I'm reading my vector calculus text when I encountered below formula.
$(\sum\limits_{i=1}^n{x_{i}})^2=(\sum\limits_{i=1}^n{x_{i}^2}+\sum\limits_{i
Is this a definition or there's a proof for above?
how is this series expansion $(\sum\limits_{i=1}^n{x_{i}})^2=(\sum\limits_{i=1}^n{x_{i}^2}+\sum\limits_{i derived?
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2Think of $(x_1+x_2)^2=x_1^2+x_2^2+2x_1 x_2$ – 2012-07-29
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0i did try that but the missing $x_{2}^2$ baffled me – 2012-07-29
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1The previous identity could be written $$(\sum_{i=1}^2 x_i)^2=\sum_{i=1}^2 x_i^2 + \sum_{i
and for $n=3$ you have $$(x_1+x_2+x_3)^2=x_1^2+x_2^2+x_3^2+2x_1 x_2 + 2x_1 x_3+2x_2 x_3,$$ that can be written as $$(\sum_{i=1}^3 x_i)^2=\sum_{i=1}^3 x_i^2 + \sum_{i – 2012-07-29 -
0ah yes, of course...earlier i took it literally as squaring the series itself to get the answer.Unfortunately I cannot choose your comment as answer to close this question since yours is the closest. – 2012-07-29
3 Answers
Consider the array
$$\left[ \begin{array}{cccc} \color{red}{x_1^2} & \color{green}{x_1x_2} & \color{blue}{x_1x_3} & x_1x_4&&& \ldots &x_1x_n \\ \color{green}{x_1 x_2} & \color{red}{x_2^2} & x_2x_3 & x_2x_4 &&&\ldots & x_2x_n \\ \color{blue}{x_1x_3} & x_2x_3 & \color{red}{x_3^2} & x_3x_4 &&&\ldots & x_3x_n \\ x_1x_4 & x_2x_4 & x_3x_4 & \color{red}{x_4^2} & x_4x_5 &&\ldots & x_4x_n \\ &&&\vdots & \color{red}{\ddots} \\ &&&\vdots \\ &&&\vdots \\ x_1x_n & x_2x_n & \ldots &&&&\ldots & \color{red}{x_n^2}\end{array}\right].$$
This is the array consisting of all terms in the expansion of $(x_1+\ldots + x_n)^2$. Notice that the array is symmetric about the diagonal. So therefore we just need to notice that
$$\begin{eqnarray*} \left(\sum_{i=1}^n x_i\right)^2 = (x_1 + \ldots+ x_n)^2 &=& \text{(sum of all terms along the diagonal)} \\ && \hspace{16mm}+ \text{($\color{red}{twice}$ the sum of all terms above the diagonal)}\\ &=& \sum_{i=1}^n x_i^2 + 2\sum_{i < j} x_ix_j.\end{eqnarray*}$$
$\hspace{6in} \square$
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1That's what I meant by *seeing* why it's true, but I was too lazy to typeset it :) – 2012-07-29
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0@Jbeuh C'était pas trop dificile.... – 2012-07-29
It's easily proved by induction using $(x_1+x_2)^2=x_1^2+2x_1x_2+x_2^2$ as the base case but it's actually more useful to "see" why it's true (replace the $\Sigma$'s with dots if it helps).
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0how about the missing $x_{2}^2$ ?I'm not sure what to made of it – 2012-07-29
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0I'm not sure what you mean. $$\left(\sum_{i=1}^{n+1} x_i\right)^2 = \left(\left(\sum_{i=1}^n x_i \right)+x_{n+1}\right)^2 = \left(\sum_{i=1}^n x_i \right)^2 + 2x_{n+1}\sum_{i=1}^n x_i + x_{n+1}^2 = \sum_{i=1}^n x_i^2 + 2\sum_{1\leq i
– 2012-07-29 -
0It's even easier to use $n=1$ as the base case (in this case $\{i
is empty). – 2012-07-29 -
0@Jbeuh turns out I took squaring of equation too literally and apply it to series. – 2012-07-29
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0@wildildildlife It's easier, but in my experience base cases which are vacuously true tend to confuse (first year) students. – 2012-07-29
You can use multinomial theorem.