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I have an engineering problem where I have to find the smallest positive real root of a polynomial in $x$: $$Ax^5+Bx^3 - C = 0$$ Instead of solving numerically, I want simple approximative formulas ("design equations") that describe the behaviour. For that matter, I split the problem into two regimes:

  • Large $x$: $\ \ \ \ x^5$ is dominant $\Longrightarrow \ \ Ax^5 \approx C \ \ \Longrightarrow \ \ x \approx \sqrt[5]{C/A} =: x_A$
  • Small $x$: $\ \ \ \ x^3$ is dominant $\Longrightarrow \ \ Bx^3 \approx C \ \ \Longrightarrow \ \ x \approx \sqrt[3]{C/B} =: x_B$

The approximations work well below/above a certain threshold on $x$, but require an ugly case distinction. In order to avoid that and having something smoother than $$x \approx \min\{x_A, x_B\}$$ I tried Pythagorean-style combination of the inverses (inspired by the resistance of parallel circuits in electrical engineering) and found that $$x \approx 1\Big/ \ \left\|\left(\begin{matrix} 1/x_A \\ 1/x_B \end{matrix}\right)\right\|_4 = 1 \Big/ \sqrt[4]{1/x_A^4 + 1/x_B^4}$$ is a really, really good approximation. Pretty much perfect. That leads me to my actual question: Is it possible to argue why that 4-norm is such an amazing approximation? Does my initial polynomial have a certain structure that explains the high accuracy of my approximation?

Since I want to present/defend that stuff, I'd appreciate some sophistication.

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  • $\begingroup$ Are $A,B,C$ positive? If so the "smallest positive real root" is the only real root. $\endgroup$ – alex.jordan Sep 2 '15 at 17:19
  • $\begingroup$ For example, the sole real root of $17x^5 + 41x^3 - 105$ is approximately $1.17624...$; the approximation gives $1.17849...$. $\endgroup$ – Unit Sep 2 '15 at 17:20
  • $\begingroup$ @alex.jordan Yes they are positive. And sure, since $f(x) = Ax^5 + Bx^3$ is strictly increasing, there is only one relevant solution. $\endgroup$ – GDumphart Sep 2 '15 at 17:21
  • $\begingroup$ What do you mean by a "really, really good approximation", and have you tried lots of values of $A$, $B$ and $C$? $\endgroup$ – David Quinn Sep 2 '15 at 17:27
  • $\begingroup$ @DavidQuinn No I haven't done exhaustive tests with all kinds of $A,B$ combos, only what my problem yielded. Here is what I mean with great approximation: i.imgur.com/7Bfl8nt.png Only ridiculous zooming exposes some error. $\endgroup$ – GDumphart Sep 2 '15 at 17:41
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Your equation can be rewritten

$$\left(\frac x{x_A}\right)^5+\left(\frac x{x_B}\right)^3-1=0.$$

With $y=\dfrac x{x_A}$ and $r=\dfrac{x_A}{x_B}$, a single parameter remains:

$$y^5+r^3y^3-1=0.$$

Then $$r=\sqrt[3]{\frac{1-y^5}{y^3}}=\frac1y\sqrt[3]{1-y^5},$$ can be compared to your approximation $$y=\frac1{\sqrt[4]{1+r^4}},$$i.e. $$r=\sqrt[4]{\frac1{y^4}-1}=\frac1y\sqrt[4]{1-y^4}.$$ The agreement is indeed excellent on a wide range [abscissa $y$, ordinate $r$]: enter image description here

For small $y$, both behaviors are identical, $\dfrac1y$. For $y$ close to $1$, behaviors are similar, approximately $\sqrt[3]{5(1-y)}$ and$\sqrt[4]{4(1-y)}$, and a blend in between.

The good agreement is explained by the fact that the functional relations are similar, with the exponent $4$ intermediate between $3$ and $5$ (but the value $4$ has nothing "magical").

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  • $\begingroup$ Using Yves's parametrization, the absolute value of the difference between the true $y$ and your approximation turns out to be at most $0.01171$. The maximum absolute difference occurs at approximately $r = 0.66145$, where the true $y \approx 0.94545$ while the approximation is $0.95716$. $\endgroup$ – Robert Israel Sep 2 '15 at 18:50
  • $\begingroup$ @RobertIsrael: you can try slightly lower values than $4$ for the exponent, such as $3.7$. $\endgroup$ – Yves Daoust Sep 2 '15 at 18:53
  • $\begingroup$ Yes, I can. The best in terms of maximum absolute difference seems to be approximately $3.85$ (i.e. approximation $(1 + r^{3.85})^{-1/3.85}$), where the maximum absolute difference is about $0.007608$. $\endgroup$ – Robert Israel Sep 2 '15 at 21:44
  • $\begingroup$ Does this mean that there's an exact solution, rather than the approximation that the OP is using $\endgroup$ – Dr Xorile Sep 2 '15 at 22:03
  • $\begingroup$ Amazing answer, very intuitive and accounts for everything I wished for, thank you. I reproduced all your steps. $\endgroup$ – GDumphart Sep 3 '15 at 6:34

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