Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$$\sum_{K=2}^{N}\sum_{L=1}^{\lfloor\frac{K}{2}\rfloor-1}$$

I want to have the $L$ summation on the outside and the $K$ summation on the inside somehow. Can this be done?

share|improve this question
    
Are these $L$ and $K$ finite sum terms? What happens when you expand the sum fully? Can you regroup the terms? –  abiessu Aug 11 '13 at 4:52
    
I tried that and couldn't find any clean way to do it –  user2175923 Aug 11 '13 at 4:53
    
For any fixed value of $L$, $K$ travels from $2L$ to $N$, So the inner sum is $\sum_{K=2L}^N$. –  André Nicolas Aug 11 '13 at 4:53
    
I tried that already; it led me to make the outer sum L=1 to floor(N/2) which is wrong –  user2175923 Aug 11 '13 at 4:58

1 Answer 1

up vote 1 down vote accepted

$$\sum_{k=2}^Na_k\sum_{j=1}^{[\frac{k}{2}]-1}b_j=a_4(b_1)+a_5(b_1)+a_6(b_1+b_2)+a_7(b_1+b_2)+a_8(b_1+b_2+b_3)...$$ $$=b_1(a_4+a_5+a_6...a_N)+b_2(a_6+a_7+a_8+...a_N)+b_3(a_8+a_9+a_{10}...+a_N)...$$ $$=\sum_{k=1}^{[\frac{N}{2}]-1}b_k\sum_{j=2(k+1)}^{N}a_j$$

So that we get,

$$\sum_{k=2}^Na_k\sum_{j=1}^{[\frac{k}{2}]-1}b_j=\sum_{k=1}^{[\frac{N}{2}]-1}b_k\sum_{j=2(k+1)}^{N}a_j$$

share|improve this answer
    
This is not correct –  user2175923 Aug 11 '13 at 5:08
    
Why is this not correct? –  Ethan Aug 11 '13 at 5:09
    
Try sample N=10, then try adding up +1 for each iteration of the inner loop. Correct one gives you 16, the rewrite gives 25 –  user2175923 Aug 11 '13 at 5:12
    
You get 16 with the new one? k=1 to 5. so j=2 to 10, 4 to 10, 6 to 10, 8 to 10, 10 to 10. That's 9+7+5+3+1=25 –  user2175923 Aug 11 '13 at 5:17
    
It looks like the mistake is on my end. My inner bound is not floor(K/2) but rather floor(K/2)-1 –  user2175923 Aug 11 '13 at 5:21

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.