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

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Discussion Overview

The discussion revolves around understanding the mathematical transformation of an equation involving the n-th power of a summation. Participants explore the steps required to derive the second line from the first in the context of a specific summation formula, focusing on the manipulation of terms and the application of summation properties.

Discussion Character

  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant presents an equation involving a summation and requests clarification on how to transition from the first to the second line of the equation.
  • Another participant attempts to clarify the notation and improve the readability of the LaTeX representation.
  • A further contribution explains a general case of raising a summation to a power, detailing how to express the result using multiple sums and the selection of terms from the original summation.
  • A participant expresses gratitude for the explanation and acknowledges the earlier issues with LaTeX formatting.

Areas of Agreement / Disagreement

Participants generally agree on the mathematical principles involved in the transformation, but there is no explicit consensus on the clarity of the initial LaTeX presentation.

Contextual Notes

Some participants note issues with LaTeX formatting that may hinder understanding, indicating a potential limitation in communication of the mathematical concepts.

duranta23
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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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Well, I can forgive it, but unfortunately I can't read it.
 
[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:
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.
 
Makes Perfect sense. Thank you so much for the explanation :D and special thanks for forgiving my latex :P
 

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