How can I find the inverse function over the interval$ [0,\infty].$?

Question

How can I find the inverse function of this function $ y = x^3 + \sqrt{x}$, could anyone give me a hint please?

Thanks!

Answer 1

$x^3 + \sqrt x = y$ becomes $x^6 - 2yx^3 - x + y^2 = 0$, whose Galois group is $S_6$, so it is not solvable by radicals.

That means, the inverse function exists, but you cannot express it using surds.

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By computation (thanks to PARI/GP), the Galois group of $x^6 - 6x^3 - x + 9 = 0$ is $S_6$. This is the same polynomial but with $y=3$.

Let $\alpha, \beta, \gamma, \delta, \varepsilon, \zeta$ be the six roots of $x^6 - 6x^3 - x + 9 = 0$ over $\Bbb Q$.

Let $y_1$ through $y_6$ be the six roots of $x^6 - 2yx^3 - x + y^2 = 0$ over $\Bbb Q(y)$.

We know that $[\Bbb Q(\alpha,\beta,\gamma,\delta,\varepsilon,\zeta) : \Bbb Q] = 720$.

It suffices to show that $[\Bbb Q(y_1,y_2,y_3,y_4,y_5,y_6) : \Bbb Q(y)] = 720$.

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Note that $\Bbb Q(y) \subseteq \Bbb Q(y_1,y_2,y_3,y_4,y_5,y_6)$ because $2y = \displaystyle \sum y_1 y_2 y_3$ (Vieta).

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We will, in particular, show that this path and the indices are correct:

$$\Bbb Q(y) \overset 6 \longrightarrow \Bbb Q(y,y_1) \overset 5 \longrightarrow \Bbb Q(y,y_1,y_2) \overset 4 \longrightarrow \Bbb Q(y,y_1,y_2,y_3) \\ \overset 3 \longrightarrow \Bbb Q(y,y_1,y_2,y_3,y_4) \overset 2 \longrightarrow \Bbb Q(y,y_1,y_2,y_3,y_4,y_5) = \Bbb Q(y_1,y_2,y_3,y_4,y_5,y_6)$$

where $\longrightarrow$ means nothing more than field extension and the number above is the degree of the extension.

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$x^6 - 2yx^3 - x + y^2$ is irreducible over $\Bbb Q(y)$ because $x^6 - 6x^3 - x + 9$ is irreducible over $\Bbb Q$. Therefore, $\Bbb Q(y,y_1)$ is $\Bbb Q(y)[X]/\langle X^6 - 2yX^3 - X + y^2 \rangle$ which is a degree $6$ extension of $\Bbb Q(y)$.

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$\displaystyle \frac {x^6 - 2yx^3 - x + y^2} {x - y_1}$ is irreducible over $\Bbb Q(y,y_1)$ because $\displaystyle \frac {x^6 - 6x^3 - x + 9} {x - \alpha}$ is irreducible over $\Bbb Q(\alpha)$. They are both degree $5$ polynomials, so indeed we have $[\Bbb Q(y,y_1,y_2):\Bbb Q(y)] = 5$.

(For reference: $$\begin{array}{cl} &\displaystyle \frac {x^6 - 2yx^3 - x + y^2} {x - y_1} \\ &\displaystyle = x^5 + y_1 x^4 + y_1^2 x^3 - (2y-y_1^3) x^2 - (2yy_1-y_1^4) x + (-1-2yy_1^2+y_1^5) \end{array}$$ because $(-1-2yy_1^2+y_1^5)(-y_1) = -y_1^6 + 2yy_1^3 + y_1 = y^2$.)

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The fact that $[\Bbb Q(y,y_1,y_2,y_3) : \Bbb Q(y,y_1,y_2)] = 4$, that $[\Bbb Q(y,y_1,y_2,y_3,y_4) : \Bbb Q(y,y_1,y_2,y_3)] = 3$, and that $[\Bbb Q(y,y_1,y_2,y_3,y_4,y_5) : \Bbb Q(y,y_1,y_2,y_3,y_4)] = 2$ follow similarly.