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 Cauchy-Schwarz Inequality proof

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spirit

spirit


Messages : 222
Date d'inscription : 25/09/2007
Age : 31

Cauchy-Schwarz Inequality proof Empty
MessageSujet: Cauchy-Schwarz Inequality proof   Cauchy-Schwarz Inequality proof EmptySam 7 Mai 2011 - 15:03

Cauchy-Schwarz Inequality

|u*v| ≤ ||u|| ||v||, such that u and v are non-zero vectors.

Proof:

If |u*v| = ||u|| ||v||, then u and v are parallel vectors, i.e.: u = kv, where k is any scalar in real number.
Let p(t) = ||tv - u||^2 ≥ 0
p(t) = (tv - u)*(tv - u) = v*v t^2 - 2t(u*v) + u*u

Let a = v*v, b = 2u*v, c = u*u

p(t) = at^2 - bt + c
p(b/2a) = a(b/2a)^2 - b(b/2a) + c = b^2 /4a - b^2 /2a + c = -b^2 /4a + c ≥ 0
∴ c ≥ b^2 /4a; 4ac ≥ b^2

With the original values:
4(v*v)(u*u) ≥ (2u*v)^2
(v*v)(u*u) ≥ (u*v)^2
||v||^2 ||u||^2 ≥ (u*v)^2
∴|u*v| ≤ ||u|| ||v|| QED


Dernière édition par spirit le Sam 7 Mai 2011 - 15:27, édité 1 fois
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Dark

Dark


Messages : 184
Date d'inscription : 13/05/2008

Cauchy-Schwarz Inequality proof Empty
MessageSujet: Re: Cauchy-Schwarz Inequality proof   Cauchy-Schwarz Inequality proof EmptySam 7 Mai 2011 - 15:09

Cauchy–Bunyakovsky–Schwarz inequality - the name is so weird. It comes from Augustin-Louis Cauchy, Viktor Bunyakovsky and Hermann Amandus Schwarz. 13
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spirit

spirit


Messages : 222
Date d'inscription : 25/09/2007
Age : 31

Cauchy-Schwarz Inequality proof Empty
MessageSujet: Re: Cauchy-Schwarz Inequality proof   Cauchy-Schwarz Inequality proof EmptySam 7 Mai 2011 - 15:11

Yeah. It's their problem that we have difficulties to pronounce it.
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spirit

spirit


Messages : 222
Date d'inscription : 25/09/2007
Age : 31

Cauchy-Schwarz Inequality proof Empty
MessageSujet: Re: Cauchy-Schwarz Inequality proof   Cauchy-Schwarz Inequality proof EmptySam 7 Mai 2011 - 15:26

Triangle inequality

||u+v|| ≤ ||u||+||v||, such that u and v are non-zero vectors.

Proof:

Consider ||u+v||^2 = (u+v)*(u+v) = u*u + 2u*v + v*v = ||u||^2 + 2u*v + ||v||^2

We know that, by the Cauchy Schwarz inequality, u*v ≤ |u*v| ≤ ||u|| ||v||

∴ ||u+v||^2 = ||u||^2 + 2u*v + ||v||^2 ≤ ||u||^2 + 2||u|| ||v|| + ||v||^2
||u+v||^2 = ||u||^2 + 2||u|| ||v|| + ||v||^2 = (||u|| + ||v||)^2
||u+v|| = ||u|| + ||v|| QED
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Shadow

Shadow


Messages : 237
Date d'inscription : 15/09/2007
Age : 32

Cauchy-Schwarz Inequality proof Empty
MessageSujet: Re: Cauchy-Schwarz Inequality proof   Cauchy-Schwarz Inequality proof EmptyDim 8 Mai 2011 - 21:08

My teacher said that the proof for Cauchy-Schwarz's inequality is simply saying that |cos(teta)| = |u*v|/(||u|| ||v||) ≤ 1
Thus, |u*v| ≤ ||u|| ||v||

Which can save you some time if you get this question on the final exam.
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spirit

spirit


Messages : 222
Date d'inscription : 25/09/2007
Age : 31

Cauchy-Schwarz Inequality proof Empty
MessageSujet: Re: Cauchy-Schwarz Inequality proof   Cauchy-Schwarz Inequality proof EmptyDim 8 Mai 2011 - 21:20

That's great. 035
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