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richyw

  • 2 years ago

Under what conditions on \(\mathbf{a}\) and \(\mathbf{b}\) is Cauchy's inequality an equality? Examine the proof

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  1. richyw
    • 2 years ago
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    The proof is done as shown. We introduce the real variable\(t\) and consider the function \[f(t)=|\mathbf{a}-t\mathbf{b}|^2=|\mathbf{a}|^2-2t\mathbf{a}\cdot\mathbf{b}+t^2|\mathbf{b}|^2\] This is a quadratic function of t. Its minimum value occurs at \[t=\frac{(\mathbf{a}\cdot\mathbf{b})}{|\mathbf{b}|^2}\] and that minimum value is \[f\left((\mathbf{a}\cdot\mathbf{b})/|\mathbf{b}|^2\right)=|\mathbf{a}|^2-\frac{(\mathbf{a}\cdot\mathbf{b})}{|\mathbf{b}|^2}\]clearly \(f(t)\geq0\,\forall \,t\) so \[|\mathbf{a}|^2-\frac{(\mathbf{a}\cdot\mathbf{b})}{|\mathbf{b}|^2}\geq 0\] which completes the proof.

  2. richyw
    • 2 years ago
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    first of all. I have no idea why they did the proof like that. How would I ever guess to do that? second, how to I examine this proof to find where cauchy's inequality an equality?

  3. wio
    • 2 years ago
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    Usually really weird stuff like that comes about by doing things backwards.

  4. wio
    • 2 years ago
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    To answer your second question... look for things like division by variables to see where holes would be.

  5. wio
    • 2 years ago
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    Like what if \(|\mathbf{b}| = 0\)

  6. experimentX
    • 2 years ago
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  7. richyw
    • 2 years ago
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    does anyone know how I would show that it is an equality if an only if a and b are colinear?

  8. joemath314159
    • 2 years ago
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    If you assume that a and b are colinear, then:\[a=tb\]for some real number t. If you plug this in the cauchy inequality it turns out to be an equality.

  9. richyw
    • 2 years ago
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    does that show that this is the only case though?

  10. richyw
    • 2 years ago
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    like the "and only" part of if and only if

  11. joemath314159
    • 2 years ago
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    Thats correct, what I posted above is only one direction of the proof (if a and b are colinear, then there is equality in Cauchy's inequality). For the other direction, you should do a proof by contradiction. Assume that there is equality, and that a and b are not colinear. Since a and b are not colinear, you have that the function:\[f(t)=|a-tb|^2\] is always strictly greater than 0 for any value of t. So using:\[|a-tb|^2>0\]and \[|(a\cdot b)|=|a|\cdot |b|\] you should end up with a contradiction.

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