Is there a good way to simplify this expression?

Question

The Short Version

Is there a way to simplify this expression?

$$ \left(\left(\left(d × \left(\frac{j}{2}\right)^2\right)^2 − \frac{1}{27} × \left(\frac{j^2}{2 s}\right)^6\right)^\frac{1}{2} + d × \left(\frac{j}{2}\right)^2\right)^\frac{1}{3} + \frac{1}{3} × \left(\frac{j^{2}}{2 s}\right)^2 × \left(\left(\left(d × \left(\frac{j}{2}\right)^2\right)^2 − \frac{1}{27} × \left(\frac{j^2}{2 s}\right)^6\right)^\frac{1}{2} + d × \left(\frac{j}{2}\right)^2\right)^\frac{−1}{3} − \frac{j^2}{2 s} $$

Specifically, I'd like to condense the cube root portion and the inverse cube root portion into a single root, if possible.

The Longer Version

We have five positive-valued variables:

• $s_{max}$
• $j_{max}$
• $a_{max}$
• $v_{max}$
• $d_{max}$

We also have the function $\operatorname{Min}(n_1,\, n_2,\, …,\, n_k)$, which returns the input with the lowest value.

Finally, we have three variables whose values are each based on combinations of the previously-defined variables:

$$ j_{limit}\,=\,{\operatorname{Min}\begin{pmatrix} \left(j_{max} × \frac{s_{max}^0}{0!}\right)^\frac{1}{1},\\ \left(a_{max} × \frac{s_{max}^1}{1!}\right)^\frac{1}{2},\\ \left(v_{max} × \frac{s_{max}^2}{2!}\right)^\frac{1}{3},\\ \left(d_{max} × \frac{s_{max}^3}{3!}\right)^\frac{1}{4} \end{pmatrix}} $$

$$ a_{limit}\,=\,{\operatorname{Min}\begin{pmatrix} a_{max},\\ \left(v_{max} × j_{limit} + \left(\frac{j_{limit}^2}{2 s_{max}}\right)^2\right)^\frac{1}{2} - \frac{j_{limit}^2}{2 s_{max}},\\ \left(\left(\left(d_{max} × \left(\frac{j_{limit}}{2}\right)^2\right)^2 − \frac{1}{27} × \left(\frac{j_{limit}^2}{2 s_{max}}\right)^6\right)^\frac{1}{2} + d_{max} × \left(\frac{j_{limit}}{2}\right)^2\right)^\frac{1}{3} + \frac{1}{3} × \left(\frac{j_{limit}^{2}}{2 s_{max}}\right)^2 × \left(\left(\left(d_{max} × \left(\frac{j_{limit}}{2}\right)^2\right)^2 − \frac{1}{27} × \left(\frac{j_{limit}^2}{2 s_{max}}\right)^6\right)^\frac{1}{2} + d_{max} × \left(\frac{j_{limit}}{2}\right)^2\right)^\frac{−1}{3} − \frac{j_{limit}^2}{2 s_{max}} \end{pmatrix}} $$

and

$$ v_{limit}\,=\,{\operatorname{Min}\begin{pmatrix} v_{max},\\ \left(d_{max} × a_{limit} + \left(\frac{a_{limit}^2}{2 j_{limit}}\right)^2\right)^\frac{1}{2} - \frac{a_{limit}^2}{2 j_{limit}} \end{pmatrix}} $$

The third option for the value of $a_{limit}$ stands out for being so much longer than all the other expressions in the values of the limit-variables. Can it be condensed any? Is there a better way to write it? As-is, it feels very unsatisfying.

Answer 1

Let $\,\displaystyle a = d \cdot \left(\frac{j}{2}\right)^2\,$ and $\,\displaystyle b = \frac{1}{3} \cdot \left(\frac{j^2}{2s}\right)^2\,$, then the expression (without the radical-free term at the end) is:

$$ \require{cancel} \begin{align} x &= \sqrt[3]{a + \sqrt{a^2 − b^3}} + \frac{b}{\sqrt[3]{a + \sqrt{a^2 − b^3}}} \cdot \color{red}{\frac{\sqrt[3]{a - \sqrt{a^2 − b^3}}}{\sqrt[3]{a - \sqrt{a^2 − b^3}}}} \\ &= \sqrt[3]{a + \sqrt{a^2 − b^3}} + \frac{ \bcancel{b}\,\sqrt[3]{a - \sqrt{a^2 − b^3}}}{\bcancel{\sqrt[3]{\cancel{a^2} - (\cancel{a^2} − b^3)}}} \\ &= \sqrt[3]{a + \sqrt{a^2 − b^3}} + \sqrt[3]{a - \sqrt{a^2 − b^3}} \end{align} $$

Let $\,u,v = \sqrt[3]{a \pm \sqrt{a^2 − b^3}}\,$, then $\,uv = b\,$, and:

$$ \begin{align} x^3 = (u+v)^3 &= u^3+v^3+3uv(u+v) \\ &= \left(a + \cancel{\sqrt{a^2 − b^3}}\right) + \left(a - \cancel{\sqrt{a^2 − b^3}}\right) + 3b(u+v) \\ &= 2a + 3 b x \end{align} $$

Therefore $\,x\,$ is a root of the depressed cubic $\,x^3 - 3bx - 2a = 0\,$, and the initial expression is in fact one form of Cardano's formula. Reverting to the original parameters, the equation is:

$$ x^3 - \cancel{3} \cdot \frac{1}{\cancel{3}} \cdot \left(\frac{j^2}{2s}\right)^2 x - 2 \cdot d \cdot \left(\frac{j}{2}\right)^2 = 0 \quad\iff\quad 4 s^2 \,x^3 - j^4 \,x - 2 d j^2s^2 = 0 $$

The latter is equivalent to the general depressed cubic, and there is no known formula simpler than the one that contains two cube roots (unless there are additional constraints on $d,j,s$).