# Need help in explaining the calculation of n-th power of a summation

1. Apr 29, 2010

### duranta23

Hi

I got to the following equation while going through a book. I can't figure out how the second line comes from the first. Can anyone please help me understand?

(1/2*\sum_{q=-Q}^Q V_s,q .H(w_q) .exp(iw_q t))^n is written as,

1/2^n * \sum_{q1=-Q}^Q \sum_{q2=-Q}^Q ....... \sum_{qn=-Q}^Q V_s,q1 . V_s,q2 .... V_s,qn .H(w_q1).H(w_q2).....H(w_qn).exp(i(w_q1 +w_q2 +....w_qn).t)

Please forgive my poor latex style writing.

2. May 2, 2010

### Martin Rattigan

Well, I can forgive it, but unfortunately I can't read it.

3. May 2, 2010

### Martin Rattigan

$(1/2*\sum_{q=-Q}^Q V_{s,q} .H(w_q) .exp(iw_q t))^n$

is written as,

$1/2^n * \sum_{q_1=-Q}^Q \sum_{q_2=-Q}^Q ....... \sum_{q_n=-Q}^Q V_{s,q_1} . V_{s,q_2} .... V_{s,q_n} .H(w_{q_1}).H(w_{q_2}).....H(w_{q_n}).exp(i(w_{q_1} +w_{q_2} +....w_{q_n}).t)$

The above is hopefully closer. I don't know if its just me, but I see quite a few posts with unprocessed Latex.

Last edited: May 2, 2010
4. May 2, 2010

### Martin Rattigan

And if you have

$(A\sum_{i=r}^s a_i)^n=A^n(a_r+a_{r+1}\dots a_s)(a_r+a_{r+1}\dots a_s)\dots (a_r+a_{r+1}\dots a_s)\text{ (to }n\text{ factors})$

you get

$A^n(\Sigma a_{i_1}a_{i_2}\dots a_{i_n}})$

where the sum on the second line runs through all different ways of selecting each $a_{i_k}$ from the corresponding bracket on the rhs of the first line.

This can be written as

$A^n\sum_{i_1=r}^s\sum_{i_2=r}^s\dots \sum_{i_n=r}^s a_{i_1}a_{i_2}\dots a_{i_n}}$

and as the indices run from r to s in each of the multiple sums, all ways of selecting $a_{i_1}a_{i_2}\dots a_{i_n}$ will occur exactly once.

With appropriate values inserted from your example this gives you the transition between the first and second lines.

5. May 3, 2010

### duranta23

Makes Perfect sense. Thank you so much for the explanation :D and special thanks for forgiving my latex :P