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Guest
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 Posted: Tue Jul 08, 2008 3:21 am Post subject: Cauchy-Schwarz inequality. |
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Hello teacher~
a = (a_1, ... , a_m)
b = (b_1, ... , b_m)
sum{i=1 to m} |(a_i).(b_i)| <= (||a||)(||b||)
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Arithmetic-geometric mean
(x-y)^2 >= 0
==> x^2 + y^2 >= 2xy
==> (x^2 + y^2)/2 >= xy
Cauchy-Schwarz inequality
x^2 + y^2 >= 2xy
Let x = |a_i| / ||a||, y = |b_i| / ||b||.
==> [ |a_i|^2 / ||a||^2 ] + [ |b_i|^2 / ||b||^2 ] >= 2.|a_i|.|
b_i| / (||a||.||b||)
==> Sum{i=1 to m} ~~
==> [ ||a||^2 / ||a||^2 ] + [ ||b||^2 / ||b||^2 ] >= sum{i=1 to m}
[ 2.|a_i|.|b_i| / (||a||.||b||) ]
==> (||a||)(||b||) >= sum{i=1 to m} |(a_i).(b_i)| |
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William Elliot
Guest
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| Posted: Tue Jul 08, 2008 11:14 am Post subject: Re: Cauchy-Schwarz inequality. |
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On Mon, 7 Jul 2008 mina_world@hanmail.net wrote:
| Quote: |
a = (a_1, ... , a_m)
b = (b_1, ... , b_m)
sum{i=1 to m} |(a_i).(b_i)| <= (||a||)(||b||)
Case a = 0: |
Case b = 0:
Case a,b /= 0: using * for dot product
0 <= (a/||a|| +- b/||b||)^2
-+2a/||a|| * b/||b|| <= a*a/|a||^2 + b*b/||b||^2 = 2
-+a*b <= ||a|| + ||b||
|a*b| <= ||a|| + ||b||
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