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

In summary, the conversation is about a mathematical equation and how it is written differently in two lines. The second line is a simplified version of the first line using multiple sums and indices. The conversation ends with gratitude for the explanation and forgiveness for the poor latex writing style.
  • #1
duranta23
3
0
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.
 
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  • #2
Well, I can forgive it, but unfortunately I can't read it.
 
  • #3
[itex](1/2*\sum_{q=-Q}^Q V_{s,q} .H(w_q) .exp(iw_q t))^n[/itex]

is written as,

[itex]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)[/itex]

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:
  • #4
And if you have

[itex](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})[/itex]

you get

[itex]A^n(\Sigma a_{i_1}a_{i_2}\dots a_{i_n}})[/itex]

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

This can be written as

[itex]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}}[/itex]

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

With appropriate values inserted from your example this gives you the transition between the first and second lines.
 
  • #5
Makes Perfect sense. Thank you so much for the explanation :D and special thanks for forgiving my latex :P
 

What is the n-th power of a summation?

The n-th power of a summation is a mathematical calculation that involves raising a summation expression, which is the addition of a series of terms, to a certain power n. This results in a new expression that represents repeated multiplication of the original summation.

How do you calculate the n-th power of a summation?

To calculate the n-th power of a summation, you first need to expand the summation expression by multiplying it by itself n times. Then, you can use the binomial theorem to simplify the expanded expression and obtain the final result.

What is the significance of calculating the n-th power of a summation?

Calculating the n-th power of a summation is useful in many areas of mathematics, such as in combinatorics, probability, and statistics. It allows us to efficiently solve complex problems that involve repeated addition or multiplication of terms.

Can you provide an example of calculating the n-th power of a summation?

Sure, for example, if we have the summation expression (2x + 3) and we want to find its square, we would expand it as (2x + 3)(2x + 3) = 4x^2 + 12x + 9. This represents the n=2 case of the n-th power of a summation.

Are there any special cases or exceptions when calculating the n-th power of a summation?

Yes, when n=0, the n-th power of a summation is equal to 1. This is because any number raised to the power of 0 is equal to 1. Additionally, when n=1, the n-th power of a summation is equal to the original summation expression, as multiplying by 1 does not change the value.

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