Spivak's Calculus (4ed) 1.19 Schwarz inequality

AI Thread Summary
The discussion centers on proving the Schwarz inequality using a quadratic equation derived from the condition that x1 and x2 cannot be expressed as a scalar multiple of y1 and y2. Participants clarify the transition from the inequality to the quadratic form, emphasizing the significance of the expression being greater than zero. Confusion arises around the interpretation of the quadratic's coefficients and the implications of the lack of real solutions, which leads to the conclusion of the Schwarz inequality. The conversation highlights the need for clarity in mathematical notation and reasoning. Ultimately, the participants work through the problem collaboratively, enhancing their understanding of the concepts involved.
swevener
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The problem
Given the Schwarz inequality, x_{1}y_{1} + x_{2}y_{2} \leq \sqrt{x_{1}^{2} + x_{2}^{2}} \sqrt{y_{1}^{2} + y_{2}^{2}}, prove that if x_{1} = \lambda y_{1} and x_{2} = \lambda y_{2} for some number \lambda \geq 0, then equality holds. Prove the same thing if y_{1} = y_{2} = 0. Now suppose that y_{1} and y_{2} are not both 0 and that there is no number \lambda such that x_{1} = \lambda y_{1} and x_{2} = \lambda y_{2}. Then

\begin{align*}<br /> 0 &amp;\lt (\lambda y_{1} - x_{1})^{2} + (\lambda y_{2} - x_{2})^{2} \\<br /> &amp;= \lambda^{2} (y_{1}^{2} + y_{2}^{2}) - 2 \lambda (x_{1} y_{1} + x_{2} y_{2}) + (x_{1}^{2} + x_{2}^{2}).<br /> \end{align*}
Use the solutions to the quadratic equation to prove the Schwarz ineq.

My confusion
I can do all the parts of this, but I'm not sure how they fit together. I can't figure out how we go from the Schwarz ineq. to the quadratic equation, so I don't know why the lack of a real solution proves the ineq. I've tried working it forward and backward and all I've got is wasted paper and a sore wrist.
 
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Hi swevener! http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif

You say you can do all the parts? Yet the problem leads you in steps to the quadratic equation, so which part can't you do?
Maybe you can't see the origin of this inequality:
\begin{align*}<br /> 0 &amp;\lt (\lambda y_{1} - x_{1})^{2} + (\lambda y_{2} - x_{2})^{2} \end{align*}
It arises because you are told there is now an "error" or difference between x1 and λy1 (and/or x2 and λy2) so add these differences together and equate result to something greater than 0. You then expand the brackets and arrive at the quadratic shown.
 
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Exactly what I was looking for. Thank you! :)
 
swevener said:
\begin{align*}<br /> 0 &amp;\lt (\lambda y_{1} - x_{1})^{2} + (\lambda y_{2} - x_{2})^{2} \\<br /> &amp;= \lambda^{2} (y_{1}^{2} + y_{2}^{2}) - 2 \lambda (x_{1} y_{1} + x_{2} y_{2}) + (x_{1}^{2} + x_{2}^{2}).<br /> \end{align*}

I'm just getting started with Spivak and math and am confused about the notation here, perhaps someone can clarify it for me. It seems to me this can be read two ways:

(a) \begin{align*}<br /> 0 &amp;\lt (\lambda y_{1} - x_{1})^{2} + (\lambda y_{2} - x_{2})^{2}) \\<br /> 0 &amp;= \lambda^{2} (y_{1}^{2} + y_{2}^{2}) - 2 \lambda (x_{1} y_{1} + x_{2} y_{2}) + (x_{1}^{2} + x_{2}^{2}).<br /> \end{align*}

or (b) \begin{align*}<br /> 0 &amp;\lt (\lambda y_{1} - x_{1})^{2} + (\lambda y_{2} - x_{2})^{2} \\<br /> &amp;= \\<br /> 0 &amp;\lt \lambda^{2} (y_{1}^{2} + y_{2}^{2}) - 2 \lambda (x_{1} y_{1} + x_{2} y_{2}) + (x_{1}^{2} + x_{2}^{2}).<br /> \end{align*}

I'm inclined to read this as (b), and that can easily be shown to hold true, but I want to make sure I understand what's going on here and am not making a mistake as I proceed.
 
Yes, (b) is what's intended. I didn't like the liberties the authors took there, either. I think it could definitely have been expressed with more rigor.

This symbol ⇔[/size][/color] would have been appropriate; though I'd be content with just plain ∴[/size][/color]
 
NascentOxygen said:
Yes, (b) is what's intended. I didn't like the liberties the authors took there, either. I think it could definitely have been expressed with more rigor.

This symbol ⇔[/size][/color] would have been appropriate; though I'd be content with just plain ∴[/size][/color]

Thanks for the prompt reply. So, I'm on the right track but there is a gap that I can't solve in this problem. I glanced after many hours at Spivak's solution and it still doesn't satisfy me. He says the equation:

\begin{align*}<br /> 0 &amp;\lt \lambda^{2} (y_{1}^{2} + y_{2}^{2}) - 2 \lambda (x_{1} y_{1} + x_{2} y_{2}) + (x_{1}^{2} + x_{2}^{2}).<br /> \end{align*}

has no solution for \lambda. That's all well and good, but then he goes on to infer that from the prior problem's relation to the quadratic equation we must have:

\begin{align*}<br /> (2(x_{1}y{1} + x_{2}y{2}/(y{1}^{2} + y_{2}^{2}))^{2} &amp;- 4(x_{1}^{2} + x_{2}^{2})/(y{1}^{2} + y_{2}^{2}) &amp;\lt 0<br /> \end{align*}

which yields the Schwarz inequality. That's well and good, and I can certainly see how this is supposed to represent the formula b^{2} - 4c \lt 0, but where does this formula even come from?

It seems to me \begin{align*}<br /> 0 &amp;\lt \lambda^{2} (y_{1}^{2} + y_{2}^{2}) - 2 \lambda (x_{1} y_{1} + x_{2} y_{2}) + (x_{1}^{2} + x_{2}^{2}).<br /> \end{align*}

Is in the: ax^{2} + bx + c = a(x^{2} +bx/a + c/a) form, which would mean if we let a = (y{1}^{2} + y_{2}^{2}) and x = -\lambda, and let c = (x_{1}^{2} + x_{2}^{2}), and let b = 2(x_{1}y{1} + x_{2}y{2}) then our formula isn't actually b^{2} - 4c \lt 0 but
b^{2}/a - 4c/a \lt 0. I'm just not seeing how to close the gap.

I hope this isn't all nonsense, and I hope there aren't too many typos (my first shot at tex).

Cheers to anyone who read,
A
 
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Likes jorgebar
Don't replace λ with -x, leave it as λ

set b = –2 (x₁y₁ + x₂y₂)
 
It just dawned on me what other moves I need to close the gap, and I picked up on a few other errors above. I don't know why I didn't see it before. Thanks for your help.
 
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