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$z=x_1^2+x_2^2$
$u=(u_1,u_2)$
$v=(v_1,v_2)$

Height of arc: $f[\theta u+(1-\theta )v]= f[\theta u_1+(1-\theta)v_1,\theta u_2 + (1-\theta)v_2 ]$ $= [\theta u_1+(1-\theta)v_1]^2 + [\theta u_2 + (1-\theta)v_2 ]^2$

Height of line segment: $\theta f(u) + (1-\theta)f(v) =\theta(u_1^2+u_2^2)+(1-\theta)(v_1^2+v_2^2)$

My book says that subtracting the height of the arc from the height of the line segment should result (after collecting terms) in the following:
$\theta(1-\theta)(u_2^2+u_2^2)+\theta(1-\theta)(v_1^2+v_2^2)-2\theta(1-\theta)(u_1v_1+u_2v_2)$

I've been trying for way too long to get that result, but I can't replicate it. Any hints?

Jordi
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  • I just checked : after developing your result and the result of your book : they coincide... with the common, simpler expression $\theta(1-\theta)((u_1-v_1)^2+(u_2-v_2)^2)$ which is maximal for $\theta=1/2$, its value being a quarter of the squared length of the considered segment. – Jean Marie Feb 09 '16 at 18:42

1 Answers1

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Subtracting the height of the arc from the height of the line segment we get

$$\theta(u_1^2+u_2^2)+(1-\theta)(v_1^2+v_2^2)- [\theta u_1+(1-\theta)v_1]^2 - [\theta u_2 + (1-\theta)v_2 ]^2$$

$$=\theta(u_1^2+u_2^2)+(1-\theta)(v_1^2+v_2^2) \\-\theta^2 u_1^2-(1-\theta)^2v_1^2-2\theta(1-\theta)u_ 1v_1-\theta^2 u_2^2+(1-\theta)^2v_2^2-2\theta(1-\theta)u_ 2v_2$$

$$=\color{red}{\theta(u_1^2+u_2^2)}+\color{blue}{(1-\theta)(v_1^2+v_2^2)}\\-{\color{red} {\theta^2 u_1^2}}-\color{blue}{(1-\theta)^2v_1^2}-\color{green}{2\theta(1-\theta)u_ 1v_1}-\color{red}{\theta^2 u_2^2}-\color{blue}{(1-\theta)^2v_2^2}-\color{green}{2\theta(1-\theta)u_ 2v_2} $$

$$=\color{red}{\theta(1-\theta)(u_1^2+u_ 2^2)}+\color{blue}{\theta(1-\theta)(v_1^2+v_ 2^2)}-\color{green}{2\theta(1-\theta)(u_1v_1+u_2v_2).}$$

mfl
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  • Ahh! I gave up too soon. Thank you! I just thought "how do I ever get rid of that $\theta^2$. I guess it's one of those days.. – Jordi Feb 09 '16 at 18:38