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Hello, I suspect this reduces to a homework problem, but I've been a bit hung up on it for the last few hours. I'm trying to minimize the (convex) function $f(x) = 1/x + ax + bx^2$ , where $x,a,b>0$. Specifically, I'm interested in the minimal objective function value as a function of $a$ and $b$. Since finding the minimizer $x^*$ is tricky (requires solving a cubic), I figured I'd try and find a lower bound using the following argument: if $b=0$, the minimizer is $x=1/\sqrt{a}$ and the minimal value is $2\sqrt{a}$. If $a=0$, the minimizer is $x=(2b)^{-1/3}$ and the minimal value is $\frac{3\cdot2^{1/3}}{2}b^{1/3}$. Therefore, one possible approximate solution is the convex combination

$(\frac{a}{a+b})\cdot2\sqrt{a} + (\frac{b}{a+b})\cdot\frac{3\cdot2^{1/3}}{2}b^{1/3}$.

Numerical simulations suggest that the above expression is a lower bound for the minimal value. Does this follow from some nice result about parameterized convex functions? It seems like it shouldn't be hard to prove. I guess in a nutshell I just want to prove that for all $x,a,b>0$ we have

$(\frac{a}{a+b})\cdot2\sqrt{a} + (\frac{b}{a+b})\cdot\frac{3\cdot2^{1/3}}{2}b^{1/3} \leq 1/x + ax + bx^2$. Thanks!

EDIT: It also appears that if I take the convex combination

$(\frac{a^{3/5}}{a^{3/5}+b^{2/5}})\cdot2\sqrt{a} + (\frac{b^{2/5}}{a^{3/5}+b^{2/5}})\cdot\frac{3\cdot2^{1/3}}{2}b^{1/3}$

then I get a tighter lower bound, and in fact the lower bound is within a factor of something like $3/2$ of the true minimal solution.

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  • $\begingroup$ can't you just split 1/x into the linear combination and use that. $\endgroup$ Apr 16, 2011 at 19:24
  • $\begingroup$ Sorry, I don't understand -- how does the "splitting" work? $\endgroup$ Apr 17, 2011 at 8:29
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    $\begingroup$ but computing the roots of your cubic don't seem to be too much more complicated than computing the lower bounds? en.wikipedia.org/wiki/Cubic_function $\endgroup$
    – Suvrit
    Apr 17, 2011 at 8:50
  • $\begingroup$ @Suvrit: is it? The expression for the roots of a cubic looks pretty complicated to me, and the lower bound is just a pair of terms. $\endgroup$ Apr 17, 2011 at 20:40

2 Answers 2

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The first inequality is true. Write $$f=\frac{a}{a+b}f_0+\frac{b}{a+b}f_1,$$ where $f_0$ and $f_1$ correspond to the case $b=0$ and $a=0$, respectively. You know that $f_0\ge2\sqrt a$ and $f_1\ge\frac{3\cdot2^{1/3}}{2}b^{1/3}$. This implies $$f\ge\frac{a}{a+b}2\sqrt a+\frac{b}{a+b}\frac{3\cdot2^{1/3}}{2}b^{1/3}.$$

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  • $\begingroup$ Is it true? We have $\frac{a}{a+b}f_0 + \frac{b}{a+b}f_1 = \frac{1}{x}+\frac{a^2}{a+b}x + \frac{b^2}{a+b}x^2$ $\endgroup$
    – Maciej S.
    May 8, 2011 at 9:10
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As Nishant Chandgotia sugessted: simply write $f(x) = \left(p\cdot \frac{1}{x} + ax\right) + \left((1-p)\frac{1}{x}+bx^2 \right)$ for some parametr $p\in[0,1]$.

For the first term, minimizer is equal to $ p^{\frac{1}{2}}a^{-\frac{1}{2}}$ and the minimal value is $p^{\frac{1}{2}} 2a^{\frac{1}{2}}$. For the second therm, minimizer is equal to $(1-p)^{\frac{1}{3}}(2b)^{-\frac{1}{3}} $ and the minimal value is $(1-p)^{\frac{2}{3}} \cdot\frac{3\cdot 2^{\frac{1}{3}} }{2} b^{\frac{1}{3}}$

The best estimate is achieved when both minimizers are equal, which means, in therms of $p$, that

\begin{equation} \left(\frac{p}{a}\right)^{3} = \left( \frac{1-p}{2b}\right)^{2} \end{equation}

Note, that this equation has a solution in interval $0 < p < 1$ by Mean-value theorem, unfortunately not expressible in nice way.

Any way, we get for any $p\in[0,1]$ the following estimate:

\begin{equation} f(x) \geqslant p^{\frac{1}{2}} \cdot 2a^{\frac{1}{2}} + (1-p)^{\frac{2}{3}} \cdot\frac{3\cdot 2^{\frac{1}{3}} }{2} b^{\frac{1}{3}} \end{equation}

Finally, we check that $p^{\frac{1}{2}} + (1-p)^{\frac{2}{3}} \geqslant 1$. This inequality implies, that estimate reamins valid after using any convex combination instead of weights $p^{\frac{1}{2}}, (1-p)^{\frac{2}{3}}$, i.e.

\begin{equation} f(x) \geqslant \alpha \cdot 2a^{\frac{1}{2}} + (1-\alpha) \cdot\frac{3\cdot 2^{\frac{1}{3}} }{2} b^{\frac{1}{3}} \end{equation}

This can be seen immediately, however, by the inequality $f \geqslant \alpha f_0 + (1-\alpha) f_1$. My goal was to complete presented ideas and to show when exact optimum is attained.

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