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This is an easy question, in a way, but I've been trying to solve it for the better part of the day and I'm getting nowhere. Specifically, I have the equation $$ f(x) := a - 2\sqrt{bx} - \sqrt{3}x^{3/4} -4x $$ where $x\geq0,a>0,b>0$. I'm trying to find (roughly) the largest value of $x$ s.t. $f(x) \geq 0$, but I don't really need an optimal solution, just anything better than $x = 0$ would make me happy at this point. I've tried solving the polynomial with the higher order quadratic formula, which leads to a god-awful mess, as well as trying to come up with workable lower-bounds for $f$, with no luck. Any ideas would be greatly appreciated.

update: Under the additional assumption that $a,b \in (0,1]$, we can conclude that $x < 1$, which implies $x^l \leq x^m$ if $l \geq m$. We then have: \begin{align} f(x) & \geq a - 2\sqrt{b}\sqrt{x} - \sqrt{3}\sqrt{x} - 4\sqrt{x}\\ &= a - \sqrt{x}(2\sqrt{b} + \sqrt{3} + 4)\\ &\geq a - \sqrt{x}(2\sqrt{b} +5.8) \end{align}

Setting $x = \frac{a^2}{(2\sqrt{b} +5.8)^2}$, then, gives a solution other than the trivial $x=0$. Although this is a rather crude attempt.

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    $\begingroup$ This isn't a polynomial (in $x$). It's a polynomial of degree $4$ in $\sqrt[4]{x}$. How rough of a solution are you willing to accept? What regime of $a, b$ do you care about (e.g. $a, b \to \infty$ or what)? $\endgroup$ – Qiaochu Yuan Jul 10 '13 at 5:46
  • $\begingroup$ @QiaochuYuan Let's say $a$ and $b$ are fixed constants in $(0,1]$, as for how rough, as I said, I'm just trying to improve on my current best $x=0$ $\endgroup$ – fairidox Jul 10 '13 at 5:49
  • $\begingroup$ Notice that $f$ is convex, which means that any tangent to the curve at a point $x$ such that $f(x) > 0$ will intercept the abscissa axis at a better approximation. The only problem is that the tangent at $x=0$ is vertical, but you can use the estimate you found. This might not be very practical, though. $\endgroup$ – zuggg Jul 10 '13 at 7:01
  • $\begingroup$ @zuggg good point, thanks, if I get you right I can take the first-order Taylor expansion of $f$ at any point $x_0$, and find the intersect with 0, to find a better minimum $x_1$. I can then repeat this over with a Taylor approximation about $x_1$, essentially doing Newton's method by hand. $\endgroup$ – fairidox Jul 10 '13 at 7:30
  • $\begingroup$ @anonymous_21321 Yes, that is correct. You would get an increasing sequence converging towards the root of $f$, but a major drawback is that the expressions quickly become ugly, so it is more of a numerical method than an analytical one. Also, you still need an initial guess for the value of $x$. $\endgroup$ – zuggg Jul 10 '13 at 7:55
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Here's something that's slightly better than the estimate you gave above. To simplify notation let $y=x^{1/4}$ and $B=\sqrt{b}$. Then $$ \begin{align} f(y) &=a-2By^2-\sqrt{3}y^3-4y^4 \\ &=2B\left(\frac{a}{2B}-y^2\left(1+\frac{\sqrt{3}}{2B}y+\frac{2}{B}y^2\right)\right)\\ &=2B\left(\frac{a}{2B}-y^2\left(1+\frac{\sqrt{3}}{4B}y\right)^2-\left(\frac{2}{B}-\frac{3}{16B^2}\right)y^4\right) \end{align} $$ If $2/B-3/16B^2\ge 0$ then $$ f(y) \ge 2B\left(\frac{a}{2B}-Y^2-\left(\frac{2}{B}-\frac{3}{16B^2}\right)Y^4\right) $$ where $Y=y+\sqrt{3}y^2/4B> y$. Then we can use the quadratic formula twice to solve for $Y^2$ as the root of the RHS, then for $y$ from the definition of $Y$.

It's kind of messy, but gives $$ x_1=\left(\frac{2\sqrt{b}}{\sqrt{3}}\left(-1+\sqrt{1+\sqrt{\frac{24}{32\sqrt{b}-3}\left(-1+\sqrt{1+\frac{(32\sqrt{b}-3)a}{8b^{3/2}}}\right)}}\right)\right)^4 $$ If $2/B-3/16B^2<0$ then instead we have $$ f(y)> 2B\left(\frac{a}{2B}-Y^2\right) $$ and the corresponding bound $$ x_2=\left(\frac{2\sqrt{b}}{\sqrt{3}} \left(-1+\sqrt{1+\sqrt{\frac{3a}{2b^{3/2}}}}\right)\right)^4 $$ Write $x_0=(a/(2\sqrt{b}+5.8))^2$ for your estimate above, then for example $$ \begin{array}{c|c|c|c|c} a & b & x_0 & x_1 & x_2 & \hat{x} \\ \hline 0.4 & 0.3 & 0.0034 & 0.0175 & 0.0398 & 0.0267 \\ 0.2 & 0.002 & 0.0012 & - & 0.0141 & 0.0220 \end{array} $$ where $\hat{x}$ is the root found numerically. If $2/B-3/16B^2>0$ then $x_2$ gives an upper bound as well; in the other case we can also get an upper bound but it requires a different choice of roots, so it will be different from $x_1$.

A possible refinement is as follows. Given bounds $l_1,u_1$ with $l_1\le \hat{x}\le u_1$, for $x\in[l_1,u_1]$ we can write $$ f(x) \ge a-\sqrt{x}(2\sqrt{b}+\sqrt{3}u_1^{1/4}+4\sqrt{u_1}) \\ l_2 = \left(\frac{a}{2\sqrt{b}+\sqrt{3}u_1^{1/4}+4\sqrt{u_1}}\right)^2 $$ and $$ f(x) \le a-\sqrt{x}(2\sqrt{b}+\sqrt{3}l_1^{1/4}+4\sqrt{l_1}) \\ u_2 = \left(\frac{a}{2\sqrt{b}+\sqrt{3}l_1^{1/4}+4\sqrt{l_1}}\right)^2 $$ then $l_2\le \hat{x}\le u_2$ and of course this can be iterated. For example, if $a=0.4,b=0.3$ as in the example above, taking $l_1=x_1=0.0175$ and $u_1=x_2=0.0398$ we can get the successive bounds $$ \begin{align} l_2=0.0225 & u_2=0.0315 \\ l_3=0.0249 & u_3=0.0286 \\ l_4=0.0259 & u_4=0.0274 \end{align} $$

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  • $\begingroup$ very nice, thanks! $\endgroup$ – fairidox Jul 25 '13 at 17:28

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