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I have the following expression:

$$ \sqrt{ 1 X^2 - \sqrt[3]{X^2 Y + \sqrt{\left(X^2 Y\right)^2 + \left(\frac{2}{3} Y\right)^3}} - \sqrt[3]{X^2 Y - \sqrt{\left(X^2 Y\right)^2 + \left(\frac{2}{3} Y\right)^3}} } + \sqrt{ 2 X^2 + \sqrt[3]{X^2 Y + \sqrt{\left(X^2 Y\right)^2 + \left(\frac{2}{3} Y\right)^3}} + \sqrt[3]{X^2 Y - \sqrt{\left(X^2 Y\right)^2 + \left(\frac{2}{3} Y\right)^3}} + \frac{2 X^3}{\sqrt{ X^2 - \sqrt[3]{X^2 Y + \sqrt{\left(X^2 Y\right)^2 + \left(\frac{2}{3} Y\right)^3}} - \sqrt[3]{X^2 Y - \sqrt{\left(X^2 Y\right)^2 + \left(\frac{2}{3} Y\right)^3}} }} } - 3 X $$

where both $X$ and $Y$ are always greater than or equal to zero.

Is there any way to simplify it so that it uses fewer nested radicals? I'd particularly like to get all the radicals out of the denominators, if possible. I'd also like to use only real numbers and avoid complex numbers, if possible.

Lawton
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2 Answers2

5

One approach to simplifying radicals is to get repeated expressions out of the way to see things more clearly.

I. Simplification

So the expression shortens to,

$$ Z =\sqrt{ X^2 -\color{blue}v } + \sqrt{2 X^2 + \color{blue}v + \frac{2 X^3}{\sqrt{ X^2 - \color{blue}v}} } - 3 X $$

where $v$ is the cubic root,

$$\color{blue}v = \sqrt[3]{X^2 Y + \sqrt{\left(X^2 Y\right)^2 + \left(\frac{2}{3} Y\right)^3}} + \sqrt[3]{X^2 Y - \sqrt{\left(X^2 Y\right)^2 + \left(\frac{2}{3} Y\right)^3}}$$

In other words, it seems the OP is trying to solve a quartic.

To re-construct the quartic, the 1st step is to multiply the four conjugates and collect the variable $Z$,

$$P(Z)= \left(Z -\Big(\sqrt{ X^2 -v } + \sqrt{2 X^2 + v + \tfrac{2 X^3}{\sqrt{ X^2 - v}}} - 3 X\Big)\right)\\ \left(Z -\Big(\sqrt{ X^2 -v } - \sqrt{2 X^2 + v + \tfrac{2 X^3}{\sqrt{ X^2 - v}}} - 3 X\Big)\right)\\ \left(Z -\Big({-\sqrt{X^2 -v}} + \sqrt{2 X^2 + v - \tfrac{2 X^3}{\sqrt{ X^2 - v}}} - 3 X\Big)\right)\\ \left(Z -\Big({-\sqrt{X^2 -v}} - \sqrt{2 X^2 + v - \tfrac{2 X^3}{\sqrt{ X^2 - v}}} - 3 X\Big)\right)=0 $$

hence,

$$P(Z) = Z^4 + 12 X Z^3 + 48 X^2 Z^2 + 64 X^3 Z + \frac{4v^3}{v-X^2} = 0$$

or simply,

$$\color{blue}{P(Z) = Z(Z+4X)^3\,(v - X^2) + 4 v^3 =0}\tag1$$

The 2nd step, re-construct its resolvent cubic,

$$P(v) = \prod_{k=0}^2\left(v-\Big(\sqrt[3]{X^2 Y + \sqrt{D}}\,\zeta_3^k+ \sqrt[3]{X^2 Y - \sqrt{D}}\,\zeta_3^{2k}\Big)\right) = 0$$

with discriminant $D=\left(X^2 Y\right)^2 + \left(\frac{2}{3} Y\right)^3$ and cube root of unity $\zeta_3 = e^{2\pi i/3}$ to get,

$$\color{blue}{P(v) = v^3 + 2 v Y - 2 X^2 Y=0}\tag2$$

The 3rd step, using resultants in Wolfram Alpha to eliminate $v$ between $(1)$ and $(2)$ yields,

$$Z^4 + 12X Z^3 + 48X^2 Z^2 + 64X^3 Z - 8 Y = 0$$

or more elegantly, the final quartic,

$$Z (Z + 4 X)^3 = 8 Y$$

II. Hypergeometric solution

Because of the special properties of the quartic,

$$Z (Z + 4 X)^3 = 8 Y$$

if a simple closed-form is wanted (and not necessarily in radicals), then it has a rather neat "cubed" generalized hypergeometric solution,

$$ Z = \frac{Y}{(2X)^3}\left({_3F_2}\Big(\tfrac14,\tfrac24,\tfrac34;\,\tfrac23,\tfrac43;\,-\tfrac{8Y}{27X^4}\Big)\right)^3$$

The OP prefers $X,Y \geq 0$ which implies the cubic discriminant $27X^4+8Y \geq 0$. So for example, let $X = 2$ and $Y = 3$, then the hypergeometric above evaluates as,

$$Z = 0.0460743\dots$$

which is the only positive real root of the quartic.

