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experimentX

  • 2 years ago

if \( a\) and \( b\) are two vectors in linear vector space, prove Cauchy-Swartz inequality \( |a|.|b| \geq |a.b|\)

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  1. phi
    • 2 years ago
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    I have seen a proof that u|dw:1346599918682:dw|ses the idea that a quadratic y= a x^2 + bx +c is always positive for all values of x if its discriminate is negative: this means complex roots

  2. experimentX
    • 2 years ago
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    I'm particularly looking for something to write in exam

  3. phi
    • 2 years ago
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    use this idea with | a + x b|^2 where a and b are vectors and x is a scalar this must always be positive for all values of x

  4. phi
    • 2 years ago
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    write the magnitude squared as a dot product | a + x b|^2 ≥0 \[|a|^2 + 2(a \cdot b) x + |b|^2 x^2 ≥ 0\] as noted above, the discriminate of this quadratic in x must be negative to guarantee all values ≥ 0 for all x: \[ 4 (a \cdot b)^2 -4 |a|^2 |b|^2 ≤ 0 \]

  5. experimentX
    • 2 years ago
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    how do you generalize it to n space?

  6. phi
    • 2 years ago
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    The vectors a and b are in n space.

  7. experimentX
    • 2 years ago
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    i see..

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