I am working on a problem that asks me to solve for x for an iteration of the Lennard-Jones equation. Specifically, I am trying to determine the equilibrium separation between two hypothetical atoms. The equation is given to me as:

PE(x) = 2.3x10^-134 Jm^12/x^12 - 6.6x10^-77 Jm^6/x^6

I'm using MapleSoft 2015 to graphically represent this equation, and I was able to use the minimize function within MapleSoft to determine that the minimum of the graph occurs when PE= -4.734782609x10^-20 J

However, I am not certain how to get the x value at that point. My graphic calculator (a TI-84 Plus) won't recognize negative exponents beyond 1E-99 or so and it just give me an error any time I try to graph this equation, so I can get at the value that way.

I'm really green with MapleSoft; I'm sure there's a way to use the software to solve for x at a given f(x) value, but I don't know how. I tried using:

solve([equation], x) and got some answers, but they're all in the form of:

(some number) - (some number)I

I have no idea what that means so it doesn't help me and at any rate MapleSoft returned a number of solutions.

What I need to know is how do I go about solving for x? Could this be done as a quadratic equation with x = x^6? Please note, I'm not asking for anyone to give me the answer; I intend to do my own homework. I just need a method of getting at the answer that gives me results rather than frustration.

Edit: Here is the string of solutions Maple returns when I right click the equation and choose "solve":

{x = 2.977618011*10^(-10)-3.978806442*10^(-16)*I}, {x = 1.488812451*10^(-10)+2.578690850*10^(-10)*I}, {x = -1.488805560*10^(-10)+2.578694829*10^(-10)*I}, {x = -2.977618011*10^(-10)+3.978806442*10^(-16)*I}, {x = -1.488812451*10^(-10)-2.578690850*10^(-10)*I}, {x = 1.488805560*10^(-10)-2.578694829*10^(-10)*I}, {x = 2.977618011*10^(-10)+3.978806442*10^(-16)*I}, {x = 1.488805560*10^(-10)+2.578694829*10^(-10)*I}, {x = -1.488812451*10^(-10)+2.578690850*10^(-10)*I}, {x = -2.977618011*10^(-10)-3.978806442*10^(-16)*I}, {x = -1.488805560*10^(-10)-2.578694829*10^(-10)*I}, {x = 1.488812451*10^(-10)-2.578690850*10^(-10)*I}

What does it mean when "*I" is in the solution? Is this saying an imaginary number is the multiplier, i.e. sqrt(-1)?


As set, when you compute the derivative and ask for its zero, you have to solve a polynomial of degree $13$ in $x$. So, I am even surprised by the fact that the post only reports $12$ solutions. Among these solutions, some of them have to be complex but not all of them.

You are basically considering $$y=\frac a {x^{12}}-\frac b {x^6}$$ So $$y'=\frac{6 b}{x^7}-\frac{12 a}{x^{13}}=\frac{6 \left(b x^6-2 a\right)}{x^{13}}$$ So, the solution corresponds to $x^6=\frac{2a}b$ and, for this value $y=-\frac{b^2}{4a}$. Using $$a=23 \times 10^{-135} J\times m^{12} \qquad , \qquad b=66 \times 10^{-78} J\times m^{12}$$ the minimum is then obtained $$x^6=\frac{23}{33}\times 10^{-57} \implies y=-\frac{1089 }{23} \times 10^{-21} \times J \times m^{12}$$ which is exactly your result.

With regard to the solutions in $x$, since you solve $bx^6=2a$, they are $$\left\{x= -\frac{\sqrt[6]{\frac{23}{33}}}{10^9 \sqrt{10}}\right\},\left\{x= \frac{\sqrt[6]{\frac{23}{33}}}{10^9 \sqrt{10}}\right\},\left\{x= -\frac{\sqrt[3]{-1} \sqrt[6]{\frac{23}{33}}}{10^9 \sqrt{10}}\right\},\left\{x= \frac{\sqrt[3]{-1} \sqrt[6]{\frac{23}{33}}}{10^9 \sqrt{10}}\right\},\left\{x=-\frac{(-1)^{2/3} \sqrt[6]{\frac{23}{33}}}{10^9 \sqrt{10}}\right\},\left\{x= \frac{(-1)^{2/3} \sqrt[6]{\frac{23}{33}}}{10^9 \sqrt{10}}\right\}$$

and, as you can see, only the first and second are real.

But, as you say, the problem would have been made more simple setting from the beginning $X=\frac 1 {x^6}$ which reduces everything to a quadratic problem. $$y=a X^2-b X \implies \frac{dy}{dx}=\frac{dy}{dX}\frac{dX}{dx}=-6(2aX-b)\frac 1 {x^7}$$

You can notice that, among the numbers you gave, two of them have an imaginary term with an exponent $10^{-16}$ while all other are $10^{-10}$. The first are due to roundoff errors (because working with decimal numbers). The othe are correct.

  • $\begingroup$ Thank you very much Claude for your excellent, insightful answer. You have been a big help. $\endgroup$ – 1John5vs7 Jul 4 '16 at 9:32
  • $\begingroup$ @1John5vs7. You are very welcome ! $\endgroup$ – Claude Leibovici Jul 4 '16 at 9:37

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