Can I see something meaningfull about this sum? Where is it equal to?
$$\left(\sum_{n=a}^{b} z_n\right)^2$$
Is it equal to: $\sum_{n=a^2}^{b^2} z_n^2$ or something else, I've no idea how to deal with it.
Thanks for the help.
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Is $z_n$ something specific or can it be general? – quid Oct 14 '15 at 17:43
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Just general!!! – Jan Eerland Oct 14 '15 at 17:43
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no its not equal to that.... however, you could use the binominal formula to 'simplify' this.... – Börge Oct 14 '15 at 17:46
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I dont think there is a general result. – vudu vucu Oct 14 '15 at 17:46
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@Börge Can you give me an example? – Jan Eerland Oct 14 '15 at 17:47
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scroll down to the anwer of quid.. he used described it more detailed – Börge Oct 14 '15 at 17:51
2 Answers
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The string $\sum_{n=a}^b z_n$ means that you take the sum of $z_a + z_{a+1}+ \dots + z_b$. When you square that you compute
$$(z_a + z_{a+1}+ \dots + z_b)(z_a + z_{a+1}+ \dots + z_b)$$
and this is not at all what you propose. To get a feeling you might want to try it out for say $a=1$ and $b=3$ spelling out everything in detail.
A general formula would be $$ \left(\sum_{n=a}^b z_n\right)^2= \sum_{n=a}^b z_n^2 + 2\sum_{a \le n < m \le b} z_nz_m.$$ Observe the special case $(z_1 +z_2)^2 = z_1^2 + z_2^2 +2z_1 z_2$ that you might have seen elsewhere.
jameselmore
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quid
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$$\left(\sum_{n=a}^b z_n\right)^2 = \left(\sum_{j=a}^b z_j\right)\left(\sum_{k=a}^b z_k\right) = \sum_{j,k=a}^b z_jz_k$$ If you want, you can pull the squares and reduce repetitive terms into the following form: $$\left(\sum_{n=a}^b z_n\right)^2 = \sum_{n=a}^b z_n^2 + 2\sum_{j=a+1}^b\left[\sum_{k=a}^{j-1} z_jz_k\right]$$ Unfortunately, there aren't many simplifications (in general) beyond this.
jameselmore
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