Consider two collections of real numbers in increasing order,
$a_1 \leq a_2 \leq a_3··· \leq a_n$ and $b_1 \leq b_2 \leq···\leq b_n$.
For any permutation $({a_1}',{a_2}',...,{a_n}')$of$(a_1,a_2,...,a_n)$, it happens that
$a_1 b_1 + a_2 b_2 + ··· + a_n b_n \geq {a_1}'b_1 + {a_2}'b_2+..+{a_n}'b_n \geq$
$\geq a_nb_1 + a_n b_2 + ··· + a_1b_n$
Rearrangement inequalty :(BOMC)
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Re: Rearrangement inequalty :(BOMC)
If \[{a_{1}}', {a_{2}}', ...{a_{n}}'\] is any permutation of \[a_{1}, a_{2}, ..., a_{n}\], how can you express \[{a_{1}}'\] in terms of \[a_{1}\]?
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Re: Rearrangement inequalty :(BOMC)
it means ${a_n}'$ can be any of $(a_1,a_2,...,a_n)$
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
Re: Rearrangement inequalty :(BOMC)
@Willpower:
note that it isnt bound to be $a_i=a_i'$
it may be $a_1=a_5'$ for example.
note that it isnt bound to be $a_i=a_i'$
it may be $a_1=a_5'$ for example.
A man is not finished when he's defeated, he's finished when he quits.
Re: Rearrangement inequalty :(BOMC)
All right! Thank you.
Everybody is a genius; but if you judge a fish on its ability to climb a tree, it will live its entire life believing that it is stupid. - Albert Einstein