The question is:
Prove that $|x| \le \sum_{i=0}^n |x^i|$
But the solution has this step:
$ (\sum_{i=0}^n |x^i|)^2 = \sum_{i=0}^n (x^i)^2 + \sum_{i\neq j} x^ix^j $
How is this true?
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@GeorgesElencwajg What about the addition?? – banana Apr 11 '17 at 05:41
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2For example, $(a+b)^2=a^2+b^2 + ab + ba$ and $(a+b+c)^2 = a^2 + b^2 + c^2 + ab + ac + bc + ba + bc + ba + bc$. – angryavian Apr 11 '17 at 05:41
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@angryavian so does i and j in what's being added represent all the possible cases depending on the variable? For example, would it represent the (+ab+ac+bc+ba+bc+ba+bc)? – banana Apr 11 '17 at 05:43
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1Yes, the "$i \ne j$" is shorthand for "all pairs $(i,j)$ such that $0 \le i \le n$ and $0 \le j \le n$ and $i \ne j$." – angryavian Apr 11 '17 at 05:46
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1Just work out an example for small n. This is a fairly standard trick so it's worth understanding in detail. BTW, I think you've left out the absolute value signs in the second sum on the RHS. – NickD Apr 11 '17 at 05:47
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@Nick Thank you! I'm having trouble coming up with an example in which $ |x| $ would be greater than, though. – banana Apr 11 '17 at 05:50
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@angryavian Thank you! – banana Apr 11 '17 at 05:51
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$|x|$ would be greater than what? – NickD Apr 11 '17 at 05:53
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@Nick I'm sorry, I meant less than $\sum_{i=0}^n|x^i|$ – banana Apr 11 '17 at 05:56
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Try $x=(1, 2, 3)$. Then $|x| = \sqrt{14}$ and $\sum_i |x^i| = 1+2 + 3 = 6$. – NickD Apr 11 '17 at 05:58