- #1

cheeez

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Any hint on how to start this?

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- Thread starter cheeez
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- #1

cheeez

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Any hint on how to start this?

- #2

mathman

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Are there any restrictions on c? As I interpret the statement, c = 1/2 will always work.

- #3

cheeez

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- #4

fzero

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You can obtain 1/2 by considering n=2. For any n, you can gather the n terms together into 2 terms.

- #5

cheeez

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i dont see how you can divide them into 2 numbers because you would need to have |z_1+..+z_i| + |z_i+1+...+z_n| >= 1 but that's not necessarily true since triangle inequality works the other way.

- #6

fzero

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- #7

cheeez

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i dont get it how are you dividing z_1,...,z_n into 2 terms so |z_1+..+z_i| + |z_i+1+...+z_n| >= 1

- #8

fzero

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i dont get it how are you dividing z_1,...,z_n into 2 terms so |z_1+..+z_i| + |z_i+1+...+z_n| >= 1

Try

[tex] \sum_k |z_k| = \left( |z_1|+\cdots+|z_i| \right) + \left( |z_{i+1}|+\cdots+|z_n| \right). [/tex]

- #9

cheeez

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- #10

fzero

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[tex] |z_1+\cdots+z_i| + |z_{i+1}+\cdots+z_n| \geq 1.

[/tex]

- #11

cheeez

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- #12

fzero

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- #13

cheeez

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yea that's what I'm trying to do, any ideas?

- #14

fzero

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- #15

cheeez

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- #16

fzero

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Those don't satisfy [tex]|z_1+z_2 +z_3| \geq 1[/tex].

- #17

cheeez

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show there is a some constant c, independent of n, s.t. if {Z_j} are complex numbers and sum of |Z_j| from 1 to n >= 1, then there's a subcollection {Z_j_k} of {Z_j} s.t. |sum of Z_j_k| >= c.

so here |z1| + |z2| + |z3| = 1/3 + 1/3 + 1/3 = 1

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