III. Euler's solution

If radicals are desired but not in the denominator, then the technique was found by Euler. Given the "depressed" form (which any quartic can be transformed to),

$$w^4+pw^2+qw+r = 0$$

then its four roots are among the 8 sign changes of,

$$w = \frac{\pm\sqrt{u_1}\pm\sqrt{u_2}\pm\sqrt{u_3}}2$$

with the $u_i$ as the three roots of the Euler resolvent cubic,

$$ u^3 + 2p u^2 + (p^2 - 4r)u - q^2 =0$$

IV. Depressing an equation

As requested by the OP, to depress an equation is to get rid of the $x^{n+1}$ term using a linear transformation. For the general quartic,

$$Z^4+aZ^3+bZ^2+cZ+d = 0$$

let $Z=w+t$, and collect the new variable $w,$

$$w^4 + (\color{blue}{a + 4 t}) w^3 + (b + 3 a t + 6 t^2) w^2 + (c + 2 b t + 3 a t^2 + 4 t^3) w +(d + c t + b t^2 + a t^3 + t^4) = 0$$

then solve $\color{blue}{a+4t = 0}$ for the variable $t$. The $w^3$ term vanishes and we get the depressed form,

$$w^4+pw^2+qw+r = 0$$

V. Example

Given,

$$w^4+3w^2+2w+1=0$$

then the Euler resolvent is,

$$u^3+6u^2+5u-4=0$$

with two conjugate roots of the quartic,

$$w_1 = \frac{+\sqrt{u_1}+\sqrt{u_2}+\sqrt{u_3}}2 \approx 0.3497 + 1.7469i$$ $$w_2 = \frac{-\sqrt{u_1}-\sqrt{u_2}+\sqrt{u_3}}2 \approx 0.3497 - 1.7469i$$

and similarly for the other pair.

Tito Piezas III
  • 60,745
  • The expression in my question does come from a sequence of equations whose previous iterations were {degree zero, degree one, degree two, degree three}, so it makes sense that this expression would be linked to a quartic. However, I haven't learned about solving quartics. Could you elaborate a little on how I can use Euler's quartic solution for my question? – Lawton Jun 19 '23 at 14:14
  • @Lawton: Ok, I'll edit my answer. All you need is a linear transformation to get the "depressed" form of any equation. – Tito Piezas III Jun 19 '23 at 14:18
  • @Lawton It is done. I trust the procedure for "depressing" an equation is adequately explained? – Tito Piezas III Jun 19 '23 at 14:43
  • Given the quartic equation $Z^4 + 12X Z^3 + 48X^2 Z^2 + 64X^3 Z - 8 Y = 0$, I calculate that $p = -6 X^2$, $q = -8 X^3$, and $r = -3 X^4 - 8 Y$, giving the depressed quartic $w^4 + (-6 X^2) w^2 + (-8 X^3) w + (-3 X^4 - 8 Y) = 0$ and the Euler resolvent cubic $u^3 - 12 X^2 u^2 + (48 X^4 + 32 Y) u - 64 X^6 = 0$. However, it appears that the resolvent cubic only has one real root. Is that expected? – Lawton Jun 19 '23 at 15:04
  • @Lawton You got everything correct. If you solve the cubic, you'll find the expression $D= 27X^4+8Y$ which is its discriminant. If $D>0$, then the cubic will have 1 real root. But if $D<0$, then the cubic will have 3 real roots. Just adjust your parameters $X,Y$ depending on what you want. – Tito Piezas III Jun 19 '23 at 15:16
  • @Lawton In fact, in your original post, the expression $(X^2Y)^2 + (2/3Y)^3$ (after removing the factor $Y^2/27$) yields the same discriminant $D$. – Tito Piezas III Jun 19 '23 at 15:23
  • Both X and Y are constrained to be positive or zero, not negative. Does that mean that the only way to get the radicals out of the denominator is to include complex numbers instead? Or is there a real-numbers-only alternative? – Lawton Jun 19 '23 at 15:30
  • @Lawton: I know at least four ways to solve the quartic: it seems three of them will not satisfy at least one of the properties you prefer. Let me check the 4th one. Update: The 4th one will also have radicals in the denominator. . – Tito Piezas III Jun 19 '23 at 15:43
  • Thank you for your help! – Lawton Jun 19 '23 at 16:08
  • One more question about this: How did you find the quartic equation in Z? I can pretty easily verify that my expression is a root of the quartic equation you give at the top of your answer, but I can't figure out how to find that quartic in the first place. – Lawton Jun 19 '23 at 19:26
  • @Lawton. I used Method 1. You expand the quartic $(Z-x_1)(Z-x_2)(Z-x_3)(Z-x_4)$ and all the radicals will cancel out. Then, using Euler's cubic resolvent, you use resultants to eliminate the variable $u$ in the quartic, so it only has $p,q,r$. P.S. Would any closed-form solution to your quartic do (not necessarily in radicals)? Because there is a simple hypergeometric solution that bypasses complex numbers. See my edited first answer. – Tito Piezas III Jun 20 '23 at 06:21
  • I don't understand what you mean by "expand the quartic $(Z − x_1) (Z − x_2) (Z − x_3) (Z − x_4)$ and all the radicals will cancel out." Just to make sure we're on the same page, I'm trying to figure out how to go from $Z = \sqrt{X^2 - v} + \sqrt{2 X^2 + v + \frac{2 X^3}{\sqrt{X^2 - v}}} - 3 X$ to $Z^4 + 12X Z^3 + 48X^2 Z^2 + 64X^3 Z - 8 Y = 0$. I've tried multiplying out the four terms in $(Z − x_1) (Z − x_2) (Z − x_3) (Z − x_4)$, setting $Z$ equal to my initial expression, and solving for $x_i$, but haven't been able to get anywhere with that. – Lawton Jun 20 '23 at 14:54
  • @Lawton: To re-construct your quartic, one has to go through three steps. I've edited my first answer, so kindly check. I trust the explanation should suffice. – Tito Piezas III Jun 20 '23 at 15:48
  • I think I understand it in this case, but it's brought up a related problem. As this comment section is getting quite long, I've created a new question for it: https://math.stackexchange.com/questions/4722491/how-do-i-find-a-cubic-equation-given-only-one-root – Lawton Jun 20 '23 at 19:49
  • @Lawton: Ok, I answered the new question. No wonder you were interested in how I derived the quartic. You realized the other members in the first family had a similar form. The second family, however, seems trickier. – Tito Piezas III Jun 21 '23 at 09:46
  • Precisely, on all counts. – Lawton Jun 21 '23 at 12:57
  • @Lawton You extrapolated $f_1(Z) \to f_2(Z) \to f_3(Z)$. So maybe just use similar reasoning to extrapolate $g_1(Z) \to g_2(Z) \to g_3(Z)$ since you have the first two equations already. – Tito Piezas III Jun 21 '23 at 14:19
2

