]+[f(i_2,j_1)+f(i_2,j_2)+...+f(i_2,j_n)]+...+[f(i_m,j_1)+f(i_m,j_2)+...+f(i_m,j_n)] =$$ $$ [f(i_1,j_1)+f(i_2,j_1)+...+f(i_m,j_1)]+[f(i_1,j_2)+f(i_2,j_2)+...+f(i_m,j_2)]+...+[f(i_1,j_n)+f(i_2,j_n)+...+f(i_m,j_n)] =$$ $$ \sum_j f(i_1,j)+\sum_j f(i_2,j)+...+\sum_j f(i_m,j) = \sum_j\sum_if(i,j).\square$$

$\endgroup$
2
  • 2
    $\begingroup$ As discussed in the comments above, this finite case is trivial. $\endgroup$
    – Simon S
    Jun 4 '15 at 11:16
  • $\begingroup$ I guess there is a misprint in the derivation above in the line "3" where you expanded in "j" not in "i". $\endgroup$
    – bruziuz
    Sep 16 at 16:25

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.