0

in the following equation

$$ax+bx^{5/3} = c$$

$a, b$, and $c$ are constant. Through numerical result I know that $x$ is near zero and positive. I want to find an analytical solution or approximation. I really appreciate if you could help me through this matter.

gt6989b
  • 54,422
amir
  • 3
  • Why are you sure one exists? E.g. substituting $u = x^{1/3}$ you get $au^3 + bu^5 = c$, a quintic, which may not generally have any analytic solutions? – gt6989b Feb 24 '20 at 16:00
  • 4
    @gt6989b The claim that quintics might not have analytic solutions is false. What you are remembering is expressible in radicals. –  Feb 24 '20 at 16:01
  • @flan interesting I just deleted my comment saying much the same as gt. I thought that this was true? – Chinny84 Feb 24 '20 at 16:03
  • @flan that's fair. still not sure you can write down a solution to an arbitrary quintic, even in terms of special functions etc. – gt6989b Feb 24 '20 at 16:03
  • 1
    @gt6989b It is well known that you can, for all degrees. In the case of the quintic you can write it in terms of the Bring radical, for example. –  Feb 24 '20 at 16:05
  • @flan +1, thanks for teaching me something new. :) appreciate it a lot – gt6989b Feb 24 '20 at 16:08
  • @gt6989b I have tried your suggestion before. I also tried to form as an exponential form and try Taylor series for $exp(bx^{2-\delta})$ but it becomes too complicated to solve. – amir Feb 24 '20 at 16:19
  • So following @flan's suggestion, perhaps look at the 2 approaches described in Wiki article on Bring radical, transforming your equation to eliminate the cubic term and possibly eventually the quadratic term – gt6989b Feb 24 '20 at 16:24
  • Thanks to @flan 's comments I am trying to find a solution. Thank you for your consideration. – amir Feb 24 '20 at 16:27
  • Welcome to the site ! . This is an interesting problem from a numerical point of view. Are $(a,b,c)$ all positive ? How large could be $b$ compared to $a$ ? – Claude Leibovici Feb 24 '20 at 16:31
  • @ClaudeLeibovici Thanks for your interest. c = 2, and $b = a \times 165.5170826624471$. For example, when a = 60.4246, b = 10001.29897874889, c = 2, I find the approximation of x^* \approx 0.00545. – amir Feb 24 '20 at 16:45

2 Answers2

1

As comments say, you can write this as $a u^3 + b u^5 = c$, where $u = x^{1/3}$. This has a real, positive root (if $a, b, c$ are positive, that is). If $u$ is near 1, one (or a few) iterations of Newton's method starting at $u_0 = 1$ should give a precise value. I.e.,

$$u_1 = \frac{2 a + 4 b + c}{3 a + 5 b}$$

(Next one gets hairy...)

vonbrand
  • 27,812
  • amir said $x$ (and therefore $u$) is near $0$, not $1$. But I agree, if all they want is a solution for a particular set of values for $a,b,c$, then Newton's method is the way to go. No "analytic solution" will give you as good a value with less calculation than Newton's method (excepting when the solution is a low-denominator rational, which is clearly not going to be true here). – Paul Sinclair Feb 25 '20 at 00:31
1

First of all, let us consider some numerical aspects which will (probably) be useful later.

I shall assume that constants $(a,b,c)$ are all positive.

We are looking for the zero of function $$f(x)=a x+bx^{5/3}-c$$ for which $$f'(x)=a+\frac{5}{3} b x^{ 2/3} \qquad \text{and} \qquad f''(x)=\frac {10b} {9x^{ 1/3}}\qquad \text{and} \qquad f'''(x)=-\frac{10 b}{27 x^{4/3}}$$

Starting at $\color{red}{x_0=0}$, the first iterate of Newton method is $\color{red}{x_1=\frac c a}$ for which $$f(x_1)=b \left(\frac{c}{a}\right)^{5/3} >0 \qquad \text{and} \qquad f''(x_1)=\frac{10 b}{9 \left(\frac{c}{a}\right)^{1/3}} >0$$ By Darboux theorem, this means that, even if we start with this probably very poor estimate, Newton method will converge without any overshoot of the solution.

As an example, using $a=60$, $b=10000$ and $c=2$ the iterates will be $$\left( \begin{array}{cc} n & x_n \\ 0 & 0.0333333 \\ 1 & 0.0140051 \\ 2 & 0.0072200 \\ 3 & 0.0055641 \\ 4 & 0.0054253 \\ 5 & 0.0054242 \end{array} \right)$$ So, there is no risk to generate a better starting point $$\color{red}{x_2=\frac{c}{a}-\frac{3 b \left(\frac{c}{a}\right)^{5/3}}{20 \left(2500 \left(\frac{c}{a}\right)^{2/3}+9\right)}}$$ at which we can compute the values $f(x_2)$, $f'(x_2)$, $f''(x_2)$, $f'''(x_2)$.

Now, let us use Householder method which will give $$\color{red}{x_3=x_2+\frac{3 f(x_2) \left(f(x_2) f''(x_2)-2 f'(x_2)^2\right)}{f(x_2)^2 f'''(x_2)+6 f'(x_2)^3-6 f(x_2) f'(x_2) f''(x_2)}}$$ For the worked example, this should give $x_3=0.0056389$. If we repeat once this last step, this should give $x_4=0.0054242$ which is the answer.

Claude Leibovici
  • 260,315