This second answer summarizes five different methods of solving the quartic, since it turns out OP's post is linked to that equation. Each method have their uses. For simplicity, we will use the depressed form,

$$x^4+px^2+qx+r = 0$$

for the first three methods and the same resolvent cubic in $u$,

$$ u^3 + 2p u^2 + (p^2 - 4r)u - q^2 =0$$

Method 1

Given one real root $u$ of the resolvent cubic, then,

$$x_{1,2} = \frac12\left(\sqrt{u}\pm\sqrt{-(2p+u)-\frac{2q}{\sqrt{u}}}\right)$$ $$x_{3,4} = \frac12\left(-\sqrt{u}\pm\sqrt{-(2p+u)+\frac{2q}{\sqrt{u}}}\right)$$

This factors the quartic into two quadratics and is the form used by Mathematica.

Method 2

Given all three roots $u_i$ of the resolvent cubic, then,

$$x_k = \frac12\Big({\pm\sqrt{u_1}}\pm\sqrt{u_2}\pm\sqrt{u_3}\,\Big)$$

where the four quartic roots $x_k$ are among the 8 sign changes. This avoids radicals in the denominator.

Method 3

By using resultants to eliminate $x$ between the depressed quartic,

$$x^4+px^2+qx+r = 0$$

and the special quadratic Tschirnhausen transformation,

$$x^2 + (q/u) x + (p/2) - y = 0$$

where $u$ is a real root of the resolvent cubic. This eliminates the $x^3,x$ terms and the quartic transforms into,

$$y^4+Ay^2+B = 0$$

so can be solved as a "quadratic". The roots of the quartic are among the 8 roots of,

$$x^2 + (q/u) x + (p/2) - y_k = 0$$

for $k=0,1,2,3$. This shows a relation between quadratic and quartic equations.

Method 4

Again, we use resultants. Eliminate $x$ between the general quartic,

$$x^4+ax^3+bx^2+cx+d = 0$$

and the quadratic Tschirnhausen transformation,

$$x^2+ex+f-y = 0$$

The two variables $e,f$ will suffice to eliminate the $x^3,x^2$ terms and are roots of quadratics. One is left with the "Bring quartic",

$$y^4+Ay+B = 0$$

This has a generalized hypergeometric solution given by,

$$ y = -\tfrac{B}{A}\;{_3F_2}\Big(\tfrac14,\tfrac24,\tfrac34;\,\tfrac23,\tfrac43;\,\tfrac{4^4B^3}{3^3A^4}\Big)$$

and is the only method in the list that does not need a cube root.

Method 5

We use resultants again. Eliminate $x$ between the general quartic and the cubic Tschirnhausen transformation,

$$x^3+ex^2+fx+g- y = 0$$

where $e,f,g$ suffice to eliminate the $x^3, x^2, x$ terms and are roots of sextics with a solvable Galois group. The quartic transforms into the binomial,

$$y^4-C = 0$$

and can be solved by a single $4$th root extraction. The roots of the quartic are among the 12 roots of,

$$x^3+ex^2+fx+g-C^{1/4}\zeta_4^k = 0$$

for $k=0,1,2,3$.

Tito Piezas III
  • 60,745