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August 28th, 2017, 09:10 PM   #1
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Double sigma

I am reading my Bulgarian textbook about Electromagnetism theory. I found an expression, which I can't understand. Here I attached the expression to my post.
Please, help me to understand this "double" sum from pure mathematical point of view. What does the expression mean?
I will be very thankful for a step-by-step explanation of this particular case of double sum.

Wish you a nice day everybody!
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 August 29th, 2017, 01:13 PM #2 Global Moderator   Joined: May 2007 Posts: 6,511 Thanks: 585 Your expression has 2 indices p and q. The double sum is simply summing over both of them. It is a standard notation, nothing to do with Bulgarian.
 August 29th, 2017, 01:56 PM #3 Senior Member   Joined: May 2016 From: USA Posts: 1,030 Thanks: 420 $\displaystyle \sum_{i=1}^3 \sum_{j=1}^3 i(2j+1) = \sum_{i=1}^3(3i + 5i +7i) = 15 * \sum_{i=1}^3i = 15(1 + 2 + 3) = 90.$ Thanks from Country Boy
 September 3rd, 2017, 12:06 PM #4 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,170 Thanks: 869 Since addition and multiplication "distribute" and addition is commutative, the order is not relevant. You can also do it as $\displaystyle \sum_{j= 1}^3 (2j+ 1)\left(\sum_{i= 1}^3 i\right)=$$\displaystyle \sum_{j=1}^3 (2j+ 1)(1+ 2+ 3)=$$\displaystyle \sum_{j=1}^3 6(2j+ 1)= 6(3)+ 6(5)+ 6(7)= 18+ 30+ 42= 90$. $\displaystyle \sum_{p= 1}^{m_p}\sum{q= 1}^{m_q} M_{pq} \frac{di_q}{dt}= \sum_{p=1}^{m_p}\left(M_{p1}\frac{di_1}{dt}+ M_{p2}\frac{di_2}{dt}+ \cdot\cdot\cdot+ M_{pm_q} M_{pm_q}\frac{di_{mq}}{dt}\right)= \left(M_{11}\frac{di_1}{dt}+ M_{21}\frac{di_1}{dt}+ \cdot\cdot\cdot+ M_{m_p1}\frac{di_1}{dt}\right)+ \left(M_{12}\frac{di_2}{dt}+ M_{22}\frac{di_2}{dt}+ \cdot\cdot\cdot+ M_{m_p2}\frac{di_2}{dt}\right)+ \cdot\cdot\cdot+ \left(M_{1m_q}\frac{di_{m_q}}{dt}+ M_{2m_q}\frac{di_{m_q}}{dt}+ \cdot\cdot\cdot+ M_{m_pm_q}\frac{di_{m_q}}{dt}\right)$

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